{"version":2,"work":"monographs-self-improving-intelligence","title":"Self-Improving Intelligence","source_revision":"cdaf825cc0547fe58973b1770b7cba2fff9bcfa4","sources":[{"repository_path":"monographs/self-improving-intelligence/CLAIMS.md","file_url":"documents/files/c18682a8352bdc094caa83d2e5c0ac1dee3bd613/CLAIMS.md","url":"read/monographs-self-improving-intelligence/claims-27a1e9babd86.html","sha256":"5f4b83fa1043bbf8b8664c8c8a558415155c73f5e0ed9aa89a1903e74f9272c3","git_blob_sha1":"c18682a8352bdc094caa83d2e5c0ac1dee3bd613","size":4495,"source_url":"read/monographs-self-improving-intelligence/claims-27a1e9babd86-source.html","passages":0,"macros":{}},{"repository_path":"monographs/self-improving-intelligence/README.md","file_url":"documents/files/a67897553701f7bc6296cf8fab5855f2f387acfa/README.md","url":"read/monographs-self-improving-intelligence/readme-d596d2856514.html","sha256":"c088357aad9adcaf79976a1bc761631091a70ef265c8fcf5f1532fa77fb1c2a5","git_blob_sha1":"a67897553701f7bc6296cf8fab5855f2f387acfa","size":5202,"source_url":"read/monographs-self-improving-intelligence/readme-d596d2856514-source.html","passages":0,"macros":{}},{"repository_path":"monographs/self-improving-intelligence/Self_Improving_Intelligence_v1.0.0.tex","file_url":"documents/files/f50a7cc7e0a8e3913a4d5c74a85d34581033e2e5/Self_Improving_Intelligence_v1.0.0.tex","url":"read/monographs-self-improving-intelligence/self-improving-intelligence-v1-0-0-63b55647d965.html","sha256":"8de5a3881f1d58e60ab5995040153bfd419de42ffdeb21926eff9b9b50ad4b00","git_blob_sha1":"f50a7cc7e0a8e3913a4d5c74a85d34581033e2e5","size":207778,"passages":641,"macros":{"headrulewidth":"0.3pt","X":"\\mathcal X","Z":"\\mathcal Z","U":"\\mathcal U","Q":"\\mathcal Q","Y":"\\mathcal Y","C":"\\mathcal C","M":"\\mathcal M","E":"\\mathbb E","Ffive":"\\mathbb F_5","TV":"d_{\\mathrm{TV}}","Eq":"\\equiv_{\\C}","supp":"\\operatorname{supp}","B":"\\mathcal B","Rcal":"\\mathcal R","Gcal":"\\mathcal G","Safe":"\\mathsf{Safe}","Post":"\\operatorname{Post}","Pareto":"\\operatorname{Pareto}"}},{"repository_path":"monographs/self-improving-intelligence/lineage/v0.6.0/README.md","file_url":"documents/files/21f9adcd2056390cdc2639b29e5e90e9b7f2ac56/README.md","url":"read/monographs-self-improving-intelligence/readme-9449377bac9d.html","sha256":"bd17034bfd59ff1b1eef46d291deaebbf65dddf81a7aa8b2198c00002f4e2da1","git_blob_sha1":"21f9adcd2056390cdc2639b29e5e90e9b7f2ac56","size":3664,"source_url":"read/monographs-self-improving-intelligence/readme-9449377bac9d-source.html","passages":0,"macros":{}}],"pdfs":[{"repository_path":"monographs/self-improving-intelligence/Self_Improving_Intelligence_v1.0.0.pdf","url":"read/monographs-self-improving-intelligence/self-improving-intelligence-v1-0-0-354d2085d7ff.html","sha256":"61660121dae183af536e565ad8e0a11e18d962807d20bcb1d55984acb1722957"}],"passages":[{"line":92,"start":8780,"end":8781,"expression":"S","context":0,"kind":"notation-table","source":2},{"line":92,"start":8785,"end":8786,"expression":"U","context":0,"kind":"notation-table","source":2},{"line":92,"start":8790,"end":8791,"expression":"L","context":0,"kind":"notation-table","source":2},{"line":93,"start":8865,"end":8868,"expression":"\\Xi","context":1,"kind":"notation-table","source":2},{"line":93,"start":8872,"end":8873,"expression":"B","context":1,"kind":"notation-table","source":2},{"line":94,"start":8931,"end":8947,"expression":"\\Rcal_\\Xi(U,L;B)","context":2,"kind":"notation-table","source":2},{"line":95,"start":9048,"end":9062,"expression":"B_0\\subseteq X","context":3,"kind":"notation-table","source":2},{"line":96,"start":9121,"end":9122,"expression":"e","context":4,"kind":"notation-table","source":2},{"line":96,"start":9126,"end":9127,"expression":"r","context":4,"kind":"notation-table","source":2},{"line":96,"start":9131,"end":9132,"expression":"P","context":4,"kind":"notation-table","source":2},{"line":97,"start":9208,"end":9216,"expression":"F_h(B_0)","context":5,"kind":"notation-table","source":2},{"line":97,"start":9296,"end":9297,"expression":"h","context":5,"kind":"notation-table","source":2},{"line":98,"start":9303,"end":9309,"expression":"\\sigma","context":6,"kind":"notation-table","source":2},{"line":98,"start":9313,"end":9317,"expression":"\\tau","context":6,"kind":"notation-table","source":2},{"line":99,"start":9389,"end":9399,"expression":"\\mathcal K","context":7,"kind":"notation-table","source":2},{"line":99,"start":9403,"end":9410,"expression":"W_{h,k}","context":7,"kind":"notation-table","source":2},{"line":99,"start":9463,"end":9464,"expression":"k","context":7,"kind":"notation-table","source":2},{"line":99,"start":9490,"end":9491,"expression":"h","context":7,"kind":"notation-table","source":2},{"line":159,"start":18665,"end":18667,"expression":"^2","context":8,"kind":"use","source":2},{"line":170,"start":20727,"end":20728,"expression":"S","context":9,"kind":"introduction-passage","source":2},{"line":170,"start":20763,"end":20764,"expression":"U","context":9,"kind":"introduction-passage","source":2},{"line":170,"start":20797,"end":20798,"expression":"L","context":9,"kind":"introduction-passage","source":2},{"line":173,"start":21124,"end":21127,"expression":"\\Xi","context":10,"kind":"definition-passage","definition_title":"Improvement contract","source":2},{"line":173,"start":21155,"end":21165,"expression":"\\mathcal H","context":10,"kind":"definition-passage","definition_title":"Improvement contract","source":2},{"line":173,"start":21307,"end":21308,"expression":"B","context":10,"kind":"definition-passage","definition_title":"Improvement contract","source":2},{"line":173,"start":21357,"end":21374,"expression":"J_\\tau(S)\\in[0,1]","context":10,"kind":"definition-passage","definition_title":"Improvement contract","source":2},{"line":176,"start":21651,"end":21664,"expression":"U(S,\\tau,L;B)","context":11,"kind":"introduction-passage","source":2},{"line":177,"start":21782,"end":21894,"expression":"\\label{eq:productivity}\n\\Rcal_\\Xi(U,L;B)=\\E_{(S,\\tau)\\sim\\mathcal H}\n\\big[J_\\tau(U(S,\\tau,L;B))-J_\\tau(S)\\big].\n","context":11,"kind":"introduction-passage","source":2},{"line":181,"start":21986,"end":21989,"expression":"\\Xi","context":11,"kind":"introduction-passage","source":2},{"line":183,"start":22235,"end":22241,"expression":"[-1,1]","context":12,"kind":"use","source":2},{"line":186,"start":22590,"end":22607,"expression":"\\Rcal_\\Xi(U',L;B)","context":13,"kind":"use","source":2},{"line":186,"start":22615,"end":22631,"expression":"\\Rcal_\\Xi(U,L;B)","context":13,"kind":"use","source":2},{"line":186,"start":22697,"end":22704,"expression":"(U',L')","context":13,"kind":"use","source":2},{"line":186,"start":22819,"end":22824,"expression":"(U,L)","context":13,"kind":"use","source":2},{"line":186,"start":22828,"end":22834,"expression":"(U',L)","context":13,"kind":"use","source":2},{"line":186,"start":22838,"end":22844,"expression":"(U,L')","context":13,"kind":"use","source":2},{"line":186,"start":22852,"end":22859,"expression":"(U',L')","context":13,"kind":"use","source":2},{"line":196,"start":23998,"end":24035,"expression":"\nZ_t=(S_t,U_t,L_t,E_t,\\Xi_t,\\rho_t),\n","context":14,"kind":"use","source":2},{"line":199,"start":24045,"end":24048,"expression":"E_t","context":14,"kind":"use","source":2},{"line":199,"start":24099,"end":24105,"expression":"\\rho_t","context":14,"kind":"use","source":2},{"line":199,"start":24233,"end":24238,"expression":"\\Xi_t","context":14,"kind":"use","source":2},{"line":204,"start":24854,"end":24872,"expression":"j\\in\\{0,\\ldots,7\\}","context":15,"kind":"use","source":2},{"line":206,"start":25076,"end":25077,"expression":"A","context":16,"kind":"use","source":2},{"line":206,"start":25092,"end":25111,"expression":"\\mathbf 1\\{j\\in A\\}","context":16,"kind":"use","source":2},{"line":206,"start":25212,"end":25215,"expression":"r_i","context":16,"kind":"use","source":2},{"line":206,"start":25240,"end":25243,"expression":"i=j","context":16,"kind":"use","source":2},{"line":208,"start":25573,"end":25582,"expression":"8(8+3)=88","context":17,"kind":"use","source":2},{"line":210,"start":25841,"end":25844,"expression":"j=0","context":18,"kind":"use","source":2},{"line":210,"start":25853,"end":25856,"expression":"j=1","context":18,"kind":"use","source":2},{"line":219,"start":27014,"end":27020,"expression":"\\theta","context":19,"kind":"use","source":2},{"line":219,"start":27039,"end":27057,"expression":"\\mathcal R(\\theta)","context":19,"kind":"use","source":2},{"line":219,"start":27252,"end":27258,"expression":"\\theta","context":19,"kind":"use","source":2},{"line":222,"start":27440,"end":27447,"expression":"P_0,P_1","context":20,"kind":"use","source":2},{"line":223,"start":27670,"end":27695,"expression":"\n\\frac{1+\\TV(P_0,P_1)}2.\n","context":21,"kind":"use","source":2},{"line":229,"start":27908,"end":27921,"expression":"\\mathcal R(0)","context":22,"kind":"use","source":2},{"line":229,"start":28195,"end":28207,"expression":"f(o)\\in[0,1]","context":23,"kind":"use","source":2},{"line":230,"start":28252,"end":28323,"expression":"\n\\tfrac12\\big(1+\\E_{P_0}f-\\E_{P_1}f\\big)\n\\leq\\tfrac12(1+\\TV(P_0,P_1)).\n","context":24,"kind":"use","source":2},{"line":242,"start":29243,"end":29271,"expression":"\\mathcal R(\\theta_1)=\\{a,c\\}","context":25,"kind":"use","source":2},{"line":242,"start":29278,"end":29306,"expression":"\\mathcal R(\\theta_2)=\\{b,c\\}","context":25,"kind":"use","source":2},{"line":242,"start":29345,"end":29346,"expression":"c","context":25,"kind":"use","source":2},{"line":242,"start":29540,"end":29547,"expression":"\\{a,b\\}","context":25,"kind":"use","source":2},{"line":242,"start":29551,"end":29558,"expression":"\\{b,c\\}","context":25,"kind":"use","source":2},{"line":242,"start":29566,"end":29573,"expression":"\\{a,c\\}","context":25,"kind":"use","source":2},{"line":247,"start":30046,"end":30047,"expression":"X","context":26,"kind":"use","source":2},{"line":247,"start":30096,"end":30106,"expression":"F_u:X\\to X","context":26,"kind":"use","source":2},{"line":247,"start":30127,"end":30131,"expression":"q(x)","context":26,"kind":"use","source":2},{"line":247,"start":30158,"end":30161,"expression":"P_0","context":26,"kind":"use","source":2},{"line":247,"start":30212,"end":30213,"expression":"q","context":26,"kind":"use","source":2},{"line":248,"start":30381,"end":30507,"expression":"\nx\\equiv_{P_{t+1}}y\n\\quad\\Longleftrightarrow\\quad\nx\\equiv_{P_t}y\\ \\text{ and }\\\nF_u(x)\\equiv_{P_t}F_u(y)\\ \\text{for every }u.\n","context":26,"kind":"use","source":2},{"line":257,"start":30800,"end":30809,"expression":"|X|-|P_0|","context":27,"kind":"use","source":2},{"line":257,"start":30882,"end":30885,"expression":"P_0","context":28,"kind":"use","source":2},{"line":257,"start":30903,"end":30906,"expression":"F_u","context":29,"kind":"use","source":2},{"line":260,"start":31136,"end":31139,"expression":"|X|","context":30,"kind":"use","source":2},{"line":260,"start":31357,"end":31358,"expression":"R","context":31,"kind":"use","source":2},{"line":260,"start":31394,"end":31397,"expression":"P_0","context":32,"kind":"use","source":2},{"line":260,"start":31404,"end":31405,"expression":"R","context":33,"kind":"use","source":2},{"line":260,"start":31416,"end":31419,"expression":"P_t","context":34,"kind":"use","source":2},{"line":260,"start":31436,"end":31437,"expression":"R","context":35,"kind":"use","source":2},{"line":260,"start":31453,"end":31454,"expression":"R","context":36,"kind":"use","source":2},{"line":260,"start":31503,"end":31506,"expression":"P_t","context":37,"kind":"use","source":2},{"line":260,"start":31544,"end":31551,"expression":"P_{t+1}","context":38,"kind":"use","source":2},{"line":260,"start":31576,"end":31577,"expression":"R","context":39,"kind":"use","source":2},{"line":260,"start":31594,"end":31597,"expression":"P_t","context":40,"kind":"use","source":2},{"line":263,"start":31817,"end":31820,"expression":"P_0","context":41,"kind":"use","source":2},{"line":271,"start":32777,"end":32778,"expression":"X","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":32806,"end":32816,"expression":"\\mathcal E","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":32844,"end":32854,"expression":"\\mathcal A","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":32869,"end":32870,"expression":"e","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":32896,"end":32910,"expression":"D_e\\subseteq X","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":32935,"end":32947,"expression":"f_e:D_e\\to X","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":32963,"end":32977,"expression":"o_e:D_e\\to O_e","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":33000,"end":33008,"expression":"c_e\\geq0","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":33038,"end":33039,"expression":"a","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":33057,"end":33065,"expression":"c_a\\geq0","context":42,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":33099,"end":33113,"expression":"G_a\\subseteq X","context":43,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":271,"start":33166,"end":33180,"expression":"B_0\\subseteq X","context":44,"kind":"definition-passage","definition_title":"Finite repair experiment model","source":2},{"line":274,"start":33285,"end":33301,"expression":"B_0\\subseteq D_e","context":45,"kind":"use","source":2},{"line":274,"start":33321,"end":33322,"expression":"o","context":45,"kind":"use","source":2},{"line":275,"start":33388,"end":33444,"expression":"\\label{eq:post}\nB_{e,o}=\\{f_e(x):x\\in B_0,\\ o_e(x)=o\\}.\n","context":45,"kind":"use","source":2},{"line":278,"start":33533,"end":33549,"expression":"B_0\\subseteq G_a","context":45,"kind":"use","source":2},{"line":278,"start":33652,"end":33653,"expression":"X","context":46,"kind":"use","source":2},{"line":282,"start":34162,"end":34170,"expression":"W(P,B_0)","context":47,"kind":"use","source":2},{"line":282,"start":34258,"end":34262,"expression":"M(P)","context":47,"kind":"use","source":2},{"line":284,"start":34396,"end":34397,"expression":"h","context":48,"kind":"use","source":2},{"line":284,"start":34405,"end":34422,"expression":"\\mathcal F_h(B_0)","context":48,"kind":"use","source":2},{"line":284,"start":34462,"end":34467,"expression":"(w,m)","context":48,"kind":"use","source":2},{"line":284,"start":34580,"end":34615,"expression":"T(B_0)=\\{(c_a,1):B_0\\subseteq G_a\\}","context":48,"kind":"use","source":2},{"line":285,"start":34636,"end":34889,"expression":"\\label{eq:pareto}\n\\mathcal F_0(B_0)&=\\Pareto T(B_0),\\\\\n\\mathcal F_h(B_0)&=\\Pareto\\left(T(B_0)\\cup\n\\bigcup_{e:B_0\\subseteq D_e}\n\\left\\{\\left(c_e+\\max_o w_o,\\ 1+\\sum_o m_o\\right):\n(w_o,m_o)\\in\\mathcal F_{h-1}(B_{e,o})\\ \\forall o\\right\\}\\right).\n\\nonumber\n","context":49,"kind":"use","source":2},{"line":296,"start":35160,"end":35163,"expression":"B_0","context":50,"kind":"use","source":2},{"line":296,"start":35179,"end":35185,"expression":"h\\geq0","context":51,"kind":"use","source":2},{"line":296,"start":35309,"end":35310,"expression":"h","context":52,"kind":"use","source":2},{"line":296,"start":35386,"end":35391,"expression":"(b,m)","context":53,"kind":"use","source":2},{"line":296,"start":35465,"end":35470,"expression":"(b,m)","context":54,"kind":"use","source":2},{"line":299,"start":35571,"end":35577,"expression":"T(B_0)","context":55,"kind":"use","source":2},{"line":299,"start":35601,"end":35604,"expression":"h-1","context":56,"kind":"use","source":2},{"line":299,"start":35637,"end":35638,"expression":"h","context":57,"kind":"use","source":2},{"line":306,"start":36994,"end":36995,"expression":"a","context":58,"kind":"use","source":2},{"line":306,"start":37010,"end":37019,"expression":"\\{0,1,2\\}","context":58,"kind":"use","source":2},{"line":306,"start":37033,"end":37034,"expression":"b","context":58,"kind":"use","source":2},{"line":306,"start":37049,"end":37054,"expression":"\\{3\\}","context":58,"kind":"use","source":2},{"line":306,"start":37085,"end":37092,"expression":"\\{0,1\\}","context":58,"kind":"use","source":2},{"line":306,"start":37100,"end":37107,"expression":"\\{2,3\\}","context":58,"kind":"use","source":2},{"line":306,"start":37135,"end":37144,"expression":"\\{0,1,2\\}","context":58,"kind":"use","source":2},{"line":306,"start":37152,"end":37157,"expression":"\\{3\\}","context":58,"kind":"use","source":2},{"line":322,"start":39336,"end":39342,"expression":"\\Theta","context":59,"kind":"use","source":2},{"line":322,"start":39414,"end":39415,"expression":"X","context":59,"kind":"use","source":2},{"line":322,"start":39480,"end":39486,"expression":"\\theta","context":59,"kind":"use","source":2},{"line":322,"start":39499,"end":39509,"expression":"D_e^\\theta","context":59,"kind":"use","source":2},{"line":322,"start":39513,"end":39523,"expression":"f_e^\\theta","context":59,"kind":"use","source":2},{"line":322,"start":39527,"end":39537,"expression":"o_e^\\theta","context":59,"kind":"use","source":2},{"line":322,"start":39545,"end":39555,"expression":"G_a^\\theta","context":59,"kind":"use","source":2},{"line":322,"start":39653,"end":39677,"expression":"B\\subseteq\\Theta\\times X","context":60,"kind":"use","source":2},{"line":325,"start":39875,"end":39988,"expression":"\\label{eq:model-support}\nB_{e,o}=\\{(\\theta,f_e^\\theta(x)):(\\theta,x)\\in B,\\ x\\in D_e^\\theta,\\ o_e^\\theta(x)=o\\}.\n","context":61,"kind":"use","source":2},{"line":328,"start":40214,"end":40215,"expression":"a","context":62,"kind":"use","source":2},{"line":328,"start":40236,"end":40251,"expression":"x\\in G_a^\\theta","context":63,"kind":"use","source":2},{"line":328,"start":40264,"end":40279,"expression":"(\\theta,x)\\in B","context":64,"kind":"use","source":2},{"line":333,"start":40780,"end":40807,"expression":"\\widetilde X=\\Theta\\times X","context":65,"kind":"use","source":2},{"line":334,"start":40814,"end":40961,"expression":"\n\\widetilde f_e(\\theta,x)=(\\theta,f_e^\\theta(x)),\\quad\n\\widetilde o_e(\\theta,x)=o_e^\\theta(x),\\quad\n\\widetilde G_a=\\{(\\theta,x):x\\in G_a^\\theta\\},\n","context":66,"kind":"use","source":2},{"line":339,"start":41198,"end":41199,"expression":"B","context":67,"kind":"use","source":2},{"line":339,"start":41330,"end":41331,"expression":"B","context":68,"kind":"use","source":2},{"line":342,"start":41965,"end":41966,"expression":"B","context":69,"kind":"use","source":2},{"line":349,"start":42703,"end":42714,"expression":"x\\in\\{0,1\\}","context":70,"kind":"use","source":2},{"line":349,"start":42754,"end":42770,"expression":"\\theta\\in\\{0,1\\}","context":70,"kind":"use","source":2},{"line":349,"start":42834,"end":42849,"expression":"o=x\\oplus\\theta","context":70,"kind":"use","source":2},{"line":349,"start":42872,"end":42875,"expression":"r_i","context":70,"kind":"use","source":2},{"line":349,"start":42900,"end":42903,"expression":"x=i","context":70,"kind":"use","source":2},{"line":349,"start":42924,"end":42930,"expression":"\\theta","context":70,"kind":"use","source":2},{"line":349,"start":43017,"end":43027,"expression":"(\\theta,x)","context":71,"kind":"use","source":2},{"line":349,"start":43079,"end":43084,"expression":"(0,0)","context":72,"kind":"use","source":2},{"line":349,"start":43091,"end":43096,"expression":"(1,1)","context":73,"kind":"use","source":2},{"line":349,"start":43124,"end":43129,"expression":"(0,1)","context":74,"kind":"use","source":2},{"line":349,"start":43136,"end":43141,"expression":"(1,0)","context":75,"kind":"use","source":2},{"line":352,"start":43389,"end":43395,"expression":"\\theta","context":76,"kind":"use","source":2},{"line":352,"start":43406,"end":43407,"expression":"x","context":76,"kind":"use","source":2},{"line":352,"start":43501,"end":43518,"expression":"r_{o\\oplus\\theta}","context":76,"kind":"use","source":2},{"line":352,"start":43801,"end":43806,"expression":"(4,7)","context":77,"kind":"use","source":2},{"line":366,"start":45671,"end":45682,"expression":"|\\Theta||X|","context":78,"kind":"use","source":2},{"line":371,"start":46464,"end":46482,"expression":"\\{U_1,\\ldots,U_N\\}","context":79,"kind":"use","source":2},{"line":371,"start":46532,"end":46544,"expression":"\\widehat R_i","context":79,"kind":"use","source":2},{"line":371,"start":46570,"end":46573,"expression":"R_i","context":79,"kind":"use","source":2},{"line":374,"start":46909,"end":46942,"expression":"|\\widehat R_i-R_i|\\leq\\varepsilon","context":80,"kind":"use","source":2},{"line":374,"start":46969,"end":46979,"expression":"\\hat\\imath","context":80,"kind":"use","source":2},{"line":374,"start":46992,"end":47004,"expression":"\\widehat R_i","context":80,"kind":"use","source":2},{"line":375,"start":47014,"end":47058,"expression":"\nR_{\\hat\\imath}\\geq \\max_iR_i-2\\varepsilon.\n","context":80,"kind":"use","source":2},{"line":378,"start":47145,"end":47164,"expression":"2\\varepsilon+\\gamma","context":80,"kind":"use","source":2},{"line":378,"start":47195,"end":47201,"expression":"\\gamma","context":80,"kind":"use","source":2},{"line":381,"start":47258,"end":47261,"expression":"i^*","context":81,"kind":"use","source":2},{"line":382,"start":47265,"end":47377,"expression":"R_{\\hat\\imath}\\geq\\widehat R_{\\hat\\imath}-\\varepsilon\n\\geq\\widehat R_{i^*}-\\varepsilon\n\\geq R_{i^*}-2\\varepsilon","context":82,"kind":"use","source":2},{"line":393,"start":48239,"end":48255,"expression":"R^*_{\\mathcal G}","context":83,"kind":"use","source":2},{"line":393,"start":48338,"end":48348,"expression":"\\mathcal G","context":83,"kind":"use","source":2},{"line":393,"start":48414,"end":48435,"expression":"R^*_{\\mathcal G}-R(U)","context":83,"kind":"use","source":2},{"line":393,"start":48509,"end":48526,"expression":"R^*_{\\mathcal G'}","context":83,"kind":"use","source":2},{"line":393,"start":48534,"end":48550,"expression":"R^*_{\\mathcal G}","context":83,"kind":"use","source":2},{"line":400,"start":49606,"end":49609,"expression":"3/8","context":84,"kind":"use","source":2},{"line":400,"start":49686,"end":49687,"expression":"1","context":84,"kind":"use","source":2},{"line":407,"start":50856,"end":50870,"expression":"\\sigma:X\\to X'","context":85,"kind":"use","source":2},{"line":407,"start":50889,"end":50918,"expression":"\\tau:\\mathcal E\\to\\mathcal E'","context":85,"kind":"use","source":2},{"line":407,"start":50949,"end":50978,"expression":"\\psi:\\mathcal A\\to\\mathcal A'","context":85,"kind":"use","source":2},{"line":407,"start":51009,"end":51035,"expression":"\\eta_e:O_e\\to O'_{\\tau(e)}","context":85,"kind":"use","source":2},{"line":408,"start":51087,"end":51255,"expression":"\\label{eq:transport}\n\\sigma(D_e)&=D'_{\\tau(e)}, &\n\\sigma f_e(x)&=f'_{\\tau(e)}\\sigma(x),\\\\\n\\eta_e(o_e(x))&=o'_{\\tau(e)}(\\sigma x), &\n\\sigma(G_a)&=G'_{\\psi(a)},\\nonumber\n","context":86,"kind":"use","source":2},{"line":417,"start":51645,"end":51648,"expression":"B_0","context":87,"kind":"use","source":2},{"line":417,"start":51670,"end":51681,"expression":"\\sigma(B_0)","context":88,"kind":"use","source":2},{"line":418,"start":51765,"end":51812,"expression":"\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n","context":89,"kind":"use","source":2},{"line":424,"start":51955,"end":51958,"expression":"B_0","context":90,"kind":"use","source":2},{"line":424,"start":51996,"end":52007,"expression":"\\sigma(B_0)","context":91,"kind":"use","source":2},{"line":425,"start":52135,"end":52176,"expression":"\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n","context":92,"kind":"use","source":2},{"line":434,"start":52998,"end":53004,"expression":"\\Delta","context":93,"kind":"use","source":2},{"line":434,"start":53010,"end":53012,"expression":"U'","context":93,"kind":"use","source":2},{"line":434,"start":53020,"end":53021,"expression":"U","context":93,"kind":"use","source":2},{"line":434,"start":53029,"end":53055,"expression":"\\widetilde U',\\widetilde U","context":93,"kind":"use","source":2},{"line":434,"start":53173,"end":53184,"expression":"\\varepsilon","context":93,"kind":"use","source":2},{"line":435,"start":53208,"end":53301,"expression":"\\label{eq:gaintransport}\n\\Rcal'(\\widetilde U')-\\Rcal'(\\widetilde U)\n\\geq\\Delta-2\\varepsilon.\n","context":93,"kind":"use","source":2},{"line":439,"start":53498,"end":53504,"expression":"[-1,1]","context":94,"kind":"use","source":2},{"line":439,"start":53549,"end":53555,"expression":"\\delta","context":95,"kind":"use","source":2},{"line":439,"start":53593,"end":53600,"expression":"2\\delta","context":96,"kind":"use","source":2},{"line":439,"start":53641,"end":53645,"expression":"\\eta","context":97,"kind":"use","source":2},{"line":439,"start":53664,"end":53688,"expression":"\\varepsilon=\\eta+2\\delta","context":98,"kind":"use","source":2},{"line":453,"start":55661,"end":55667,"expression":"d\\geq0","context":99,"kind":"use","source":2},{"line":453,"start":55689,"end":55695,"expression":"s\\geq0","context":100,"kind":"use","source":2},{"line":453,"start":55734,"end":55739,"expression":"(b,m)","context":101,"kind":"use","source":2},{"line":453,"start":55795,"end":55804,"expression":"(b+d,m+s)","context":102,"kind":"use","source":2},{"line":473,"start":58788,"end":58794,"expression":"n\\geq1","context":103,"kind":"use","source":2},{"line":473,"start":59090,"end":59091,"expression":"n","context":103,"kind":"use","source":2},{"line":477,"start":59421,"end":59445,"expression":"\n1+\\lceil\\log_2 n\\rceil\n","context":104,"kind":"use","source":2},{"line":480,"start":59470,"end":59474,"expression":"2n-1","context":105,"kind":"use","source":2},{"line":483,"start":59954,"end":59955,"expression":"n","context":106,"kind":"use","source":2},{"line":483,"start":60004,"end":60005,"expression":"d","context":107,"kind":"use","source":2},{"line":483,"start":60020,"end":60023,"expression":"2^d","context":108,"kind":"use","source":2},{"line":483,"start":60037,"end":60062,"expression":"d\\geq\\lceil\\log_2 n\\rceil","context":109,"kind":"use","source":2},{"line":483,"start":60128,"end":60129,"expression":"l","context":110,"kind":"use","source":2},{"line":483,"start":60164,"end":60171,"expression":"i_1,i_2","context":111,"kind":"use","source":2},{"line":483,"start":60258,"end":60278,"expression":"i_1+2i_2=i_1+i_2+l-1","context":112,"kind":"use","source":2},{"line":483,"start":60288,"end":60295,"expression":"i_2=l-1","context":113,"kind":"use","source":2},{"line":483,"start":60317,"end":60333,"expression":"2l-1+i_1\\geq2n-1","context":114,"kind":"use","source":2},{"line":508,"start":64122,"end":64127,"expression":"n=2^k","context":115,"kind":"use","source":2},{"line":508,"start":64135,"end":64141,"expression":"k\\geq1","context":116,"kind":"use","source":2},{"line":508,"start":64163,"end":64164,"expression":"q","context":117,"kind":"use","source":2},{"line":508,"start":64260,"end":64261,"expression":"k","context":118,"kind":"use","source":2},{"line":508,"start":64374,"end":64375,"expression":"q","context":119,"kind":"use","source":2},{"line":509,"start":64564,"end":64625,"expression":"\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n","context":120,"kind":"use","source":2},{"line":512,"start":64701,"end":64702,"expression":"k","context":121,"kind":"use","source":2},{"line":515,"start":64888,"end":64890,"expression":"nq","context":122,"kind":"use","source":2},{"line":515,"start":64915,"end":64916,"expression":"k","context":123,"kind":"use","source":2},{"line":515,"start":64968,"end":64969,"expression":"n","context":124,"kind":"use","source":2},{"line":515,"start":65008,"end":65010,"expression":"nq","context":125,"kind":"use","source":2},{"line":515,"start":65051,"end":65052,"expression":"k","context":126,"kind":"use","source":2},{"line":515,"start":65147,"end":65150,"expression":"2nk","context":127,"kind":"use","source":2},{"line":522,"start":66287,"end":66293,"expression":"N\\geq1","context":128,"kind":"use","source":2},{"line":522,"start":66624,"end":66625,"expression":"N","context":129,"kind":"use","source":2},{"line":525,"start":66746,"end":66747,"expression":"j","context":130,"kind":"use","source":2},{"line":535,"start":68617,"end":68625,"expression":"10{,}256","context":131,"kind":"use","source":2},{"line":535,"start":68650,"end":68658,"expression":"10{,}380","context":131,"kind":"use","source":2},{"line":535,"start":68699,"end":68702,"expression":"488","context":131,"kind":"use","source":2},{"line":538,"start":69197,"end":69225,"expression":"(\\text{work},\\text{storage})","context":132,"kind":"use","source":2},{"line":539,"start":69252,"end":69257,"expression":"(2,3)","context":133,"kind":"use","source":2},{"line":540,"start":69277,"end":69282,"expression":"(3,7)","context":134,"kind":"use","source":2},{"line":541,"start":69305,"end":69311,"expression":"(4,15)","context":135,"kind":"use","source":2},{"line":542,"start":69339,"end":69345,"expression":"(5,31)","context":136,"kind":"use","source":2},{"line":547,"start":69849,"end":69856,"expression":"2{,}000","context":137,"kind":"use","source":2},{"line":572,"start":73784,"end":73822,"expression":"D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}","context":138,"kind":"use","source":2},{"line":572,"start":73861,"end":73877,"expression":"2\\min\\{D_0,D_1\\}","context":139,"kind":"use","source":2},{"line":572,"start":73959,"end":73965,"expression":"c\\geq0","context":140,"kind":"use","source":2},{"line":572,"start":74035,"end":74056,"expression":"2(1+c)\\min\\{D_0,D_1\\}","context":141,"kind":"use","source":2},{"line":575,"start":74098,"end":74100,"expression":"2d","context":142,"kind":"use","source":2},{"line":575,"start":74161,"end":74162,"expression":"d","context":143,"kind":"use","source":2},{"line":575,"start":74188,"end":74189,"expression":"d","context":144,"kind":"use","source":2},{"line":594,"start":76130,"end":76137,"expression":"2{,}000","context":145,"kind":"use","source":2},{"line":614,"start":79283,"end":79284,"expression":"m","context":146,"kind":"use","source":2},{"line":614,"start":79386,"end":79387,"expression":"a","context":146,"kind":"use","source":2},{"line":614,"start":79434,"end":79435,"expression":"b","context":146,"kind":"use","source":2},{"line":614,"start":79465,"end":79470,"expression":"m-a-b","context":146,"kind":"use","source":2},{"line":614,"start":79530,"end":79536,"expression":"d\\geq1","context":146,"kind":"use","source":2},{"line":615,"start":79592,"end":79603,"expression":"\nC=2^{d-1}\n","context":147,"kind":"use","source":2},{"line":618,"start":79692,"end":79699,"expression":"m\\leq2C","context":148,"kind":"use","source":2},{"line":620,"start":79900,"end":79907,"expression":"(a,b,m)","context":149,"kind":"use","source":2},{"line":620,"start":79920,"end":79921,"expression":"C","context":149,"kind":"use","source":2},{"line":620,"start":79997,"end":80012,"expression":"(a,b,m)=(0,3,4)","context":149,"kind":"use","source":2},{"line":620,"start":80123,"end":80128,"expression":"(1,3)","context":149,"kind":"use","source":2},{"line":623,"start":80345,"end":80355,"expression":"\\mathcal P","context":150,"kind":"use","source":2},{"line":623,"start":80398,"end":80415,"expression":"r:\\mathcal P\\to Z","context":151,"kind":"use","source":2},{"line":623,"start":80458,"end":80483,"expression":"\\ell:\\mathcal P\\to\\{0,1\\}","context":152,"kind":"use","source":2},{"line":623,"start":80539,"end":80552,"expression":"f:Z\\to\\{0,1\\}","context":152,"kind":"use","source":2},{"line":623,"start":80566,"end":80573,"expression":"fr=\\ell","context":152,"kind":"use","source":2},{"line":623,"start":80598,"end":80602,"expression":"\\ell","context":152,"kind":"use","source":2},{"line":623,"start":80643,"end":80644,"expression":"r","context":152,"kind":"use","source":2},{"line":626,"start":80788,"end":80789,"expression":"f","context":153,"kind":"use","source":2},{"line":626,"start":80842,"end":80843,"expression":"f","context":154,"kind":"use","source":2},{"line":626,"start":81021,"end":81022,"expression":"r","context":155,"kind":"use","source":2},{"line":626,"start":81111,"end":81112,"expression":"f","context":156,"kind":"use","source":2},{"line":629,"start":81363,"end":81370,"expression":"(a,b,m)","context":157,"kind":"use","source":2},{"line":629,"start":81466,"end":81467,"expression":"C","context":157,"kind":"use","source":2},{"line":629,"start":81531,"end":81532,"expression":"C","context":158,"kind":"use","source":2},{"line":632,"start":81776,"end":81795,"expression":"(a,b,m,C)=(0,3,4,2)","context":159,"kind":"use","source":2},{"line":632,"start":81802,"end":81811,"expression":"(0,3,4,4)","context":159,"kind":"use","source":2},{"line":632,"start":81842,"end":81849,"expression":"(a,b,m)","context":159,"kind":"use","source":2},{"line":638,"start":82844,"end":82853,"expression":"(a,b,m,C)","context":160,"kind":"use","source":2},{"line":638,"start":82903,"end":82914,"expression":"a,b,m,C,0,1","context":160,"kind":"use","source":2},{"line":638,"start":82954,"end":82958,"expression":"\\max","context":160,"kind":"use","source":2},{"line":638,"start":82962,"end":82966,"expression":"\\min","context":160,"kind":"use","source":2},{"line":638,"start":83009,"end":83024,"expression":">,<,\\geq,\\leq,=","context":160,"kind":"use","source":2},{"line":645,"start":84266,"end":84279,"expression":"C\\in\\{1,2,4\\}","context":161,"kind":"use","source":2},{"line":645,"start":84283,"end":84296,"expression":"2\\leq m\\leq2C","context":161,"kind":"use","source":2},{"line":645,"start":84304,"end":84312,"expression":"a,b\\geq0","context":161,"kind":"use","source":2},{"line":645,"start":84316,"end":84325,"expression":"a+b\\leq m","context":161,"kind":"use","source":2},{"line":645,"start":84516,"end":84517,"expression":"C","context":161,"kind":"use","source":2},{"line":650,"start":85205,"end":85215,"expression":"\\mathcal G","context":162,"kind":"use","source":2},{"line":650,"start":85284,"end":85294,"expression":"\\mathcal P","context":163,"kind":"use","source":2},{"line":650,"start":85325,"end":85329,"expression":"\\ell","context":164,"kind":"use","source":2},{"line":650,"start":85440,"end":85444,"expression":"\\ell","context":165,"kind":"use","source":2},{"line":650,"start":85480,"end":85490,"expression":"\\mathcal P","context":166,"kind":"use","source":2},{"line":650,"start":85656,"end":85666,"expression":"\\mathcal P","context":167,"kind":"use","source":2},{"line":650,"start":85696,"end":85706,"expression":"\\mathcal G","context":168,"kind":"use","source":2},{"line":650,"start":85743,"end":85755,"expression":"|\\mathcal G|","context":169,"kind":"use","source":2},{"line":653,"start":86132,"end":86136,"expression":"\\ell","context":170,"kind":"use","source":2},{"line":653,"start":86149,"end":86159,"expression":"\\mathcal P","context":171,"kind":"use","source":2},{"line":653,"start":86221,"end":86233,"expression":"|\\mathcal G|","context":172,"kind":"use","source":2},{"line":659,"start":87017,"end":87020,"expression":"a>a","context":173,"kind":"use","source":2},{"line":659,"start":87091,"end":87094,"expression":"b>a","context":174,"kind":"use","source":2},{"line":659,"start":87155,"end":87158,"expression":"b>C","context":175,"kind":"use","source":2},{"line":660,"start":87245,"end":87310,"expression":"\\label{eq:learned-prune}\n\\rho(a,b,m,C)=\\mathbf 1\\{\\max(a,b)>C\\}.