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bigcup","display":"\\bigcup","kind":"use","occurrences":[567]},{"id":"symbol-c5604e7103d9198f","tex":"\\bigcup_{j\\in J}","aliases":"⋃_j∈ J bigcup in","display":"\\bigcup_{j\\in J}","kind":"use","occurrences":[567]},{"id":"symbol-437de0b3cc0e5a5f","tex":"\\langle","aliases":"⟨ langle","display":"\\langle","kind":"use","occurrences":[53]},{"id":"symbol-7b52d298f1764e49","tex":"\\rangle","aliases":"⟩ rangle","display":"\\rangle","kind":"use","occurrences":[54]}],"contexts":["ary}\n\\label{sec:publication-evidence}\nThe authoritative formal baseline for this edition is \\path{AVOC_Formal_v0.41.0.zip}, with SHA-256\n\\begin{quote}\\small\\ttfamily\n\\detokenize{04efab87010b3b88c9a119275e190423a2465da32ba1e526edb7e6d2e0cd077e}.\n\\end{quote}\nThe original archive is preserved intact in the accompanying package. Its internal payload manifest was freshly checked for this edition: all $16{,}912$ listed payload digests match. The archive contains the mathematical source, interfaces, experiments, review records, failed attempts, and reproduction instructions. It excludes the Lean runtime and compiled project/dependency objects.","\\subsection{Archived formal evidence}\nThe v0.41.0 release record reports $222$ positive build targets, all passing under warnings-as-errors, and $2{,}268$ exact named theorem/lemma axiom reports. The allowed axiom union is \\path{Classical.choice}, \\path{Quot.sound}, and \\path{propext}. There are $195$ non-audit mathematical modules; the latest block adds four modules and $69$ named theorem/lemma declarations. The $191$ inherited mathematical modules were checked byte-identical to the preceding release. The archive also records $135$ substantive negative proof rejections.","ed formal evidence}\nThe v0.41.0 release record reports $222$ positive build targets, all passing under warnings-as-errors, and $2{,}268$ exact named theorem/lemma axiom reports. The allowed axiom union is \\path{Classical.choice}, \\path{Quot.sound}, and \\path{propext}. There are $195$ non-audit mathematical modules; the latest block adds four modules and $69$ named theorem/lemma declarations. The $191$ inherited mathematical modules were checked byte-identical to the preceding release. The archive also records $135$ substantive negative proof rejections.","rrors, and $2{,}268$ exact named theorem/lemma axiom reports. The allowed axiom union is \\path{Classical.choice}, \\path{Quot.sound}, and \\path{propext}. There are $195$ non-audit mathematical modules; the latest block adds four modules and $69$ named theorem/lemma declarations. The $191$ inherited mathematical modules were checked byte-identical to the preceding release. The archive also records $135$ substantive negative proof rejections.","\\subsection{Fresh checks for this edition}\nA fresh source inventory matches the actual $195$ mathematical modules and $2{,}268$ named declarations to the delivered registry. It checks the archived declaration-source digests and recorded source locations. Fresh exact Python computations reproduce the declared examples and finite recorder comparisons. The detailed tested families, assertions, input digests, outputs, and commands are included in the publication's \\path{reviews} directory.","\\fig{10_completion_order_four_cases.pdf}{Every pair of attempted outcomes is logged and resolved, yet successes finishing first make the first reported bit successful with probability $3/4$. 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{₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad /\\quad REALIZATION CALCULUS\\par}\n \\vspace{17mm}{\\Huge\\bfseries AVOC\\par}\n \\vspace{9mm}{\\LARGE\\bfseries Certified Artifact Planning,","\\eta$}}1 {ν}{{$\\nu$}}1 {χ}{{$\\chi$}}1 {θ}{{$\\theta$}}1 {ρ}{{$\\rho$}}1 {Γ}{{$\\Gamma$}}1 {Σ}{{$\\Sigma$}}1 {∧}{{$\\land$}}1 {∨}{{$\\lor$}}1 {¬}{{$\\neg$}}1 {×}{{$\\times$}}1 {⊆}{{$\\subseteq$}}1 {⊂}{{$\\subset$}}1 {⟨}{{$\\langle$}}1 {⟩}{{$\\rangle$}}1 {∑}{{$\\sum$}}1 {∏}{{$\\prod$}}1 {∣}{{$\\mid$}}1 {∘}{{$\\circ$}}1 {↑}{{$\\uparrow$}}1 {↓}{{$\\downarrow$}}1 {▸}{{$\\triangleright$}}1 {₀}{{$_0$}}1 {₁}{{$_1$}}1 {₂}{{$_2$}}1 {₃}{{$_3$}}1 {₄}{{$_4$}}1 {₅}{{$_5$}}1 {₆}{{$_6$}}1 {₇}{{$_7$}}1 {₈}{{$_8$}}1 {₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad 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{∏}{{$\\prod$}}1 {∣}{{$\\mid$}}1 {∘}{{$\\circ$}}1 {↑}{{$\\uparrow$}}1 {↓}{{$\\downarrow$}}1 {▸}{{$\\triangleright$}}1 {₀}{{$_0$}}1 {₁}{{$_1$}}1 {₂}{{$_2$}}1 {₃}{{$_3$}}1 {₄}{{$_4$}}1 {₅}{{$_5$}}1 {₆}{{$_6$}}1 {₇}{{$_7$}}1 {₈}{{$_8$}}1 {₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad /\\quad REALIZATION CALCULUS\\par}\n \\vspace{17mm}{\\Huge\\bfseries AVOC\\par}\n \\vspace{9mm}{\\LARGE\\bfseries Certified Artifact Planning,","}1 {ρ}{{$\\rho$}}1 {Γ}{{$\\Gamma$}}1 {Σ}{{$\\Sigma$}}1 {∧}{{$\\land$}}1 {∨}{{$\\lor$}}1 {¬}{{$\\neg$}}1 {×}{{$\\times$}}1 {⊆}{{$\\subseteq$}}1 {⊂}{{$\\subset$}}1 {⟨}{{$\\langle$}}1 {⟩}{{$\\rangle$}}1 {∑}{{$\\sum$}}1 {∏}{{$\\prod$}}1 {∣}{{$\\mid$}}1 {∘}{{$\\circ$}}1 {↑}{{$\\uparrow$}}1 {↓}{{$\\downarrow$}}1 {▸}{{$\\triangleright$}}1 {₀}{{$_0$}}1 {₁}{{$_1$}}1 {₂}{{$_2$}}1 {₃}{{$_3$}}1 {₄}{{$_4$}}1 {₅}{{$_5$}}1 {₆}{{$_6$}}1 {₇}{{$_7$}}1 {₈}{{$_8$}}1 {₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad /\\quad REALIZATION CALCULUS\\par}\n \\vspace{17mm}{\\Huge\\bfseries AVOC\\par}\n \\vspace{9mm}{\\LARGE\\bfseries Certified Artifact Planning,","$}}1 {Γ}{{$\\Gamma$}}1 {Σ}{{$\\Sigma$}}1 {∧}{{$\\land$}}1 {∨}{{$\\lor$}}1 {¬}{{$\\neg$}}1 {×}{{$\\times$}}1 {⊆}{{$\\subseteq$}}1 {⊂}{{$\\subset$}}1 {⟨}{{$\\langle$}}1 {⟩}{{$\\rangle$}}1 {∑}{{$\\sum$}}1 {∏}{{$\\prod$}}1 {∣}{{$\\mid$}}1 {∘}{{$\\circ$}}1 {↑}{{$\\uparrow$}}1 {↓}{{$\\downarrow$}}1 {▸}{{$\\triangleright$}}1 {₀}{{$_0$}}1 {₁}{{$_1$}}1 {₂}{{$_2$}}1 {₃}{{$_3$}}1 {₄}{{$_4$}}1 {₅}{{$_5$}}1 {₆}{{$_6$}}1 {₇}{{$_7$}}1 {₈}{{$_8$}}1 {₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad /\\quad REALIZATION CALCULUS\\par}\n \\vspace{17mm}{\\Huge\\bfseries AVOC\\par}\n \\vspace{9mm}{\\LARGE\\bfseries Certified Artifact Planning,","amma$}}1 {Σ}{{$\\Sigma$}}1 {∧}{{$\\land$}}1 {∨}{{$\\lor$}}1 {¬}{{$\\neg$}}1 {×}{{$\\times$}}1 {⊆}{{$\\subseteq$}}1 {⊂}{{$\\subset$}}1 {⟨}{{$\\langle$}}1 {⟩}{{$\\rangle$}}1 {∑}{{$\\sum$}}1 {∏}{{$\\prod$}}1 {∣}{{$\\mid$}}1 {∘}{{$\\circ$}}1 {↑}{{$\\uparrow$}}1 {↓}{{$\\downarrow$}}1 {▸}{{$\\triangleright$}}1 {₀}{{$_0$}}1 {₁}{{$_1$}}1 {₂}{{$_2$}}1 {₃}{{$_3$}}1 {₄}{{$_4$}}1 {₅}{{$_5$}}1 {₆}{{$_6$}}1 {₇}{{$_7$}}1 {₈}{{$_8$}}1 {₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad /\\quad