\n","context":176,"kind":"use","source":2},{"line":673,"start":89224,"end":89231,"expression":"C,m,a,b","context":177,"kind":"use","source":2},{"line":673,"start":89251,"end":89257,"expression":"C\\geq1","context":178,"kind":"use","source":2},{"line":673,"start":89261,"end":89274,"expression":"2\\leq m\\leq2C","context":179,"kind":"use","source":2},{"line":673,"start":89278,"end":89286,"expression":"a,b\\geq0","context":180,"kind":"use","source":2},{"line":673,"start":89294,"end":89303,"expression":"a+b\\leq m","context":181,"kind":"use","source":2},{"line":673,"start":89344,"end":89349,"expression":"m-a-b","context":182,"kind":"use","source":2},{"line":673,"start":89398,"end":89399,"expression":"C","context":183,"kind":"use","source":2},{"line":673,"start":89424,"end":89439,"expression":"\\max(a,b)\\leq C","context":184,"kind":"use","source":2},{"line":673,"start":89477,"end":89488,"expression":"\\max(a,b)>C","context":185,"kind":"use","source":2},{"line":676,"start":89699,"end":89702,"expression":"a>C","context":186,"kind":"use","source":2},{"line":676,"start":89708,"end":89711,"expression":"b>C","context":187,"kind":"use","source":2},{"line":676,"start":89774,"end":89783,"expression":"a,b\\leq C","context":188,"kind":"use","source":2},{"line":676,"start":89791,"end":89798,"expression":"r=m-a-b","context":189,"kind":"use","source":2},{"line":676,"start":89812,"end":89815,"expression":"C-a","context":190,"kind":"use","source":2},{"line":676,"start":89858,"end":89861,"expression":"C-b","context":191,"kind":"use","source":2},{"line":676,"start":89921,"end":89933,"expression":"2C-a-b\\geq r","context":192,"kind":"use","source":2},{"line":676,"start":89952,"end":89953,"expression":"r","context":193,"kind":"use","source":2},{"line":679,"start":90332,"end":90341,"expression":"C=2^{d-1}","context":194,"kind":"use","source":2},{"line":683,"start":90846,"end":90854,"expression":"a,b\\leq4","context":195,"kind":"use","source":2},{"line":683,"start":90921,"end":90929,"expression":"a+b\\leq8","context":195,"kind":"use","source":2},{"line":687,"start":91342,"end":91353,"expression":"\\max(a,b)>C","context":196,"kind":"use","source":2},{"line":690,"start":91794,"end":91813,"expression":"\\max(a,b)>\\min(C,4)","context":197,"kind":"use","source":2},{"line":690,"start":91941,"end":91961,"expression":"(a,b,m,C)=(0,5,16,8)","context":197,"kind":"use","source":2},{"line":690,"start":92019,"end":92024,"expression":"(8,8)","context":197,"kind":"use","source":2},{"line":696,"start":92892,"end":92919,"expression":"d_0=\\lceil\\log_2|B_0|\\rceil","context":198,"kind":"use","source":2},{"line":698,"start":93428,"end":93431,"expression":"d_0","context":199,"kind":"use","source":2},{"line":698,"start":93464,"end":93467,"expression":"d_0","context":199,"kind":"use","source":2},{"line":703,"start":94677,"end":94678,"expression":"C","context":200,"kind":"use","source":2},{"line":710,"start":95995,"end":95996,"expression":"C","context":201,"kind":"use","source":2},{"line":710,"start":96130,"end":96131,"expression":"C","context":201,"kind":"use","source":2},{"line":710,"start":96158,"end":96159,"expression":"C","context":201,"kind":"use","source":2},{"line":716,"start":96888,"end":96895,"expression":"\\{0,1\\}","context":202,"kind":"use","source":2},{"line":716,"start":96902,"end":96909,"expression":"\\{2,3\\}","context":202,"kind":"use","source":2},{"line":716,"start":96981,"end":96984,"expression":"2/2","context":202,"kind":"use","source":2},{"line":722,"start":97714,"end":97715,"expression":"B","context":203,"kind":"use","source":2},{"line":722,"start":97770,"end":97784,"expression":"G_r\\subseteq B","context":203,"kind":"use","source":2},{"line":722,"start":97841,"end":97850,"expression":"\\kappa(B)","context":203,"kind":"use","source":2},{"line":722,"start":97921,"end":97922,"expression":"B","context":203,"kind":"use","source":2},{"line":722,"start":97930,"end":97946,"expression":"\\kappa(B)=\\infty","context":203,"kind":"use","source":2},{"line":725,"start":98297,"end":98298,"expression":"d","context":204,"kind":"use","source":2},{"line":725,"start":98314,"end":98323,"expression":"\\kappa(B)","context":205,"kind":"use","source":2},{"line":725,"start":98363,"end":98379,"expression":"\\kappa(B)\\leq2^d","context":206,"kind":"use","source":2},{"line":725,"start":98445,"end":98457,"expression":"2\\kappa(B)-1","context":207,"kind":"use","source":2},{"line":728,"start":98752,"end":98753,"expression":"B","context":208,"kind":"use","source":2},{"line":728,"start":98793,"end":98802,"expression":"\\kappa(B)","context":209,"kind":"use","source":2},{"line":728,"start":98829,"end":98830,"expression":"d","context":210,"kind":"use","source":2},{"line":728,"start":98845,"end":98848,"expression":"2^d","context":211,"kind":"use","source":2},{"line":728,"start":99081,"end":99092,"expression":"\\kappa(B)=4","context":212,"kind":"use","source":2},{"line":728,"start":99099,"end":99102,"expression":"d=2","context":213,"kind":"use","source":2},{"line":731,"start":99264,"end":99277,"expression":"\\kappa(B)=|B|","context":214,"kind":"use","source":2},{"line":731,"start":99339,"end":99350,"expression":"\\kappa(B)=1","context":214,"kind":"use","source":2},{"line":763,"start":103190,"end":103193,"expression":"w>0","context":215,"kind":"use","source":2},{"line":764,"start":103307,"end":103383,"expression":"\n\\Delta(w)=w\\,(R_{\\mathrm{full}}-R_{\\mathrm{stream}})-N_{\\mathrm{compare}}.\n","context":215,"kind":"use","source":2},{"line":767,"start":103427,"end":103439,"expression":"w>14209/8287","context":215,"kind":"use","source":2},{"line":767,"start":103446,"end":103449,"expression":"w=1","context":215,"kind":"use","source":2},{"line":769,"start":103799,"end":103800,"expression":"D","context":216,"kind":"use","source":2},{"line":769,"start":103873,"end":103884,"expression":"\\Delta(w)>0","context":216,"kind":"use","source":2},{"line":769,"start":103888,"end":103889,"expression":"N","context":216,"kind":"use","source":2},{"line":769,"start":103930,"end":103942,"expression":"N\\Delta(w)>D","context":216,"kind":"use","source":2},{"line":788,"start":106962,"end":106973,"expression":"\\max(a,b)>C","context":217,"kind":"use","source":2},{"line":793,"start":108092,"end":108118,"expression":"\\Gcal=(p_0,\\ldots,p_{N-1})","context":218,"kind":"use","source":2},{"line":793,"start":108209,"end":108215,"expression":"z\\in D","context":218,"kind":"use","source":2},{"line":793,"start":108237,"end":108241,"expression":"f(z)","context":218,"kind":"use","source":2},{"line":793,"start":108347,"end":108348,"expression":"D","context":218,"kind":"use","source":2},{"line":798,"start":109129,"end":109130,"expression":"+","context":219,"kind":"use","source":2},{"line":798,"start":109134,"end":109138,"expression":"\\min","context":219,"kind":"use","source":2},{"line":798,"start":109142,"end":109146,"expression":"\\max","context":219,"kind":"use","source":2},{"line":798,"start":109170,"end":109171,"expression":"<","context":219,"kind":"use","source":2},{"line":798,"start":109175,"end":109178,"expression":"\\le","context":219,"kind":"use","source":2},{"line":798,"start":109182,"end":109183,"expression":">","context":219,"kind":"use","source":2},{"line":798,"start":109187,"end":109190,"expression":"\\ge","context":219,"kind":"use","source":2},{"line":798,"start":109198,"end":109199,"expression":"=","context":219,"kind":"use","source":2},{"line":798,"start":109268,"end":109271,"expression":"x,y","context":219,"kind":"use","source":2},{"line":798,"start":109409,"end":109428,"expression":"(x,y)\\in\\{1,2,3\\}^2","context":220,"kind":"use","source":2},{"line":805,"start":110431,"end":110444,"expression":"x\\in\\{1,2,3\\}","context":221,"kind":"use","source":2},{"line":805,"start":110458,"end":110467,"expression":"\\max(x,1)","context":221,"kind":"use","source":2},{"line":805,"start":110474,"end":110475,"expression":"x","context":221,"kind":"use","source":2},{"line":805,"start":110488,"end":110491,"expression":"x=0","context":221,"kind":"use","source":2},{"line":805,"start":110580,"end":110591,"expression":"\\max(x,0)=x","context":221,"kind":"use","source":2},{"line":814,"start":111906,"end":111931,"expression":"\\llbracket p\\rrbracket(z)","context":222,"kind":"introduction-passage","source":2},{"line":814,"start":111971,"end":111972,"expression":"z","context":222,"kind":"introduction-passage","source":2},{"line":814,"start":111980,"end":111984,"expression":"n(p)","context":222,"kind":"introduction-passage","source":2},{"line":815,"start":112106,"end":112231,"expression":"\n n(p)=n(q)\\quad\\Longrightarrow\\quad\n \\llbracket p\\rrbracket(z)=\\llbracket q\\rrbracket(z)\n \\quad\\text{for every admitted }z.\n","context":222,"kind":"introduction-passage","source":2},{"line":823,"start":112760,"end":112763,"expression":"p_i","context":223,"kind":"use","source":2},{"line":824,"start":112809,"end":112844,"expression":"\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n","context":224,"kind":"use","source":2},{"line":827,"start":112853,"end":112865,"expression":"\\rho(i)\\le i","context":225,"kind":"use","source":2},{"line":827,"start":112869,"end":112872,"expression":"p_i","context":226,"kind":"use","source":2},{"line":827,"start":112879,"end":112890,"expression":"p_{\\rho(i)}","context":227,"kind":"use","source":2},{"line":827,"start":112974,"end":112979,"expression":"\\Gcal","context":228,"kind":"use","source":2},{"line":830,"start":113489,"end":113496,"expression":"\\rho(i)","context":229,"kind":"use","source":2},{"line":830,"start":113508,"end":113509,"expression":"i","context":230,"kind":"use","source":2},{"line":830,"start":113534,"end":113546,"expression":"\\rho(i)\\le i","context":231,"kind":"use","source":2},{"line":839,"start":115223,"end":115226,"expression":"p_i","context":232,"kind":"use","source":2},{"line":839,"start":115260,"end":115263,"expression":"p_j","context":233,"kind":"use","source":2},{"line":839,"start":115301,"end":115304,"expression":"p_j","context":234,"kind":"use","source":2},{"line":839,"start":115325,"end":115328,"expression":"p_i","context":235,"kind":"use","source":2},{"line":839,"start":115369,"end":115372,"expression":"p_j","context":236,"kind":"use","source":2},{"line":839,"start":115472,"end":115475,"expression":"p_i","context":237,"kind":"use","source":2},{"line":839,"start":115539,"end":115542,"expression":"p_i","context":238,"kind":"use","source":2},{"line":839,"start":115585,"end":115588,"expression":"p_i","context":239,"kind":"use","source":2},{"line":847,"start":117667,"end":117680,"expression":"\\{0,1,3,5\\}^4","context":240,"kind":"use","source":2},{"line":847,"start":117724,"end":117737,"expression":"\\{0,1,2,3\\}^4","context":241,"kind":"use","source":2},{"line":873,"start":119995,"end":120010,"expression":"C_{\\rm prepare}","context":242,"kind":"use","source":2},{"line":873,"start":120144,"end":120147,"expression":"d_t","context":242,"kind":"use","source":2},{"line":873,"start":120183,"end":120184,"expression":"t","context":242,"kind":"use","source":2},{"line":874,"start":120261,"end":120298,"expression":"\n \\sum_{t=1}^{T}d_t>C_{\\rm prepare}.\n","context":242,"kind":"use","source":2},{"line":883,"start":121255,"end":121260,"expression":"x+0=x","context":243,"kind":"use","source":2},{"line":883,"start":121367,"end":121391,"expression":"a\\mathbin{\\oplus}b=a+b+1","context":243,"kind":"use","source":2},{"line":883,"start":121400,"end":121423,"expression":"x\\mathbin{\\oplus}0\\ne x","context":243,"kind":"use","source":2},{"line":889,"start":122149,"end":122150,"expression":"X","context":244,"kind":"use","source":2},{"line":889,"start":122193,"end":122207,"expression":"S_r\\subseteq X","context":244,"kind":"use","source":2},{"line":889,"start":122239,"end":122240,"expression":"r","context":244,"kind":"use","source":2},{"line":889,"start":122280,"end":122283,"expression":"w_x","context":244,"kind":"use","source":2},{"line":890,"start":122312,"end":122403,"expression":"\\label{eq:repair-weights}\n \\sum_{x\\in S_r}w_x\\le1\\quad\\text{for every permitted repair }r.\n","context":244,"kind":"use","source":2},{"line":896,"start":122801,"end":122802,"expression":"B","context":245,"kind":"use","source":2},{"line":896,"start":122813,"end":122814,"expression":"L","context":246,"kind":"use","source":2},{"line":896,"start":122858,"end":122859,"expression":"h","context":247,"kind":"use","source":2},{"line":897,"start":122897,"end":122928,"expression":"\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n","context":248,"kind":"use","source":2},{"line":900,"start":122963,"end":122968,"expression":"h\\ge1","context":249,"kind":"use","source":2},{"line":900,"start":123073,"end":123080,"expression":"2^{h-1}","context":250,"kind":"use","source":2},{"line":903,"start":123332,"end":123333,"expression":"B","context":251,"kind":"use","source":2},{"line":903,"start":123605,"end":123606,"expression":"h","context":252,"kind":"use","source":2},{"line":903,"start":123621,"end":123624,"expression":"2^h","context":253,"kind":"use","source":2},{"line":903,"start":123651,"end":123652,"expression":"h","context":254,"kind":"use","source":2},{"line":903,"start":123700,"end":123703,"expression":"h-1","context":255,"kind":"use","source":2},{"line":916,"start":126370,"end":126377,"expression":"\\{1,2\\}","context":256,"kind":"use","source":2},{"line":916,"start":126381,"end":126388,"expression":"\\{2,3\\}","context":256,"kind":"use","source":2},{"line":916,"start":126396,"end":126403,"expression":"\\{1,3\\}","context":256,"kind":"use","source":2},{"line":916,"start":126499,"end":126514,"expression":"w_1=w_2=w_3=1/2","context":256,"kind":"use","source":2},{"line":916,"start":126527,"end":126530,"expression":"3/2","context":256,"kind":"use","source":2},{"line":921,"start":127191,"end":127193,"expression":"2g","context":257,"kind":"use","source":2},{"line":921,"start":127231,"end":127232,"expression":"g","context":257,"kind":"use","source":2},{"line":921,"start":127248,"end":127249,"expression":"g","context":257,"kind":"use","source":2},{"line":921,"start":127518,"end":127519,"expression":"g","context":258,"kind":"use","source":2},{"line":921,"start":127531,"end":127538,"expression":"\\log_2g","context":259,"kind":"use","source":2},{"line":921,"start":127618,"end":127627,"expression":"1+\\log_2g","context":260,"kind":"use","source":2},{"line":921,"start":127661,"end":127665,"expression":"2g-1","context":261,"kind":"use","source":2},{"line":952,"start":131282,"end":131283,"expression":"K","context":262,"kind":"use","source":2},{"line":952,"start":131314,"end":131318,"expression":"A,A'","context":263,"kind":"use","source":2},{"line":952,"start":131356,"end":131360,"expression":"D,D'","context":264,"kind":"use","source":2},{"line":953,"start":131433,"end":131452,"expression":"\n (A-A')+(D-D')>K.\n","context":265,"kind":"use","source":2},{"line":959,"start":131858,"end":131861,"expression":"A+D","context":266,"kind":"use","source":2},{"line":959,"start":131890,"end":131897,"expression":"K+A'+D'","context":267,"kind":"use","source":2},{"line":959,"start":132006,"end":132007,"expression":"K","context":268,"kind":"use","source":2},{"line":977,"start":135069,"end":135087,"expression":"a=(a_1,\\ldots,a_d)","context":269,"kind":"use","source":2},{"line":977,"start":135317,"end":135318,"expression":"v","context":269,"kind":"use","source":2},{"line":977,"start":135371,"end":135391,"expression":"\\operatorname{pa}(v)","context":269,"kind":"use","source":2},{"line":977,"start":135458,"end":135484,"expression":"D_v\\subseteq\\{1,\\ldots,d\\}","context":270,"kind":"use","source":2},{"line":977,"start":135567,"end":135570,"expression":"w_v","context":271,"kind":"use","source":2},{"line":979,"start":135749,"end":135752,"expression":"D_v","context":272,"kind":"use","source":2},{"line":981,"start":136192,"end":136193,"expression":"a","context":273,"kind":"use","source":2},{"line":981,"start":136199,"end":136201,"expression":"a'","context":273,"kind":"use","source":2},{"line":981,"start":136293,"end":136299,"expression":"\\Delta","context":273,"kind":"use","source":2},{"line":981,"start":136427,"end":136428,"expression":"v","context":273,"kind":"use","source":2},{"line":982,"start":136441,"end":136533,"expression":"\n a|_{D_v}\\ne a'|_{D_v}\n \\quad\\text{or}\\quad\n \\operatorname{pa}(v)\\cap\\Delta\\ne\\varnothing.\n","context":273,"kind":"use","source":2},{"line":990,"start":137115,"end":137121,"expression":"\\Delta","context":274,"kind":"use","source":2},{"line":990,"start":137186,"end":137192,"expression":"\\Delta","context":275,"kind":"use","source":2},{"line":993,"start":137577,"end":137583,"expression":"\\Delta","context":276,"kind":"use","source":2},{"line":993,"start":137631,"end":137637,"expression":"\\Delta","context":277,"kind":"use","source":2},{"line":999,"start":138875,"end":138876,"expression":"u","context":278,"kind":"use","source":2},{"line":999,"start":138883,"end":138884,"expression":"v","context":278,"kind":"use","source":2},{"line":999,"start":138943,"end":138944,"expression":"u","context":278,"kind":"use","source":2},{"line":999,"start":139027,"end":139028,"expression":"v","context":278,"kind":"use","source":2},{"line":999,"start":139055,"end":139056,"expression":"u","context":278,"kind":"use","source":2},{"line":1038,"start":145966,"end":145971,"expression":"d\\ge1","context":279,"kind":"use","source":2},{"line":1038,"start":145997,"end":146019,"expression":"K\\in\\mathbb R_{\\ge0}^d","context":279,"kind":"use","source":2},{"line":1038,"start":146042,"end":146059,"expression":"s_i\\in\\mathbb R^d","context":279,"kind":"use","source":2},{"line":1038,"start":146122,"end":146123,"expression":"i","context":279,"kind":"use","source":2},{"line":1038,"start":146135,"end":146138,"expression":"n_i","context":279,"kind":"use","source":2},{"line":1039,"start":146221,"end":146275,"expression":"\\label{eq:vector-saving}\n \\Delta(n)=\\sum_i n_i s_i-K.\n","context":279,"kind":"use","source":2},{"line":1042,"start":146306,"end":146328,"expression":"w\\in\\mathbb R_{\\ge0}^d","context":280,"kind":"use","source":2},{"line":1042,"start":146419,"end":146434,"expression":"w\\cdot\\Delta(n)","context":281,"kind":"use","source":2},{"line":1045,"start":146666,"end":146685,"expression":"w\\cdot\\Delta(n)\\ge0","context":282,"kind":"use","source":2},{"line":1045,"start":146723,"end":146724,"expression":"w","context":283,"kind":"use","source":2},{"line":1045,"start":146742,"end":146751,"expression":"\\Delta(n)","context":284,"kind":"use","source":2},{"line":1045,"start":146851,"end":146854,"expression":"W_0","context":285,"kind":"use","source":2},{"line":1046,"start":146864,"end":146951,"expression":"\n \\min_{w\\in\\operatorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n","context":286,"kind":"use","source":2},{"line":1053,"start":147280,"end":147281,"expression":"j","context":287,"kind":"use","source":2},{"line":1053,"start":147301,"end":147302,"expression":"j","context":288,"kind":"use","source":2},{"line":1060,"start":148326,"end":148350,"expression":"\n (1552,\\ 1552,\\ -1868)\n","context":289,"kind":"use","source":2},{"line":1063,"start":148544,"end":148551,"expression":"(1,1,1)","context":290,"kind":"use","source":2},{"line":1063,"start":148591,"end":148598,"expression":"(1,0,1)","context":291,"kind":"use","source":2},{"line":1063,"start":148638,"end":148645,"expression":"(0,0,1)","context":292,"kind":"use","source":2},{"line":1066,"start":149057,"end":149064,"expression":"(1,1,1)","context":293,"kind":"use","source":2},{"line":1077,"start":150861,"end":150877,"expression":"\\mathcal H_{t-1}","context":294,"kind":"use","source":2},{"line":1077,"start":150924,"end":150925,"expression":"t","context":295,"kind":"use","source":2},{"line":1077,"start":150947,"end":150950,"expression":"A_t","context":296,"kind":"use","source":2},{"line":1077,"start":150966,"end":150967,"expression":"t","context":297,"kind":"use","source":2},{"line":1078,"start":151015,"end":151084,"expression":"\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n","context":298,"kind":"use","source":2},{"line":1082,"start":151106,"end":151119,"expression":"\\delta_t\\geq0","context":298,"kind":"use","source":2},{"line":1082,"start":151127,"end":151158,"expression":"\\sum_{t\\geq1}\\delta_t\\leq\\delta","context":298,"kind":"use","source":2},{"line":1083,"start":151168,"end":151212,"expression":"\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n","context":299,"kind":"use","source":2},{"line":1089,"start":151378,"end":151398,"expression":"\\Pr(A_t)\\leq\\delta_t","context":300,"kind":"use","source":2},{"line":1092,"start":151919,"end":151943,"expression":"\\delta_t=\\delta/[t(t+1)]","context":301,"kind":"use","source":2},{"line":1095,"start":152158,"end":152161,"expression":"n_t","context":302,"kind":"use","source":2},{"line":1095,"start":152282,"end":152294,"expression":"Y_i\\in[-1,1]","context":303,"kind":"use","source":2},{"line":1095,"start":152302,"end":152310,"expression":"\\Delta_t","context":304,"kind":"use","source":2},{"line":1095,"start":152343,"end":152359,"expression":"\\widehat\\Delta_t","context":305,"kind":"use","source":2},{"line":1096,"start":152430,"end":152521,"expression":"\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n","context":306,"kind":"use","source":2},{"line":1100,"start":152557,"end":152575,"expression":"\\Delta_t\\geq\\gamma","context":307,"kind":"use","source":2},{"line":1101,"start":152603,"end":152676,"expression":"\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n","context":308,"kind":"use","source":2},{"line":1104,"start":152824,"end":152827,"expression":"n_t","context":309,"kind":"use","source":2},{"line":1104,"start":152859,"end":152862,"expression":"n_t","context":310,"kind":"use","source":2},{"line":1109,"start":153607,"end":153608,"expression":"N","context":311,"kind":"use","source":2},{"line":1120,"start":155648,"end":155660,"expression":"Y_t\\in[-1,1]","context":312,"kind":"use","source":2},{"line":1120,"start":155796,"end":155810,"expression":"\\gamma\\in[0,1)","context":312,"kind":"use","source":2},{"line":1123,"start":155852,"end":155950,"expression":"\\label{eq:anytime-null}\n\\E[Y_t\\mid\\mathcal H_{t-1}]\\leq\\gamma\n\\quad\\text{for every audit step }t,\n","context":313,"kind":"use","source":2},{"line":1127,"start":155972,"end":155988,"expression":"\\mathcal H_{t-1}","context":313,"kind":"use","source":2},{"line":1127,"start":156121,"end":156127,"expression":"\\gamma","context":313,"kind":"use","source":2},{"line":1129,"start":156305,"end":156314,"expression":"\\lambda_t","context":314,"kind":"use","source":2},{"line":1129,"start":156334,"end":156350,"expression":"\\mathcal H_{t-1}","context":314,"kind":"use","source":2},{"line":1129,"start":156359,"end":156384,"expression":"\\lambda_t\\leq1/(1+\\gamma)","context":314,"kind":"use","source":2},{"line":1130,"start":156410,"end":156498,"expression":"\\label{eq:anytime-capital}\nE_0=1,\\qquad E_t=E_{t-1}\\bigl(1+\\lambda_t(Y_t-\\gamma)\\bigr).\n","context":314,"kind":"use","source":2},{"line":1133,"start":156527,"end":156530,"expression":"E_t","context":314,"kind":"use","source":2},{"line":1136,"start":156857,"end":156860,"expression":"E_t","context":315,"kind":"use","source":2},{"line":1136,"start":156907,"end":156921,"expression":"\\delta\\in(0,1)","context":316,"kind":"use","source":2},{"line":1137,"start":156926,"end":156983,"expression":"\n\\Pr\\left(\\sup_{t\\geq0}E_t\\geq1/\\delta\\right)\\leq\\delta.\n","context":317,"kind":"use","source":2},{"line":1140,"start":157061,"end":157069,"expression":"1/\\delta","context":318,"kind":"use","source":2},{"line":1143,"start":157286,"end":157311,"expression":"Y_t-\\gamma\\geq-(1+\\gamma)","context":319,"kind":"use","source":2},{"line":1144,"start":157423,"end":157534,"expression":"\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n","context":320,"kind":"use","source":2},{"line":1149,"start":157542,"end":157546,"expression":"\\tau","context":321,"kind":"use","source":2},{"line":1149,"start":157595,"end":157596,"expression":"N","context":322,"kind":"use","source":2},{"line":1149,"start":157623,"end":157635,"expression":"\\tau\\wedge N","context":323,"kind":"use","source":2},{"line":1149,"start":157733,"end":157747,"expression":"\\{\\tau\\leq N\\}","context":324,"kind":"use","source":2},{"line":1149,"start":157772,"end":157780,"expression":"1/\\delta","context":325,"kind":"use","source":2},{"line":1149,"start":157829,"end":157854,"expression":"\\Pr(\\tau\\leq N)\\leq\\delta","context":326,"kind":"use","source":2},{"line":1149,"start":157866,"end":157867,"expression":"N","context":327,"kind":"use","source":2},{"line":1156,"start":158521,"end":158524,"expression":"Y_t","context":328,"kind":"use","source":2},{"line":1156,"start":158557,"end":158565,"expression":"\\gamma=0","context":328,"kind":"use","source":2},{"line":1156,"start":158576,"end":158587,"expression":"\\lambda_t=1","context":328,"kind":"use","source":2},{"line":1156,"start":158670,"end":158673,"expression":"E_t","context":328,"kind":"use","source":2},{"line":1156,"start":158855,"end":158860,"expression":"11/16","context":329,"kind":"use","source":2},{"line":1156,"start":158895,"end":158898,"expression":"1/4","context":330,"kind":"use","source":2},{"line":1156,"start":158950,"end":158955,"expression":"[0,1]","context":331,"kind":"use","source":2},{"line":1164,"start":159858,"end":159859,"expression":"j","context":332,"kind":"use","source":2},{"line":1164,"start":159966,"end":159977,"expression":"E_0^{(j)}=1","context":332,"kind":"use","source":2},{"line":1164,"start":160001,"end":160009,"expression":"\\delta_j","context":332,"kind":"use","source":2},{"line":1164,"start":160016,"end":160040,"expression":"\\sum_j\\delta_j\\leq\\delta","context":332,"kind":"use","source":2},{"line":1164,"start":160140,"end":160146,"expression":"\\delta","context":332,"kind":"use","source":2},{"line":1174,"start":161619,"end":161718,"expression":"\nC=(C_{\\rm model},C_{\\rm search},C_{\\rm validate},C_{\\rm install},\nC_{\\rm execute},C_{\\rm store}).\n","context":333,"kind":"use","source":2},{"line":1180,"start":162400,"end":162412,"expression":"n(n+\\log_2n)","context":334,"kind":"use","source":2},{"line":1183,"start":162855,"end":162861,"expression":"C\\geq0","context":335,"kind":"use","source":2},{"line":1183,"start":162958,"end":162961,"expression":"d>0","context":335,"kind":"use","source":2},{"line":1183,"start":162978,"end":162979,"expression":"N","context":335,"kind":"use","source":2},{"line":1183,"start":163073,"end":163077,"expression":"Nd-C","context":336,"kind":"use","source":2},{"line":1183,"start":163128,"end":163133,"expression":"N>C/d","context":337,"kind":"use","source":2},{"line":1186,"start":163227,"end":163229,"expression":"Nd","context":338,"kind":"use","source":2},{"line":1189,"start":163550,"end":163551,"expression":"d","context":339,"kind":"use","source":2},{"line":1193,"start":163842,"end":163843,"expression":"N","context":340,"kind":"use","source":2},{"line":1193,"start":163925,"end":163931,"expression":"B\\geq1","context":340,"kind":"use","source":2},{"line":1193,"start":164046,"end":164057,"expression":"1-(1-1/N)^B","context":340,"kind":"use","source":2},{"line":1203,"start":165456,"end":165457,"expression":"Z","context":341,"kind":"use","source":2},{"line":1205,"start":165872,"end":165873,"expression":"z","context":342,"kind":"use","source":2},{"line":1205,"start":165916,"end":165920,"expression":"A(z)","context":342,"kind":"use","source":2},{"line":1205,"start":165954,"end":165964,"expression":"\\Post(z,a)","context":342,"kind":"use","source":2},{"line":1205,"start":166105,"end":166121,"expression":"\\Safe\\subseteq Z","context":342,"kind":"use","source":2},{"line":1205,"start":166313,"end":166314,"expression":"Z","context":343,"kind":"use","source":2},{"line":1208,"start":166383,"end":166467,"expression":"\nK_0=\\Safe,\\qquad\nK_{n+1}=\\{z\\in K_n:\\exists a\\in A(z),\\ \\Post(z,a)\\subseteq K_n\\}.\n","context":344,"kind":"use","source":2},{"line":1212,"start":166524,"end":166534,"expression":"\\mathcal K","context":344,"kind":"use","source":2},{"line":1214,"start":166737,"end":166756,"expression":"p(z,a,z')\\in\\{0,1\\}","context":345,"kind":"use","source":2},{"line":1215,"start":166972,"end":167235,"expression":"\\label{eq:progress}\nW_{0,0}&=\\mathcal K, & W_{0,k}&=\\varnothing\\quad(k>0),\\\\\nW_{h,k}&=\\{z\\in\\mathcal K:\\exists a\\in A(z),\\ \\Post(z,a)\\subseteq\\mathcal K,\\nonumber\\\\[-2pt]\n&\\hspace{34mm} z'\\in W_{h-1,\\max\\{0,k-p(z,a,z')\\}}\n\\quad\\forall z'\\in\\Post(z,a)\\}.\\nonumber\n","context":346,"kind":"use","source":2},{"line":1221,"start":167253,"end":167254,"expression":"h","context":347,"kind":"use","source":2},{"line":1221,"start":167297,"end":167298,"expression":"k","context":347,"kind":"use","source":2},{"line":1224,"start":167549,"end":167556,"expression":"W_{h,k}","context":348,"kind":"use","source":2},{"line":1224,"start":167636,"end":167646,"expression":"\\mathcal K","context":349,"kind":"use","source":2},{"line":1224,"start":167673,"end":167674,"expression":"k","context":350,"kind":"use","source":2},{"line":1224,"start":167708,"end":167709,"expression":"h","context":351,"kind":"use","source":2},{"line":1224,"start":167793,"end":167798,"expression":"(h,k)","context":352,"kind":"use","source":2},{"line":1227,"start":168027,"end":168037,"expression":"\\mathcal K","context":353,"kind":"use","source":2},{"line":1227,"start":168109,"end":168112,"expression":"h-1","context":354,"kind":"use","source":2},{"line":1227,"start":168128,"end":168129,"expression":"z","context":355,"kind":"use","source":2},{"line":1227,"start":168232,"end":168242,"expression":"\\mathcal K","context":356,"kind":"use","source":2},{"line":1227,"start":168304,"end":168317,"expression":"\\max\\{0,k-p\\}","context":357,"kind":"use","source":2},{"line":1236,"start":169907,"end":169908,"expression":"h","context":358,"kind":"use","source":2},{"line":1236,"start":169933,"end":169934,"expression":"h","context":359,"kind":"use","source":2},{"line":1236,"start":170008,"end":170009,"expression":"h","context":360,"kind":"use","source":2},{"line":1240,"start":170497,"end":170502,"expression":"\\Safe","context":361,"kind":"use","source":2},{"line":1248,"start":171612,"end":171623,"expression":"R_t\\in[a,b]","context":362,"kind":"use","source":2},{"line":1248,"start":171682,"end":171690,"expression":"\\gamma>0","context":362,"kind":"use","source":2},{"line":1249,"start":171807,"end":171853,"expression":"\n\\left\\lfloor\\frac{b-a}{\\gamma}\\right\\rfloor.\n","context":362,"kind":"use","source":2},{"line":1255,"start":172032,"end":172033,"expression":"k","context":363,"kind":"use","source":2},{"line":1255,"start":172086,"end":172119,"expression":"R_t\\geq R_0+k\\gamma\\geq a+k\\gamma","context":364,"kind":"use","source":2},{"line":1255,"start":172129,"end":172138,"expression":"R_t\\leq b","context":365,"kind":"use","source":2},{"line":1255,"start":172220,"end":172226,"expression":"2^{-t}","context":366,"kind":"use","source":2},{"line":1255,"start":172233,"end":172239,"expression":"t\\geq1","context":367,"kind":"use","source":2},{"line":1260,"start":172775,"end":172778,"expression":"g_t","context":368,"kind":"use","source":2},{"line":1260,"start":172812,"end":172817,"expression":"c_t>0","context":368,"kind":"use","source":2},{"line":1260,"start":172937,"end":172944,"expression":"g_t/c_t","context":368,"kind":"use","source":2},{"line":1260,"start":172956,"end":172959,"expression":"g_t","context":368,"kind":"use","source":2},{"line":1260,"start":172999,"end":173002,"expression":"c_t","context":368,"kind":"use","source":2},{"line":1270,"start":174324,"end":174330,"expression":"r_\\phi","context":369,"kind":"use","source":2},{"line":1270,"start":174360,"end":174372,"expression":"G_{\\theta,a}","context":369,"kind":"use","source":2},{"line":1270,"start":174405,"end":174415,"expression":"\\pi_\\omega","context":369,"kind":"use","source":2},{"line":1270,"start":174458,"end":174464,"expression":"p_\\psi","context":369,"kind":"use","source":2},{"line":1270,"start":174549,"end":174572,"expression":"\\phi,\\theta,\\omega,\\psi","context":369,"kind":"use","source":2},{"line":1277,"start":176031,"end":176032,"expression":"D","context":370,"kind":"use","source":2},{"line":1277,"start":176077,"end":176080,"expression":"F_a","context":370,"kind":"use","source":2},{"line":1277,"start":176088,"end":176096,"expression":"q:D\\to Z","context":370,"kind":"use","source":2},{"line":1278,"start":176113,"end":176140,"expression":"\nqF_a=G_aq\\quad\\text{on }D\n","context":370,"kind":"use","source":2},{"line":1281,"start":176163,"end":176164,"expression":"a","context":370,"kind":"use","source":2},{"line":1281,"start":176201,"end":176210,"expression":"y=\\bar yq","context":370,"kind":"use","source":2},{"line":1281,"start":176216,"end":176217,"expression":"D","context":370,"kind":"use","source":2},{"line":1281,"start":176258,"end":176259,"expression":"w","context":370,"kind":"use","source":2},{"line":1282,"start":176264,"end":176299,"expression":"\nqF_w=G_wq,\\qquad yF_w=\\bar yG_wq.\n","context":370,"kind":"use","source":2},{"line":1287,"start":176482,"end":176505,"expression":"qF_aF_w=G_aqF_w=G_aG_wq","context":371,"kind":"use","source":2},{"line":1287,"start":176569,"end":176575,"expression":"\\bar y","context":372,"kind":"use","source":2},{"line":1290,"start":176692,"end":176693,"expression":"q","context":373,"kind":"use","source":2},{"line":1290,"start":176699,"end":176702,"expression":"G_a","context":373,"kind":"use","source":2},{"line":1307,"start":179623,"end":179628,"expression":"n=2^k","context":374,"kind":"use","source":2},{"line":1307,"start":179709,"end":179710,"expression":"k","context":374,"kind":"use","source":2},{"line":1310,"start":180151,"end":180152,"expression":"k","context":375,"kind":"use","source":2},{"line":1310,"start":180241,"end":180245,"expression":"2n-1","context":376,"kind":"use","source":2},{"line":1310,"start":180338,"end":180341,"expression":"n-1","context":377,"kind":"use","source":2},{"line":1310,"start":180517,"end":180520,"expression":"k+1","context":378,"kind":"use","source":2},{"line":1310,"start":180527,"end":180528,"expression":"n","context":379,"kind":"use","source":2},{"line":1313,"start":180593,"end":180594,"expression":"k","context":380,"kind":"use","source":2},{"line":1313,"start":180612,"end":180617,"expression":"2^k=n","context":381,"kind":"use","source":2},{"line":1313,"start":180766,"end":180767,"expression":"k","context":382,"kind":"use","source":2},{"line":1313,"start":180787,"end":180788,"expression":"k","context":383,"kind":"use","source":2},{"line":1313,"start":180864,"end":180865,"expression":"n","context":384,"kind":"use","source":2},{"line":1313,"start":180879,"end":180882,"expression":"n-1","context":385,"kind":"use","source":2},{"line":1313,"start":181022,"end":181025,"expression":"n-1","context":386,"kind":"use","source":2},{"line":1313,"start":181223,"end":181227,"expression":"2n-1","context":387,"kind":"use","source":2},{"line":1318,"start":181505,"end":181508,"expression":"n-1","context":388,"kind":"use","source":2},{"line":1318,"start":181570,"end":181578,"expression":"\\log_2 n","context":388,"kind":"use","source":2},{"line":1318,"start":181655,"end":181659,"expression":"2n-1","context":388,"kind":"use","source":2},{"line":1322,"start":182052,"end":182055,"expression":"n=8","context":389,"kind":"use","source":2},{"line":1327,"start":182985,"end":182988,"expression":"3/8","context":390,"kind":"use","source":2},{"line":1400,"start":195964,"end":195967,"expression":"n=8","context":391,"kind":"use","source":2},{"line":1400,"start":195988,"end":195991,"expression":"j=5","context":391,"kind":"use","source":2},{"line":1400,"start":196019,"end":196022,"expression":"101","context":391,"kind":"use","source":2},{"line":1401,"start":196067,"end":196131,"expression":"\nA_0=\\{1,3,5,7\\},\\qquad A_1=\\{2,3,6,7\\},\\qquad A_2=\\{4,5,6,7\\}.\n","context":391,"kind":"use","source":2},{"line":1405,"start":196236,"end":196322,"expression":"\n\\{0,\\ldots,7\\}\\xrightarrow{1}\\{1,3,5,7\\}\n\\xrightarrow{0}\\{1,5\\}\\xrightarrow{1}\\{5\\}.\n","context":391,"kind":"use","source":2},{"line":1409,"start":196349,"end":196352,"expression":"r_5","context":392,"kind":"use","source":2},{"line":1413,"start":196882,"end":196902,"expression":"\\sigma(i)=3i+1\\pmod8","context":393,"kind":"use","source":2},{"line":1413,"start":196934,"end":196945,"expression":"\\sigma(5)=0","context":393,"kind":"use","source":2},{"line":1413,"start":196968,"end":196979,"expression":"\\sigma(A_r)","context":393,"kind":"use","source":2},{"line":1413,"start":197014,"end":197028,"expression":"r'_{\\sigma(5)}","context":393,"kind":"use","source":2},{"line":1416,"start":197404,"end":197413,"expression":"\\{0,1,2\\}","context":394,"kind":"use","source":2},{"line":1416,"start":197430,"end":197431,"expression":"a","context":394,"kind":"use","source":2},{"line":1416,"start":197668,"end":197675,"expression":"\\{2,3\\}","context":394,"kind":"use","source":2}],"entries":[{"id":"symbol-df7e70e5021544f4","tex":"B","aliases":"B","display":"B","kind":"notation-table","occurrences":[4,5,6,10,24,95,26,27,30,31,96,98,99,101,104,106,107,108,113,131,132,135,138,139,140,180,181,182,183,184,185,324,334,335,336,337,338,340,341,342,343,344,347,349,350,414,417,420,549,550]},{"id":"symbol-3ea71b67efc35822","tex":"B_0","aliases":"B_0","display":"B_0","kind":"notation-table","occurrences":[6,10,95,96,98,99,101,104,106,107,108,113,180,181,182,183,184,324]},{"id":"symbol-3f79bb7b435b0532","tex":"e","aliases":"e","display":"e","kind":"notation-table","occurrences":[7,87,88,89,90,91,96,98,107,127,128,129,132,137,178,179,185,402,405,539]},{"id":"symbol-f67ab10ad4e4c531","tex":"F","aliases":"F","display":"F","kind":"notation-table","occurrences":[10,64,68,71,596,598,603,604]},{"id":"symbol-e982a068c0aac4ff","tex":"F_h","aliases":"F_h","display":"F_h","kind":"notation-table","occurrences":[10]},{"id":"symbol-aaa9402664f1a41f","tex":"h","aliases":"h","display":"h","kind":"notation-table","occurrences":[10,11,15,17,103,104,107,109,110,114,115,182,416,417,418,419,421,422,423,424,539,560,561,563,566,567,569,573,574,575]},{"id":"symbol-9da7de32ab512a2d","tex":"\\mathcal K","aliases":"K mathcal","display":"\\mathcal K","kind":"notation-table","occurrences":[14,558,560,564,568,571]},{"id":"symbol-8254c329a92850f6","tex":"k","aliases":"k","display":"k","kind":"notation-table","occurrences":[15,16,214,215,217,219,220,222,225,226,560,562,563,565,567,572,580,581,608,609,610,613,615,616,617,618]},{"id":"symbol-72dfcfb0c470ac25","tex":"L","aliases":"L","display":"L","kind":"notation-table","occurrences":[2,5,21,26,27,30,31,32,33,34,35,36,37,415,417]},{"id":"symbol-5c62e091b8c0565f","tex":"P","aliases":"P","display":"P","kind":"notation-table","occurrences":[9,52,53,56,66,68,69,70,74,76,79,80,82,83,101,102]},{"id":"symbol-454349e422f05297","tex":"r","aliases":"r","display":"r","kind":"notation-table","occurrences":[8,44,144,154,259,262,264,267,312,315,316,335,402,405,410,411,413,498,539,590,633,636,637]},{"id":"symbol-3bce33d4a711b894","tex":"\\Rcal","aliases":"Rcal R mathcal","display":"{\\mathcal R}","kind":"notation-table","occurrences":[5,27,30,31,191]},{"id":"symbol-bdc7ac27c66dc39c","tex":"\\Rcal_\\Xi","aliases":"Rcal_Ξ R_Ξ Rcal Xi mathcal Xi","display":"{\\mathcal R}_\\Xi","kind":"notation-table","occurrences":[5,27,30,31]},{"id":"symbol-8de0b3c47f112c59","tex":"S","aliases":"S","display":"S","kind":"notation-table","occurrences":[0,25,19,26,27,37,410,413]},{"id":"symbol-a25513c7e0f6eaa8","tex":"U","aliases":"U","display":"U","kind":"notation-table","occurrences":[1,5,20,26,27,30,31,32,33,34,35,36,37,157,170,187,188]},{"id":"symbol-fcb5f40df9be6bae","tex":"W","aliases":"W","display":"W","kind":"notation-table","occurrences":[15,101,475,476,560,563]},{"id":"symbol-67fbcb7a77b5ea1b","tex":"W_{h,k}","aliases":"W_h,k","display":"W_{h,k}","kind":"notation-table","occurrences":[15,560,563]},{"id":"symbol-4b68ab3847feda7d","tex":"X","aliases":"X","display":"X","kind":"notation-table","occurrences":[6,84,88,89,94,95,63,64,69,72,100,125,131,136,156,175,409,410]},{"id":"symbol-c645657d025cc351","tex":"\\Xi","aliases":"Ξ Xi","display":"\\Xi","kind":"notation-table","occurrences":[3,5,22,27,28,30,31,37,40]},{"id":"symbol-9c3245dfb4ac54c1","tex":"\\sigma","aliases":"σ sigma","display":"\\sigma","kind":"notation-table","occurrences":[12,175,179,181,182,184,185,634,635,636,637]},{"id":"symbol-55d1823545b5b4c3","tex":"\\tau","aliases":"τ tau","display":"\\tau","kind":"notation-table","occurrences":[13,25,26,27,176,178,179,185,520,522,523,525]},{"id":"symbol-39c3d5a5ec4b06af","tex":"\\subseteq","aliases":"⊆ subseteq","display":"\\subseteq","kind":"notation-table","occurrences":[6,88,94,95,96,99,106,107,131,335,410,447,555,557,560]},{"id":"symbol-09b0b2df35128655","tex":"\\mathcal A","aliases":"A mathcal","display":"\\mathcal A","kind":"definition-passage","occurrences":[86,177]},{"id":"symbol-ca978112ca1bbdca","tex":"a","aliases":"a","display":"a","kind":"definition-passage","occurrences":[92,93,94,57,60,62,99,106,116,130,133,134,137,179,248,250,254,256,269,272,274,275,276,282,283,296,297,299,300,303,304,305,307,308,309,311,312,313,315,318,319,320,321,322,359,402,405,407,444,450,451,454,539,554,557,559,560,577,579,581,591,596,598,599,604,607,639]},{"id":"symbol-2e7d2c03a9507ae2","tex":"c","aliases":"c","display":"c","kind":"definition-passage","occurrences":[91,93,57,58,59,61,62,106,107,241,242,539,586,587,589]},{"id":"symbol-d9494e21b365d43b","tex":"c_a","aliases":"c_a","display":"c_a","kind":"definition-passage","occurrences":[93,106]},{"id":"symbol-2a62ce5b7390bdd7","tex":"c_e","aliases":"c_e","display":"c_e","kind":"definition-passage","occurrences":[91,107]},{"id":"symbol-3f39d5c348e5b79d","tex":"D","aliases":"D","display":"D","kind":"definition-passage","occurrences":[88,89,90,96,107,127,132,179,239,240,242,355,358,361,363,439,440,441,442,447,449,454,595,597,598,601]},{"id":"symbol-3b858f3e8665ea65","tex":"D_e","aliases":"D_e","display":"D_e","kind":"definition-passage","occurrences":[88,89,90,96,107,179]},{"id":"symbol-060ae96635fef1f6","tex":"\\mathcal E","aliases":"E mathcal","display":"\\mathcal E","kind":"definition-passage","occurrences":[85,176]},{"id":"symbol-252f10c83610ebca","tex":"f","aliases":"f","display":"f","kind":"definition-passage","occurrences":[89,55,56,98,128,132,137,179,261,262,265,266,268,362]},{"id":"symbol-48dc92a98cff75ec","tex":"f_e","aliases":"f_e","display":"f_e","kind":"definition-passage","occurrences":[89,98,179]},{"id":"symbol-333e0a1e27815d0c","tex":"G","aliases":"G","display":"G","kind":"definition-passage","occurrences":[94,99,106,130,134,137,179,335,591,598,603,604,607]},{"id":"symbol-8d428b98976a03b3","tex":"G_a","aliases":"G_a","display":"G_a","kind":"definition-passage","occurrences":[94,99,106,179,598,604,607]},{"id":"symbol-a03ffaed013c4d19","tex":"\\mathcal H","aliases":"H mathcal","display":"\\mathcal H","kind":"definition-passage","occurrences":[23,27,484,488,498,506,507,510,519]},{"id":"symbol-6da43b944e494e88","tex":"J","aliases":"J","display":"J","kind":"definition-passage","occurrences":[25,27]},{"id":"symbol-ac50e45a798d4049","tex":"J_\\tau","aliases":"J_τ tau","display":"J_\\tau","kind":"definition-passage","occurrences":[25,27]},{"id":"symbol-c4694f2e93d5c4e7","tex":"O","aliases":"O","display":"O","kind":"definition-passage","occurrences":[90,178]},{"id":"symbol-65c74c15a686187b","tex":"o","aliases":"o","display":"o","kind":"definition-passage","occurrences":[90,55,97,98,107,129,132,137,143,154,179,185,539]},{"id":"symbol-3554f05e6897e490","tex":"O_e","aliases":"O_e","display":"O_e","kind":"definition-passage","occurrences":[90,178]},{"id":"symbol-b2bdbcfb9c9cfc41","tex":"o_e","aliases":"o_e","display":"o_e","kind":"definition-passage","occurrences":[90,98,179]},{"id":"symbol-1e6eb42c12591791","tex":"\\to","aliases":"→ to","display":"\\to","kind":"definition-passage","occurrences":[89,90,64,175,176,177,178,259,260,261,597]},{"id":"symbol-729f2cd8398e9960","tex":"\\in","aliases":"∈ in","display":"\\in","kind":"definition-passage","occurrences":[25,41,43,55,98,107,132,134,135,137,141,142,239,280,361,373,374,413,417,465,466,470,476,495,504,505,515,557,559,560,577]},{"id":"symbol-2dd02f064b00c6d5","tex":"\\geq","aliases":"≥ geq","display":"\\geq","kind":"definition-passage","occurrences":[91,93,109,163,167,191,197,198,201,208,213,215,227,241,251,279,282,301,303,315,489,490,491,498,499,500,516,518,541,549,581,584]},{"id":"symbol-380918b946a52664","tex":"=","aliases":"=","display":"=","kind":"introduction-passage","occurrences":[27,382,37,45,46,47,48,57,58,98,106,107,132,136,137,143,145,179,182,185,196,211,212,214,219,252,256,262,272,279,299,312,317,322,324,338,347,348,349,350,352,354,360,371,377,378,384,405,406,407,428,444,469,476,493,512,519,528,529,535,539,557,560,598,600,603,604,608,616,626,628,629,631,634,635]},{"id":"symbol-fb7427a95ad8c5b4","tex":"\\E","aliases":"E mathbb","display":"{\\mathbb E}","kind":"introduction-passage","occurrences":[27,56,506,519]},{"id":"symbol-08ef244830e2cec4","tex":"\\E_{(S,\\tau)\\sim\\mathcal H}","aliases":"E_(S,τ)∼ H E tau sim mathcal mathbb tau sim mathcal","display":"{\\mathbb E}_{(S,\\tau)\\sim\\mathcal H}","kind":"introduction-passage","occurrences":[27]},{"id":"symbol-1e8b482a0cdcae31","tex":"\\llbracket","aliases":"llbracket","display":null,"kind":"introduction-passage","occurrences":[379,382]},{"id":"symbol-e0987528a0170db9","tex":"\\Longrightarrow","aliases":"Longrightarrow","display":"\\Longrightarrow","kind":"introduction-passage","occurrences":[382]},{"id":"symbol-1b16b1df538ba12d","tex":"n","aliases":"n","display":"n","kind":"introduction-passage","occurrences":[381,382,201,202,203,204,205,208,213,214,219,221,223,224,226,384,468,469,471,472,474,476,494,498,500,501,502,539,540,557,608,611,612,614,616,619,620,621,622,623,624,625,626,628]},{"id":"symbol-148de9c5a7a44d19","tex":"p","aliases":"p","display":"p","kind":"introduction-passage","occurrences":[379,381,382,360,383,384,386,387,392,393,394,395,396,397,398,399,402,405,559,560,572,593]},{"id":"symbol-8e35c2cd3bf6641b","tex":"q","aliases":"q","display":"q","kind":"introduction-passage","occurrences":[382,65,67,216,218,219,221,224,597,598,600,603,604,606]},{"id":"symbol-cbe1f992df7dd288","tex":"\\rrbracket","aliases":"rrbracket","display":null,"kind":"introduction-passage","occurrences":[379,382]},{"id":"symbol-594e519ae499312b","tex":"z","aliases":"z","display":"z","kind":"introduction-passage","occurrences":[379,380,382,361,362,552,553,554,557,559,560,570]},{"id":"symbol-616522c60d7b122c","tex":"\\sim","aliases":"∼ sim","display":"\\sim","kind":"introduction-passage","occurrences":[27]},{"id":"symbol-684888c0ebb17f37","tex":"*","aliases":"*","display":"*","kind":"use","occurrences":[166,167,168,170,171,172]},{"id":"symbol-a318c24216defe20","tex":"+","aliases":"+","display":"+","kind":"use","occurrences":[46,53,56,68,80,107,164,196,200,203,211,213,219,242,283,304,319,364,406,407,435,440,441,442,493,511,512,518,519,540,557,581,613,634]},{"id":"symbol-c0eca90b4aa4b1d1","tex":"\\mathbf1","aliases":"1 mathbf","display":"\\mathbf1","kind":"use","occurrences":[43,299]},{"id":"symbol-559aead08264d579","tex":"A","aliases":"A","display":"A","kind":"use","occurrences":[42,43,438,440,441,442,486,488,491,492,553,557,560,631,636]},{"id":"symbol-06f7344d1b73ecc4","tex":"A'","aliases":"A'","display":"A'","kind":"use","occurrences":[438,440,442]},{"id":"symbol-7226e2825e376081","tex":"\\mathcal A'","aliases":"A' mathcal","display":"\\mathcal A'","kind":"use","occurrences":[177]},{"id":"symbol-a80c8a675a0e6f40","tex":"a'","aliases":"a'","display":"a'","kind":"use","occurrences":[451,454]},{"id":"symbol-993220b6cfa76019","tex":"A_0","aliases":"A_0","display":"A_0","kind":"use","occurrences":[631]},{"id":"symbol-47b34af59b56e555","tex":"A_1","aliases":"A_1","display":"A_1","kind":"use","occurrences":[631]},{"id":"symbol-1426674c44e8876f","tex":"a_1","aliases":"a_1","display":"a_1","kind":"use","occurrences":[444]},{"id":"symbol-f75dd322f11ad636","tex":"A_2","aliases":"A_2","display":"A_2","kind":"use","occurrences":[631]},{"id":"symbol-5a8e57a002dc32f4","tex":"a_d","aliases":"a_d","display":"a_d","kind":"use","occurrences":[444]},{"id":"symbol-da62b0f97d94856e","tex":"A_r","aliases":"A_r","display":"A_r","kind":"use","occurrences":[636]},{"id":"symbol-82821bc2124f1658","tex":"A_t","aliases":"A_t","display":"A_t","kind":"use","occurrences":[486,488,491,492]},{"id":"symbol-3e23e8160039594a","tex":"b","aliases":"b","display":"b","kind":"use","occurrences":[58,60,61,111,112,118,199,200,249,250,254,256,269,272,274,275,276,282,283,297,298,299,300,303,304,305,307,308,310,311,312,314,315,318,319,320,321,322,359,407,577,579,582]},{"id":"symbol-0071df57536df27e","tex":"B'_{\\tau(e),\\eta_e(o)}","aliases":"B'_τ(e),η_e(o) tau eta","display":"B'_{\\tau(e),\\eta_e(o)}","kind":"use","occurrences":[185]},{"id":"symbol-f94c2a70b1030a1a","tex":"B_{e,o}","aliases":"B_e,o","display":"B_{e,o}","kind":"use","occurrences":[98,107,132,185]},{"id":"symbol-fcd1d016a0b9e4bf","tex":"\\bar y","aliases":"bar y bar","display":"\\bar y","kind":"use","occurrences":[600,603,605]},{"id":"symbol-6b23c0d5f35d1b11","tex":"C","aliases":"C","display":"C","kind":"use","occurrences":[252,253,255,270,271,272,275,276,280,281,284,298,299,300,301,302,306,307,308,309,310,311,313,314,315,317,320,321,322,327,328,329,330,359,402,405,539,541,544,545]},{"id":"symbol-ebd1b51c8a60881c","tex":"C_{\\rm execute}","aliases":"C_rm execute rm","display":"C_{\\rm execute}","kind":"use","occurrences":[539]},{"id":"symbol-f1ce996935a45a50","tex":"C_{\\rm install}","aliases":"C_rm install rm","display":"C_{\\rm install}","kind":"use","occurrences":[539]},{"id":"symbol-b1fc30ad0cce6444","tex":"C_{\\rm model}","aliases":"C_rm model rm","display":"C_{\\rm model}","kind":"use","occurrences":[539]},{"id":"symbol-c3ebead36d730927","tex":"C_{\\rm prepare}","aliases":"C_rm prepare rm","display":"C_{\\rm prepare}","kind":"use","occurrences":[402,405]},{"id":"symbol-21a5ce52f6366e9c","tex":"C_{\\rm search}","aliases":"C_rm search rm","display":"C_{\\rm search}","kind":"use","occurrences":[539]},{"id":"symbol-7a5ef302e1b9bf69","tex":"C_{\\rm store}","aliases":"C_rm store rm","display":"C_{\\rm store}","kind":"use","occurrences":[539]},{"id":"symbol-7792cf3103d96f1c","tex":"C_{\\rm validate}","aliases":"C_rm validate rm","display":"C_{\\rm validate}","kind":"use","occurrences":[539]},{"id":"symbol-c99be13da294dc50","tex":"c_t","aliases":"c_t","display":"c_t","kind":"use","occurrences":[586,587,589]},{"id":"symbol-2eff86b42c380025","tex":"\\mathrm{compare}","aliases":"compare mathrm","display":"\\mathrm{compare}","kind":"use","occurrences":[352]},{"id":"symbol-491630d7018dba28","tex":"\\operatorname{conv}","aliases":"conv operatorname","display":"\\operatorname{conv}","kind":"use","occurrences":[476]},{"id":"symbol-18ac3e7343f01689","tex":"d","aliases":"d","display":"d","kind":"use","occurrences":[197,200,206,207,208,243,244,245,251,252,317,324,325,326,339,341,345,346,348,403,405,444,447,464,465,466,470,539,542,544,545,546,547]},{"id":"symbol-27513713e8c146a0","tex":"D'","aliases":"D'","display":"D'","kind":"use","occurrences":[439,440,442]},{"id":"symbol-32f4f8a168125618","tex":"D'_{\\tau(e)}","aliases":"D'_τ(e) tau","display":"D'_{\\tau(e)}","kind":"use","occurrences":[179]},{"id":"symbol-04edc0a682114fdd","tex":"D_0","aliases":"D_0","display":"D_0","kind":"use","occurrences":[239,240,242]},{"id":"symbol-86bc40a9dfa0a3bf","tex":"d_0","aliases":"d_0","display":"d_0","kind":"use","occurrences":[324,325,326]},{"id":"symbol-99e5f8e078f9fe31","tex":"D_1","aliases":"D_1","display":"D_1","kind":"use","occurrences":[239,240,242]},{"id":"symbol-70f89149d18a83e8","tex":"D_e^\\theta","aliases":"D_e^θ theta","display":"D_e^\\theta","kind":"use","occurrences":[127,132]},{"id":"symbol-b377ac1e0c05ff0f","tex":"d_t","aliases":"d_t","display":"d_t","kind":"use","occurrences":[403,405]},{"id":"symbol-ef97823489e8f12d","tex":"D_v","aliases":"D_v","display":"D_v","kind":"use","occurrences":[447,449,454]},{"id":"symbol-a9f51566bd6705f7","tex":"E","aliases":"E","display":"E","kind":"use","occurrences":[37,38,512,513,514,516,519,530,535]},{"id":"symbol-7d11c517fc1de2b4","tex":"\\mathcal E'","aliases":"E' mathcal","display":"\\mathcal E'","kind":"use","occurrences":[176]},{"id":"symbol-535f17963fb5c472","tex":"E_0","aliases":"E_0","display":"E_0","kind":"use","occurrences":[512]},{"id":"symbol-67371d243352fe13","tex":"E_0^{(j)}","aliases":"E_0^(j)","display":"E_0^{(j)}","kind":"use","occurrences":[535]},{"id":"symbol-5bbddd50e7cf1221","tex":"\\E_{P_0}","aliases":"E_P_0 E mathbb","display":"{\\mathbb E}_{P_0}","kind":"use","occurrences":[56]},{"id":"symbol-ff556bf7d9b30168","tex":"\\E_{P_1}","aliases":"E_P_1 E mathbb","display":"{\\mathbb E}_{P_1}","kind":"use","occurrences":[56]},{"id":"symbol-f75d15c864ad3cef","tex":"E_t","aliases":"E_t","display":"E_t","kind":"use","occurrences":[37,38,512,513,514,516,519,530]},{"id":"symbol-2c7a3501c3e771c4","tex":"E_{t-1}","aliases":"E_t-1","display":"E_{t-1}","kind":"use","occurrences":[512,519]},{"id":"symbol-a8adfd562576492b","tex":"\\exp","aliases":"exp","display":"\\exp","kind":"use","occurrences":[498]},{"id":"symbol-77dd4d71ed008fa6","tex":"\\mathcal F","aliases":"F mathcal","display":"\\mathcal F","kind":"use","occurrences":[104,107,182]},{"id":"symbol-cace9a962804366a","tex":"\\mathcal F'_h","aliases":"F'_h mathcal","display":"\\mathcal F'_h","kind":"use","occurrences":[182]},{"id":"symbol-d6012f1edaba837f","tex":"f'_{\\tau(e)}","aliases":"f'_τ(e) tau","display":"f'_{\\tau(e)}","kind":"use","occurrences":[179]},{"id":"symbol-313137accdec2e68","tex":"\\mathcal F_0","aliases":"F_0 mathcal","display":"\\mathcal F_0","kind":"use","occurrences":[107]},{"id":"symbol-ff53476e6fc124f9","tex":"F_a","aliases":"F_a","display":"F_a","kind":"use","occurrences":[596,598,604]},{"id":"symbol-109ec22b64c2454d","tex":"f_e^\\theta","aliases":"f_e^θ theta","display":"f_e^\\theta","kind":"use","occurrences":[128,132,137]},{"id":"symbol-5e3d805240078a53","tex":"\\mathcal F_h","aliases":"F_h mathcal","display":"\\mathcal F_h","kind":"use","occurrences":[104,107,182]},{"id":"symbol-874f13ac2a9c06a1","tex":"\\mathcal F_{h-1}","aliases":"F_h-1 mathcal","display":"\\mathcal F_{h-1}","kind":"use","occurrences":[107]},{"id":"symbol-4a14d2ab0a464e3b","tex":"F_u","aliases":"F_u","display":"F_u","kind":"use","occurrences":[64,68,71]},{"id":"symbol-52bb9700d630070f","tex":"F_w","aliases":"F_w","display":"F_w","kind":"use","occurrences":[603,604]},{"id":"symbol-00c37c61b4205056","tex":"\\mathrm{full}","aliases":"full mathrm","display":"\\mathrm{full}","kind":"use","occurrences":[352]},{"id":"symbol-15e2ee7354a2849e","tex":"\\mathcal G","aliases":"G mathcal","display":"\\mathcal G","kind":"use","occurrences":[168,169,170,171,172,285,291,292,295]},{"id":"symbol-cd0aa9856147b6c5","tex":"g","aliases":"g","display":"g","kind":"use","occurrences":[430,431,432,433,434,435,436,585,587,588]},{"id":"symbol-abeca96c3125e618","tex":"\\mathcal G'","aliases":"G' mathcal","display":"\\mathcal G'","kind":"use","occurrences":[171]},{"id":"symbol-3dfdbb430c5e1740","tex":"G'_{\\psi(a)}","aliases":"G'_ψ(a) psi","display":"G'_{\\psi(a)}","kind":"use","occurrences":[179]},{"id":"symbol-9eb17f719597fdbe","tex":"G_a^\\theta","aliases":"G_a^θ theta","display":"G_a^\\theta","kind":"use","occurrences":[130,134,137]},{"id":"symbol-3f8720ad3f121275","tex":"G_r","aliases":"G_r","display":"G_r","kind":"use","occurrences":[335]},{"id":"symbol-1481d5cc6f09169d","tex":"g_t","aliases":"g_t","display":"g_t","kind":"use","occurrences":[585,587,588]},{"id":"symbol-6d584c042dd0c359","tex":"G_w","aliases":"G_w","display":"G_w","kind":"use","occurrences":[603,604]},{"id":"symbol-5d4698720d0499ca","tex":"G_{\\theta,a}","aliases":"G_θ,a theta","display":"G_{\\theta,a}","kind":"use","occurrences":[591]},{"id":"symbol-8c918c284a1665cd","tex":"\\Gcal","aliases":"Gcal G mathcal","display":"{\\mathcal G}","kind":"use","occurrences":[360,388]},{"id":"symbol-edcbd44645488c8a","tex":"\\mathcal H_{t-1}","aliases":"H_t-1 mathcal","display":"\\mathcal H_{t-1}","kind":"use","occurrences":[484,488,498,506,507,510,519]},{"id":"symbol-518159fee1404da8","tex":"\\hat\\imath","aliases":"hatimath hat imath","display":"\\hat\\imath","kind":"use","occurrences":[161,163,167]},{"id":"symbol-de7d1b721a1e0632","tex":"i","aliases":"i","display":"i","kind":"use","occurrences":[44,45,144,145,158,159,160,162,163,166,167,210,211,212,213,383,384,385,386,387,389,390,391,392,395,397,398,399,466,467,468,469,495,539,634]},{"id":"symbol-1eba5c75dddfe29c","tex":"i^*","aliases":"i^*","display":"i^*","kind":"use","occurrences":[166,167]},{"id":"symbol-291335565d353600","tex":"i_1","aliases":"i_1","display":"i_1","kind":"use","occurrences":[210,211,213]},{"id":"symbol-0475b2ac7d225af5","tex":"i_2","aliases":"i_2","display":"i_2","kind":"use","occurrences":[210,211,212]},{"id":"symbol-189f40034be7a199","tex":"j","aliases":"j","display":"j","kind":"use","occurrences":[41,43,45,47,48,229,384,393,394,396,477,478,534,535,536,537,629]},{"id":"symbol-86be9a55762d316a","tex":"K","aliases":"K","display":"K","kind":"use","occurrences":[437,440,442,443,465,469,557]},{"id":"symbol-f3c143179694d5bc","tex":"K_0","aliases":"K_0","display":"K_0","kind":"use","occurrences":[557]},{"id":"symbol-bb9a538f5e406f30","tex":"K_n","aliases":"K_n","display":"K_n","kind":"use","occurrences":[557]},{"id":"symbol-8a89bee060b13402","tex":"K_{n+1}","aliases":"K_n+1","display":"K_{n+1}","kind":"use","occurrences":[557]},{"id":"symbol-acac86c0e609ca90","tex":"l","aliases":"l","display":"l","kind":"use","occurrences":[209,211,212,213,539]},{"id":"symbol-34c05dc8f5073293","tex":"L'","aliases":"L'","display":"L'","kind":"use","occurrences":[32,35,36]},{"id":"symbol-49e6cb003fe8b536","tex":"L_t","aliases":"L_t","display":"L_t","kind":"use","occurrences":[37]},{"id":"symbol-abd805d5765e70d2","tex":"\\lceil","aliases":"lceil","display":"\\lceil","kind":"use","occurrences":[203,208,324]},{"id":"symbol-316e91d8c0b468a8","tex":"\\lfloor","aliases":"lfloor","display":null,"kind":"use","occurrences":[579]},{"id":"symbol-bfbb661daa93662b","tex":"\\log","aliases":"log","display":"\\log","kind":"use","occurrences":[203,208,324,434,435,500,540,624]},{"id":"symbol-356d34f32f81e480","tex":"\\log_2","aliases":"log_2 log","display":"\\log_2","kind":"use","occurrences":[203,208,324,434,435,540,624]},{"id":"symbol-584e1f4fc366a43f","tex":"\\Longleftrightarrow","aliases":"Longleftrightarrow","display":"\\Longleftrightarrow","kind":"use","occurrences":[68]},{"id":"symbol-08f271887ce94707","tex":"M","aliases":"M","display":"M","kind":"use","occurrences":[102]},{"id":"symbol-62c66a7a5dd70c31","tex":"m","aliases":"m","display":"m","kind":"use","occurrences":[105,107,111,112,199,200,247,250,253,254,256,269,272,274,275,276,281,283,299,300,302,304,305,312,322,539,560]},{"id":"symbol-7f54f92432b51276","tex":"m_o","aliases":"m_o","display":"m_o","kind":"use","occurrences":[107]},{"id":"symbol-51b7ccf8b09eaeca","tex":"\\mathbin","aliases":"mathbin","display":null,"kind":"use","occurrences":[407,408]},{"id":"symbol-e481658e9c6ce30a","tex":"\\max","aliases":"max","display":"\\max","kind":"use","occurrences":[107,163,277,299,307,308,320,321,359,366,375,378,560,572]},{"id":"symbol-7be0073546f6c028","tex":"\\max_i","aliases":"max_i max","display":"\\max_i","kind":"use","occurrences":[163]},{"id":"symbol-d441c86815baa4be","tex":"\\max_o","aliases":"max_o max","display":"\\max_o","kind":"use","occurrences":[107]},{"id":"symbol-19f3073c143a3a98","tex":"\\mid","aliases":"mid","display":null,"kind":"use","occurrences":[488,498,506,519]},{"id":"symbol-4d95ab92bdb1dbe9","tex":"\\min","aliases":"min","display":"\\min","kind":"use","occurrences":[240,242,278,321,365,384,476]},{"id":"symbol-cc8961cbd7cc9df9","tex":"\\min_{w\\in W_0}","aliases":"min_w∈ W_0 min in","display":"\\min_{w\\in W_0}","kind":"use","occurrences":[476]},{"id":"symbol-7e18358242f4b0dc","tex":"\\min_{w\\in\\operatorname{conv}(W_0)}","aliases":"min_w∈conv(W_0) min in operatorname","display":"\\min_{w\\in\\operatorname{conv}(W_0)}","kind":"use","occurrences":[476]},{"id":"symbol-8ce86a6ae65d3692","tex":"N","aliases":"N","display":"N","kind":"use","occurrences":[157,227,228,352,357,358,360,503,521,522,523,525,526,543,544,545,546,548,550]},{"id":"symbol-3f620354ed5db565","tex":"\\mathbb N","aliases":"N mathbb","display":"\\mathbb N","kind":"use","occurrences":[239]},{"id":"symbol-f8c1643801c1b1fe","tex":"\\mathbb N_{>0}","aliases":"N_>0 mathbb","display":"\\mathbb N_{>0}","kind":"use","occurrences":[239]},{"id":"symbol-e66911ccd27e5fbc","tex":"N_{\\mathrm{compare}}","aliases":"N_compare mathrm","display":"N_{\\mathrm{compare}}","kind":"use","occurrences":[352]},{"id":"symbol-ae50884201122a48","tex":"n_i","aliases":"n_i","display":"n_i","kind":"use","occurrences":[468,469]},{"id":"symbol-ae4704a81fdf75d2","tex":"n_t","aliases":"n_t","display":"n_t","kind":"use","occurrences":[494,498,500,501,502]},{"id":"symbol-76e48dcd185aa356","tex":"O'_{\\tau(e)}","aliases":"O'_τ(e) tau","display":"O'_{\\tau(e)}","kind":"use","occurrences":[178]},{"id":"symbol-990c002ed2409e69","tex":"o'_{\\tau(e)}","aliases":"o'_τ(e) tau","display":"o'_{\\tau(e)}","kind":"use","occurrences":[179]},{"id":"symbol-706818517abf30a8","tex":"o_e^\\theta","aliases":"o_e^θ theta","display":"o_e^\\theta","kind":"use","occurrences":[129,132,137]},{"id":"symbol-c33eb003dc4eec09","tex":"\\mathcal P","aliases":"P mathcal","display":"\\mathcal P","kind":"use","occurrences":[258,259,260,286,289,290,294]},{"id":"symbol-e50b5dbfe7add137","tex":"P_0","aliases":"P_0","display":"P_0","kind":"use","occurrences":[52,53,56,66,69,70,74,83]},{"id":"symbol-db6b2ac0be28aba6","tex":"p_0","aliases":"p_0","display":"p_0","kind":"use","occurrences":[360]},{"id":"symbol-d1641bdc80ea783c","tex":"P_1","aliases":"P_1","display":"P_1","kind":"use","occurrences":[52,53,56]},{"id":"symbol-6158f77c3833268f","tex":"p_i","aliases":"p_i","display":"p_i","kind":"use","occurrences":[383,384,386,392,395,397,398,399]},{"id":"symbol-85687051228b1e6b","tex":"p_j","aliases":"p_j","display":"p_j","kind":"use","occurrences":[384,393,394,396]},{"id":"symbol-cc9e0dabacaadc8b","tex":"p_{N-1}","aliases":"p_N-1","display":"p_{N-1}","kind":"use","occurrences":[360]},{"id":"symbol-12b33428971a165a","tex":"P_t","aliases":"P_t","display":"P_t","kind":"use","occurrences":[68,76,79,82]},{"id":"symbol-fa1ae76a27248a1e","tex":"P_{t+1}","aliases":"P_t+1","display":"P_{t+1}","kind":"use","occurrences":[68,80]},{"id":"symbol-fbd55c78e5bd8f12","tex":"p_{\\rho(i)}","aliases":"p_ρ(i) rho","display":"p_{\\rho(i)}","kind":"use","occurrences":[387]},{"id":"symbol-7f6ddefca2507bfe","tex":"p_\\psi","aliases":"p_ψ psi","display":"p_\\psi","kind":"use","occurrences":[593]},{"id":"symbol-4bc9c56157a62dc5","tex":"\\operatorname{pa}","aliases":"pa operatorname","display":"\\operatorname{pa}","kind":"use","occurrences":[446,454]},{"id":"symbol-dfea44c8abc940ec","tex":"\\Pareto","aliases":"Pareto operatorname","display":"{\\operatorname{Pareto}}","kind":"use","occurrences":[107]},{"id":"symbol-1cc1c79952eb4cf2","tex":"\\pmod","aliases":"pmod","display":"\\pmod","kind":"use","occurrences":[634]},{"id":"symbol-f1dd2bd1079bcf86","tex":"\\Post","aliases":"Post operatorname","display":"{\\operatorname{Post}}","kind":"use","occurrences":[554,557,560]},{"id":"symbol-6c033af8552f2dad","tex":"\\Pr","aliases":"Pr","display":"\\Pr","kind":"use","occurrences":[488,491,492,498,516,525]},{"id":"symbol-8c2574892063f995","tex":"R","aliases":"R","display":"R","kind":"use","occurrences":[73,75,77,78,81,159,160,163,167,168,170,171,172,219,352,577,581,582]},{"id":"symbol-97a4ded34746a4f6","tex":"\\mathbb R","aliases":"R mathbb","display":"\\mathbb R","kind":"use","occurrences":[465,466,470]},{"id":"symbol-1f4ac7c68b8abd53","tex":"\\mathcal R","aliases":"R mathcal","display":"\\mathcal R","kind":"use","occurrences":[50,54,57,58]},{"id":"symbol-d41fe2809fb29445","tex":"r'_{\\sigma(5)}","aliases":"r'_σ(5) sigma","display":"r'_{\\sigma(5)}","kind":"use","occurrences":[637]},{"id":"symbol-1b221660c5e1ad9a","tex":"R^*_{\\mathcal G}","aliases":"R^*_ G mathcal","display":"R^*_{\\mathcal G}","kind":"use","occurrences":[168,170,172]},{"id":"symbol-0faa5db6de8d543f","tex":"R^*_{\\mathcal G'}","aliases":"R^*_ G' mathcal","display":"R^*_{\\mathcal G'}","kind":"use","occurrences":[171]},{"id":"symbol-55271cfa55ee85c8","tex":"r^2","aliases":"r^2","display":"r^2","kind":"use","occurrences":[498]},{"id":"symbol-855b51431c459bf3","tex":"\\mathbb R^d","aliases":"R^d mathbb","display":"\\mathbb R^d","kind":"use","occurrences":[466]},{"id":"symbol-9eb12de9c970740e","tex":"R_0","aliases":"R_0","display":"R_0","kind":"use","occurrences":[581]},{"id":"symbol-b22adc2a54ec6353","tex":"r_5","aliases":"r_5","display":"r_5","kind":"use","occurrences":[633]},{"id":"symbol-148f6d1b72f67e4b","tex":"R_{\\mathrm{full}}","aliases":"R_full mathrm","display":"R_{\\mathrm{full}}","kind":"use","occurrences":[352]},{"id":"symbol-ff25d16c5ae0dc58","tex":"R_{\\hat\\imath}","aliases":"R_hatimath hat imath","display":"R_{\\hat\\imath}","kind":"use","occurrences":[163,167]},{"id":"symbol-00c00f8247bf728a","tex":"R_i","aliases":"R_i","display":"R_i","kind":"use","occurrences":[159,160,163]},{"id":"symbol-f231812574787070","tex":"r_i","aliases":"r_i","display":"r_i","kind":"use","occurrences":[44,144]},{"id":"symbol-d05f49cca47725be","tex":"R_{i^*}","aliases":"R_i^*","display":"R_{i^*}","kind":"use","occurrences":[167]},{"id":"symbol-07c29c38146ae7ad","tex":"r_{o\\oplus\\theta}","aliases":"r_o⊕θ oplus theta","display":"r_{o\\oplus\\theta}","kind":"use","occurrences":[154]},{"id":"symbol-1ba3cd8a0856a76d","tex":"R_{\\mathrm{shortcut}}","aliases":"R_shortcut mathrm","display":"R_{\\mathrm{shortcut}}","kind":"use","occurrences":[219]},{"id":"symbol-8bbe0b8d06054149","tex":"R_{\\mathrm{stream}}","aliases":"R_stream mathrm","display":"R_{\\mathrm{stream}}","kind":"use","occurrences":[352]},{"id":"symbol-49121c793f36cbb5","tex":"R_t","aliases":"R_t","display":"R_t","kind":"use","occurrences":[577,581,582]},{"id":"symbol-e4b6d938c63d0d7a","tex":"r_\\phi","aliases":"r_ϕ phi","display":"r_\\phi","kind":"use","occurrences":[590]},{"id":"symbol-43dcd88270900239","tex":"\\mathbb R_{\\ge0}^d","aliases":"R_≥0^d mathbb ge","display":"\\mathbb R_{\\ge0}^d","kind":"use","occurrences":[465,470]},{"id":"symbol-fcf0d1ec782295a3","tex":"\\Rcal'","aliases":"Rcal' R' Rcal mathcal","display":"{\\mathcal R}'","kind":"use","occurrences":[191]},{"id":"symbol-87db0642ef898b44","tex":"\\rceil","aliases":"rceil","display":"\\rceil","kind":"use","occurrences":[203,208,324]},{"id":"symbol-452acf9313788e80","tex":"\\rfloor","aliases":"rfloor","display":null,"kind":"use","occurrences":[579]},{"id":"symbol-e402978cefc6e89c","tex":"\\rm","aliases":"rm","display":null,"kind":"use","occurrences":[402,405,539]},{"id":"symbol-043a718774c572bd","tex":"s","aliases":"s","display":"s","kind":"use","occurrences":[198,200,466,469,539]},{"id":"symbol-c8ea458b12d6729a","tex":"s_i","aliases":"s_i","display":"s_i","kind":"use","occurrences":[466,469]},{"id":"symbol-b37e9083bef6f614","tex":"S_r","aliases":"S_r","display":"S_r","kind":"use","occurrences":[410,413]},{"id":"symbol-87807c21edc049cd","tex":"S_t","aliases":"S_t","display":"S_t","kind":"use","occurrences":[37]},{"id":"symbol-a354a58aabc0fe1c","tex":"\\Safe","aliases":"Safe mathsf","display":"{\\mathsf{Safe}}","kind":"use","occurrences":[555,557,576]},{"id":"symbol-f020c9fbd9af5930","tex":"\\mathrm{shortcut}","aliases":"shortcut mathrm","display":"\\mathrm{shortcut}","kind":"use","occurrences":[219]},{"id":"symbol-d6008072c780cd95","tex":"\\mathrm{stream}","aliases":"stream mathrm","display":"\\mathrm{stream}","kind":"use","occurrences":[352]},{"id":"symbol-729e41c77e53a59d","tex":"\\sup","aliases":"sup","display":"\\sup","kind":"use","occurrences":[516]},{"id":"symbol-53bc99b07a603a4f","tex":"\\sup_{t\\geq0}","aliases":"sup_t≥0 sup geq","display":"\\sup_{t\\geq0}","kind":"use","occurrences":[516]},{"id":"symbol-e632b7095b0bf32c","tex":"T","aliases":"T","display":"T","kind":"use","occurrences":[106,107,113,405]},{"id":"symbol-e3b98a4da31a127d","tex":"t","aliases":"t","display":"t","kind":"use","occurrences":[37,38,39,40,68,76,79,80,82,403,404,405,484,485,486,487,488,489,490,491,492,493,494,496,497,498,499,500,501,502,504,506,507,509,510,511,512,513,514,516,518,519,527,529,530,539,560,577,581,582,583,584,585,586,587,588,589]},{"id":"symbol-b3f5e05f76a79869","tex":"\\TV","aliases":"TV d_TV mathrm","display":"{d_{\\mathrm{TV}}}","kind":"use","occurrences":[53,56]},{"id":"symbol-0bfe935e70c321c7","tex":"u","aliases":"u","display":"u","kind":"use","occurrences":[64,68,71,459,461,463,539]},{"id":"symbol-ec61c8382b2d3710","tex":"U'","aliases":"U'","display":"U'","kind":"use","occurrences":[30,32,34,36,187]},{"id":"symbol-98de45ac415e9b4a","tex":"U_1","aliases":"U_1","display":"U_1","kind":"use","occurrences":[157]},{"id":"symbol-9fcfc7136a869616","tex":"U_N","aliases":"U_N","display":"U_N","kind":"use","occurrences":[157]},{"id":"symbol-6ae21ddbb7e958a2","tex":"U_t","aliases":"U_t","display":"U_t","kind":"use","occurrences":[37]},{"id":"symbol-4c94485e0c21ae6c","tex":"v","aliases":"v","display":"v","kind":"use","occurrences":[445,446,447,448,449,453,454,460,462,539]},{"id":"symbol-50e721e49c013f00","tex":"w","aliases":"w","display":"w","kind":"use","occurrences":[105,107,351,352,353,354,356,358,412,413,417,428,448,470,471,472,473,476,602,603,604]},{"id":"symbol-b9d7c18b6334d33d","tex":"W_0","aliases":"W_0","display":"W_0","kind":"use","occurrences":[475,476]},{"id":"symbol-c5480dc96ac56ca2","tex":"W_{0,0}","aliases":"W_0,0","display":"W_{0,0}","kind":"use","occurrences":[560]},{"id":"symbol-3aded8f2b3589e32","tex":"W_{0,k}","aliases":"W_0,k","display":"W_{0,k}","kind":"use","occurrences":[560]},{"id":"symbol-438757f12dd8c3fc","tex":"w_1","aliases":"w_1","display":"w_1","kind":"use","occurrences":[428]},{"id":"symbol-3106fac3e3c8f992","tex":"w_2","aliases":"w_2","display":"w_2","kind":"use","occurrences":[428]},{"id":"symbol-9763a163e7ff6591","tex":"w_3","aliases":"w_3","display":"w_3","kind":"use","occurrences":[428]},{"id":"symbol-7c3094fc743065b1","tex":"W_{h-1,\\max\\{0,k-p(z,a,z')\\}}","aliases":"W_h-1,max{0,k-p(z,a,z')} max","display":"W_{h-1,\\max\\{0,k-p(z,a,z')\\}}","kind":"use","occurrences":[560]},{"id":"symbol-a24f305d6325f526","tex":"w_o","aliases":"w_o","display":"w_o","kind":"use","occurrences":[107]},{"id":"symbol-e6e9e4d147adf393","tex":"w_v","aliases":"w_v","display":"w_v","kind":"use","occurrences":[448]},{"id":"symbol-5b4ce5956eb97972","tex":"w_x","aliases":"w_x","display":"w_x","kind":"use","occurrences":[412,413,417]},{"id":"symbol-cb66caadea9f666a","tex":"\\wedge","aliases":"wedge","display":"\\wedge","kind":"use","occurrences":[522]},{"id":"symbol-967c63ebf4cd78be","tex":"\\widehat R","aliases":"widehat R widehat","display":"\\widehat R","kind":"use","occurrences":[158,160,162,167]},{"id":"symbol-e0887913ce9f0cef","tex":"\\widehat R_{\\hat\\imath}","aliases":"widehat R_hatimath widehat hat imath","display":"\\widehat R_{\\hat\\imath}","kind":"use","occurrences":[167]},{"id":"symbol-f3e1cec7580875b3","tex":"\\widehat R_i","aliases":"widehat R_i widehat","display":"\\widehat R_i","kind":"use","occurrences":[158,160,162]},{"id":"symbol-19029b9c979ec627","tex":"\\widehat R_{i^*}","aliases":"widehat R_i^* widehat","display":"\\widehat R_{i^*}","kind":"use","occurrences":[167]},{"id":"symbol-5df53739c784d094","tex":"\\widehat\\Delta","aliases":"widehatΔ widehat Delta","display":"\\widehat\\Delta","kind":"use","occurrences":[497,498,500]},{"id":"symbol-0ba62b2ff576b8eb","tex":"\\widehat\\Delta_t","aliases":"widehatΔ_t widehat Delta","display":"\\widehat\\Delta_t","kind":"use","occurrences":[497,498,500]},{"id":"symbol-d74a2e10811483dc","tex":"\\widetilde f","aliases":"widetilde f widetilde","display":"\\widetilde f","kind":"use","occurrences":[137]},{"id":"symbol-128fa659bfedffe5","tex":"\\widetilde f_e","aliases":"widetilde f_e widetilde","display":"\\widetilde f_e","kind":"use","occurrences":[137]},{"id":"symbol-929c170d8af8ade4","tex":"\\widetilde G","aliases":"widetilde G widetilde","display":"\\widetilde G","kind":"use","occurrences":[137]},{"id":"symbol-008a0a204981576f","tex":"\\widetilde G_a","aliases":"widetilde G_a widetilde","display":"\\widetilde G_a","kind":"use","occurrences":[137]},{"id":"symbol-756f2ac70a75c1ed","tex":"\\widetilde o","aliases":"widetilde o widetilde","display":"\\widetilde o","kind":"use","occurrences":[137]},{"id":"symbol-15c79c23846df118","tex":"\\widetilde o_e","aliases":"widetilde o_e widetilde","display":"\\widetilde o_e","kind":"use","occurrences":[137]},{"id":"symbol-aba1aecba944a16d","tex":"\\widetilde U","aliases":"widetilde U widetilde","display":"\\widetilde U","kind":"use","occurrences":[189,191]},{"id":"symbol-72746316ee7d6770","tex":"\\widetilde U'","aliases":"widetilde U' widetilde","display":"\\widetilde U'","kind":"use","occurrences":[189,191]},{"id":"symbol-ea2a8d1df2130cc8","tex":"\\widetilde X","aliases":"widetilde X widetilde","display":"\\widetilde X","kind":"use","occurrences":[136]},{"id":"symbol-2d711642b726b044","tex":"x","aliases":"x","display":"x","kind":"use","occurrences":[65,68,98,132,134,135,137,141,143,145,147,153,179,372,373,374,375,376,377,378,406,408,412,413,417,539]},{"id":"symbol-f5857cac0ddcea35","tex":"X'","aliases":"X'","display":"X'","kind":"use","occurrences":[175]},{"id":"symbol-c8031fd0bb708709","tex":"\\xrightarrow","aliases":"xrightarrow","display":null,"kind":"use","occurrences":[632]},{"id":"symbol-18f5384d58bcb1bb","tex":"Y","aliases":"Y","display":"Y","kind":"use","occurrences":[495,504,506,512,518,519,527]},{"id":"symbol-a1fce4363854ff88","tex":"y","aliases":"y","display":"y","kind":"use","occurrences":[68,372,373,600,603]},{"id":"symbol-8e1b422661cf5616","tex":"Y_i","aliases":"Y_i","display":"Y_i","kind":"use","occurrences":[495]},{"id":"symbol-50e74e1b36da6948","tex":"Y_t","aliases":"Y_t","display":"Y_t","kind":"use","occurrences":[504,506,512,518,519,527]},{"id":"symbol-bbeebd879e1dff69","tex":"Z","aliases":"Z","display":"Z","kind":"use","occurrences":[37,259,261,551,555,556,597]},{"id":"symbol-1479ca52dd340112","tex":"z'","aliases":"z'","display":"z'","kind":"use","occurrences":[559,560]},{"id":"symbol-3480de748bc9723a","tex":"Z_t","aliases":"Z_t","display":"Z_t","kind":"use","occurrences":[37]},{"id":"symbol-cbe5cfdf7c2118a9","tex":"|","aliases":"|","display":"|","kind":"use","occurrences":[69,72,156,160,292,295,324,349,454]},{"id":"symbol-0044552e1459e501","tex":"|_{D_v}","aliases":"|_D_v","display":"|_{D_v}","kind":"use","occurrences":[454]},{"id":"symbol-8779068734acaece","tex":"\\cdot","aliases":"· cdot","display":"\\cdot","kind":"use","occurrences":[471,472,476]},{"id":"symbol-349ba76d907adfab","tex":"\\times","aliases":"× times","display":"\\times","kind":"use","occurrences":[131,136]},{"id":"symbol-67610b0632683369","tex":"\\gamma","aliases":"γ gamma","display":"\\gamma","kind":"use","occurrences":[164,165,499,500,505,506,508,511,512,518,519,528,578,579,581]},{"id":"symbol-f1941b975ffcc891","tex":"\\Delta","aliases":"Δ Delta","display":"\\Delta","kind":"use","occurrences":[186,191,352,356,358,452,454,455,456,457,458,469,471,472,474,476,496,498,499]},{"id":"symbol-d203ba01eef4198c","tex":"\\delta","aliases":"δ delta","display":"\\delta","kind":"use","occurrences":[193,194,196,488,489,490,491,492,493,500,515,516,517,524,525,536,537,538]},{"id":"symbol-e76cde478bccee9e","tex":"\\delta_j","aliases":"δ_j delta","display":"\\delta_j","kind":"use","occurrences":[536,537]},{"id":"symbol-c0cfccdc70abb5b6","tex":"\\Delta_t","aliases":"Δ_t Delta","display":"\\Delta_t","kind":"use","occurrences":[496,498,499]},{"id":"symbol-4abc637c4be411ae","tex":"\\delta_t","aliases":"δ_t delta","display":"\\delta_t","kind":"use","occurrences":[488,489,490,492,493,500]},{"id":"symbol-daec86669d42946c","tex":"\\varepsilon","aliases":"ε varepsilon","display":"\\varepsilon","kind":"use","occurrences":[160,163,164,167,190,191,196]},{"id":"symbol-59a94416a2a319ca","tex":"\\eta","aliases":"η eta","display":"\\eta","kind":"use","occurrences":[178,179,185,195,196]},{"id":"symbol-26849087606e21ff","tex":"\\eta_e","aliases":"η_e eta","display":"\\eta_e","kind":"use","occurrences":[178,179,185]},{"id":"symbol-c7b34ab03e848fc4","tex":"\\Theta","aliases":"Θ Theta","display":"\\Theta","kind":"use","occurrences":[124,131,136,156]},{"id":"symbol-2b4af44cb17349bb","tex":"\\theta","aliases":"θ theta","display":"\\theta","kind":"use","occurrences":[49,50,51,57,58,126,127,128,129,130,132,134,135,137,142,143,146,147,152,154,591,594]},{"id":"symbol-9f146cf9b0e1580e","tex":"\\theta_1","aliases":"θ_1 theta","display":"\\theta_1","kind":"use","occurrences":[57]},{"id":"symbol-d74d781424901ae1","tex":"\\theta_2","aliases":"θ_2 theta","display":"\\theta_2","kind":"use","occurrences":[58]},{"id":"symbol-055de4ec3d8e4886","tex":"\\kappa","aliases":"κ kappa","display":"\\kappa","kind":"use","occurrences":[336,338,340,341,342,344,347,349,350]},{"id":"symbol-583cf6290785d72c","tex":"\\lambda","aliases":"λ lambda","display":"\\lambda","kind":"use","occurrences":[509,511,512,519,529]},{"id":"symbol-29b14a34343e28e4","tex":"\\lambda_t","aliases":"λ_t lambda","display":"\\lambda_t","kind":"use","occurrences":[509,511,512,519,529]},{"id":"symbol-93b36ed48fba09a5","tex":"\\Xi_t","aliases":"Ξ_t Xi","display":"\\Xi_t","kind":"use","occurrences":[37,40]},{"id":"symbol-f1cf04093d08b6dc","tex":"\\pi","aliases":"π pi","display":"\\pi","kind":"use","occurrences":[592]},{"id":"symbol-587890bf36e41b2d","tex":"\\pi_\\omega","aliases":"π_ω pi