REALIZATION CALCULUS\\par}\n \\vspace{17mm}{\\Huge\\bfseries AVOC\\par}\n \\vspace{9mm}{\\LARGE\\bfseries Certified Artifact Planning,","{$\\Sigma$}}1 {∧}{{$\\land$}}1 {∨}{{$\\lor$}}1 {¬}{{$\\neg$}}1 {×}{{$\\times$}}1 {⊆}{{$\\subseteq$}}1 {⊂}{{$\\subset$}}1 {⟨}{{$\\langle$}}1 {⟩}{{$\\rangle$}}1 {∑}{{$\\sum$}}1 {∏}{{$\\prod$}}1 {∣}{{$\\mid$}}1 {∘}{{$\\circ$}}1 {↑}{{$\\uparrow$}}1 {↓}{{$\\downarrow$}}1 {▸}{{$\\triangleright$}}1 {₀}{{$_0$}}1 {₁}{{$_1$}}1 {₂}{{$_2$}}1 {₃}{{$_3$}}1 {₄}{{$_4$}}1 {₅}{{$_5$}}1 {₆}{{$_6$}}1 {₇}{{$_7$}}1 {₈}{{$_8$}}1 {₉}{{$_9$}}1}\n\\lstset{style=LeanCode}\n\\newcommand{\\avoccover}[2]{\\hypersetup{pageanchor=false}\\begin{titlepage}\\color{navy}\n \\vspace*{16mm}{\\large KOMOREBI\\quad /\\quad REALIZATION CALCULUS\\par}\n \\vspace{17mm}{\\Huge\\bfseries AVOC\\par}\n \\vspace{9mm}{\\LARGE\\bfseries Certified Artifact Planning,","\\begin{definition}[Deterministic refinement and task sufficiency]\nLet $X$ be a raw state space and let $\\phi:X\\to Z$ and $T:X\\to Y$ be functions. We write\n\\[\n \\phi\\succeq T\n \\quad\\Longleftrightarrow\\quad\n \\exists d:Z\\to Y\\ \\ \\forall x\\in X,\\ d(\\phi(x))=T(x).\n\\]\nMore generally, $\\phi\\succeq\\psi$ means that $\\psi$ factors through $\\phi$. The relation is called refinement. The representation $\\phi$ is task sufficient for $T$ when $\\phi\\succeq T$. Two representations are information equivalent when each refines the other.\n\\end{definition}","rministic refinement and task sufficiency]\nLet $X$ be a raw state space and let $\\phi:X\\to Z$ and $T:X\\to Y$ be functions. We write\n\\[\n \\phi\\succeq T\n \\quad\\Longleftrightarrow\\quad\n \\exists d:Z\\to Y\\ \\ \\forall x\\in X,\\ d(\\phi(x))=T(x).\n\\]\nMore generally, $\\phi\\succeq\\psi$ means that $\\psi$ factors through $\\phi$. The relation is called refinement. The representation $\\phi$ is task sufficient for $T$ when $\\phi\\succeq T$. Two representations are information equivalent when each refines the other.\n\\end{definition}","refinement and task sufficiency]\nLet $X$ be a raw state space and let $\\phi:X\\to Z$ and $T:X\\to Y$ be functions. We write\n\\[\n \\phi\\succeq T\n \\quad\\Longleftrightarrow\\quad\n \\exists d:Z\\to Y\\ \\ \\forall x\\in X,\\ d(\\phi(x))=T(x).\n\\]\nMore generally, $\\phi\\succeq\\psi$ means that $\\psi$ factors through $\\phi$. The relation is called refinement. The representation $\\phi$ is task sufficient for $T$ when $\\phi\\succeq T$. Two representations are information equivalent when each refines the other.\n\\end{definition}","\\begin{theorem}[Exact fiber criterion]\n\\label{thm:fiber-criterion}\nSuppose $Y$ is nonempty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states.","\\begin{theorem}[Exact fiber criterion]\n\\label{thm:fiber-criterion}\nSuppose $Y$ is nonempty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses cl","\\begin{theorem}[Exact fiber criterion]\n\\label{thm:fiber-criterion}\nSuppose $Y$ is nonempty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the","\\begin{theorem}[Exact fiber criterion]\n\\label{thm:fiber-criterion}\nSuppose $Y$ is nonempty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","er criterion]\n\\label{thm:fiber-criterion}\nSuppose $Y$ is nonempty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","criterion}\nSuppose $Y$ is nonempty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","ty. Then $\\phi$ is sufficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","fficient for $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","r $T$ if and only if\n\\[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","[\n \\phi(x)=\\phi(y)\\quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","quad\\Longrightarrow\\quad T(x)=T(y)\n \\qquad(x,y\\in X).