omega","display":"\\pi_\\omega","kind":"use","occurrences":[592]},{"id":"symbol-dabec2cdc19c3167","tex":"\\rho","aliases":"ρ rho","display":"\\rho","kind":"use","occurrences":[37,39,299,384,385,387,389,391]},{"id":"symbol-faf2d9e8110151a4","tex":"\\rho_t","aliases":"ρ_t rho","display":"\\rho_t","kind":"use","occurrences":[37,39]},{"id":"symbol-10ce0200b450ea95","tex":"\\phi","aliases":"ϕ phi","display":"\\phi","kind":"use","occurrences":[590,594]},{"id":"symbol-f301f970eb09c336","tex":"\\psi","aliases":"ψ psi","display":"\\psi","kind":"use","occurrences":[177,179,593,594]},{"id":"symbol-11baa595827a4e0f","tex":"\\omega","aliases":"ω omega","display":"\\omega","kind":"use","occurrences":[592,594]},{"id":"symbol-f0dc60cd5b6e9148","tex":"\\ldots","aliases":"… ldots","display":"\\ldots","kind":"use","occurrences":[41,157,360,444,447,632]},{"id":"symbol-5b86eb1897c44804","tex":"\\ell","aliases":"ℓ ell","display":"\\ell","kind":"use","occurrences":[260,262,263,287,288,293]},{"id":"symbol-4ec4830bef761b7a","tex":"\\forall","aliases":"∀ forall","display":"\\forall","kind":"use","occurrences":[107,560]},{"id":"symbol-b161bf60d6b2d19d","tex":"\\exists","aliases":"∃ exists","display":"\\exists","kind":"use","occurrences":[491,557,560]},{"id":"symbol-4bab02d584adcb84","tex":"\\varnothing","aliases":"∅ varnothing","display":"\\varnothing","kind":"use","occurrences":[454,560]},{"id":"symbol-56dadbdf217ad92b","tex":"\\sum","aliases":"∑ sum","display":"\\sum","kind":"use","occurrences":[107,405,413,417,469,490,537]},{"id":"symbol-91e6cb6007134ac7","tex":"\\sum_i","aliases":"∑_i sum","display":"\\sum_i","kind":"use","occurrences":[469]},{"id":"symbol-f9e2dbe88f36e12f","tex":"\\sum_j","aliases":"∑_j sum","display":"\\sum_j","kind":"use","occurrences":[537]},{"id":"symbol-ccbd66bdaed18a2b","tex":"\\sum_o","aliases":"∑_o sum","display":"\\sum_o","kind":"use","occurrences":[107]},{"id":"symbol-702bc9ceff3fca47","tex":"\\sum_{t=1}^{T}","aliases":"∑_t=1^T sum","display":"\\sum_{t=1}^{T}","kind":"use","occurrences":[405]},{"id":"symbol-075f5b1599c3c4ed","tex":"\\sum_{t\\geq1}","aliases":"∑_t≥1 sum geq","display":"\\sum_{t\\geq1}","kind":"use","occurrences":[490]},{"id":"symbol-7fb07bdba495887b","tex":"\\sum_{x\\in B}","aliases":"∑_x∈ B sum in","display":"\\sum_{x\\in B}","kind":"use","occurrences":[417]},{"id":"symbol-2fa0bdda5c9e6c7e","tex":"\\sum_{x\\in S_r}","aliases":"∑_x∈ S_r sum in","display":"\\sum_{x\\in S_r}","kind":"use","occurrences":[413]},{"id":"symbol-db8e84427c2e1e02","tex":"\\infty","aliases":"∞ infty","display":"\\infty","kind":"use","occurrences":[239,338]},{"id":"symbol-795a720997b72208","tex":"\\cap","aliases":"∩ cap","display":"\\cap","kind":"use","occurrences":[454]},{"id":"symbol-d2ecc6688baa55b4","tex":"\\cup","aliases":"∪ cup","display":"\\cup","kind":"use","occurrences":[107,239]},{"id":"symbol-4d3a285fd7941bf1","tex":"\\ne","aliases":"≠ ne","display":"\\ne","kind":"use","occurrences":[408,454]},{"id":"symbol-ea7ef77bf7563068","tex":"\\equiv","aliases":"≡ equiv","display":"\\equiv","kind":"use","occurrences":[68]},{"id":"symbol-b4f882cbfabf90e0","tex":"\\equiv_{P_t}","aliases":"≡_P_t equiv","display":"\\equiv_{P_t}","kind":"use","occurrences":[68]},{"id":"symbol-e3fa6485b9bac2a9","tex":"\\equiv_{P_{t+1}}","aliases":"≡_P_t+1 equiv","display":"\\equiv_{P_{t+1}}","kind":"use","occurrences":[68]},{"id":"symbol-7a03384e6e8b8519","tex":"\\le","aliases":"≤ le","display":"\\le","kind":"use","occurrences":[368,385,391,413,417]},{"id":"symbol-b92a69b030a52379","tex":"\\leq","aliases":"≤ leq","display":"\\leq","kind":"use","occurrences":[56,160,253,279,281,283,302,304,307,311,318,319,341,488,490,491,492,498,506,511,516,519,523,525,537,582]},{"id":"symbol-d38f53421f824a97","tex":"\\ge","aliases":"≥ ge","display":"\\ge","kind":"use","occurrences":[370,418,464,465,470,472]},{"id":"symbol-0b887c3f37e371d2","tex":"\\oplus","aliases":"⊕ oplus","display":"\\oplus","kind":"use","occurrences":[143,154,407,408]},{"id":"symbol-036096cd7e9ecc74","tex":"\\bigcup","aliases":"⋃ bigcup","display":"\\bigcup","kind":"use","occurrences":[107]},{"id":"symbol-4042331604425568","tex":"\\bigcup_{e:B_0\\subseteq D_e}","aliases":"⋃_e:B_0⊆ D_e bigcup subseteq","display":"\\bigcup_{e:B_0\\subseteq D_e}","kind":"use","occurrences":[107]}],"contexts":["midrule\n$S$, $U$, $L$ & A target solver, an improvement procedure, and a reusable library.","$\\Xi$, $B$ & An evaluation contract and its resource allowance.","$\\Rcal_\\Xi(U,L;B)$ & Expected improvement produced by the procedure, not an oracle over its unselected candidates.","$B_0\\subseteq X$ & Possible current states of a finite repair problem.","$e$, $r$, $P$ & An experiment, a terminal repair, and an executable diagnostic tree.","$F_h(B_0)$ & Nondominated execution/storage pairs for successful trees of depth at most $h$.","$\\sigma$, $\\tau$ & Correspondences of states and operations in a declared transfer.","$\\mathcal K$, $W_{h,k}$ & A safe continuation set and states guaranteeing $k$ progress events within $h$ attempts.","The September 2026 AIDE$^2$ preprint studies repeated modification of a research agent's harness and reports transfer beyond its selection tasks~\\cite{aide}. Its separate comparison of a discovered agent as an outer-loop improver is explicitly inconclusive relative to the strong reference. This distinction is directly relevant: improvements in research performance and improvements in driving the next improvement loop are related hypotheses, not interchangeable observations. These public works are research context. Their empirical findings are not treated as evidence for the architecture proposed here.","\\section{A protected comparison for improvement procedures}\\label{sec:contract}\nLet $S$ denote a solver implementation, $U$ an improvement procedure, and $L$ the library available to that procedure. The distinction is functional rather than physical. They may occupy one executable program. Their different roles are retained so that changes can be attributed rather than hidden in a single checkpoint name.","\\begin{definition}[Improvement contract]\\label{def:contract}\nA contract $\\Xi$ specifies a distribution $\\mathcal H$ of starting-solver/task pairs, an initial information and library boundary, permitted observations and modifications, a resource allowance $B$, protected predicates, and an evaluation rule $J_\\tau(S)\\in[0,1]$. The improvement procedure returns one selected descendant or the unchanged baseline. Selection must use only the admitted development information. The final evaluation is not available as an oracle for choosing retrospectively among all candidates.\n\\end{definition}","Write $U(S,\\tau,L;B)$ for the selected result, including the procedure's randomness where applicable. Its productivity is\n\\begin{equation}\\label{eq:productivity}\n\\Rcal_\\Xi(U,L;B)=\\E_{(S,\\tau)\\sim\\mathcal H}\n\\big[J_\\tau(U(S,\\tau,L;B))-J_\\tau(S)\\big].\n\\end{equation}\nInvalid execution and out-of-budget behavior receive the treatment fixed by $\\Xi$, not an exclusion chosen after seeing the outcome. The score may be a primary coordinate of a larger profile. Resource, validity, and group-specific constraints need not be folded into one scalar.","The value in~\\eqref{eq:productivity} lies in $[-1,1]$. A procedure that may return the baseline need not achieve a nonnegative true value: noisy selection can still choose a worse descendant. A nonnegative certificate requires additional evidence. The mere existence of a fallback does not make every selector good at using it.","\\subsection{Fixed-library and full-system comparisons}\nA comparison of $\\Rcal_\\Xi(U',L;B)$ with $\\Rcal_\\Xi(U,L;B)$ isolates process change at a common library. A comparison with $(U',L')$ measures the combined descendant. Both are useful, but they answer different questions. A crossed comparison of $(U,L)$, $(U',L)$, $(U,L')$, and $(U',L')$ can expose interaction between procedure and library. There is no general additive decomposition that makes their contributions independent.","\\subsection{What persists across a recursive step}\nA lineage state can be written schematically as\n\\[\nZ_t=(S_t,U_t,L_t,E_t,\\Xi_t,\\rho_t),\n\\]\nwhere $E_t$ records evidence and rejected alternatives, and $\\rho_t$ records remaining resources and scope. A proposed modification may target any admitted subset. A comparison across different $\\Xi_t$ needs a declared bridge. It cannot be justified merely because both scores are called accuracy.","\\section{A small world with reusable diagnostic structure}\\label{sec:world}\nConsider eight cells with exactly one latent defect, indexed by $j\\in\\{0,\\ldots,7\\}$. The solver must repair that cell while preserving the others. Its old representation retains only the number of defects, so every episode initially looks identical.","A diagnostic query names a subset $A$ and returns $\\mathbf 1\\{j\\in A\\}$. The reference language contains eight singleton queries and three bit-coordinate queries. Repair $r_i$ succeeds exactly when $i=j$. Because repair is irreversible within the episode, a procedure may commit only when every surviving possibility justifies its choice. Otherwise it may continue investigating or abstain.","The model is fully declared. Setting each known cell as the defect on a calibration copy would permit learning the complete query table in $8(8+3)=88$ observation calls. The code instead generates that table from the analytical specification and records it as supplied information. It does not claim to have executed those calibrations or inferred a hidden environment.","Equality-first testing asks whether $j=0$, then $j=1$, and so on. Its worst case needs seven queries, identifying the last cell by elimination. The three bit-coordinate queries need only three. Each procedure then performs one repair. Both complete binary trees have eight leaves and fifteen stored records. Faster execution has not made the controller free.","Let a hidden case $\\theta$ determine a set $\\mathcal R(\\theta)$ of successful repairs. The set may have several members. Diagnosis need only obtain enough information to choose a repair in the appropriate set. It need not identify every hidden property of $\\theta$.","\\begin{proposition}[Feedback indistinguishability limits repair]\\label{prop:indistinguishable}\nSuppose two cases have disjoint successful-repair sets and induce transcript laws $P_0,P_1$ under a fixed admitted investigation protocol. A repair selector observes only the transcript and independent private randomness. Under equal prior probabilities, its average probability of successful repair is at most\n\\[\n\\frac{1+\\TV(P_0,P_1)}2.\n\\]\nIn particular, identical transcript laws rule out certain success in both cases.\n\\end{proposition}\n\\begin{proof}\nFrom the selected repair construct a binary decision: report case zero if the repair belongs to $\\mathcal R(0)$, and case one otherwise. Success in case zero implies a correct decision. Success in case one also implies a correct decision because the two repair sets are disjoint. Thus repair success is at most the success of this binar","roposition}[Feedback indistinguishability limits repair]\\label{prop:indistinguishable}\nSuppose two cases have disjoint successful-repair sets and induce transcript laws $P_0,P_1$ under a fixed admitted investigation protocol. A repair selector observes only the transcript and independent private randomness. Under equal prior probabilities, its average probability of successful repair is at most\n\\[\n\\frac{1+\\TV(P_0,P_1)}2.\n\\]\nIn particular, identical transcript laws rule out certain success in both cases.\n\\end{proposition}\n\\begin{proof}\nFrom the selected repair construct a binary decision: report case zero if the repair belongs to $\\mathcal R(0)$, and case one otherwise. Success in case zero implies a correct decision. Success in case one also implies a correct decision because the two repair sets are disjoint. Thus repair success is at most the success of this binary test. For a randomized test with probability $f(o)\\in[0,1]$ of reporting zero, the latter success is\n\\[\n\\tfrac12\\big(1+\\E_{P_0}f-\\E_{P_1}f\\big)\n\\leq\\tfrac12(1+\\TV(P_0,P_1)).\n\\]\nThe inequality follows by summing the positi…","ector observes only the transcript and independent private randomness. Under equal prior probabilities, its average probability of successful repair is at most\n\\[\n\\frac{1+\\TV(P_0,P_1)}2.\n\\]\nIn particular, identical transcript laws rule out certain success in both cases.\n\\end{proposition}\n\\begin{proof}\nFrom the selected repair construct a binary decision: report case zero if the repair belongs to $\\mathcal R(0)$, and case one otherwise. Success in case zero implies a correct decision. Success in case one also implies a correct decision because the two repair sets are disjoint. Thus repair success is at most the success of this binary test. For a randomized test with probability $f(o)\\in[0,1]$ of reporting zero, the latter success is\n\\[\n\\tfrac12\\big(1+\\E_{P_0}f-\\E_{P_1}f\\big)\n\\leq\\tfrac12(1+\\TV(P_0,P_1)).\n\\]\nThe inequality follows by summing the positive differences of the two finite probability laws. The same argument applies to any measurable transcript space with the usual total-variation definition.\n\\end{proof}","}\n\\begin{proof}\nFrom the selected repair construct a binary decision: report case zero if the repair belongs to $\\mathcal R(0)$, and case one otherwise. Success in case zero implies a correct decision. Success in case one also implies a correct decision because the two repair sets are disjoint. Thus repair success is at most the success of this binary test. For a randomized test with probability $f(o)\\in[0,1]$ of reporting zero, the latter success is\n\\[\n\\tfrac12\\big(1+\\E_{P_0}f-\\E_{P_1}f\\big)\n\\leq\\tfrac12(1+\\TV(P_0,P_1)).\n\\]\nThe inequality follows by summing the positive differences of the two finite probability laws. The same argument applies to any measurable transcript space with the usual total-variation definition.\n\\end{proof}","ry decision: report case zero if the repair belongs to $\\mathcal R(0)$, and case one otherwise. Success in case zero implies a correct decision. Success in case one also implies a correct decision because the two repair sets are disjoint. Thus repair success is at most the success of this binary test. For a randomized test with probability $f(o)\\in[0,1]$ of reporting zero, the latter success is\n\\[\n\\tfrac12\\big(1+\\E_{P_0}f-\\E_{P_1}f\\big)\n\\leq\\tfrac12(1+\\TV(P_0,P_1)).\n\\]\nThe inequality follows by summing the positive differences of the two finite probability laws. The same argument applies to any measurable transcript space with the usual total-variation definition.\n\\end{proof}","\\subsection{Repair sufficiency is not full identification}\nSuppose $\\mathcal R(\\theta_1)=\\{a,c\\}$ and $\\mathcal R(\\theta_2)=\\{b,c\\}$. The cases are distinct, yet repair $c$ works in both. A method can stop without learning which case occurred. Conversely, pairwise overlap of repair sets does not guarantee one repair for an entire uncertainty set. The three sets $\\{a,b\\}$, $\\{b,c\\}$, and $\\{a,c\\}$ intersect pairwise and have empty total intersection. The stopping condition must refer to the whole surviving set.","\\subsection{Repairing a failed representation}\nLet $X$ be a finite set with deterministic operations $F_u:X\\to X$ and output labels $q(x)$. Begin with a partition $P_0$ that preserves both the old retained labels and $q$. Define the next partition by splitting a block whenever two states in it have successors in different current blocks under some permitted operation. Equivalently,\n\\[\nx\\equiv_{P_{t+1}}y\n\\quad\\Longleftrightarrow\\quad\nx\\equiv_{P_t}y\\ \\text{ and }\\\nF_u(x)\\equiv_{P_t}F_u(y)\\ \\text{for every }u.\n\\]\nThe construction uses a complete transition description. Having a few observed contradictions does not give a learner that table for free.","\\begin{proposition}[Least stable refinement of a retained representation]\\label{prop:refinement}\nThe splitting construction stabilizes after at most $|X|-|P_0|$ strict refinements. Its final partition is the coarsest refinement of $P_0$ on which every $F_u$ has a well-defined quotient operation. It preserves every finite continuation of the retained labels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ b","\\begin{proposition}[Least stable refinement of a retained representation]\\label{prop:refinement}\nThe splitting construction stabilizes after at most $|X|-|P_0|$ strict refinements. Its final partition is the coarsest refinement of $P_0$ on which every $F_u$ has a well-defined quotient operation. It preserves every finite continuation of the retained labels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refine","\\begin{proposition}[Least stable refinement of a retained representation]\\label{prop:refinement}\nThe splitting construction stabilizes after at most $|X|-|P_0|$ strict refinements. Its final partition is the coarsest refinement of $P_0$ on which every $F_u$ has a well-defined quotient operation. It preserves every finite continuation of the retained labels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hen","efinement}\nThe splitting construction stabilizes after at most $|X|-|P_0|$ strict refinements. Its final partition is the coarsest refinement of $P_0$ on which every $F_u$ has a well-defined quotient operation. It preserves every finite continuation of the retained labels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","es every finite continuation of the retained labels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","etained labels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","bels and outputs.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","puts.\n\\end{proposition}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","on}\n\\begin{proof}\nA strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","A strict refinement increases the number of nonempty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","pty blocks, which is bounded by $|X|$. At a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","a fixed point, equivalent states have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","s have equivalent successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","successors, so the quotient operations are well-defined. Induction on word length then preserves all retained labels after every finite word. For minimality, let $R$ be any stable partition refining $P_0$. If $R$ refines $P_t$, stability of $R$ implies that $R$-equivalent states have successors in the same $P_t$ blocks, so they remain together in $P_{t+1}$. Induction shows that $R$ refines every $P_t$ and hence the terminal partition.\n\\end{proof}","This is a standard partition-refinement argument expressed as a repair of an old representation. It is not necessarily the smallest representation for the outputs alone: $P_0$ may retain distinctions that are no longer needed. Discarding those distinctions requires a different contract. The continuation and cost perspectives are developed more fully in \\emph{Machine Understanding}~\\cite{understanding}.","\\begin{definition}[Finite repair experiment model]\\label{def:repairmodel}\nA model has a finite nonempty state set $X$, a finite experiment set $\\mathcal E$, and a finite repair set $\\mathcal A$. Experiment $e$ is allowed on a domain $D_e\\subseteq X$, changes the state by $f_e:D_e\\to X$, and reports $o_e:D_e\\to O_e$. Its execution cost $c_e\\geq0$ is fixed and known. Repair $a$ has fixed cost $c_a\\geq0$ and succeeds on a declared set $G_a\\subseteq X$. All costs are in one stated unit. A nonempty set $B_0\\subseteq X$ records the possible current states.\n\\end{definition}","riment model]\\label{def:repairmodel}\nA model has a finite nonempty state set $X$, a finite experiment set $\\mathcal E$, and a finite repair set $\\mathcal A$. Experiment $e$ is allowed on a domain $D_e\\subseteq X$, changes the state by $f_e:D_e\\to X$, and reports $o_e:D_e\\to O_e$. Its execution cost $c_e\\geq0$ is fixed and known. Repair $a$ has fixed cost $c_a\\geq0$ and succeeds on a declared set $G_a\\subseteq X$. All costs are in one stated unit. A nonempty set $B_0\\subseteq X$ records the possible current states.\n\\end{definition}","state set $X$, a finite experiment set $\\mathcal E$, and a finite repair set $\\mathcal A$. Experiment $e$ is allowed on a domain $D_e\\subseteq X$, changes the state by $f_e:D_e\\to X$, and reports $o_e:D_e\\to O_e$. Its execution cost $c_e\\geq0$ is fixed and known. Repair $a$ has fixed cost $c_a\\geq0$ and succeeds on a declared set $G_a\\subseteq X$. All costs are in one stated unit. A nonempty set $B_0\\subseteq X$ records the possible current states.\n\\end{definition}","A chosen experiment must be uniformly allowed: $B_0\\subseteq D_e$. After observing $o$, the possible \\emph{post-experiment} states are\n\\begin{equation}\\label{eq:post}\nB_{e,o}=\\{f_e(x):x\\in B_0,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are retained. A terminal repair is justified when $B_0\\subseteq G_a$. If the original state or an earlier obligation matters, it must be retained as an immutable tag in $X$. Merely changing the current state into a known state is not recovery of its history.","ment must be uniformly allowed: $B_0\\subseteq D_e$. After observing $o$, the possible \\emph{post-experiment} states are\n\\begin{equation}\\label{eq:post}\nB_{e,o}=\\{f_e(x):x\\in B_0,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are retained. A terminal repair is justified when $B_0\\subseteq G_a$. If the original state or an earlier obligation matters, it must be retained as an immutable tag in $X$. Merely changing the current state into a known state is not recovery of its history.","Let $W(P,B_0)$ be the largest execution cost of a complete path, including its terminal repair. Let $M(P)$ be the number of stored tree records. These costs are not independently minimized. A single tree has to realize both.","For horizon $h$, let $\\mathcal F_h(B_0)$ be the nondominated attainable pairs $(w,m)$. A pair dominates another when it is no larger in either coordinate and strictly smaller in at least one. Let $T(B_0)=\\{(c_a,1):B_0\\subseteq G_a\\}$. Then\n\\begin{align}\\label{eq:pareto}\n\\mathcal F_0(B_0)&=\\Pareto T(B_0),","For horizon $h$, let $\\mathcal F_h(B_0)$ be the nondominated attainable pairs $(w,m)$. A pair dominates another when it is no larger in either coordinate and strictly smaller in at least one. Let $T(B_0)=\\{(c_a,1):B_0\\subseteq G_a\\}$. Then\n\\begin{align}\\label{eq:pareto}\n\\mathcal F_0(B_0)&=\\Pareto T(B_0),\\\\\n\\mathcal F_h(B_0)&=\\Pareto\\left(T(B_0)\\cup\n\\bigcup_{e:B_0\\subseteq D_e}\n\\left\\{\\left(c_e+\\max_o w_o,\\ 1+\\sum_o m_o\\right):\n(w_o,m_o)\\in\\mathcal F_{h-1}(B_{e,o})\\ \\forall o\\right\\}\\right).\n\\nonumber\n\\end{align}\nAn empty child frontier makes that experiment unavailable for a successful complete tree. An empty union is not a zero-cost solution.","\\begin{theorem}[Exact finite repair synthesis with compiled storage]\\label{thm:synthesis}\nFor every finite model, nonempty $B_0$, and integer $h\\geq0$, recurrence~\\eqref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the","\\begin{theorem}[Exact finite repair synthesis with compiled storage]\\label{thm:synthesis}\nFor every finite model, nonempty $B_0$, and integer $h\\geq0$, recurrence~\\eqref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The","\\begin{theorem}[Exact finite repair synthesis with compiled storage]\\label{thm:synthesis}\nFor every finite model, nonempty $B_0$, and integer $h\\geq0$, recurrence~\\eqref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The storage is the root record plus the sum over every child, including children not traversed in a particular execution. Conver","\\begin{theorem}[Exact finite repair synthesis with compiled storage]\\label{thm:synthesis}\nFor every finite model, nonempty $B_0$, and integer $h\\geq0$, recurrence~\\eqref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The storage is the root record plus the sum over every child, including children not traversed in a particular execution. Conversely, choosing a successful child for each report constructs a successful root tr","repair synthesis with compiled storage]\\label{thm:synthesis}\nFor every finite model, nonempty $B_0$, and integer $h\\geq0$, recurrence~\\eqref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The storage is the root record plus the sum over every child, including children not traversed in a particular execution. Conversely, choosing a successful child for each report constructs a successful root tree with those costs. Replacing a dominated child by a dominating one never…","nteger $h\\geq0$, recurrence~\\eqref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The storage is the root record plus the sum over every child, including children not traversed in a particular execution. Conversely, choosing a successful child for each report constructs a successful root tree with those costs. Replacing a dominated child by a dominating one never increases the root maximum or sum. Thus Pareto pruning loses no feasible allowance. Induction proves both…","qref{eq:pareto} yields exactly the nondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The storage is the root record plus the sum over every child, including children not traversed in a particular execution. Conversely, choosing a successful child for each report constructs a successful root tree with those costs. Replacing a dominated child by a dominating one never increases the root maximum or sum. Thus Pareto pruning loses no feasible allowance. Induction proves both completeness and witness cons…","ondominated cost pairs of successful diagnostic trees of depth at most $h$. Every retained pair has an executable witness. A work/storage allowance $(b,m)$ is feasible if and only if some retained pair is componentwise at most $(b,m)$.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero only a justified terminal repair can succeed, giving $T(B_0)$. Assume the claim at $h-1$. A successful tree at horizon $h$ either stops immediately or has one uniformly allowed first experiment. Its child at each possible report is successful on exactly the post-experiment support in~\\eqref{eq:post}. The execution cost is the root cost plus the worst child cost. The storage is the root record plus the sum over every child, including children not traversed in a particular execution. Conversely, choosing a successful child for each report constructs a successful root tree with those costs. Replacing a dominated child by a dominating one never increases the root maximum or sum. Thus Pareto pruning loses no feasible allowance. Induction proves both completeness and witness construction.\n\\end{proof}","\\subsection{A more informative question can be a worse repair question}\n\\begin{counterexample}[Entropy reduction need not minimize repair work]\\label{ex:entropy}\nThere are four equally likely cases. Repair $a$ succeeds in $\\{0,1,2\\}$ and repair $b$ succeeds in $\\{3\\}$. A balanced query separates $\\{0,1\\}$ from $\\{2,3\\}$. Another query separates $\\{0,1,2\\}$ from $\\{3\\}$. Both cost one unit and leave the state unchanged. The balanced query has the larger entropy reduction, but its second branch still needs another observation. The unbalanced query supports a correct repair after one observation in every case. Maximizing information about the entire hidden case is not the same objective as minimizing the cost of a successful correction.\n\\end{counterexample}","A useful first extension retains several explicit possibilities rather than replacing uncertainty by one convenient estimate. Let $\\Theta$ be a finite nonempty set of candidate models with a common state set $X$, experiment names, repair names, and nonnegative costs. Model $\\theta$ specifies $D_e^\\theta$, $f_e^\\theta$, $o_e^\\theta$, and $G_a^\\theta$. The unknown model is fixed during an investigation. The initial information is a nonempty set $B\\subseteq\\Theta\\times X$. The procedure observes reports, not the model label or the hidden physical state.","t possibilities rather than replacing uncertainty by one convenient estimate. Let $\\Theta$ be a finite nonempty set of candidate models with a common state set $X$, experiment names, repair names, and nonnegative costs. Model $\\theta$ specifies $D_e^\\theta$, $f_e^\\theta$, $o_e^\\theta$, and $G_a^\\theta$. The unknown model is fixed during an investigation. The initial information is a nonempty set $B\\subseteq\\Theta\\times X$. The procedure observes reports, not the model label or the hidden physical state.","An experiment is allowed only when it is allowed under every pair still possible. Its update is\n\\begin{equation}\\label{eq:model-support}\nB_{e,o}=\\{(\\theta,f_e^\\theta(x)):(\\theta,x)\\in B,\\ x\\in D_e^\\theta,\\ o_e^\\theta(x)=o\\}.\n\\end{equation}\nThe model tag remains unchanged. Observations can exclude a model, but the algorithm cannot switch the underlying model between steps merely because doing so produces a convenient worst case. A terminal repair $a$ is justified when $x\\in G_a^\\theta$ for every $(\\theta,x)\\in B$.","der every pair still possible. Its update is\n\\begin{equation}\\label{eq:model-support}\nB_{e,o}=\\{(\\theta,f_e^\\theta(x)):(\\theta,x)\\in B,\\ x\\in D_e^\\theta,\\ o_e^\\theta(x)=o\\}.\n\\end{equation}\nThe model tag remains unchanged. Observations can exclude a model, but the algorithm cannot switch the underlying model between steps merely because doing so produces a convenient worst case. A terminal repair $a$ is justified when $x\\in G_a^\\theta$ for every $(\\theta,x)\\in B$.","ossible. Its update is\n\\begin{equation}\\label{eq:model-support}\nB_{e,o}=\\{(\\theta,f_e^\\theta(x)):(\\theta,x)\\in B,\\ x\\in D_e^\\theta,\\ o_e^\\theta(x)=o\\}.\n\\end{equation}\nThe model tag remains unchanged. Observations can exclude a model, but the algorithm cannot switch the underlying model between steps merely because doing so produces a convenient worst case. A terminal repair $a$ is justified when $x\\in G_a^\\theta$ for every $(\\theta,x)\\in B$.","n{equation}\\label{eq:model-support}\nB_{e,o}=\\{(\\theta,f_e^\\theta(x)):(\\theta,x)\\in B,\\ x\\in D_e^\\theta,\\ o_e^\\theta(x)=o\\}.\n\\end{equation}\nThe model tag remains unchanged. Observations can exclude a model, but the algorithm cannot switch the underlying model between steps merely because doing so produces a convenient worst case. A terminal repair $a$ is justified when $x\\in G_a^\\theta$ for every $(\\theta,x)\\in B$.","\\begin{theorem}[Repair synthesis under a fixed unknown model]\\label{thm:uncertain-synthesis}\nConstruct a finite repair model on $\\widetilde X=\\Theta\\times X$ by\n\\[\n\\widetilde f_e(\\theta,x)=(\\theta,f_e^\\theta(x)),\\quad\n\\widetilde o_e(\\theta,x)=o_e^\\theta(x),\\quad\n\\widetilde G_a=\\{(\\theta,x):x\\in G_a^\\theta\\},\n\\]\nwith the corresponding tagged domains and the common action costs. Applying Theorem~\\ref{thm:synthesis} to this model gives exactly the nondominated work/storage pairs of report-dependent trees that succeed for every initial pair in $B$, within the declared horizon. Every retained tree is valid for the actual system whenever its model and initial state belong to $B$.\n\\end{theorem}\n\\begin{proof}\nA tree receives the same experiment names and report alphabet in both descriptions. For each experiment and report, its possible tagged successor","\\begin{theorem}[Repair synthesis under a fixed unknown model]\\label{thm:uncertain-synthesis}\nConstruct a finite repair model on $\\widetilde X=\\Theta\\times X$ by\n\\[\n\\widetilde f_e(\\theta,x)=(\\theta,f_e^\\theta(x)),\\quad\n\\widetilde o_e(\\theta,x)=o_e^\\theta(x),\\quad\n\\widetilde G_a=\\{(\\theta,x):x\\in G_a^\\theta\\},\n\\]\nwith the corresponding tagged domains and the common action costs. Applying Theorem~\\ref{thm:synthesis} to this model gives exactly the nondominated work/storage pairs of report-dependent trees that succeed for every initial pair in $B$, within the declared horizon. Every retained tree is valid for the actual system whenever its model and initial state belong to $B$.\n\\end{theorem}\n\\begin{proof}\nA tree receives the same experiment names and report alphabet in both descriptions. For each experiment and report, its possible tagged successor states are exactly~\\eqref{eq:model-support}. Its terminal repair is valid on the tagged support exactly when it is valid for every retained model/state p","a\\times X$ by\n\\[\n\\widetilde f_e(\\theta,x)=(\\theta,f_e^\\theta(x)),\\quad\n\\widetilde o_e(\\theta,x)=o_e^\\theta(x),\\quad\n\\widetilde G_a=\\{(\\theta,x):x\\in G_a^\\theta\\},\n\\]\nwith the corresponding tagged domains and the common action costs. Applying Theorem~\\ref{thm:synthesis} to this model gives exactly the nondominated work/storage pairs of report-dependent trees that succeed for every initial pair in $B$, within the declared horizon. Every retained tree is valid for the actual system whenever its model and initial state belong to $B$.\n\\end{theorem}\n\\begin{proof}\nA tree receives the same experiment names and report alphabet in both descriptions. For each experiment and report, its possible tagged successor states are exactly~\\eqref{eq:model-support}. Its terminal repair is valid on the tagged support exactly when it is valid for every retained model/state pair. Consequently the report histories, path costs, branch counts, and stopping conditions agree node by node. Induction on tree depth gives a cost-preserving correspondence of all successful trees. The finite synthesis theorem then supp…","{(\\theta,x):x\\in G_a^\\theta\\},\n\\]\nwith the corresponding tagged domains and the common action costs. Applying Theorem~\\ref{thm:synthesis} to this model gives exactly the nondominated work/storage pairs of report-dependent trees that succeed for every initial pair in $B$, within the declared horizon. Every retained tree is valid for the actual system whenever its model and initial state belong to $B$.\n\\end{theorem}\n\\begin{proof}\nA tree receives the same experiment names and report alphabet in both descriptions. For each experiment and report, its possible tagged successor states are exactly~\\eqref{eq:model-support}. Its terminal repair is valid on the tagged support exactly when it is valid for every retained model/state pair. Consequently the report histories, path costs, branch counts, and stopping conditions agree node by node. Induction on tree depth gives a cost-preserving correspondence of all successful trees. The finite synthesis theorem then supplies the frontier and its witnesses. If the actual pair starts in $B$, each actual observation leaves its actual successor in the up…","repair is valid on the tagged support exactly when it is valid for every retained model/state pair. Consequently the report histories, path costs, branch counts, and stopping conditions agree node by node. Induction on tree depth gives a cost-preserving correspondence of all successful trees. The finite synthesis theorem then supplies the frontier and its witnesses. If the actual pair starts in $B$, each actual observation leaves its actual successor in the updated support, so the executed terminal repair is valid for that pair.\n\\end{proof}","\\subsection{One hidden bit and two sensor conventions}\n\\begin{counterexample}[Known-model repair without a common unknown-model policy]\\label{ex:orientation}\nLet $x\\in\\{0,1\\}$ be the cell requiring repair and let $\\theta\\in\\{0,1\\}$ be a fixed sensor convention. A nondestructive query reports $o=x\\oplus\\theta$ at cost one. Repair $r_i$ succeeds exactly when $x=i$ and costs one. If $\\theta$ is known, one query followed by the appropriate repair always succeeds. If all four $(\\theta,x)$ pairs are initially possible, report zero leaves $(0,0)$ and $(1,1)$, while report one leaves $(0,1)$ and $(1,0)$. Repeating the same query gives no further information. No finite tree using only this query can guarantee a repair, although each known model admits a two-unit procedure.\n\\end{counterexample}","[Known-model repair without a common unknown-model policy]\\label{ex:orientation}\nLet $x\\in\\{0,1\\}$ be the cell requiring repair and let $\\theta\\in\\{0,1\\}$ be a fixed sensor convention. A nondestructive query reports $o=x\\oplus\\theta$ at cost one. Repair $r_i$ succeeds exactly when $x=i$ and costs one. If $\\theta$ is known, one query followed by the appropriate repair always succeeds. If all four $(\\theta,x)$ pairs are initially possible, report zero leaves $(0,0)$ and $(1,1)$, while report one leaves $(0,1)$ and $(1,0)$. Repeating the same query gives no further information. No finite tree using only this query can guarantee a repair, although each known model admits a two-unit procedure.\n\\end{counterexample}","el{ex:orientation}\nLet $x\\in\\{0,1\\}$ be the cell requiring repair and let $\\theta\\in\\{0,1\\}$ be a fixed sensor convention. A nondestructive query reports $o=x\\oplus\\theta$ at cost one. Repair $r_i$ succeeds exactly when $x=i$ and costs one. If $\\theta$ is known, one query followed by the appropriate repair always succeeds. If all four $(\\theta,x)$ pairs are initially possible, report zero leaves $(0,0)$ and $(1,1)$, while report one leaves $(0,1)$ and $(1,0)$. Repeating the same query gives no further information. No finite tree using only this query can guarantee a repair, although each known model admits a two-unit procedure.\n\\end{counterexample}","ation}\nLet $x\\in\\{0,1\\}$ be the cell requiring repair and let $\\theta\\in\\{0,1\\}$ be a fixed sensor convention. A nondestructive query reports $o=x\\oplus\\theta$ at cost one. Repair $r_i$ succeeds exactly when $x=i$ and costs one. If $\\theta$ is known, one query followed by the appropriate repair always succeeds. If all four $(\\theta,x)$ pairs are initially possible, report zero leaves $(0,0)$ and $(1,1)$, while report one leaves $(0,1)$ and $(1,0)$. Repeating the same query gives no further information. No finite tree using only this query can guarantee a repair, although each known model admits a two-unit procedure.\n\\end{counterexample}","ell requiring repair and let $\\theta\\in\\{0,1\\}$ be a fixed sensor convention. A nondestructive query reports $o=x\\oplus\\theta$ at cost one. Repair $r_i$ succeeds exactly when $x=i$ and costs one. If $\\theta$ is known, one query followed by the appropriate repair always succeeds. If all four $(\\theta,x)$ pairs are initially possible, report zero leaves $(0,0)$ and $(1,1)$, while report one leaves $(0,1)$ and $(1,0)$. Repeating the same query gives no further information. No finite tree using only this query can guarantee a repair, although each known model admits a two-unit procedure.\n\\end{counterexample}","g repair and let $\\theta\\in\\{0,1\\}$ be a fixed sensor convention. A nondestructive query reports $o=x\\oplus\\theta$ at cost one. Repair $r_i$ succeeds exactly when $x=i$ and costs one. If $\\theta$ is known, one query followed by the appropriate repair always succeeds. If all four $(\\theta,x)$ pairs are initially possible, report zero leaves $(0,0)$ and $(1,1)$, while report one leaves $(0,1)$ and $(1,0)$. Repeating the same query gives no further information. No finite tree using only this query can guarantee a repair, although each known model admits a two-unit procedure.\n\\end{counterexample}","Add a permitted reference measurement that reports $\\theta$, leaves $x$ unchanged, and costs two. The procedure can calibrate, make the ordinary query, and choose $r_{o\\oplus\\theta}$. Its worst-case work is four and its complete binary tree has seven records. Either measurement alone leaves both repairs possible. The two measurements are therefore both necessary in this example, regardless of their order. The executable reference construction returns exactly $(4,7)$ at horizons at least two.","ves $x$ unchanged, and costs two. The procedure can calibrate, make the ordinary query, and choose $r_{o\\oplus\\theta}$. Its worst-case work is four and its complete binary tree has seven records. Either measurement alone leaves both repairs possible. The two measurements are therefore both necessary in this example, regardless of their order. The executable reference construction returns exactly $(4,7)$ at horizons at least two.","The finite construction also makes the cost of caution visible. The tagged state set can have $|\\Theta||X|$ members, and the support search can be substantially larger than a search under one assumed model. Representation refinement, model acquisition, and compiler improvement are consequently coupled research problems. A more reliable model can increase the computational burden of planning until a better discovery procedure exploits its structure.","For a fixed finite collection $\\{U_1,\\ldots,U_N\\}$, suppose a declared assessment gives estimates $\\widehat R_i$ of true productivities $R_i$. This assessment can be exact on a fully specified finite distribution or statistical under additional assumptions. Choosing a candidate must use the assessment actually available at selection time, not a hidden final score inspected afterward.","\\begin{proposition}[Selection under uniform assessment error]\\label{prop:selection}\nIf $|\\widehat R_i-R_i|\\leq\\varepsilon$ for every candidate and $\\hat\\imath$ maximizes $\\widehat R_i$, then\n\\[\nR_{\\hat\\imath}\\geq \\max_iR_i-2\\varepsilon.\n\\]\nIf a baseline is included and the selected candidate's estimated advantage exceeds $2\\varepsilon+\\gamma$, its true advantage exceeds $\\gamma$.\n\\end{proposition}\n\\begin{proof}\nFor a true maximizer $i^*$,\n$R_{\\hat\\imath}\\geq\\widehat R_{\\hat\\imath}-\\varepsilon\n\\geq\\widehat R_{i^*}-\\varepsilon\n\\geq R_{i^*}-2\\varepsilon$.\nApply the same two error bounds to the selected candidate and the baseline for the second assertion.\n\\end{proof}","niform assessment error]\\label{prop:selection}\nIf $|\\widehat R_i-R_i|\\leq\\varepsilon$ for every candidate and $\\hat\\imath$ maximizes $\\widehat R_i$, then\n\\[\nR_{\\hat\\imath}\\geq \\max_iR_i-2\\varepsilon.\n\\]\nIf a baseline is included and the selected candidate's estimated advantage exceeds $2\\varepsilon+\\gamma$, its true advantage exceeds $\\gamma$.\n\\end{proposition}\n\\begin{proof}\nFor a true maximizer $i^*$,\n$R_{\\hat\\imath}\\geq\\widehat R_{\\hat\\imath}-\\varepsilon\n\\geq\\widehat R_{i^*}-\\varepsilon\n\\geq R_{i^*}-2\\varepsilon$.\nApply the same two error bounds to the selected candidate and the baseline for the second assertion.\n\\end{proof}","assessment error]\\label{prop:selection}\nIf $|\\widehat R_i-R_i|\\leq\\varepsilon$ for every candidate and $\\hat\\imath$ maximizes $\\widehat R_i$, then\n\\[\nR_{\\hat\\imath}\\geq \\max_iR_i-2\\varepsilon.\n\\]\nIf a baseline is included and the selected candidate's estimated advantage exceeds $2\\varepsilon+\\gamma$, its true advantage exceeds $\\gamma$.\n\\end{proposition}\n\\begin{proof}\nFor a true maximizer $i^*$,\n$R_{\\hat\\imath}\\geq\\widehat R_{\\hat\\imath}-\\varepsilon\n\\geq\\widehat R_{i^*}-\\varepsilon\n\\geq R_{i^*}-2\\varepsilon$.\nApply the same two error bounds to the selected candidate and the baseline for the second assertion.\n\\end{proof}","For a fixed budget, let $R^*_{\\mathcal G}$ be the largest productivity among procedures expressible in a declared language $\\mathcal G$, with installed data and interpreter costs held fixed. The gap $R^*_{\\mathcal G}-R(U)$ is search or selection headroom \\emph{within that language}. Comparing $R^*_{\\mathcal G'}$ with $R^*_{\\mathcal G}$ concerns expressibility and resources. Neither upper envelope is automatically available to the learner.","Under a cap of three diagnostic calls and one repair call, the equality-first procedure certifies and repairs the first three cells, then abstains on the other five. On the uniform finite case distribution, its productivity from a zero-success baseline is $3/8$. The depth-three procedure certifies all eight cases, giving productivity $1$. These are exact properties of stipulated procedures. The compiler selected a better procedure within a supplied finite language. It did not invent the language, learn a general visual representation, or improve the compiler that performed this search.","Consider source and target finite repair models. Let $\\sigma:X\\to X'$ be a bijection, $\\tau:\\mathcal E\\to\\mathcal E'$ a bijection of experiments, $\\psi:\\mathcal A\\to\\mathcal A'$ a bijection of repairs, and $\\eta_e:O_e\\to O'_{\\tau(e)}$ a bijection of report labels. Assume\n\\begin{align}\\label{eq:transport}\n\\sigma(D_e)&=D'_{\\tau(e)}, &\n\\sigma f_e(x)&=f'_{\\tau(e)}\\sigma(x),","Consider source and target finite repair models. Let $\\sigma:X\\to X'$ be a bijection, $\\tau:\\mathcal E\\to\\mathcal E'$ a bijection of experiments, $\\psi:\\mathcal A\\to\\mathcal A'$ a bijection of repairs, and $\\eta_e:O_e\\to O'_{\\tau(e)}$ a bijection of report labels. Assume\n\\begin{align}\\label{eq:transport}\n\\sigma(D_e)&=D'_{\\tau(e)}, &\n\\sigma f_e(x)&=f'_{\\tau(e)}\\sigma(x),\\\\\n\\eta_e(o_e(x))&=o'_{\\tau(e)}(\\sigma x), &\n\\sigma(G_a)&=G'_{\\psi(a)},\\nonumber\n\\end{align}\nand matching execution costs. These assumptions concern supplied semantics. Their discovery or learned approximation is an additional problem.","\\begin{theorem}[Exact transport of diagnostic procedures]\\label{thm:transport}\nUnder~\\eqref{eq:transport}, relabeling every experiment, branch, and repair transports successful diagnostic trees bijectively between the source support $B_0$ and target support $\\sigma(B_0)$. Execution cost, complete tree storage, and depth are preserved. In particular,\n\\[\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n\\]\nThe same conclusion holds for every selected tree, not only a cost-optimal tree.\n\\end{theorem}\n\\begin{proof}\nA terminal repair is valid on $B_0$ exactly when its image is valid on $\\sigma(B_0)$. For an internal experiment, domain preservation gives uniform admissibility. Transition and observation preservation yield\n\\[\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n\\]\nApply induction to every child. The report bijection preserves the number of children, the maximum execution cost, and the sum of stored child records. Inverse maps give","\\begin{theorem}[Exact transport of diagnostic procedures]\\label{thm:transport}\nUnder~\\eqref{eq:transport}, relabeling every experiment, branch, and repair transports successful diagnostic trees bijectively between the source support $B_0$ and target support $\\sigma(B_0)$. Execution cost, complete tree storage, and depth are preserved. In particular,\n\\[\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n\\]\nThe same conclusion holds for every selected tree, not only a cost-optimal tree.\n\\end{theorem}\n\\begin{proof}\nA terminal repair is valid on $B_0$ exactly when its image is valid on $\\sigma(B_0)$. For an internal experiment, domain preservation gives uniform admissibility. Transition and observation preservation yield\n\\[\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n\\]\nApply induction to every child. The report bijection preserves the number of children, the maximum execution cost, and the sum of stored child records. Inverse maps give the converse correspondence. The","\\begin{theorem}[Exact transport of diagnostic procedures]\\label{thm:transport}\nUnder~\\eqref{eq:transport}, relabeling every experiment, branch, and repair transports successful diagnostic trees bijectively between the source support $B_0$ and target support $\\sigma(B_0)$. Execution cost, complete tree storage, and depth are preserved. In particular,\n\\[\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n\\]\nThe same conclusion holds for every selected tree, not only a cost-optimal tree.\n\\end{theorem}\n\\begin{proof}\nA terminal repair is valid on $B_0$ exactly when its image is valid on $\\sigma(B_0)$. For an internal experiment, domain preservation gives uniform admissibility. Transition and observation preservation yield\n\\[\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n\\]\nApply induction to every child. The report bijection preserves the number of children, the maximum execution cost, and the sum of stored child records. Inverse maps give the converse correspondence. The equality of frontiers follows because costs and dominance are unchanged.\n\\end{proof}","and repair transports successful diagnostic trees bijectively between the source support $B_0$ and target support $\\sigma(B_0)$. Execution cost, complete tree storage, and depth are preserved. In particular,\n\\[\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n\\]\nThe same conclusion holds for every selected tree, not only a cost-optimal tree.\n\\end{theorem}\n\\begin{proof}\nA terminal repair is valid on $B_0$ exactly when its image is valid on $\\sigma(B_0)$. For an internal experiment, domain preservation gives uniform admissibility. Transition and observation preservation yield\n\\[\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n\\]\nApply induction to every child. The report bijection preserves the number of children, the maximum execution cost, and the sum of stored child records. Inverse maps give the converse correspondence. The equality of frontiers follows because costs and dominance are unchanged.\n\\end{proof}","ic trees bijectively between the source support $B_0$ and target support $\\sigma(B_0)$. Execution cost, complete tree storage, and depth are preserved. In particular,\n\\[\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n\\]\nThe same conclusion holds for every selected tree, not only a cost-optimal tree.\n\\end{theorem}\n\\begin{proof}\nA terminal repair is valid on $B_0$ exactly when its image is valid on $\\sigma(B_0)$. For an internal experiment, domain preservation gives uniform admissibility. Transition and observation preservation yield\n\\[\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n\\]\nApply induction to every child. The report bijection preserves the number of children, the maximum execution cost, and the sum of stored child records. Inverse maps give the converse correspondence. The equality of frontiers follows because costs and dominance are unchanged.\n\\end{proof}","e preserved. In particular,\n\\[\n\\mathcal F_h(B_0)=\\mathcal F'_h(\\sigma(B_0)).\n\\]\nThe same conclusion holds for every selected tree, not only a cost-optimal tree.\n\\end{theorem}\n\\begin{proof}\nA terminal repair is valid on $B_0$ exactly when its image is valid on $\\sigma(B_0)$. For an internal experiment, domain preservation gives uniform admissibility. Transition and observation preservation yield\n\\[\n\\sigma(B_{e,o})=B'_{\\tau(e),\\eta_e(o)}.\n\\]\nApply induction to every child. The report bijection preserves the number of children, the maximum execution cost, and the sum of stored child records. Inverse maps give the converse correspondence. The equality of frontiers follows because costs and dominance are unchanged.\n\\end{proof}","\\subsection{Approximate performance transfer}\nSuppose a source comparison gives productivity advantage $\\Delta$ to $U'$ over $U$. Let $\\widetilde U',\\widetilde U$ be actual transported target procedures. If each procedure's productivity differs from its source value by at most $\\varepsilon$, then\n\\begin{equation}\\label{eq:gaintransport}\n\\Rcal'(\\widetilde U')-\\Rcal'(\\widetilde U)\n\\geq\\Delta-2\\varepsilon.\n\\end{equation}\nThis follows by subtracting the two inequalities. A common source of the premise is an exact task correspondence plus a bounded evaluation-law shift. For any fixed gain function in $[-1,1]$, changing the task law by total variation $\\delta$ changes its expectation by at most $2\\delta$. A further pointwise gain discrepancy $\\eta$ therefore gives $\\varepsilon=\\eta+2\\delta$.","If each procedure's productivity differs from its source value by at most $\\varepsilon$, then\n\\begin{equation}\\label{eq:gaintransport}\n\\Rcal'(\\widetilde U')-\\Rcal'(\\widetilde U)\n\\geq\\Delta-2\\varepsilon.\n\\end{equation}\nThis follows by subtracting the two inequalities. A common source of the premise is an exact task correspondence plus a bounded evaluation-law shift. For any fixed gain function in $[-1,1]$, changing the task law by total variation $\\delta$ changes its expectation by at most $2\\delta$. A further pointwise gain discrepancy $\\eta$ therefore gives $\\varepsilon=\\eta+2\\delta$.","ource value by at most $\\varepsilon$, then\n\\begin{equation}\\label{eq:gaintransport}\n\\Rcal'(\\widetilde U')-\\Rcal'(\\widetilde U)\n\\geq\\Delta-2\\varepsilon.\n\\end{equation}\nThis follows by subtracting the two inequalities. A common source of the premise is an exact task correspondence plus a bounded evaluation-law shift. For any fixed gain function in $[-1,1]$, changing the task law by total variation $\\delta$ changes its expectation by at most $2\\delta$. A further pointwise gain discrepancy $\\eta$ therefore gives $\\varepsilon=\\eta+2\\delta$.","begin{equation}\\label{eq:gaintransport}\n\\Rcal'(\\widetilde U')-\\Rcal'(\\widetilde U)\n\\geq\\Delta-2\\varepsilon.\n\\end{equation}\nThis follows by subtracting the two inequalities. A common source of the premise is an exact task correspondence plus a bounded evaluation-law shift. For any fixed gain function in $[-1,1]$, changing the task law by total variation $\\delta$ changes its expectation by at most $2\\delta$. A further pointwise gain discrepancy $\\eta$ therefore gives $\\varepsilon=\\eta+2\\delta$.","widetilde U')-\\Rcal'(\\widetilde U)\n\\geq\\Delta-2\\varepsilon.\n\\end{equation}\nThis follows by subtracting the two inequalities. A common source of the premise is an exact task correspondence plus a bounded evaluation-law shift. For any fixed gain function in $[-1,1]$, changing the task law by total variation $\\delta$ changes its expectation by at most $2\\delta$. A further pointwise gain discrepancy $\\eta$ therefore gives $\\varepsilon=\\eta+2\\delta$.","idetilde U)\n\\geq\\Delta-2\\varepsilon.\n\\end{equation}\nThis follows by subtracting the two inequalities. A common source of the premise is an exact task correspondence plus a bounded evaluation-law shift. For any fixed gain function in $[-1,1]$, changing the task law by total variation $\\delta$ changes its expectation by at most $2\\delta$. A further pointwise gain discrepancy $\\eta$ therefore gives $\\varepsilon=\\eta+2\\delta$.","\\begin{proposition}[The resource price of conservative extension]\\label{prop:extension}\nSuppose an embedding preserves every old procedure's successful behavior and adds at most $d\\geq0$ execution work and $s\\geq0$ stored records. Every old allowance $(b,m)$ is then feasible in the new language under allowance $(b+d,m+s)$. Feasibility at the unchanged allowance follows only when the embedded procedure fits that allowance. Semantic conservativity alone supplies no such guarantee.\n\\end{proposition}\n\\begin{proof}\nEmbed a witness for the old allowance. By hypothesis its behavior is preserved and its costs are bounded by the shifted pair. The unchanged-budget assertion requires the corresponding inequalities. For the final claim, an old one-unit procedure can be embedded with identical output into a new interpreter requiring two units of startup work. It remains semantically av","\\begin{proposition}[The resource price of conservative extension]\\label{prop:extension}\nSuppose an embedding preserves every old procedure's successful behavior and adds at most $d\\geq0$ execution work and $s\\geq0$ stored records. Every old allowance $(b,m)$ is then feasible in the new language under allowance $(b+d,m+s)$. Feasibility at the unchanged allowance follows only when the embedded procedure fits that allowance. Semantic conservativity alone supplies no such guarantee.\n\\end{proposition}\n\\begin{proof}\nEmbed a witness for the old allowance. By hypothesis its behavior is preserved and its costs are bounded by the shifted pair. The unchanged-budget assertion requires the corresponding inequalities. For the final claim, an old one-unit procedure can be embedded with identical output into a new interpreter requiring two units of startup work. It remains semantically available but cannot execute w","\\begin{proposition}[The resource price of conservative extension]\\label{prop:extension}\nSuppose an embedding preserves every old procedure's successful behavior and adds at most $d\\geq0$ execution work and $s\\geq0$ stored records. Every old allowance $(b,m)$ is then feasible in the new language under allowance $(b+d,m+s)$. Feasibility at the unchanged allowance follows only when the embedded procedure fits that allowance. Semantic conservativity alone supplies no such guarantee.\n\\end{proposition}\n\\begin{proof}\nEmbed a witness for the old allowance. By hypothesis its behavior is preserved and its costs are bounded by the shifted pair. The unchanged-budget assertion requires the corresponding inequalities. For the final claim, an old one-unit procedure can be embedded with identical output into a new interpreter requiring two units of startup work. It remains semantically available but cannot execute within the old one-unit allowance.\n\\end{proof","\\begin{proposition}[The resource price of conservative extension]\\label{prop:extension}\nSuppose an embedding preserves every old procedure's successful behavior and adds at most $d\\geq0$ execution work and $s\\geq0$ stored records. Every old allowance $(b,m)$ is then feasible in the new language under allowance $(b+d,m+s)$. Feasibility at the unchanged allowance follows only when the embedded procedure fits that allowance. Semantic conservativity alone supplies no such guarantee.\n\\end{proposition}\n\\begin{proof}\nEmbed a witness for the old allowance. By hypothesis its behavior is preserved and its costs are bounded by the shifted pair. The unchanged-budget assertion requires the corresponding inequalities. For the final claim, an old one-unit procedure can be embedded with identical output into a new interpreter requiring two units of startup work. It remains semantically available but cannot execute within the old one-unit allowance.\n\\end{proof}","\\subsection{The scope in which a short proof is decisive}\nConsider $n\\geq1$ possible immutable cases. Every admitted query is total, nondestructive, binary-valued, and costs one unit. Every available repair costs one unit and succeeds on exactly one case. At least one repair is available for each case. The target is certain successful repair from the complete set of $n$ possibilities. A literal controller stores one record per query or repair node. Redundant unary query nodes are permitted but have their full costs.","\\begin{theorem}[Checked lower-bound closure of a diagnostic search]\\label{thm:optimality-stop}\nEvery successful controller in this scope has worst-case execution work at least\n\\[\n1+\\lceil\\log_2 n\\rceil\n\\]\nand storage at least $2n-1$. If an independently checked candidate attains both bounds and respects the requested horizon, its cost pair is the entire nondominated frontier. Returning this candidate is therefore equivalent in the protected work/storage objective to completing the exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number","\\begin{theorem}[Checked lower-bound closure of a diagnostic search]\\label{thm:optimality-stop}\nEvery successful controller in this scope has worst-case execution work at least\n\\[\n1+\\lceil\\log_2 n\\rceil\n\\]\nand storage at least $2n-1$. If an independently checked candidate attains both bounds and respects the requested horizon, its cost pair is the entire nondominated frontier. Returning this candidate is therefore equivalent in the protected work/storage objective to completing the exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ c","quested horizon, its cost pair is the entire nondominated frontier. Returning this candidate is therefore equivalent in the protected work/storage objective to completing the exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","minated frontier. Returning this candidate is therefore equivalent in the protected work/storage objective to completing the exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}",". Returning this candidate is therefore equivalent in the protected work/storage objective to completing the exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","candidate is therefore equivalent in the protected work/storage objective to completing the exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","exhaustive synthesis.\n\\end{theorem}\n\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","\\begin{proof}\nEvery case produces one deterministic query history. Two different cases cannot terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","terminate at the same repair leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","leaf because no repair succeeds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","ds in both. Thus there are at least $n$ nonempty leaves. A tree of binary query depth $d$ has at most $2^d$ leaves, so $d\\geq\\lceil\\log_2 n\\rceil$. Every path also executes a unit-cost repair. For storage, let $l$ be the number of leaves and let $i_1,i_2$ count unary and binary internal nodes. Counting edges in a finite rooted tree gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\geq2n-1$. A successful candidate attaining both coordinatewise lower bounds dominates every other attainable cost pair. Its independent validity check and horizon check make it a feasible witness, so the Pareto frontier contains only that pair.\n\\end{proof}","\\begin{proposition}[Read cost of the certified binary-family shortcut]\\label{prop:compiler-work}\nLet $n=2^k$ with $k\\geq1$. Suppose there are $q$ total unit-cost binary immutable queries, and the balanced proposer produces a perfect depth-$k$ tree. In the stipulated implementation, the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the","\\begin{proposition}[Read cost of the certified binary-family shortcut]\\label{prop:compiler-work}\nLet $n=2^k$ with $k\\geq1$. Suppose there are $q$ total unit-cost binary immutable queries, and the balanced proposer produces a perfect depth-$k$ tree. In the stipulated implementation, the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\","\\begin{proposition}[Read cost of the certified binary-family shortcut]\\label{prop:compiler-work}\nLet $n=2^k$ with $k\\geq1$. Suppose there are $q$ total unit-cost binary immutable queries, and the balanced proposer produces a perfect depth-$k$ tree. In the stipulated implementation, the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin","\\begin{proposition}[Read cost of the certified binary-family shortcut]\\label{prop:compiler-work}\nLet $n=2^k$ with $k\\geq1$. Suppose there are $q$ total unit-cost binary immutable queries, and the balanced proposer produces a perfect depth-$k$ tree. In the stipulated implementation, the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partitio","\\begin{proposition}[Read cost of the certified binary-family shortcut]\\label{prop:compiler-work}\nLet $n=2^k$ with $k\\geq1$. Suppose there are $q$ total unit-cost binary immutable queries, and the balanced proposer produces a perfect depth-$k$ tree. In the stipulated implementation, the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the su","$ total unit-cost binary immutable queries, and the balanced proposer produces a perfect depth-$k$ tree. In the stipulated implementation, the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coord…",", the semantic guard reads every query row once, the proposer scans all $q$ queries at each internal support, and verification performs one support reconstruction and one complete case replay. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus …","play. The total number of transition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus the balancing rule chooses an equal split until singleton leaves are reached. When a two-case support admits a singleton tie, either split reaches singleton leaves. Relabeling the cases o…","ransition/report record reads is\n\\begin{equation}\\label{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus the balancing rule chooses an equal split until singleton leaves are reached. When a two-case support admits a singleton tie, either split reaches singleton leaves. Relabeling the cases or reversing report names pr…","el{eq:compiler-reads}\nR_{\\mathrm{shortcut}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus the balancing rule chooses an equal split until singleton leaves are reached. When a two-case support admits a singleton tie, either split reaches singleton leaves. Relabeling the cases or reversing report names preserves all branch cardinalities and this argument.\n\\…","t}}=nq(k+1)+2nk.\n\\end{equation}\nThe usual hidden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus the balancing rule chooses an equal split until singleton leaves are reached. When a two-case support admits a singleton tie, either split reaches singleton leaves. Relabeling the cases or reversing report names preserves all branch cardinalities and this argument.\n\\end{proof}","idden-cell family with all singleton queries and $k$ coordinate queries satisfies the premise, including after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus the balancing rule chooses an equal split until singleton leaves are reached. When a two-case support admits a singleton tie, either split reaches singleton leaves. Relabeling the cases or reversing report names preserves all branch cardinalities and this argument.\n\\end{proof}","ncluding after arbitrary cell relabeling, report-bit flips, or reordering of the query list.\n\\end{proposition}\n\\begin{proof}\nThe guard uses $nq$ reads. At each of the $k$ query levels, the current supports partition the $n$ cases, so scanning all queries uses $nq$ reads per level. Every case traverses $k$ query nodes in the support reconstruction and again in the independent replay, contributing $2nk$. Summing gives~\\eqref{eq:compiler-reads}. In the hidden-cell family, every nonsingleton support reached by coordinate splits is a subcube. A remaining coordinate divides it equally, while a singleton query cannot give a smaller maximum branch. Thus the balancing rule chooses an equal split until singleton leaves are reached. When a two-case support admits a singleton tie, either split reaches singleton leaves. Relabeling the cases or reversing report names preserves all branch cardinalities and this argument.\n\\end{proof}","\\subsection{Some verification work cannot disappear}\n\\begin{proposition}[A table-access floor for universal scope certification]\\label{prop:table-floor}\nSuppose a deterministic validator receives an arbitrary table of $N\\geq1$ independently variable records only through individual record reads. It must accept every table whose records all satisfy a fixed per-record predicate and reject every table with a violating record. Assume each record position admits both a satisfying and a violating value. On every accepted table, the validator must read all $N$ positions at least once.\n\\end{proposition}\n\\begin{proof}\nSuppose it accepts a satisfying table while leaving position $j$ unread. Change only that position to a violating value. Every record the validator reads has the same value, so determinism gives the same sequence of reads and the same acceptance. This contradicts rejection of every table containing a violati","pose a deterministic validator receives an arbitrary table of $N\\geq1$ independently variable records only through individual record reads. It must accept every table whose records all satisfy a fixed per-record predicate and reject every table with a violating record. Assume each record position admits both a satisfying and a violating value. On every accepted table, the validator must read all $N$ positions at least once.\n\\end{proposition}\n\\begin{proof}\nSuppose it accepts a satisfying table while leaving position $j$ unread. Change only that position to a violating value. Every record the validator reads has the same value, so determinism gives the same sequence of reads and the same acceptance. This contradicts rejection of every table containing a violation. Therefore no accepted table can have an unread position.\n\\end{proof}","ual record reads. It must accept every table whose records all satisfy a fixed per-record predicate and reject every table with a violating record. Assume each record position admits both a satisfying and a violating value. On every accepted table, the validator must read all $N$ positions at least once.\n\\end{proposition}\n\\begin{proof}\nSuppose it accepts a satisfying table while leaving position $j$ unread. Change only that position to a violating value. Every record the validator reads has the same value, so determinism gives the same sequence of reads and the same acceptance. This contradicts rejection of every table containing a violation. Therefore no accepted table can have an unread position.\n\\end{proof}","The developmental totals are $10{,}256$ reads for the parent, $10{,}380$ for the guarded first-query rule, and $488$ for the guarded balanced rule. The selector therefore installs the balanced rule. These totals concern a fixed, fully enumerated development catalogue. They do not imply a statistical generalization guarantee over an unknown problem distribution. The selected artifact changes the control flow of future compilations without containing the hidden state of a future repair episode.","\\begin{center}\\begin{tabular}{@{}rrrr@{}}\\toprule\nCases & Parent compiler reads & Descendant reads & Returned $(\\text{work},\\text{storage})$","midrule\n2 & 16 & 16 & $(2,3)$","4 & 352 & 88 & $(3,7)$","8 & 9,904 & 400 & $(4,15)$","16 & 576,128 & 1,728 & $(5,31)$","Under a common cap of $2{,}000$ table reads, the descendant returns a checked optimal sixteen-cell controller. The parent reaches the cap without returning its complete frontier. Its outcome is recorded as \\emph{incomplete within the allowance}, not as a proof that no controller exists. Without that cap, the parent also returns the same optimal cost pair. The difference lies in discovery work rather than repair validity or online diagnostic depth.","lowdown from a certified two-worker race]\\label{prop:race}\nSuppose a parent and a shortcut worker can be paused after each unit of a common work measure and resumed without losing progress. Each may announce completion only after producing a result certified for the same target. Alternate one unit of each worker and stop at the first certified completion. If their standalone completion costs are $D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","nd a shortcut worker can be paused after each unit of a common work measure and resumed without losing progress. Each may announce completion only after producing a result certified for the same target. Alternate one unit of each worker and stop at the first certified completion. If their standalone completion costs are $D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","sing progress. Each may announce completion only after producing a result certified for the same target. Alternate one unit of each worker and stop at the first certified completion. If their standalone completion costs are $D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","rtified for the same target. Alternate one unit of each worker and stop at the first certified completion. If their standalone completion costs are $D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","and stop at the first certified completion. If their standalone completion costs are $D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","completion costs are $D_0,D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","D_1\\in\\mathbb N_{>0}\\cup\\{\\infty\\}$ and at least one is finite, at most $2\\min\\{D_0,D_1\\}$ worker units are executed. If each dispatched unit costs at most an additional $c\\geq0$ scheduler units in the same measure, total charged work is at most $2(1+c)\\min\\{D_0,D_1\\}$.\n\\end{proposition}\n\\begin{proof}\nAfter $2d$ dispatched worker units, each worker has received exactly $d$ units of progress. Set $d$ to the smaller finite standalone completion cost. That worker has completed its certification by this point, so the race has stopped no later. Multiplying the number of dispatches by the maximum combined work per dispatch gives the overhead bound.\n\\end{proof}","bottomrule\n\\end{tabular}\\end{center}\nThe separate common input-shape validation consumes another 320 records before the measured compilation begins. It is not included in the $2{,}000$-read compiler allowance. Counting that shared setup on both sides gives 2,048 reads for the descendant and 576,448 for the parent. No particular implementation is permitted to hide its extra guard inside an uncharged setup step.","\\subsection{The information available inside a search}\nA compiler is considering a binary query on a support of $m$ possible immutable cases. It reads the query's public model rows in a fixed order. After a prefix, $a$ of the inspected cases have report zero and $b$ have report one. There are $m-a-b$ rows left to read. If this query is the first of at most $d\\geq1$ remaining queries, either child can contain at most\n\\[\nC=2^{d-1}\n\\]\ncases when distinct cases need distinct repair leaves. The local cardinality premise $m\\leq2C$ is checked before a root is investigated. An overfull support cannot fit the remaining binary depth, even before any rows of the proposed query are read.","ation available inside a search}\nA compiler is considering a binary query on a support of $m$ possible immutable cases. It reads the query's public model rows in a fixed order. After a prefix, $a$ of the inspected cases have report zero and $b$ have report one. There are $m-a-b$ rows left to read. If this query is the first of at most $d\\geq1$ remaining queries, either child can contain at most\n\\[\nC=2^{d-1}\n\\]\ncases when distinct cases need distinct repair leaves. The local cardinality premise $m\\leq2C$ is checked before a root is investigated. An overfull support cannot fit the remaining binary depth, even before any rows of the proposed query are read.","le immutable cases. It reads the query's public model rows in a fixed order. After a prefix, $a$ of the inspected cases have report zero and $b$ have report one. There are $m-a-b$ rows left to read. If this query is the first of at most $d\\geq1$ remaining queries, either child can contain at most\n\\[\nC=2^{d-1}\n\\]\ncases when distinct cases need distinct repair leaves. The local cardinality premise $m\\leq2C$ is checked before a root is investigated. An overfull support cannot fit the remaining binary depth, even before any rows of the proposed query are read.","The initial search representation contains $(a,b,m)$ but omits $C$. This can lose a distinction relevant to the search itself. For example, $(a,b,m)=(0,3,4)$ cannot have a completion whose two branch sizes are at most two. It can have a completion with branch sizes $(1,3)$ when the allowed child capacity is four. The read prefix is the same. The permitted future investigation is different.","\\begin{proposition}[Feature sufficiency for a search decision]\\label{prop:search-feature}\nLet $\\mathcal P$ be a finite set of search contexts, let $r:\\mathcal P\\to Z$ be their available feature map, and let $\\ell:\\mathcal P\\to\\{0,1\\}$ be the required decision label. A deterministic rule $f:Z\\to\\{0,1\\}$ satisfying $fr=\\ell$ exists if and only if $\\ell$ is constant on each nonempty fibre of $r$. A pair in one fibre with different labels rules out every such rule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A","\\begin{proposition}[Feature sufficiency for a search decision]\\label{prop:search-feature}\nLet $\\mathcal P$ be a finite set of search contexts, let $r:\\mathcal P\\to Z$ be their available feature map, and let $\\ell:\\mathcal P\\to\\{0,1\\}$ be the required decision label. A deterministic rule $f:Z\\to\\{0,1\\}$ satisfying $fr=\\ell$ exists if and only if $\\ell$ is constant on each nonempty fibre of $r$. A pair in one fibre with different labels rules out every such rule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A conflicting pair violates necessity for every possible $f$.","\\begin{proposition}[Feature sufficiency for a search decision]\\label{prop:search-feature}\nLet $\\mathcal P$ be a finite set of search contexts, let $r:\\mathcal P\\to Z$ be their available feature map, and let $\\ell:\\mathcal P\\to\\{0,1\\}$ be the required decision label. A deterministic rule $f:Z\\to\\{0,1\\}$ satisfying $fr=\\ell$ exists if and only if $\\ell$ is constant on each nonempty fibre of $r$. A pair in one fibre with different labels rules out every such rule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A conflicting pair violates necessity for every possible $f$.\n\\end{proof}","xts, let $r:\\mathcal P\\to Z$ be their available feature map, and let $\\ell:\\mathcal P\\to\\{0,1\\}$ be the required decision label. A deterministic rule $f:Z\\to\\{0,1\\}$ satisfying $fr=\\ell$ exists if and only if $\\ell$ is constant on each nonempty fibre of $r$. A pair in one fibre with different labels rules out every such rule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A conflicting pair violates necessity for every possible $f$.\n\\end{proof}","e map, and let $\\ell:\\mathcal P\\to\\{0,1\\}$ be the required decision label. A deterministic rule $f:Z\\to\\{0,1\\}$ satisfying $fr=\\ell$ exists if and only if $\\ell$ is constant on each nonempty fibre of $r$. A pair in one fibre with different labels rules out every such rule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A conflicting pair violates necessity for every possible $f$.\n\\end{proof}","ch nonempty fibre of $r$. A pair in one fibre with different labels rules out every such rule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A conflicting pair violates necessity for every possible $f$.\n\\end{proof}","ule, irrespective of its program length.\n\\end{proposition}\n\\begin{proof}\nIf $f$ exists, equal features have the same output under $f$ and hence the same label. Conversely, assign to each attained feature value its common label. This is well defined when the fibre condition holds. Values outside the image of $r$ can be assigned arbitrarily. A conflicting pair violates necessity for every possible $f$.\n\\end{proof}","The criterion is elementary factorization, not a new general learning theorem. Its role is to distinguish insufficient information from inadequate search effort. In the finite construction, exhaustive search for a larger formula over $(a,b,m)$ cannot fix the conflict. The supplied feature registry contains the remaining-depth capacity $C$. Exposing it resolves this particular obstruction. Computing $C$ from an already known depth is not a physical experiment and does not identify the next episode's hidden defect.","criterion is elementary factorization, not a new general learning theorem. Its role is to distinguish insufficient information from inadequate search effort. In the finite construction, exhaustive search for a larger formula over $(a,b,m)$ cannot fix the conflict. The supplied feature registry contains the remaining-depth capacity $C$. Exposing it resolves this particular obstruction. Computing $C$ from an already known depth is not a physical experiment and does not identify the next episode's hidden defect.","\\begin{counterexample}[The same prefix requires different rejection decisions]\\label{ex:missing-capacity}\nFor the two contexts $(a,b,m,C)=(0,3,4,2)$ and $(0,3,4,4)$, the initial feature vector $(a,b,m)$ is identical. In the first, three cases already occupy a child with capacity two. In the second, assigning the remaining case to the other child gives a capacity-feasible completion. Thus the predicate ``no capacity-feasible completion exists'' cannot be computed from the initial features alone. Adding the capacity makes this specific decision representable. It does not establish that the actual query library can distinguish every resulting child.\n\\end{counterexample}","\\section{Composing a rule from primitive syntax}\\label{sec:grammar-learning}\nThe enlarged feature vector is $(a,b,m,C)$. The reference language has six integer atoms, $a,b,m,C,0,1$, three binary arithmetic operations, $\\max$, $\\min$, and addition, and the five comparisons $>,<,\\geq,\\leq,=$. A candidate is a Boolean comparison with at most five syntax nodes. No general program execution is admitted. The interpreter evaluates only these finite arithmetic trees.","\\subsection{What the teacher knows}\nDevelopment uses the 198 count-prefix contexts with $C\\in\\{1,2,4\\}$, $2\\leq m\\leq2C$, and $a,b\\geq0$, $a+b\\leq m$. For each context, the teacher enumerates the possible numbers of report-one cases among the unread rows. It labels the context as reject precisely when no such completion can put at most $C$ cases in either branch. The teacher therefore solves a small finite feasibility problem. It is not reading a label generated by the final formula.","\\begin{theorem}[Finite counterexample refinement within a grammar]\\label{thm:grammar-refinement}\nLet $\\mathcal G$ be a finite nonempty set of deterministic candidate programs, let $\\mathcal P$ be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.","\\begin{theorem}[Finite counterexample refinement within a grammar]\\label{thm:grammar-refinement}\nLet $\\mathcal G$ be a finite nonempty set of deterministic candidate programs, let $\\mathcal P$ be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program becau","\\begin{theorem}[Finite counterexample refinement within a grammar]\\label{thm:grammar-refinement}\nLet $\\mathcal G$ be a finite nonempty set of deterministic candidate programs, let $\\mathcal P$ be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the const","\\begin{theorem}[Finite counterexample refinement within a grammar]\\label{thm:grammar-refinement}\nLet $\\mathcal G$ be a finite nonempty set of deterministic candidate programs, let $\\mathcal P$ be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all","\\begin{theorem}[Finite counterexample refinement within a grammar]\\label{thm:grammar-refinement}\nLet $\\mathcal G$ be a finite nonempty set of deterministic candidate programs, let $\\mathcal P$ be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid c","ic candidate programs, let $\\mathcal P$ be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid counterexample. There are only $|\\mathcal G|$ candidates, so after at most that many distinct proposals the teacher has accepted one or all candidates are excluded. If…","be a finite domain, and let $\\ell$ be its specified labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid counterexample. There are only $|\\mathcal G|$ candidates, so after at most that many distinct proposals the teacher has accepted one or all candidates are excluded. If a returned program were preceded by a g…","fied labeling. Suppose a teacher either returns a point where the proposed program differs from $\\ell$ or certifies agreement on all of $\\mathcal P$. A learner that proposes only candidates consistent with every retained counterexample rejects each proposed program at most once. It returns a correct program on $\\mathcal P$, or certifies that none in $\\mathcal G$ meets the labeling, after at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid counterexample. There are only $|\\mathcal G|$ candidates, so after at most that many distinct proposals the teacher has accepted one or all candidates are excluded. If a returned program were preceded by a globally correct program in the fixed order, the…","r at most $|\\mathcal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid counterexample. There are only $|\\mathcal G|$ candidates, so after at most that many distinct proposals the teacher has accepted one or all candidates are excluded. If a returned program were preceded by a globally correct program in the fixed order, the earlier program would remain consistent and would have been selected instead. This proves the ordering assertion.\n\\end{proof}","cal G|$ teacher calls. With a fixed total order and the earliest-consistent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid counterexample. There are only $|\\mathcal G|$ candidates, so after at most that many distinct proposals the teacher has accepted one or all candidates are excluded. If a returned program were preceded by a globally correct program in the fixed order, the earlier program would remain consistent and would have been selected instead. This proves the ordering assertion.\n\\end{proof}","ent selection rule, any returned program is the earliest one that agrees on the entire domain.\n\\end{theorem}\n\\begin{proof}\nA counterexample removes the current program because it must be included in the constraints for every later proposal. Hence that program cannot be proposed again. A program agreeing with $\\ell$ on all of $\\mathcal P$ is never removed by a valid counterexample. There are only $|\\mathcal G|$ candidates, so after at most that many distinct proposals the teacher has accepted one or all candidates are excluded. If a returned program were preceded by a globally correct program in the fixed order, the earlier program would remain consistent and would have been selected instead. This proves the ordering assertion.\n\\end{proof}","\\subsection{The synthesized expression and its provenance}\nThe reference learner encounters three disagreements. Its first candidate, $a>a$, fails on a prefix with too many ones. The next proposed candidate, $b>a$, rejects a prefix that can still be completed. The third, $b>C$, misses a prefix with too many zeros. The fourth teacher proposal is\n\\begin{equation}\\label{eq:learned-prune}\n\\rho(a,b,m,C)=\\mathbf 1\\{\\max(a,b)>C\\}.\n\\end{equation}\nThe formula uses five syntax nodes and passes all 198 developmental contexts. The enumeration generates all 3,960 AST roots and considers 190 of them, including candidates rejected directly by retained examples. There are 282 retained-example checks, 210 full-domain teacher comparisons, 344 completion-count checks to construct the labels, and 1,898 interpreted syntax-node visits in the learne","\\subsection{The synthesized expression and its provenance}\nThe reference learner encounters three disagreements. Its first candidate, $a>a$, fails on a prefix with too many ones. The next proposed candidate, $b>a$, rejects a prefix that can still be completed. The third, $b>C$, misses a prefix with too many zeros. The fourth teacher proposal is\n\\begin{equation}\\label{eq:learned-prune}\n\\rho(a,b,m,C)=\\mathbf 1\\{\\max(a,b)>C\\}.\n\\end{equation}\nThe formula uses five syntax nodes and passes all 198 developmental contexts. The enumeration generates all 3,960 AST roots and considers 190 of them, including candidates rejected directly by retained examples. There are 282 retained-example checks, 210 full-domain teacher comparisons, 344 completion-count checks to construct the labels, and 1,898 interpreted syntax-node visits in the learner's recorded comparisons. The full grammar has 19,440 literal node occurre","\\subsection{The synthesized expression and its provenance}\nThe reference learner encounters three disagreements. Its first candidate, $a>a$, fails on a prefix with too many ones. The next proposed candidate, $b>a$, rejects a prefix that can still be completed. The third, $b>C$, misses a prefix with too many zeros. The fourth teacher proposal is\n\\begin{equation}\\label{eq:learned-prune}\n\\rho(a,b,m,C)=\\mathbf 1\\{\\max(a,b)>C\\}.\n\\end{equation}\nThe formula uses five syntax nodes and passes all 198 developmental contexts. The enumeration generates all 3,960 AST roots and considers 190 of them, including candidates rejected directly by retained examples. There are 282 retained-example checks, 210 full-domain teacher comparisons, 344 completion-count checks to construct the labels, and 1,898 interpreted syntax-node visits in the learner's recorded comparisons. The full grammar has 19,440 literal node occurrences. Shared in-memory subexpressions can reduce actual storage,","\\subsection{The synthesized expression and its provenance}\nThe reference learner encounters three disagreements. Its first candidate, $a>a$, fails on a prefix with too many ones. The next proposed candidate, $b>a$, rejects a prefix that can still be completed. The third, $b>C$, misses a prefix with too many zeros. The fourth teacher proposal is\n\\begin{equation}\\label{eq:learned-prune}\n\\rho(a,b,m,C)=\\mathbf 1\\{\\max(a,b)>C\\}.\n\\end{equation}\nThe formula uses five syntax nodes and passes all 198 developmental contexts. The enumeration generates all 3,960 AST roots and considers 190 of them, including candidates rejected directly by retained examples. There are 282 retained-example checks, 210 full-domain teacher comparisons, 344 completion-count checks to construct the labels, and 1,898 interpreted syntax-node visits in the learner's recorded comparisons. The full grammar has 19,440 literal node occurrences. Shared in-memory subexpressions can reduce actual storage, so this last quantity is not a measured byte count.","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ re","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases bet","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these t","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, fi","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and put","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the re","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns","\\begin{proposition}[Exact obstruction to a capacity-feasible completion]\\label{prop:prefix-capacity}\nLet $C,m,a,b$ be integers with $C\\geq1$, $2\\leq m\\leq2C$, $a,b\\geq0$, and $a+b\\leq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an inst","eq m$. A binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","binary completion of the remaining $m-a-b$ reports with both final branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","branch counts at most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","most $C$ exists if and only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","nd only if $\\max(a,b)\\leq C$. Consequently, a prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","prefix satisfying $\\max(a,b)>C$ can be rejected before its remaining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","ining rows are read, without rejecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","jecting any completion that satisfies these two capacity constraints.\n\\end{proposition}\n\\begin{proof}\nCounts do not decrease as rows are read, so $a>C$ or $b>C$ makes a feasible completion impossible. Conversely, suppose $a,b\\leq C$. Let $r=m-a-b$. There are $C-a$ available places in the zero branch and $C-b$ in the one branch. Both are nonnegative and their sum is $2C-a-b\\geq r$. Distribute the $r$ remaining cases between these two capacities, filling either branch up to its capacity and putting any remainder in the other. This gives the required completion. The argument concerns possible report assignments, not whether an installed query has that assignment.\n\\end{proof}","The theorem applies to arbitrary positive integer capacities. The connection to query depth uses $C=2^{d-1}$. An actual search separately checks that both observed branches are nonempty, since a redundant unary query cannot be part of a controller attaining the minimal literal storage. It also verifies all later subtrees. Capacity feasibility is necessary, not sufficient, for actual repair solvability.","\\begin{figure}[!ht]\\centering\n\\includegraphics[width=.90\\linewidth]{figures/prefix_capacity.pdf}\n\\caption{An exact count-prefix model with eight cases and child capacity four. A prefix outside the square $a,b\\leq4$ cannot be completed within that capacity. Prefixes also satisfy $a+b\\leq8$. Surviving the count test does not guarantee that the installed query library can separate the children. The dotted arrow is one unfinished read that can stop immediately after its fifth equal report.}\\label{fig:prefix-capacity}\n\\end{figure}","\\subsection{A deliberately narrow certificate checker}\nThe executable gate accepts only ASTs that normalize by swapping a comparison's sides or maximum's arguments to $\\max(a,b)>C$. It rejects other finite-consistent formulas rather than claim to prove arbitrary arithmetic identities. Thus the universal justification is the preceding prose proof, coupled to a project-local structural checker and a runtime scope guard. This is not a Lean certificate or an independent theorem-proving service.","\\begin{counterexample}[Agreement on training prefixes does not establish scope]\\label{ex:prune-extrapolation}\nThe formula $\\max(a,b)>\\min(C,4)$ agrees with~\\eqref{eq:learned-prune} on every developmental context because every developmental capacity is at most four. At $(a,b,m,C)=(0,5,16,8)$ it rejects, while a completion with final branch sizes $(8,8)$ exists. A larger input has not merely changed the frequency of familiar cases. It has reached a scope on which the bounded-cap formula was never justified. The structural gate does not accept that formula.\n\\end{counterexample}","\\subsection{The same search with fewer model-row reads}\nThe reference compiler first checks the immutable, unit-cost, binary-query and singleton-repair scope of Theorem~\\ref{thm:optimality-stop}. It then conducts an ordered depth-first feasibility search at the information lower depth $d_0=\\lceil\\log_2|B_0|\\rceil$. The full-scan version reads every current row of a candidate query before rejecting an overfull child. The streaming version consults the certified prefix rule after each read. Both use the same query order, report order, remaining depth, and stopping rules. If a root has capacity-feasible children but a child cannot be completed using the actual library, the search backtracks to the next root.","A successful controller is separately checked and attains both lower bounds. When no controller exists at $d_0$, a requested horizon equal to $d_0$ permits a genuine finite infeasibility result. A longer requested horizon instead invokes the unchanged general compiler. Failure of the initial scope guard also invokes that compiler. Exhaustion of a model-read allowance returns incomplete, never infeasible merely because the allowance ran out.","Within a root scan, after a prior prefix has passed, the predicate can be evaluated by comparing only the newly incremented count with $C$. This lowering preserves every rejection decision.\n\\end{theorem}\n\\begin{proof}\nBy Proposition~\\ref{prop:prefix-capacity}, every root rejected early would be rejected after its full scan. A root not rejected early has its complete rows read and produces the same two child supports in the same order. Induction on remaining depth therefore gives the same child searches, backtracking choices, first completed controller, and lower-depth failure. Nonempty binary splits strictly reduce support, and the depth decreases, so the finite search terminates. Any successful tree can omit redundant unary roots without losing case information in this immutable unit-cost scope. The search consequently cover","For the lowering, both prior counts are at most $C$, since otherwise the scan would already have stopped. One report increments exactly one count. The unchanged count remains at most $C$, so the maximum exceeds $C$ exactly when the incremented count does. This implication requires the stated loop invariant.\n\\end{proof}","\\begin{counterexample}[Local capacity without a completing query library]\\label{ex:local-capacity}\nFour cases admit singleton repairs. The only available query divides them into $\\{0,1\\}$ and $\\{2,3\\}$ and leaves the state unchanged. At depth two, the root satisfies the $2/2$ capacity test. Repeating the only query cannot separate either pair, so no successful depth-two tree exists. Accepting the first capacity-feasible root as a finished certificate would be wrong. The streaming compiler searches its children, records the failure, and concludes infeasibility only after exhausting the actual library.\n\\end{counterexample}","Let $B$ be a finite nonempty immutable-case support and let $G_r\\subseteq B$ be each repair's success set restricted to it. Define $\\kappa(B)$ to be the smallest number of repair-success sets whose union covers $B$. Set $\\kappa(B)=\\infty$ when no such cover exists. This number concerns the repaired objective, not the number of hidden states. Computing it can itself be expensive.","\\begin{proposition}[Repair-relative leaf requirements]\\label{prop:repair-cover}\nFor finite immutable cases with binary deterministic queries, any certain successful diagnostic tree of query depth at most $d$ has at least $\\kappa(B)$ nonempty repair leaves. Consequently $\\kappa(B)\\leq2^d$ is necessary for success, and literal tree storage is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records,","\\begin{proposition}[Repair-relative leaf requirements]\\label{prop:repair-cover}\nFor finite immutable cases with binary deterministic queries, any certain successful diagnostic tree of query depth at most $d$ has at least $\\kappa(B)$ nonempty repair leaves. Consequently $\\kappa(B)\\leq2^d$ is necessary for success, and literal tree storage is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes add","\\begin{proposition}[Repair-relative leaf requirements]\\label{prop:repair-cover}\nFor finite immutable cases with binary deterministic queries, any certain successful diagnostic tree of query depth at most $d$ has at least $\\kappa(B)$ nonempty repair leaves. Consequently $\\kappa(B)\\leq2^d$ is necessary for success, and literal tree storage is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has","\\begin{proposition}[Repair-relative leaf requirements]\\label{prop:repair-cover}\nFor finite immutable cases with binary deterministic queries, any certain successful diagnostic tree of query depth at most $d$ has at least $\\kappa(B)$ nonempty repair leaves. Consequently $\\kappa(B)\\leq2^d$ is necessary for success, and literal tree storage is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking","sequently $\\kappa(B)\\leq2^d$ is necessary for success, and literal tree storage is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking a completing tree. Thus the necessary conditions are not sufficient.\n\\end{proof}","for success, and literal tree storage is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking a completing tree. Thus the necessary conditions are not sufficient.\n\\end{proof}","ge is at least $2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking a completing tree. Thus the necessary conditions are not sufficient.\n\\end{proof}","2\\kappa(B)-1$. These conditions are not sufficient for success under a specified query library.\n\\end{proposition}\n\\begin{proof}\nEvery initial case reaches one repair leaf. The set of cases reaching that leaf is contained in its repair's success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking a completing tree. Thus the necessary conditions are not sufficient.\n\\end{proof}","success set. The success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking a completing tree. Thus the necessary conditions are not sufficient.\n\\end{proof}","success sets of the attained leaves therefore cover $B$, so the number of leaves is at least $\\kappa(B)$. A binary tree of depth $d$ has at most $2^d$ leaves. The edge-count argument from Theorem~\\ref{thm:optimality-stop} gives at least twice the number of nonempty leaves minus one literal records, with any unary nodes adding further storage. The preceding one-query example has $\\kappa(B)=4$ and $d=2$, satisfying both numerical lower bounds while lacking a completing tree. Thus the necessary conditions are not sufficient.\n\\end{proof}","For singleton repairs, $\\kappa(B)=|B|$, recovering the earlier argument. With a universal repair, $\\kappa(B)=1$ even when many cases remain unknown. The executable streaming path retains the easier singleton guard rather than quietly substituting a hard cover computation whose cost would be uncharged. The broader proposition indicates how a different pruning contract could be developed, not a hidden extension of the present code.","If a model-row read has declared price $w>0$ and a new comparison has price one, the additional per-use saving under this conservative two-unit account is\n\\[\n\\Delta(w)=w\\,(R_{\\mathrm{full}}-R_{\\mathrm{stream}})-N_{\\mathrm{compare}}.\n\\]\nFor the 64-case row it is positive when $w>14209/8287$. At $w=1$ it is negative. The price is a modeling choice, not a measured statement about Python or a physical machine. Interpreted AST visits, installed code, grammar construction, labels, and development search remain additional recorded coordinates. No weighted sum is presented as a universal cost of intelligence.","If development has total charged cost $D$ in a chosen common accounting unit and the same future use has saving $\\Delta(w)>0$, $N$ repetitions repay that cost only when $N\\Delta(w)>D$. This is the amortization argument already used earlier, applied to the learned compiler rule. It does not authorize ignoring development because deployment might eventually repeat often.","The new object is neither a completed diagnostic tree nor the target expression $\\max(a,b)>C$. It is a library of checked equations and an ordered map from grammar expressions to earlier equivalent representatives. Learning the library does not require the labels of a later synthesis task. Once the representation has been constructed and validated, subsequent tasks still begin with empty task-specific counterexample sets.","\\subsection{Four objects and two meanings of success}\nLet $\\Gcal=(p_0,\\ldots,p_{N-1})$ be a finite ordered grammar of pure, total expressions. A target task supplies contexts $z\\in D$ and desired values $f(z)$. An ordinary learner uses counterexamples to find the first grammar expression that agrees with all of $D$. A separate equational learner observes expressions on a development probe set and proposes equations. A checker admits only equations supported by a derivation in the declared arithmetic theory.","\\section{Learning equations without confusing probes with proofs}\\label{sec:equation-learning}\nThe reference language uses exact integers, the total operations $+$, $\\min$, $\\max$, and the comparisons $<$, $\\le$, $>$, $\\ge$, and $=$. A finite schema language contains 135 terms built from variables $x,y$, small integer literals, Boolean literals, and at most one binary operation. The proposer evaluates those schemas on the nine contexts $(x,y)\\in\\{1,2,3\\}^2$.","sing probes with proofs}\\label{sec:equation-learning}\nThe reference language uses exact integers, the total operations $+$, $\\min$, $\\max$, and the comparisons $<$, $\\le$, $>$, $\\ge$, and $=$. A finite schema language contains 135 terms built from variables $x,y$, small integer literals, Boolean literals, and at most one binary operation. The proposer evaluates those schemas on the nine contexts $(x,y)\\in\\{1,2,3\\}^2$.","\\begin{counterexample}[A probe-equivalent rule that fails on a new input]\\label{ex:probe-equation}\nOn the development inputs $x\\in\\{1,2,3\\}$, the terms $\\max(x,1)$ and $x$ agree. At $x=0$ they differ. The proposed equation is not admitted by the proof checker. By contrast, $\\max(x,0)=x$ holds on all nonnegative integers but not on all integers. A checker using the stronger integer-wide contract may fail to admit a rule that would be valid under a narrower natural-number contract. Rejection, counterexample, and scope restriction remain different outcomes.\n\\end{counterexample}","\\section{A certified quotient preserves the investigation's evidence}\\label{sec:ordered-quotient}\nWrite $\\llbracket p\\rrbracket(z)$ for an expression's value in context $z$. Let $n(p)$ be a deterministic representation produced by finitely many checked substitutions of valid equations. We only require\n\\[\n n(p)=n(q)\\quad\\Longrightarrow\\quad\n \\llbracket p\\rrbracket(z)=\\llbracket q\\rrbracket(z)\n \\quad\\text{for every admitted }z.\n\\]\nThe reverse implication is unnecessary. Distinct normal forms may denote the same function. Missing such an equivalence wastes an optimization opportunity but does not remove a valid candidate.","\\begin{theorem}[Sound ordered quotienting from checked equations]\\label{thm:certified-quotient}\nAssume the expression language is pure and total, every admitted schema equation is valid for all well-typed substitutions in the common semantic domain, and each normalization terminates after valid contextual replacements. For each $p_i$, keep its earliest grammar representative\n\\[\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n\\]\nThen $\\rho(i)\\le i$, $p_i$ and $p_{\\rho(i)}$ agree on every admitted context, and every pointwise specification realizable in $\\Gcal$ is realizable in its retained representatives. No confluence or complete semantic-equivalence decision procedure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivi","\\begin{theorem}[Sound ordered quotienting from checked equations]\\label{thm:certified-quotient}\nAssume the expression language is pure and total, every admitted schema equation is valid for all well-typed substitutions in the common semantic domain, and each normalization terminates after valid contextual replacements. For each $p_i$, keep its earliest grammar representative\n\\[\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n\\]\nThen $\\rho(i)\\le i$, $p_i$ and $p_{\\rho(i)}$ agree on every admitted context, and every pointwise specification realizable in $\\Gcal$ is realizable in its retained representatives. No confluence or complete semantic-equivalence decision procedure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence …","dered quotienting from checked equations]\\label{thm:certified-quotient}\nAssume the expression language is pure and total, every admitted schema equation is valid for all well-typed substitutions in the common semantic domain, and each normalization terminates after valid contextual replacements. For each $p_i$, keep its earliest grammar representative\n\\[\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n\\]\nThen $\\rho(i)\\le i$, $p_i$ and $p_{\\rho(i)}$ agree on every admitted context, and every pointwise specification realizable in $\\Gcal$ is realizable in its retained representatives. No confluence or complete semantic-equivalence decision procedure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Repl…","g from checked equations]\\label{thm:certified-quotient}\nAssume the expression language is pure and total, every admitted schema equation is valid for all well-typed substitutions in the common semantic domain, and each normalization terminates after valid contextual replacements. For each $p_i$, keep its earliest grammar representative\n\\[\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n\\]\nThen $\\rho(i)\\le i$, $p_i$ and $p_{\\rho(i)}$ agree on every admitted context, and every pointwise specification realizable in $\\Gcal$ is realizable in its retained representatives. No confluence or complete semantic-equivalence decision procedure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Replacing a satisfyi…","cked equations]\\label{thm:certified-quotient}\nAssume the expression language is pure and total, every admitted schema equation is valid for all well-typed substitutions in the common semantic domain, and each normalization terminates after valid contextual replacements. For each $p_i$, keep its earliest grammar representative\n\\[\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n\\]\nThen $\\rho(i)\\le i$, $p_i$ and $p_{\\rho(i)}$ agree on every admitted context, and every pointwise specification realizable in $\\Gcal$ is realizable in its retained representatives. No confluence or complete semantic-equivalence decision procedure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Replacing a satisfying express…","every admitted schema equation is valid for all well-typed substitutions in the common semantic domain, and each normalization terminates after valid contextual replacements. For each $p_i$, keep its earliest grammar representative\n\\[\n \\rho(i)=\\min\\{j:n(p_j)=n(p_i)\\}.\n\\]\nThen $\\rho(i)\\le i$, $p_i$ and $p_{\\rho(i)}$ agree on every admitted context, and every pointwise specification realizable in $\\Gcal$ is realizable in its retained representatives. No confluence or complete semantic-equivalence decision procedure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Replacing a satisfying expression by its representative preserves every pointwise requirement. The argument uses the soundness…","dure is required.\n\\end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Replacing a satisfying expression by its representative preserves every pointwise requirement. The argument uses the soundness of each actual derivation, not uniqueness or completeness of normalization.\n\\end{proof}","end{theorem}\n\\begin{proof}\nA schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Replacing a satisfying expression by its representative preserves every pointwise requirement. The argument uses the soundness of each actual derivation, not uniqueness or completeness of normalization.\n\\end{proof}","A schema identity remains true under a well-typed substitution. Purity and totality make replacement inside any expression context preserve its value. Induction over a finite derivation gives equality between an expression and its normalized result. Two expressions with the same result therefore have equal values by transitivity. The defining set for $\\rho(i)$ contains $i$, giving existence and $\\rho(i)\\le i$. Replacing a satisfying expression by its representative preserves every pointwise requirement. The argument uses the soundness of each actual derivation, not uniqueness or completeness of normalization.\n\\end{proof}","e same submitted ASTs, counterexamples, and final accepted AST or grammar-exhaustion outcome. With the original AST interpreter retained, it performs no more retained-example evaluations or interpreter-node visits. Its teacher calls and teacher-evaluation work are unchanged. Quotient construction, storage, and dispatch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation w…","s, and final accepted AST or grammar-exhaustion outcome. With the original AST interpreter retained, it performs no more retained-example evaluations or interpreter-node visits. Its teacher calls and teacher-evaluation work are unchanged. Quotient construction, storage, and dispatch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and …","ustion outcome. With the original AST interpreter retained, it performs no more retained-example evaluations or interpreter-node visits. Its teacher calls and teacher-evaluation work are unchanged. Quotient construction, storage, and dispatch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and their evaluations remain identical. If th…","original AST interpreter retained, it performs no more retained-example evaluations or interpreter-node visits. Its teacher calls and teacher-evaluation work are unchanged. Quotient construction, storage, and dispatch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and their evaluations remain identical. If the finite grammar is exhau…","rms no more retained-example evaluations or interpreter-node visits. Its teacher calls and teacher-evaluation work are unchanged. Quotient construction, storage, and dispatch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and their evaluations remain identical. If the finite grammar is exhausted, both learners have rejected the same …","uation work are unchanged. Quotient construction, storage, and dispatch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and their evaluations remain identical. If the finite grammar is exhausted, both learners have rejected the same possible behaviors.\n\\end{proof}","atch costs are separate.\n\\end{theorem}\n\\begin{proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and their evaluations remain identical. If the finite grammar is exhausted, both learners have rejected the same possible behaviors.\n\\end{proof}","proof}\nConsider a removed expression $p_i$ and its earlier representative $p_j$. If the reference learner accepted $p_j$, it never reaches $p_i$. Otherwise a retained example refuted $p_j$, either before its teacher call or as a result of that call. That example is still retained when $p_i$ is reached. Semantic equality means the same example refutes $p_i$, so the reference learner never submits $p_i$ to the teacher. Deleting that unsuccessful consistency check leaves the sequence of proposals unchanged. Induction over the ordered scan gives the same candidate AST at every teacher call. The deterministic teacher then supplies the same first counterexample or acceptance. Removed candidates contribute only nonnegative evaluation work, while submitted expressions and their evaluations remain identical. If the finite grammar is exhausted, both learners have rejected the same possible behaviors.\n\\end{proof}","The frozen quotient is next used for nine finite targets involving sums, extrema, comparisons, and equality, followed by a parity target that is not expressible on its complete declared domain. These are changed target functions in the same arithmetic language, not a statistical sample of unrelated research domains. The first nine use all 256 contexts in $\\{0,1,3,5\\}^4$. The parity control uses all contexts in $\\{0,1,2,3\\}^4$, so more than one even value is present.","ient is next used for nine finite targets involving sums, extrema, comparisons, and equality, followed by a parity target that is not expressible on its complete declared domain. These are changed target functions in the same arithmetic language, not a statistical sample of unrelated research domains. The first nine use all 256 contexts in $\\{0,1,3,5\\}^4$. The parity control uses all contexts in $\\{0,1,2,3\\}^4$, so more than one even value is present.","Let $C_{\\rm prepare}$ be the complete preparation cost in a declared common resource model, including failed proposals and certificate checking, and let $d_t$ be the subsequent saving on task $t$ after including dispatch. Reuse is beneficial in that model exactly when\n\\[\n \\sum_{t=1}^{T}d_t>C_{\\rm prepare}.\n\\]\nThis is the earlier amortization identity applied to acquisition itself. The reference does not convert unlike counters into a universal wall-clock claim. Even one setup coordinate, 518,140 schema matches, exceeds the 522 interpreter-node visits saved in one cold capacity-rule run. A short workload or an expensive matcher can make the full modification undesirable.","\\begin{counterexample}[The same operator spelling with changed semantics]\\label{ex:semantic-cache}\nThe identity $x+0=x$ is valid for ordinary addition. Suppose a revised interpreter uses the same printed symbol but executes $a\\mathbin{\\oplus}b=a+b+1$. Then $x\\mathbin{\\oplus}0\\ne x$. A cache certified for the former interpreter cannot be imported merely because its expression strings still parse. A semantic identifier helps detect ordinary version mismatch, but an identifier falsely attached to a changed implementation is not itself proof of faithful execution.\n\\end{counterexample}","Let $X$ be a finite immutable state set and let $S_r\\subseteq X$ be the success set of repair $r$. Choose nonnegative rational weights $w_x$ satisfying\n\\begin{equation}\\label{eq:repair-weights}\n \\sum_{x\\in S_r}w_x\\le1\\quad\\text{for every permitted repair }r.\n\\end{equation}\nThis is a feasible dual weight assignment for unweighted set cover. We use only its elementary weak-duality property, not an assumption that the best assignment has been found~\\cite{williamson}.","\\begin{proposition}[Repair-relative weight and prefix obstruction]\\label{prop:weighted-cover}\nSuppose a total immutable deterministic binary diagnostic tree succeeds on nonempty support $B$ and has $L$ reachable leaves and maximum query depth $h$. Under~\\eqref{eq:repair-weights},\n\\[\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n\\]\nAt a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give wei","\\begin{proposition}[Repair-relative weight and prefix obstruction]\\label{prop:weighted-cover}\nSuppose a total immutable deterministic binary diagnostic tree succeeds on nonempty support $B$ and has $L$ reachable leaves and maximum query depth $h$. Under~\\eqref{eq:repair-weights},\n\\[\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n\\]\nAt a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most","\\begin{proposition}[Repair-relative weight and prefix obstruction]\\label{prop:weighted-cover}\nSuppose a total immutable deterministic binary diagnostic tree succeeds on nonempty support $B$ and has $L$ reachable leaves and maximum query depth $h$. Under~\\eqref{eq:repair-weights},\n\\[\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n\\]\nAt a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the firs","\\begin{proposition}[Repair-relative weight and prefix obstruction]\\label{prop:weighted-cover}\nSuppose a total immutable deterministic binary diagnostic tree succeeds on nonempty support $B$ and has $L$ reachable leaves and maximum query depth $h$. Under~\\eqref{eq:repair-weights},\n\\[\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n\\]\nAt a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ le","\\begin{proposition}[Repair-relative weight and prefix obstruction]\\label{prop:weighted-cover}\nSuppose a total immutable deterministic binary diagnostic tree succeeds on nonempty support $B$ and has $L$ reachable leaves and maximum query depth $h$. Under~\\eqref{eq:repair-weights},\n\\[\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n\\]\nAt a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidat","uction]\\label{prop:weighted-cover}\nSuppose a total immutable deterministic binary diagnostic tree succeeds on nonempty support $B$ and has $L$ reachable leaves and maximum query depth $h$. Under~\\eqref{eq:repair-weights},\n\\[\n \\sum_{x\\in B}w_x\\le L\\le2^h.\n\\]\nAt a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidate root, each child has at most $h-1$ query levels. Unread cases can only add nonnegative weight to an out…","t a candidate query root with $h\\ge1$ remaining query levels, if the cases already read into either outcome branch have weight greater than $2^{h-1}$, no assignment of its unread reports can make that branch repairable within the remaining depth. This implication is valid even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidate root, each child has at most $h-1$ query levels. Unread cases can only add nonnegative weight to an outcome branch, so a prefix already exceeding its leaf capacity cannot be rescued by any completion.\n\\end{proof}","even when one repair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidate root, each child has at most $h-1$ query levels. Unread cases can only add nonnegative weight to an outcome branch, so a prefix already exceeding its leaf capacity cannot be rescued by any completion.\n\\end{proof}","pair succeeds in several states.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidate root, each child has at most $h-1$ query levels. Unread cases can only add nonnegative weight to an outcome branch, so a prefix already exceeding its leaf capacity cannot be rescued by any completion.\n\\end{proof}","s.\n\\end{proposition}\n\\begin{proof}\nImmutable deterministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidate root, each child has at most $h-1$ query levels. Unread cases can only add nonnegative weight to an outcome branch, so a prefix already exceeding its leaf capacity cannot be rescued by any completion.\n\\end{proof}","rministic execution partitions $B$ into its nonempty reachable leaf supports. Each leaf support is contained in the success set of its chosen repair. Nonnegativity and~\\eqref{eq:repair-weights} give weight at most one for every leaf, so summing gives the first inequality. A binary tree of depth at most $h$ has at most $2^h$ leaves, by induction on $h$. At a candidate root, each child has at most $h-1$ query levels. Unread cases can only add nonnegative weight to an outcome branch, so a prefix already exceeding its leaf capacity cannot be rescued by any completion.\n\\end{proof}","For example, on three cases with success sets $\\{1,2\\}$, $\\{2,3\\}$, and $\\{1,3\\}$, the ordered greedy constructor returns total weight one. The feasible fractional assignment $w_1=w_2=w_3=1/2$ has total $3/2$ and proves that at least two leaves are required. Thus the supplied constructor does not optimize the bound. Both assignments are valid and neither creates a missing experiment.","\\subsection{Paired repairs change the task, not just its notation}\nConsider $2g$ possible hidden cases divided into $g$ pairs, where $g$ is a power of two. One repair succeeds on both members of each pair. The query language contains equality questions and bits of the pair index. The greedy weight constructor gives weight one to the first member of each pair and zero to the second, with total weight $g$. A depth $\\log_2g$ tree is both necessary and sufficient, so the diagnostic-plus-repair work is $1+\\log_2g$ and a full binary witness uses $2g-1$ records.","section{Paired repairs change the task, not just its notation}\nConsider $2g$ possible hidden cases divided into $g$ pairs, where $g$ is a power of two. One repair succeeds on both members of each pair. The query language contains equality questions and bits of the pair index. The greedy weight constructor gives weight one to the first member of each pair and zero to the second, with total weight $g$. A depth $\\log_2g$ tree is both necessary and sufficient, so the diagnostic-plus-repair work is $1+\\log_2g$ and a full binary witness uses $2g-1$ records.","d repairs change the task, not just its notation}\nConsider $2g$ possible hidden cases divided into $g$ pairs, where $g$ is a power of two. One repair succeeds on both members of each pair. The query language contains equality questions and bits of the pair index. The greedy weight constructor gives weight one to the first member of each pair and zero to the second, with total weight $g$. A depth $\\log_2g$ tree is both necessary and sufficient, so the diagnostic-plus-repair work is $1+\\log_2g$ and a full binary witness uses $2g-1$ records.","ivided into $g$ pairs, where $g$ is a power of two. One repair succeeds on both members of each pair. The query language contains equality questions and bits of the pair index. The greedy weight constructor gives weight one to the first member of each pair and zero to the second, with total weight $g$. A depth $\\log_2g$ tree is both necessary and sufficient, so the diagnostic-plus-repair work is $1+\\log_2g$ and a full binary witness uses $2g-1$ records.","of two. One repair succeeds on both members of each pair. The query language contains equality questions and bits of the pair index. The greedy weight constructor gives weight one to the first member of each pair and zero to the second, with total weight $g$. A depth $\\log_2g$ tree is both necessary and sufficient, so the diagnostic-plus-repair work is $1+\\log_2g$ and a full binary witness uses $2g-1$ records.","and the original and optimized compilers satisfy the common-domain premises of Theorem~\\ref{thm:streaming-pruning}. If all stages complete, quotient-guided rule acquisition followed by optimized compilation returns the same learned expression and the same diagnostic controller as ordinary rule acquisition followed by the reference ordered compilation. In a common additive resource model, writing $K$ for extra preparation cost, $A,A'$ for the two acquisition costs, and $D,D'$ for the two compilation costs, the composite saves work exactly when\n\\[\n (A-A')+(D-D')>K.\n\\]\nThe semantic conclusion does not imply the cost inequality, and a budget interruption does not imply infeasibility.\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither s…","ompilers satisfy the common-domain premises of Theorem~\\ref{thm:streaming-pruning}. If all stages complete, quotient-guided rule acquisition followed by optimized compilation returns the same learned expression and the same diagnostic controller as ordinary rule acquisition followed by the reference ordered compilation. In a common additive resource model, writing $K$ for extra preparation cost, $A,A'$ for the two acquisition costs, and $D,D'$ for the two compilation costs, the composite saves work exactly when\n\\[\n (A-A')+(D-D')>K.\n\\]\nThe semantic conclusion does not imply the cost inequality, and a budget interruption does not imply infeasibility.\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither semantic theorem bounds $K$ or as…","s of Theorem~\\ref{thm:streaming-pruning}. If all stages complete, quotient-guided rule acquisition followed by optimized compilation returns the same learned expression and the same diagnostic controller as ordinary rule acquisition followed by the reference ordered compilation. In a common additive resource model, writing $K$ for extra preparation cost, $A,A'$ for the two acquisition costs, and $D,D'$ for the two compilation costs, the composite saves work exactly when\n\\[\n (A-A')+(D-D')>K.\n\\]\nThe semantic conclusion does not imply the cost inequality, and a budget interruption does not imply infeasibility.\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither semantic theorem bounds $K$ or assigns exchange rates among distinct resour…","ided rule acquisition followed by optimized compilation returns the same learned expression and the same diagnostic controller as ordinary rule acquisition followed by the reference ordered compilation. In a common additive resource model, writing $K$ for extra preparation cost, $A,A'$ for the two acquisition costs, and $D,D'$ for the two compilation costs, the composite saves work exactly when\n\\[\n (A-A')+(D-D')>K.\n\\]\nThe semantic conclusion does not imply the cost inequality, and a budget interruption does not imply infeasibility.\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither semantic theorem bounds $K$ or assigns exchange rates among distinct resource units. An interrupted finite search has not exhausted its candidates and t…","semantic conclusion does not imply the cost inequality, and a budget interruption does not imply infeasibility.\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither semantic theorem bounds $K$ or assigns exchange rates among distinct resource units. An interrupted finite search has not exhausted its candidates and therefore cannot use completeness to infer infeasibility.\n\\end{proof}","ply the cost inequality, and a budget interruption does not imply infeasibility.\n\\end{proposition}\n\\begin{proof}\nTheorem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither semantic theorem bounds $K$ or assigns exchange rates among distinct resource units. An interrupted finite search has not exhausted its candidates and therefore cannot use completeness to infer infeasibility.\n\\end{proof}","orem~\\ref{thm:cegis-quotient} gives the same acquired AST and teacher transcript. Passing the unchanged scope checker places that expression in the stated compiler-preservation theorem, which gives the same returned controller. The reference total is $A+D$ and the modified total is $K+A'+D'$. Subtracting gives the displayed necessary and sufficient cost condition. Neither semantic theorem bounds $K$ or assigns exchange rates among distinct resource units. An interrupted finite search has not exhausted its candidates and therefore cannot use completeness to infer infeasibility.\n\\end{proof}","\\subsection{A declared dependency graph}\nLet an artifact state be $a=(a_1,\\ldots,a_d)$. Its components may include operation semantics, a grammar and its order, an equation library, experiment tables, repair-success sets, a returned controller, and a resource model. A finite directed acyclic graph has one vertex $v$ for each claimed guarantee. Its incoming vertices $\\operatorname{pa}(v)$ identify guarantees used as premises. Its direct dependency set $D_v\\subseteq\\{1,\\ldots,d\\}$ identifies the artifact components used directly. Each vertex carries a witness $w_v$ and a local verification rule.","state be $a=(a_1,\\ldots,a_d)$. Its components may include operation semantics, a grammar and its order, an equation library, experiment tables, repair-success sets, a returned controller, and a resource model. A finite directed acyclic graph has one vertex $v$ for each claimed guarantee. Its incoming vertices $\\operatorname{pa}(v)$ identify guarantees used as premises. Its direct dependency set $D_v\\subseteq\\{1,\\ldots,d\\}$ identifies the artifact components used directly. Each vertex carries a witness $w_v$ and a local verification rule.","quation library, experiment tables, repair-success sets, a returned controller, and a resource model. A finite directed acyclic graph has one vertex $v$ for each claimed guarantee. Its incoming vertices $\\operatorname{pa}(v)$ identify guarantees used as premises. Its direct dependency set $D_v\\subseteq\\{1,\\ldots,d\\}$ identifies the artifact components used directly. Each vertex carries a witness $w_v$ and a local verification rule.","The dependency declaration must be \\emph{semantically complete}: the meaning of the claim and the local rule is determined by the components in $D_v$ and the stated parent claims. Hidden environment assumptions cannot be left outside this declaration merely because the implementation does not know how to check them. They can instead be recorded as explicit external premises. Verifier soundness, the interpretation of primitive arithmetic, and the faithfulness of the supplied model to an external process are not established by naming them in a registry.","Suppose a revision replaces $a$ by $a'$. Mark a vertex directly stale whenever one of its declared component values differs. Let $\\Delta$ be the union of those vertices and all their descendants in the dependency graph. Equivalently, in a topological order, mark $v$ whenever\n\\[\n a|_{D_v}\\ne a'|_{D_v}\n \\quad\\text{or}\\quad\n \\operatorname{pa}(v)\\cap\\Delta\\ne\\varnothing.\n\\]\nHere equality refers to the declared artifact value, not merely an unchanged version label. This procedure is conservative. A changed premise may leave the conclusion true, but the old evidence is not reused without a further argument.","\\begin{theorem}[Dependency-complete reuse across a revision]\\label{thm:certificate-reuse}\nAssume a finite acyclic dependency graph, semantically complete direct dependencies, and sound local verification rules. Suppose the old witnesses establish every old claim from the declared external premises. After a revision, the old claims outside $\\Delta$ remain established by their old witnesses. If every vertex in $\\Delta$ is re-established in topological order using the revised artifact values and already established parent claims, all revised claims hold. If an affected vertex is not re-established, no dependent claim is justified by this reuse rule alone. These statements are conditional on the external premises applicable to the revised realization.\n\\end{theorem}\n\\begin{proof}\nA vertex outside $\\Delta$ has unchanged direct inputs and no parent in $\\Delta$. Inducting through a topological order, its parents remain established with the same meanings, so dependency completeness leaves its old verification argument applicable. For an affe","m}[Dependency-complete reuse across a revision]\\label{thm:certificate-reuse}\nAssume a finite acyclic dependency graph, semantically complete direct dependencies, and sound local verification rules. Suppose the old witnesses establish every old claim from the declared external premises. After a revision, the old claims outside $\\Delta$ remain established by their old witnesses. If every vertex in $\\Delta$ is re-established in topological order using the revised artifact values and already established parent claims, all revised claims hold. If an affected vertex is not re-established, no dependent claim is justified by this reuse rule alone. These statements are conditional on the external premises applicable to the revised realization.\n\\end{theorem}\n\\begin{proof}\nA vertex outside $\\Delta$ has unchanged direct inputs and no parent in $\\Delta$. Inducting through a topological order, its parents remain established with the same meanings, so dependency completeness leaves its old verification argument applicable. For an affected vertex, assume its parents have already been established in …","rtex in $\\Delta$ is re-established in topological order using the revised artifact values and already established parent claims, all revised claims hold. If an affected vertex is not re-established, no dependent claim is justified by this reuse rule alone. These statements are conditional on the external premises applicable to the revised realization.\n\\end{theorem}\n\\begin{proof}\nA vertex outside $\\Delta$ has unchanged direct inputs and no parent in $\\Delta$. Inducting through a topological order, its parents remain established with the same meanings, so dependency completeness leaves its old verification argument applicable. For an affected vertex, assume its parents have already been established in the revised realization. Soundness of the newly checked local rule then establishes its revised claim. A second induction yields the result when every affected vertex is re-established. Without that step, the rule has an unsupported premise and supplies no conclusion along that dependency. This is a limitation of the reuse argument, not a proof that the dependent claim is false.\n\\end{proo…","r using the revised artifact values and already established parent claims, all revised claims hold. If an affected vertex is not re-established, no dependent claim is justified by this reuse rule alone. These statements are conditional on the external premises applicable to the revised realization.\n\\end{theorem}\n\\begin{proof}\nA vertex outside $\\Delta$ has unchanged direct inputs and no parent in $\\Delta$. Inducting through a topological order, its parents remain established with the same meanings, so dependency completeness leaves its old verification argument applicable. For an affected vertex, assume its parents have already been established in the revised realization. Soundness of the newly checked local rule then establishes its revised claim. A second induction yields the result when every affected vertex is re-established. Without that step, the rule has an unsupported premise and supplies no conclusion along that dependency. This is a limitation of the reuse argument, not a proof that the dependent claim is false.\n\\end{proof}","\\begin{counterexample}[An unchanged name hides a changed premise]\\label{ex:incomplete-dependency}\nA stored claim asserts that two binary values $u$ and $v$ agree. Initially both are zero. A registry records only $u$ as a dependency and continues to validate that its stored value is zero. Change $v$ to one without changing $u$. The declared check still passes, but the claimed equality is false. Recording and replaying the incomplete registry cannot repair its omitted premise. Dependency completeness is a mathematical assumption, not a consequence of serialization or a checksum.\n\\end{counterexample}","\\subsection{Which gains survive a change of resource prices?}\nFix an additive ledger with $d\\ge1$ named event types. Let $K\\in\\mathbb R_{\\ge0}^d$ be extra setup, let $s_i\\in\\mathbb R^d$ be the old-minus-new per-use event vector for workload type $i$, and let $n_i$ be its nonnegative integer multiplicity. The net event saving is\n\\begin{equation}\\label{eq:vector-saving}\n \\Delta(n)=\\sum_i n_i s_i-K.\n\\end{equation}\nA price vector $w\\in\\mathbb R_{\\ge0}^d$ specifies conversions into one declared additive resource measure. Its priced saving is $w\\cdot\\Delta(n)$. The prices are modeling assumptions, not learned hardware laws inferred from the event counts.","a change of resource prices?}\nFix an additive ledger with $d\\ge1$ named event types. Let $K\\in\\mathbb R_{\\ge0}^d$ be extra setup, let $s_i\\in\\mathbb R^d$ be the old-minus-new per-use event vector for workload type $i$, and let $n_i$ be its nonnegative integer multiplicity. The net event saving is\n\\begin{equation}\\label{eq:vector-saving}\n \\Delta(n)=\\sum_i n_i s_i-K.\n\\end{equation}\nA price vector $w\\in\\mathbb R_{\\ge0}^d$ specifies conversions into one declared additive resource measure. Its priced saving is $w\\cdot\\Delta(n)$. The prices are modeling assumptions, not learned hardware laws inferred from the event counts.","$ be extra setup, let $s_i\\in\\mathbb R^d$ be the old-minus-new per-use event vector for workload type $i$, and let $n_i$ be its nonnegative integer multiplicity. The net event saving is\n\\begin{equation}\\label{eq:vector-saving}\n \\Delta(n)=\\sum_i n_i s_i-K.\n\\end{equation}\nA price vector $w\\in\\mathbb R_{\\ge0}^d$ specifies conversions into one declared additive resource measure. Its priced saving is $w\\cdot\\Delta(n)$. The prices are modeling assumptions, not learned hardware laws inferred from the event counts.","\\begin{proposition}[Resource dominance and declared price uncertainty]\\label{prop:resource-prices}\nFor a fixed ledger and workload, $w\\cdot\\Delta(n)\\ge0$ for every nonnegative price vector $w$ if and only if $\\Delta(n)$ is componentwise nonnegative. If admissible prices form the convex hull of a nonempty finite set $W_0$, then\n\\[\n \\min_{w\\in\\operatorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving","\\begin{proposition}[Resource dominance and declared price uncertainty]\\label{prop:resource-prices}\nFor a fixed ledger and workload, $w\\cdot\\Delta(n)\\ge0$ for every nonnegative price vector $w$ if and only if $\\Delta(n)$ is componentwise nonnegative. If admissible prices form the convex hull of a nonempty finite set $W_0$, then\n\\[\n \\min_{w\\in\\operatorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving the converse. Every point of the specif","\\begin{proposition}[Resource dominance and declared price uncertainty]\\label{prop:resource-prices}\nFor a fixed ledger and workload, $w\\cdot\\Delta(n)\\ge0$ for every nonnegative price vector $w$ if and only if $\\Delta(n)$ is componentwise nonnegative. If admissible prices form the convex hull of a nonempty finite set $W_0$, then\n\\[\n \\min_{w\\in\\operatorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving the converse. Every point of the specified convex hull is a convex","\\begin{proposition}[Resource dominance and declared price uncertainty]\\label{prop:resource-prices}\nFor a fixed ledger and workload, $w\\cdot\\Delta(n)\\ge0$ for every nonnegative price vector $w$ if and only if $\\Delta(n)$ is componentwise nonnegative. If admissible prices form the convex hull of a nonempty finite set $W_0$, then\n\\[\n \\min_{w\\in\\operatorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving the converse. Every point of the specified convex hull is a convex combination of its listed points. Linearity makes its priced saving the same combination of their savi","\\begin{proposition}[Resource dominance and declared price uncertainty]\\label{prop:resource-prices}\nFor a fixed ledger and workload, $w\\cdot\\Delta(n)\\ge0$ for every nonnegative price vector $w$ if and only if $\\Delta(n)$ is componentwise nonnegative. If admissible prices form the convex hull of a nonempty finite set $W_0$, then\n\\[\n \\min_{w\\in\\operatorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving the converse. Every point of the specified convex hull is a convex combination of its listed points. Linearity makes its priced saving the same combination of their savings, which cannot be smaller than their minimum. A listed minimizer attains tha…","ratorname{conv}(W_0)}w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving the converse. Every point of the specified convex hull is a convex combination of its listed points. Linearity makes its priced saving the same combination of their savings, which cannot be smaller than their minimum. A listed minimizer attains that minimum, proving the equality and the strict criterion.\n\\end{proof}","w\\cdot\\Delta(n)\n =\\min_{w\\in W_0}w\\cdot\\Delta(n).\n\\]\nThus strict improvement throughout that price set holds exactly when every listed vertex gives a strictly positive margin. The claims concern the same workflow, workload, and complete additive ledger.\n\\end{proposition}\n\\begin{proof}\nComponentwise nonnegativity makes every nonnegative weighted sum nonnegative. If coordinate $j$ is negative, the $j$th coordinate vector is a nonnegative price vector giving a negative sum, proving the converse. Every point of the specified convex hull is a convex combination of its listed points. Linearity makes its priced saving the same combination of their savings, which cannot be smaller than their minimum. A listed minimizer attains that minimum, proving the equality and the strict criterion.\n\\end{proof}","\\begin{counterexample}[One valid controller, opposite resource conclusions]\\label{ex:price-change}\nFor the 32-case weighted-search comparison, the old-minus-new search-event vector is\n\\[\n (1552,\\ 1552,\\ -1868)\n\\]\nin model-record reads, rational additions, and capacity comparisons. Scope checking and final witness checking are equal for both procedures and cancel in this particular difference. Prices $(1,1,1)$ give a saving of 1,236 units. Prices $(1,0,1)$ give a loss of 316 units, and prices $(0,0,1)$ give a loss of 1,868 units. The identical returned controller remains valid in every comparison. None of the price choices was measured as a universal property of a processor.\n\\end{counterexample}","in{counterexample}[One valid controller, opposite resource conclusions]\\label{ex:price-change}\nFor the 32-case weighted-search comparison, the old-minus-new search-event vector is\n\\[\n (1552,\\ 1552,\\ -1868)\n\\]\nin model-record reads, rational additions, and capacity comparisons. Scope checking and final witness checking are equal for both procedures and cancel in this particular difference. Prices $(1,1,1)$ give a saving of 1,236 units. Prices $(1,0,1)$ give a loss of 316 units, and prices $(0,0,1)$ give a loss of 1,868 units. The identical returned controller remains valid in every comparison. None of the price choices was measured as a universal property of a processor.\n\\end{counterexample}","te resource conclusions]\\label{ex:price-change}\nFor the 32-case weighted-search comparison, the old-minus-new search-event vector is\n\\[\n (1552,\\ 1552,\\ -1868)\n\\]\nin model-record reads, rational additions, and capacity comparisons. Scope checking and final witness checking are equal for both procedures and cancel in this particular difference. Prices $(1,1,1)$ give a saving of 1,236 units. Prices $(1,0,1)$ give a loss of 316 units, and prices $(0,0,1)$ give a loss of 1,868 units. The identical returned controller remains valid in every comparison. None of the price choices was measured as a universal property of a processor.\n\\end{counterexample}","For the 32-case weighted-search comparison, the old-minus-new search-event vector is\n\\[\n (1552,\\ 1552,\\ -1868)\n\\]\nin model-record reads, rational additions, and capacity comparisons. Scope checking and final witness checking are equal for both procedures and cancel in this particular difference. Prices $(1,1,1)$ give a saving of 1,236 units. Prices $(1,0,1)$ give a loss of 316 units, and prices $(0,0,1)$ give a loss of 1,868 units. The identical returned controller remains valid in every comparison. None of the price choices was measured as a universal property of a processor.\n\\end{counterexample}","The zero setup in that isolated execution comparison is not a statement that preparing the whole improvement was free. For an explicitly hypothetical additional setup charge of 2,000 comparison units and prices $(1,1,1)$, one use loses 764 units while two uses save 472. The purpose of this small calculation is to expose the dependence on the declared ledger. The actual equation-discovery, proof-search, quotient-construction, and revision-management records must be included when assessing the entire first-edition workflow.","\\begin{proposition}[Conditional error budgets across an adaptive lineage]\\label{prop:lineage}\nLet $\\mathcal H_{t-1}$ be the information available before attempt $t$. Suppose the event $A_t$ that attempt $t$ accepts a false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-de","\\begin{proposition}[Conditional error budgets across an adaptive lineage]\\label{prop:lineage}\nLet $\\mathcal H_{t-1}$ be the information available before attempt $t$. Suppose the event $A_t$ that attempt $t$ accepts a false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-dependent stop can only reduce the set of attempts","\\begin{proposition}[Conditional error budgets across an adaptive lineage]\\label{prop:lineage}\nLet $\\mathcal H_{t-1}$ be the information available before attempt $t$. Suppose the event $A_t$ that attempt $t$ accepts a false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-dependent stop can only reduce the set of attempts whose acceptance events","\\begin{proposition}[Conditional error budgets across an adaptive lineage]\\label{prop:lineage}\nLet $\\mathcal H_{t-1}$ be the information available before attempt $t$. Suppose the event $A_t$ that attempt $t$ accepts a false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-dependent stop can only reduce the set of attempts whose acceptance events occur.\n\\end{proof","\\begin{proposition}[Conditional error budgets across an adaptive lineage]\\label{prop:lineage}\nLet $\\mathcal H_{t-1}$ be the information available before attempt $t$. Suppose the event $A_t$ that attempt $t$ accepts a false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-dependent stop can only reduce the set of attempts whose acceptance events occur.\n\\end{proof}","{proposition}[Conditional error budgets across an adaptive lineage]\\label{prop:lineage}\nLet $\\mathcal H_{t-1}$ be the information available before attempt $t$. Suppose the event $A_t$ that attempt $t$ accepts a false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-dependent stop can only reduce the set of attempts whose acceptance events occur.\n\\end{proof}","false improvement claim satisfies\n\\[\n\\Pr(A_t\\mid\\mathcal H_{t-1})\\leq\\delta_t\n\\quad\\text{almost surely},\n\\]\nfor deterministic $\\delta_t\\geq0$ with $\\sum_{t\\geq1}\\delta_t\\leq\\delta$. Then\n\\[\n\\Pr\\big(\\exists t\\geq1:A_t\\big)\\leq\\delta.\n\\]\nThe candidate-generating process and the time at which attempts stop may depend on the earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_t)\\leq\\delta_t$. The union bound gives the result for any finite number of attempts. The infinite statement follows from continuity of probability for the increasing finite unions. A history-dependent stop can only reduce the set of attempts whose acceptance events occur.\n\\end{proof}","No independence between attempts is needed. The conditional premise is stronger than a marginal guarantee for a candidate considered before adaptive selection. It must be justified by the actual sampling or verification mechanism. A budget such as $\\delta_t=\\delta/[t(t+1)]$ is summable, but writing that schedule does not make an invalid test conditionally valid.","\\subsection{One sufficient fresh-audit construction}\nFreeze a candidate and its comparator using the prior history. Draw $n_t$ fresh evaluation cases conditionally independently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_","\\subsection{One sufficient fresh-audit construction}\nFreeze a candidate and its comparator using the prior history. Draw $n_t$ fresh evaluation cases conditionally independently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-un","\\subsection{One sufficient fresh-audit construction}\nFreeze a candidate and its comparator using the prior history. Draw $n_t$ fresh evaluation cases conditionally independently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or","\\subsection{One sufficient fresh-audit construction}\nFreeze a candidate and its comparator using the prior history. Draw $n_t$ fresh evaluation cases conditionally independently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or a separately justified spending rule. Time-unifor","\\subsection{One sufficient fresh-audit construction}\nFreeze a candidate and its comparator using the prior history. Draw $n_t$ fresh evaluation cases conditionally independently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or a separately justified spending rule. Time-uniform concentration is an established subject~\\cite{howard}.","$n_t$ fresh evaluation cases conditionally independently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or a separately justified spending rule. Time-uniform concentration is an established subject~\\cite{howard}.","ependently from the declared distribution and record paired score differences $Y_i\\in[-1,1]$. Let $\\Delta_t$ be their conditional mean and $\\widehat\\Delta_t$ their average. The bounded-variable concentration inequality gives\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or a separately justified spending rule. Time-uniform concentration is an established subject~\\cite{howard}.","ves\n\\[\n\\Pr\\left(\\widehat\\Delta_t-\\Delta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or a separately justified spending rule. Time-uniform concentration is an established subject~\\cite{howard}.","elta_t\\geq r\\mid\\mathcal H_{t-1}\\right)\n\\leq \\exp(-n_tr^2/2).\n\\]\nConsequently, accepting a claim $\\Delta_t\\geq\\gamma$ only when\n\\begin{equation}\\label{eq:audit}\n\\widehat\\Delta_t-\\sqrt{2\\log(1/\\delta_t)/n_t}\\geq\\gamma\n\\end{equation}\nmeets the preceding conditional error requirement for that claim. This is a fixed-sample sufficient construction. The candidate and $n_t$ may depend on past data, but $n_t$ is fixed before this audit is examined. Inspecting the same sample repeatedly and stopping at a favorable value requires a time-uniform method or a separately justified spending rule. Time-uniform concentration is an established subject~\\cite{howard}.","\\begin{counterexample}[Reused evidence can reward memorization]\\label{ex:reuse}\nA task has $N$ equally likely names and an externally fixed binary answer for each. After the answers on a development subset become available, a candidate stores them and gives the wrong answer on every other name. It is perfect on that subset and has true accuracy equal only to the subset's fraction of the task universe. An outer loop that keeps selecting by the same subset can improve its reported score without improving beyond the information it has exposed. Renaming the subset a holdout does not restore independence.\n\\end{counterexample}","We give a small time-uniform alternative whose finite arithmetic is directly inspectable. Before an audit begins, freeze the candidate procedure, its comparator, their installed information, the task law, the common resource cap, and a paired score $Y_t\\in[-1,1]$. Timeouts and invalid outcomes must have a stipulated score rather than disappear from the sample. Fix a proposed improvement margin $\\gamma\\in[0,1)$.","The null condition is\n\\begin{equation}\\label{eq:anytime-null}\n\\E[Y_t\\mid\\mathcal H_{t-1}]\\leq\\gamma\n\\quad\\text{for every audit step }t,\n\\end{equation}\nwhere $\\mathcal H_{t-1}$ includes all information available before the next score. Conditionally independent cases from a fixed task law with mean at most $\\gamma$ satisfy this condition. An auditor cannot change the task distribution toward favorable cases and retain that interpretation without a new argument.","Choose a nonnegative bet $\\lambda_t$, measurable from $\\mathcal H_{t-1}$, with $\\lambda_t\\leq1/(1+\\gamma)$. Define\n\\begin{equation}\\label{eq:anytime-capital}\nE_0=1,\\qquad E_t=E_{t-1}\\bigl(1+\\lambda_t(Y_t-\\gamma)\\bigr).\n\\end{equation}\nThe variable $E_t$ is an evidence process, not a posterior probability that the candidate is better. Predictability means choosing the bet before observing the score to which it is applied.","\\begin{theorem}[An anytime-valid paired improvement audit]\\label{thm:anytime}\nUnder~\\eqref{eq:anytime-null} and the stated predictable bet constraints, $E_t$ is a nonnegative supermartingale. For every $\\delta\\in(0,1)$,\n\\[\n\\Pr\\left(\\sup_{t\\geq0}E_t\\geq1/\\delta\\right)\\leq\\delta.\n\\]\nConsequently an audit may reject the stated null at its first crossing of $1/\\delta$, while choosing its stopping time from the observed audit history. The result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first","\\begin{theorem}[An anytime-valid paired improvement audit]\\label{thm:anytime}\nUnder~\\eqref{eq:anytime-null} and the stated predictable bet constraints, $E_t$ is a nonnegative supermartingale. For every $\\delta\\in(0,1)$,\n\\[\n\\Pr\\left(\\sup_{t\\geq0}E_t\\geq1/\\delta\\right)\\leq\\delta.\n\\]\nConsequently an audit may reject the stated null at its first crossing of $1/\\delta$, while choosing its stopping time from the observed audit history. The result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at","\\begin{theorem}[An anytime-valid paired improvement audit]\\label{thm:anytime}\nUnder~\\eqref{eq:anytime-null} and the stated predictable bet constraints, $E_t$ is a nonnegative supermartingale. For every $\\delta\\in(0,1)$,\n\\[\n\\Pr\\left(\\sup_{t\\geq0}E_t\\geq1/\\delta\\right)\\leq\\delta.\n\\]\nConsequently an audit may reject the stated null at its first crossing of $1/\\delta$, while choosing its stopping time from the observed audit history. The result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induc","\\begin{theorem}[An anytime-valid paired improvement audit]\\label{thm:anytime}\nUnder~\\eqref{eq:anytime-null} and the stated predictable bet constraints, $E_t$ is a nonnegative supermartingale. For every $\\delta\\in(0,1)$,\n\\[\n\\Pr\\left(\\sup_{t\\geq0}E_t\\geq1/\\delta\\right)\\leq\\delta.\n\\]\nConsequently an audit may reject the stated null at its first crossing of $1/\\delta$, while choosing its stopping time from the observed audit history. The result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at leas","rtingale. For every $\\delta\\in(0,1)$,\n\\[\n\\Pr\\left(\\sup_{t\\geq0}E_t\\geq1/\\delta\\right)\\leq\\delta.\n\\]\nConsequently an audit may reject the stated null at its first crossing of $1/\\delta$, while choosing its stopping time from the observed audit history. The result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","stated null at its first crossing of $1/\\delta$, while choosing its stopping time from the observed audit history. The result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","result does not require a predetermined number of cases or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","es or independence beyond the conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","e conditional-mean premise.\n\\end{theorem}\n\\begin{proof}\nSince $Y_t-\\gamma\\geq-(1+\\gamma)$, every factor in~\\eqref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","ref{eq:anytime-capital} is nonnegative. Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","Conditional expectation and predictability give\n\\[\n\\E[E_t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","t\\mid\\mathcal H_{t-1}]\n=E_{t-1}\\bigl(1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","1+\\lambda_t(\\E[Y_t\\mid\\mathcal H_{t-1}]-\\gamma)\\bigr)\n\\leq E_{t-1}.\n\\]\nLet $\\tau$ be the first threshold crossing. For a finite $N$, the stopped process at $\\tau\\wedge N$ still has expected value at most one, by induction on the stopped conditional expectations. On $\\{\\tau\\leq N\\}$ its value is at least $1/\\delta$, while it is nonnegative otherwise. Therefore $\\Pr(\\tau\\leq N)\\leq\\delta$. Letting $N$ increase proves the result for any crossing time.\n\\end{proof}","\\subsection{Why seeing the outcome before choosing the bet fails}\n\\begin{counterexample}[Hindsight betting creates false evidence]\\label{ex:hindsight-bet}\nLet $Y_t$ be independent fair signs and $\\gamma=0$. Choose $\\lambda_t=1$ when the current sign is positive and zero otherwise. This invalid rule doubles $E_t$ on a positive score and leaves it unchanged on a negative one. Within four scores it crosses the threshold four whenever at least two scores are positive, an event of probability $11/16$. This exceeds the nominal bound $1/4$. Each individual bet lies in the numerical range $[0,1]$. What fails is predictability.\n\\end{counterexample}","ight betting creates false evidence]\\label{ex:hindsight-bet}\nLet $Y_t$ be independent fair signs and $\\gamma=0$. Choose $\\lambda_t=1$ when the current sign is positive and zero otherwise. This invalid rule doubles $E_t$ on a positive score and leaves it unchanged on a negative one. Within four scores it crosses the threshold four whenever at least two scores are positive, an event of probability $11/16$. This exceeds the nominal bound $1/4$. Each individual bet lies in the numerical range $[0,1]$. What fails is predictability.\n\\end{counterexample}","el{ex:hindsight-bet}\nLet $Y_t$ be independent fair signs and $\\gamma=0$. Choose $\\lambda_t=1$ when the current sign is positive and zero otherwise. This invalid rule doubles $E_t$ on a positive score and leaves it unchanged on a negative one. Within four scores it crosses the threshold four whenever at least two scores are positive, an event of probability $11/16$. This exceeds the nominal bound $1/4$. Each individual bet lies in the numerical range $[0,1]$. What fails is predictability.\n\\end{counterexample}","s and $\\gamma=0$. Choose $\\lambda_t=1$ when the current sign is positive and zero otherwise. This invalid rule doubles $E_t$ on a positive score and leaves it unchanged on a negative one. Within four scores it crosses the threshold four whenever at least two scores are positive, an event of probability $11/16$. This exceeds the nominal bound $1/4$. Each individual bet lies in the numerical range $[0,1]$. What fails is predictability.\n\\end{counterexample}","\\subsection{Evidence across recursive generations}\nFor attempt $j$, condition on the complete earlier lineage, freeze its candidate, and initialize a fresh valid audit at $E_0^{(j)}=1$ with error allowance $\\delta_j$. If $\\sum_j\\delta_j\\leq\\delta$, Proposition~\\ref{prop:lineage} bounds the probability of ever accepting a false margin claim by $\\delta$. Each audit can stop sequentially, and the next candidate may depend on completed earlier audits. The premise remains conditional validity for each actual attempt.","Use a resource vector rather than an unexplained token total:\n\\[\nC=(C_{\\rm model},C_{\\rm search},C_{\\rm validate},C_{\\rm install},\nC_{\\rm execute},C_{\\rm store}).\n\\]\nThese coordinates need not share units. Model construction may involve environment observations. Search and validation may involve processor work. Installation may require conversion or compilation. Execution work concerns one deployment. Storage concerns the retained library and complete controller. A scalar comparison needs declared conversion weights or a common physical budget, not an arbitrary sum of unlike quantities.","For the hidden-cell instrument, a compiled tree has fifteen records even though a successful depth-three run visits only four. The bit-query API is installed for both procedures in the matched comparison. Its table representation, if used, occupies $n(n+\\log_2n)$ Boolean entries for the chosen power-of-two family. The reported three-versus-seven diagnostic comparison does not imply that inspecting a large subset costs one machine instruction in every implementation. It uses one sensor call as its stipulated unit.","\\begin{proposition}[Amortization of a reusable process improvement]\\label{prop:amortization}\nIn a fixed additive resource unit, suppose installing a new procedure has incremental cost $C\\geq0$ relative to the old one, and its expected per-use saving on an unchanged task distribution is $d>0$. If there are $N$ uses and no additional costs or changes in the task law, its expected cumulative saving is $Nd-C$. It is strictly resource-improving exactly when $N>C/d$.\n\\end{proposition}\n\\begin{proof}\nLinearity of expectation gives expected total use savings $Nd$, without requiring independence between uses when each has the stated expectation. Subtract the fixed incremental cost and solve the strict inequality.\n\\end{proof}","gin{proposition}[Amortization of a reusable process improvement]\\label{prop:amortization}\nIn a fixed additive resource unit, suppose installing a new procedure has incremental cost $C\\geq0$ relative to the old one, and its expected per-use saving on an unchanged task distribution is $d>0$. If there are $N$ uses and no additional costs or changes in the task law, its expected cumulative saving is $Nd-C$. It is strictly resource-improving exactly when $N>C/d$.\n\\end{proposition}\n\\begin{proof}\nLinearity of expectation gives expected total use savings $Nd$, without requiring independence between uses when each has the stated expectation. Subtract the fixed incremental cost and solve the strict inequality.\n\\end{proof}","rovement]\\label{prop:amortization}\nIn a fixed additive resource unit, suppose installing a new procedure has incremental cost $C\\geq0$ relative to the old one, and its expected per-use saving on an unchanged task distribution is $d>0$. If there are $N$ uses and no additional costs or changes in the task law, its expected cumulative saving is $Nd-C$. It is strictly resource-improving exactly when $N>C/d$.\n\\end{proposition}\n\\begin{proof}\nLinearity of expectation gives expected total use savings $Nd$, without requiring independence between uses when each has the stated expectation. Subtract the fixed incremental cost and solve the strict inequality.\n\\end{proof}","edure has incremental cost $C\\geq0$ relative to the old one, and its expected per-use saving on an unchanged task distribution is $d>0$. If there are $N$ uses and no additional costs or changes in the task law, its expected cumulative saving is $Nd-C$. It is strictly resource-improving exactly when $N>C/d$.\n\\end{proposition}\n\\begin{proof}\nLinearity of expectation gives expected total use savings $Nd$, without requiring independence between uses when each has the stated expectation. Subtract the fixed incremental cost and solve the strict inequality.\n\\end{proof}","The formula is not a prediction of future demand. Distribution drift, maintenance, growing context, retrieval overhead, or a changed interface can alter $d$ or add costs. The value of the repair library depends on whether later problems actually admit its operations.","\\subsection{Larger search spaces can lower productivity}\n\\begin{counterexample}[More expressibility with worse fixed-budget discovery]\\label{ex:dilution}\nA search language has $N$ candidates, exactly one useful. A procedure samples uniformly with replacement $B\\geq1$ times and recognizes usefulness perfectly after each sample. Its probability of finding the useful candidate is $1-(1-1/N)^B$. Adding an additional useless candidate preserves every old expression and strictly decreases this probability. The best attainable repair is unchanged, while the implemented discovery procedure becomes worse under the same sampling budget.\n\\end{counterexample}","Let $Z$ be a finite set of complete lineage configurations. A configuration includes the currently selected solver and improver, retained libraries and evidence, the active comparison contract, and relevant resource state. The following theorem assumes this configuration is observed. Unknown configurations require belief-state modeling or a different synthesis result. No unrestricted introspection is inferred.","Each $z$ has a finite set of authorized attempts $A(z)$. An attempt has a nonempty set $\\Post(z,a)$ of possible successor configurations. This can represent all outcomes of a stochastic proposal or an abstract uncertainty set. A safe set $\\Safe\\subseteq Z$ encodes the declared semantic, evidence, authority, and resource conditions. A pause or terminal state may have a zero-progress self-loop. Actual expensive attempts must debit resources in $Z$, or the model must explicitly state their replenishment.","s $A(z)$. An attempt has a nonempty set $\\Post(z,a)$ of possible successor configurations. This can represent all outcomes of a stochastic proposal or an abstract uncertainty set. A safe set $\\Safe\\subseteq Z$ encodes the declared semantic, evidence, authority, and resource conditions. A pause or terminal state may have a zero-progress self-loop. Actual expensive attempts must debit resources in $Z$, or the model must explicitly state their replenishment.","Define\n\\[\nK_0=\\Safe,\\qquad\nK_{n+1}=\\{z\\in K_n:\\exists a\\in A(z),\\ \\Post(z,a)\\subseteq K_n\\}.\n\\]\nThe descending finite sequence reaches a fixed point $\\mathcal K$. It is the set from which some policy keeps every successor safe indefinitely. This can include doing nothing forever. Safety alone is not recursive improvement.","For each transition, a progress mark $p(z,a,z')\\in\\{0,1\\}$ is supplied by a checked criterion. A mark of one means the specified process-improvement claim has been established under the same comparison. It is not assigned merely because a file changed. Define\n\\begin{align}\\label{eq:progress}\nW_{0,0}&=\\mathcal K, & W_{0,k}&=\\varnothing\\quad(k>0),","For each transition, a progress mark $p(z,a,z')\\in\\{0,1\\}$ is supplied by a checked criterion. A mark of one means the specified process-improvement claim has been established under the same comparison. It is not assigned merely because a file changed. Define\n\\begin{align}\\label{eq:progress}\nW_{0,0}&=\\mathcal K, & W_{0,k}&=\\varnothing\\quad(k>0),\\\\\nW_{h,k}&=\\{z\\in\\mathcal K:\\exists a\\in A(z),\\ \\Post(z,a)\\subseteq\\mathcal K,\\nonumber\\\\[-2pt]\n&\\hspace{34mm} z'\\in W_{h-1,\\max\\{0,k-p(z,a,z')\\}}\n\\quad\\forall z'\\in\\Post(z,a)\\}.\\nonumber\n\\end{align}\nHere $h$ is the number of remaining attempts and $k$ the number of still-required progress events. A pause action lets a policy that finishes early use the remaining horizon without changing the result.","[-2pt]\n&\\hspace{34mm} z'\\in W_{h-1,\\max\\{0,k-p(z,a,z')\\}}\n\\quad\\forall z'\\in\\Post(z,a)\\}.\\nonumber\n\\end{align}\nHere $h$ is the number of remaining attempts and $k$ the number of still-required progress events. A pause action lets a policy that finishes early use the remaining horizon without changing the result.","\\begin{theorem}[Common-policy recursive progress]\\label{thm:progress}\nA configuration belongs to $W_{h,k}$ if and only if there exists one authorized policy that keeps all outcomes in $\\mathcal K$ and guarantees at least $k$ checked progress events within $h$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must adm","\\begin{theorem}[Common-policy recursive progress]\\label{thm:progress}\nA configuration belongs to $W_{h,k}$ if and only if there exists one authorized policy that keeps all outcomes in $\\mathcal K$ and guarantees at least $k$ checked progress events within $h$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditi","\\begin{theorem}[Common-policy recursive progress]\\label{thm:progress}\nA configuration belongs to $W_{h,k}$ if and only if there exists one authorized policy that keeps all outcomes in $\\mathcal K$ and guarantees at least $k$ checked progress events within $h$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the membersh","\\begin{theorem}[Common-policy recursive progress]\\label{thm:progress}\nA configuration belongs to $W_{h,k}$ if and only if there exists one authorized policy that keeps all outcomes in $\\mathcal K$ and guarantees at least $k$ checked progress events within $h$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Convers","\\begin{theorem}[Common-policy recursive progress]\\label{thm:progress}\nA configuration belongs to $W_{h,k}$ if and only if there exists one authorized policy that keeps all outcomes in $\\mathcal K$ and guarantees at least $k$ checked progress events within $h$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Conversely, choose a witnessing first action and a witnessing continuation for each observed suc","omes in $\\mathcal K$ and guarantees at least $k$ checked progress events within $h$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Conversely, choose a witnessing first action and a witnessing continuation for each observed successor. The resulting policy works on every path. Induction proves necessity, sufficiency, and the stated memory requirement. Finiteness makes the construction terminating.\n\\end{proof}","$ attempts. A policy depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Conversely, choose a witnessing first action and a witnessing continuation for each observed successor. The resulting policy works on every path. Induction proves necessity, sufficiency, and the stated memory requirement. Finiteness makes the construction terminating.\n\\end{proof}","y depending on the current configuration and the remaining pair $(h,k)$ suffices. The finite recurrence gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Conversely, choose a witnessing first action and a witnessing continuation for each observed successor. The resulting policy works on every path. Induction proves necessity, sufficiency, and the stated memory requirement. Finiteness makes the construction terminating.\n\\end{proof}","gives a witness when feasible and a complete negative answer within the stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Conversely, choose a witnessing first action and a witnessing continuation for each observed successor. The resulting policy works on every path. Induction proves necessity, sufficiency, and the stated memory requirement. Finiteness makes the construction terminating.\n\\end{proof}","stated model when infeasible.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, zero remaining progress is possible exactly in $\\mathcal K$, and positive remaining progress is impossible. Assume the claim for $h-1$. A policy at $z$ must choose one initial action before its outcome is known. Every possible successor must remain in $\\mathcal K$ and must admit a continuation meeting the remaining target $\\max\\{0,k-p\\}$. By induction these conditions are exactly the memberships in~\\eqref{eq:progress}. Conversely, choose a witnessing first action and a witnessing continuation for each observed successor. The resulting policy works on every path. Induction proves necessity, sufficiency, and the stated memory requirement. Finiteness makes the construction terminating.\n\\end{proof}","\\subsection{Safe rejection can prevent a worst-case progress guarantee}\n\\begin{counterexample}[Repeated opportunity is not bounded sure progress]\\label{ex:retry}\nAt a safe configuration, an attempt either produces a genuine improvement or rejects the candidate and returns to the same configuration. Both outcomes are possible at every attempt. A safe pause is also available. For every finite $h$, there is a path with $h$ consecutive rejections, so no policy guarantees one improvement within $h$ attempts. With an independent positive success probability, eventual success can nevertheless occur with probability one. The worst-case finite theorem and that probabilistic statement answer different questions.\n\\end{counterexample}","ction can prevent a worst-case progress guarantee}\n\\begin{counterexample}[Repeated opportunity is not bounded sure progress]\\label{ex:retry}\nAt a safe configuration, an attempt either produces a genuine improvement or rejects the candidate and returns to the same configuration. Both outcomes are possible at every attempt. A safe pause is also available. For every finite $h$, there is a path with $h$ consecutive rejections, so no policy guarantees one improvement within $h$ attempts. With an independent positive success probability, eventual success can nevertheless occur with probability one. The worst-case finite theorem and that probabilistic statement answer different questions.\n\\end{counterexample}","epeated opportunity is not bounded sure progress]\\label{ex:retry}\nAt a safe configuration, an attempt either produces a genuine improvement or rejects the candidate and returns to the same configuration. Both outcomes are possible at every attempt. A safe pause is also available. For every finite $h$, there is a path with $h$ consecutive rejections, so no policy guarantees one improvement within $h$ attempts. With an independent positive success probability, eventual success can nevertheless occur with probability one. The worst-case finite theorem and that probabilistic statement answer different questions.\n\\end{counterexample}","\\subsection{Why local improvement does not imply recursive viability}\nA second finite example has a proposal that raises the current score but deletes the only retained witness needed to validate any subsequent proposal. Its destination lies outside $\\Safe$ when witness availability is protected. If witness availability is omitted, it may remain safe but have no productive continuation. The result depends on the contract, not the flattering name assigned to the edit.","\\begin{proposition}[A fixed-margin ceiling on a fixed score]\\label{prop:ceiling}\nSuppose a lineage has a fixed measure $R_t\\in[a,b]$, every accepted progress event increases it by at least $\\gamma>0$, and all other transitions are nondecreasing in that same measure. Then the number of progress events is at most\n\\[\n\\left\\lfloor\\frac{b-a}{\\gamma}\\right\\rfloor.\n\\]\nAn infinite sequence of positive increments is compatible with bounded total improvement when no common positive lower bound is imposed.\n\\end{proposition}\n\\begin{proof}\nAfter $k$ progress events, telescoping and nondecrease give $R_t\\geq R_0+k\\gamma\\geq a+k\\gamma$. Since $R_t\\leq b$, the bound follows. For the second assertion, start at zero and use increments $2^{-t}$ for $t\\geq1$. Every increment is positive and their partial sums remain below one.\n\\end{proof}","accepted progress event increases it by at least $\\gamma>0$, and all other transitions are nondecreasing in that same measure. Then the number of progress events is at most\n\\[\n\\left\\lfloor\\frac{b-a}{\\gamma}\\right\\rfloor.\n\\]\nAn infinite sequence of positive increments is compatible with bounded total improvement when no common positive lower bound is imposed.\n\\end{proposition}\n\\begin{proof}\nAfter $k$ progress events, telescoping and nondecrease give $R_t\\geq R_0+k\\gamma\\geq a+k\\gamma$. Since $R_t\\leq b$, the bound follows. For the second assertion, start at zero and use increments $2^{-t}$ for $t\\geq1$. Every increment is positive and their partial sums remain below one.\n\\end{proof}","ma>0$, and all other transitions are nondecreasing in that same measure. Then the number of progress events is at most\n\\[\n\\left\\lfloor\\frac{b-a}{\\gamma}\\right\\rfloor.\n\\]\nAn infinite sequence of positive increments is compatible with bounded total improvement when no common positive lower bound is imposed.\n\\end{proposition}\n\\begin{proof}\nAfter $k$ progress events, telescoping and nondecrease give $R_t\\geq R_0+k\\gamma\\geq a+k\\gamma$. Since $R_t\\leq b$, the bound follows. For the second assertion, start at zero and use increments $2^{-t}$ for $t\\geq1$. Every increment is positive and their partial sums remain below one.\n\\end{proof}","reasing in that same measure. Then the number of progress events is at most\n\\[\n\\left\\lfloor\\frac{b-a}{\\gamma}\\right\\rfloor.\n\\]\nAn infinite sequence of positive increments is compatible with bounded total improvement when no common positive lower bound is imposed.\n\\end{proposition}\n\\begin{proof}\nAfter $k$ progress events, telescoping and nondecrease give $R_t\\geq R_0+k\\gamma\\geq a+k\\gamma$. Since $R_t\\leq b$, the bound follows. For the second assertion, start at zero and use increments $2^{-t}$ for $t\\geq1$. Every increment is positive and their partial sums remain below one.\n\\end{proof}","\\frac{b-a}{\\gamma}\\right\\rfloor.\n\\]\nAn infinite sequence of positive increments is compatible with bounded total improvement when no common positive lower bound is imposed.\n\\end{proposition}\n\\begin{proof}\nAfter $k$ progress events, telescoping and nondecrease give $R_t\\geq R_0+k\\gamma\\geq a+k\\gamma$. Since $R_t\\leq b$, the bound follows. For the second assertion, start at zero and use increments $2^{-t}$ for $t\\geq1$. Every increment is positive and their partial sums remain below one.\n\\end{proof}","amma}\\right\\rfloor.\n\\]\nAn infinite sequence of positive increments is compatible with bounded total improvement when no common positive lower bound is imposed.\n\\end{proposition}\n\\begin{proof}\nAfter $k$ progress events, telescoping and nondecrease give $R_t\\geq R_0+k\\gamma\\geq a+k\\gamma$. Since $R_t\\leq b$, the bound follows. For the second assertion, start at zero and use increments $2^{-t}$ for $t\\geq1$. Every increment is positive and their partial sums remain below one.\n\\end{proof}","A simple accounting model can express the additional conditions for acceleration. Let $g_t$ be an expected useful gain and $c_t>0$ the complete cost of the corresponding discovery and acceptance cycle. Improvement productivity per unit resource is $g_t/c_t$. Raising $g_t$ alone does not raise that ratio when $c_t$ rises faster. A larger reachable repair set is similarly insufficient when proposal quality falls or validation becomes the limiting cost.","\\subsection{What would learn}\nLet $r_\\phi$ encode the retained state, $G_{\\theta,a}$ describe proposed operations, $\\pi_\\omega$ choose observations or experiments, and $p_\\psi$ propose candidate programs or parameter updates. A development episode can modify $\\phi,\\theta,\\omega,\\psi$ through a declared learning rule. A persistent program library can also change by adding, replacing, or removing checked operators. Temporary deliberation without a retained modification is not called a new learned generation here.","\\begin{proposition}[Local operation agreement preserves finite continuation]\\label{prop:continuation}\nLet $D$ be invariant under every source operation $F_a$, let $q:D\\to Z$, and suppose\n\\[\nqF_a=G_aq\\quad\\text{on }D\n\\]\nfor every admitted $a$. If the required output satisfies $y=\\bar yq$ on $D$, then for every finite operation word $w$,\n\\[\nqF_w=G_wq,\\qquad yF_w=\\bar yG_wq.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe empty word gives the identity. Assume the relation holds for a prefix. Domain invariance allows application of the next local equation, giving $qF_aF_w=G_aqF_w=G_aG_wq$. Induction establishes the first identity and composing with $\\bar y$ gives the second.\n\\end{proof}","let $q:D\\to Z$, and suppose\n\\[\nqF_a=G_aq\\quad\\text{on }D\n\\]\nfor every admitted $a$. If the required output satisfies $y=\\bar yq$ on $D$, then for every finite operation word $w$,\n\\[\nqF_w=G_wq,\\qquad yF_w=\\bar yG_wq.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe empty word gives the identity. Assume the relation holds for a prefix. Domain invariance allows application of the next local equation, giving $qF_aF_w=G_aqF_w=G_aG_wq$. Induction establishes the first identity and composing with $\\bar y$ gives the second.\n\\end{proof}","the required output satisfies $y=\\bar yq$ on $D$, then for every finite operation word $w$,\n\\[\nqF_w=G_wq,\\qquad yF_w=\\bar yG_wq.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe empty word gives the identity. Assume the relation holds for a prefix. Domain invariance allows application of the next local equation, giving $qF_aF_w=G_aqF_w=G_aG_wq$. Induction establishes the first identity and composing with $\\bar y$ gives the second.\n\\end{proof}","The proposition is a specification for meaningful reuse, not a method for learning $q$ or $G_a$. Invariance matters because an unchecked operation can leave the domain where the equations hold. Output preservation matters because commuting hidden states need not retain the task variable. Actual causal use matters because a probe can decode an appropriate state that the solver does not use.","For $n=2^k$ cells, assume a single defective cell, binary reports, singleton queries, and $k$ bit-coordinate queries. All queries leave the cell unchanged and cost one diagnostic call. A correct final repair costs one additional call. Distinct cells require distinct terminal repairs. There are no false repair commits. A literal tree record contains one experiment or repair identifier plus its child references.","\\begin{theorem}[Optimal diagnosis in the hidden-cell family]\\label{thm:cells}\nEvery certain diagnosis needs at least $k$ binary queries in the worst case. The bit-query procedure attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary","\\begin{theorem}[Optimal diagnosis in the hidden-cell family]\\label{thm:cells}\nEvery certain diagnosis needs at least $k$ binary queries in the worst case. The bit-query procedure attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer","\\begin{theorem}[Optimal diagnosis in the hidden-cell family]\\label{thm:cells}\nEvery certain diagnosis needs at least $k$ binary queries in the worst case. The bit-query procedure attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singl","certain diagnosis needs at least $k$ binary queries in the worst case. The bit-query procedure attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again tota…","agnosis needs at least $k$ binary queries in the worst case. The bit-query procedure attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1…","it-query procedure attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work count…","attains this bound and uses $2n-1$ records in a full binary tree. Any procedure using singleton queries alone needs at least $n-1$ queries in the worst case, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","se, and equality-first testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","t testing attains that bound with the same number of tree records. Thus their worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","worst total execution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","ecution costs, including repair, are $k+1$ and $n$.\n\\end{theorem}\n\\begin{proof}\nA binary tree of depth less than $k$ has fewer than $2^k=n$ leaves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","ves. Every successful leaf can identify at most one cell because its repair must be correct for every remaining case. Therefore depth at least $k$ is required. The $k$ coordinate bits identify every cell and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","ll and give a complete binary tree with $n$ leaves and $n-1$ internal nodes. For singleton tests, an adversary can answer no on each tested cell while at least two untested cells remain. Fewer than $n-1$ distinct singleton queries therefore cannot guarantee identification. Equality-first testing uses one leaf for each positive answer and one final leaf for the last remaining cell, again totaling $2n-1$ records. Adding the terminal repair gives the claimed work counts.\n\\end{proof}","\\begin{figure}[!ht]\\centering\n\\includegraphics[width=.92\\linewidth]{figures/diagnostic_cost.pdf}\n\\caption{Exact total execution costs for the power-of-two cell family. Singleton testing requires $n-1$ diagnostic queries, while the admitted bit queries require $\\log_2 n$. A final repair adds one call to each plotted curve. Both full trees have $2n-1$ stored records. The curves omit separately charged model acquisition, compilation, and adapters.}\\label{fig:cost}\n\\end{figure}","\\subsection{What is actually compiled}\nThe reference compiler receives an observation table and the set of successful terminal repairs. It computes the frontier in Theorem~\\ref{thm:synthesis} and returns an optimal tree under a chosen work/storage allowance. At $n=8$ it returns a four-call, fifteen-record witness. A separately coded interpreter checks every initial cell and every branch. The reference comparison constructs the equality-first baseline independently, rather than comparing the compiler to a weakened version of its own output.","\\subsection{The measured object is a procedure}\nUnder the earlier four-call cap, the equality-first improver repairs $3/8$ of the uniform eight-cell cases and otherwise abstains. The compiled improver repairs all cases. The repaired solvers themselves are simple: each now knows the correct cell-specific operation. What improved is the procedure that acquires that case-specific correction. The library and sensor language are unchanged.","\\appendix\n\\section{A complete diagnostic trace}\\label{app:trace}\nTake $n=8$ and hidden defect $j=5$, whose binary digits are $101$. One optimal tree uses the three queries\n\\[\nA_0=\\{1,3,5,7\\},\\qquad A_1=\\{2,3,6,7\\},\\qquad A_2=\\{4,5,6,7\\}.\n\\]\nThe precise order may differ among equally optimal compiled trees. For this order, the supports are\n\\[\n\\{0,\\ldots,7\\}\\xrightarrow{1}\\{1,3,5,7\\}\n\\xrightarrow{0}\\{1,5\\}\\xrightarrow{1}\\{5\\}.\n\\]\nThe terminal repair is $r_5$. The path uses three diagnostic calls and one repair call. The complete tree stores all eight repair leaves and seven query nodes, not merely this path.","p:trace}\nTake $n=8$ and hidden defect $j=5$, whose binary digits are $101$. One optimal tree uses the three queries\n\\[\nA_0=\\{1,3,5,7\\},\\qquad A_1=\\{2,3,6,7\\},\\qquad A_2=\\{4,5,6,7\\}.\n\\]\nThe precise order may differ among equally optimal compiled trees. For this order, the supports are\n\\[\n\\{0,\\ldots,7\\}\\xrightarrow{1}\\{1,3,5,7\\}\n\\xrightarrow{0}\\{1,5\\}\\xrightarrow{1}\\{5\\}.\n\\]\nThe terminal repair is $r_5$. The path uses three diagnostic calls and one repair call. The complete tree stores all eight repair leaves and seven query nodes, not merely this path.","For a target relabeling $\\sigma(i)=3i+1\\pmod8$, the defect has target label $\\sigma(5)=0$. Each query becomes $\\sigma(A_r)$ and the terminal repair becomes $r'_{\\sigma(5)}$. Reports and tree shape agree with the source, but the label adapter is part of the transported interface. Applying the old numeric repair label five directly in the target would be incorrect. This is why syntactic resemblance does not replace a semantic map.","\\subsection{A repair problem that should stop before full identification}\nIn the four-case example, the support $\\{0,1,2\\}$ admits repair $a$ immediately. A compiler that insists on identifying the precise case would spend unnecessary work. The successful stopping condition is the existence of one action whose success set contains the whole support. Conversely, the support $\\{2,3\\}$ admits no common repair in that example. A leaf chosen there without another observation is invalid even though either repair succeeds in one of its cases."]}