\n\\]\nIn particular, a single pair with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","ir with $\\phi(x)=\\phi(y)$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}",")$ and $T(x)\\ne T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","T(y)$ proves insufficiency.\n\\end{theorem}\n\\begin{proof}\nIf $T=d\\circ\\phi$, equality of the two representations gives equality after applying $d$. Conversely, fix a default $y_0\\in Y$. For every reachable $z\\in\\phi(X)$ choose $x_z$ with $\\phi(x_z)=z$ and set $d(z)=T(x_z)$. Set $d(z)=y_0$ on unreachable representation states. The fiber hypothesis ensures that $T(x_z)=T(x)$ whenever $z=\\phi(x)$, so $d(\\phi(x))=T(x)$. The nonempty-codomain hypothesis supplies the default on unreachable states. The formal proof uses classical choice for the representatives.\n\\end{proof}","The target $T$ itself is sufficient for $T$. Every sufficient representation refines $T$, by definition. Consequently $T$ is a canonical coarsest sufficient representation in this information preorder. A sufficient representation is task minimal when $T$ also refines it, and hence task minimality is exactly information equivalence with $T$. This is a statement about retained distinctions, rather than a theorem that the target can be computed cheaply or stored in a shortest file.","\\begin{proposition}[Finite data processing]\nFor a finite sample $D\\subseteq X$, let $c_D(\\phi)=|\\phi(D)|$ denote the number of reached representation states. If $\\phi\\succeq\\psi$, then\n\\[\n c_D(\\psi)\\le c_D(\\phi).\n\\]\n\\end{proposition}\n\\begin{proof}\nChoose $d$ with $\\psi=d\\circ\\phi$. Then $\\psi(D)=d(\\phi(D))$, and the image of a finite set under a function has cardinality at most that of the original set.\n\\end{proof}","The finite sample serves only to count reached states. Sufficiency in Theorem~\\ref{thm:fiber-criterion} is quantified over all of $X$. Agreement on a sample is therefore a different kind of evidence unless that sample exhausts the domain.","\\begin{example}[A task must be declared before compression]\nLet $X=\\{0,1\\}^3$ and let $T(x_1,x_2,x_3)=x_1\\wedge x_2$. The representation $\\phi(x)=T(x)$ is sufficient for this target, although it discards $x_3$ and many distinctions among the first two coordinates. The representation $\\psi(x)=x_1\\oplus x_2$ is insufficient: $(0,0,0)$ and $(1,1,0)$ have the same $\\psi$ value and different $T$ values. No decoder can repair this collision. Conversely, retaining all three coordinates refines $T$ but retains more information than this task requires.\n\\end{example}","x_2,x_3)=x_1\\wedge x_2$. The representation $\\phi(x)=T(x)$ is sufficient for this target, although it discards $x_3$ and many distinctions among the first two coordinates. The representation $\\psi(x)=x_1\\oplus x_2$ is insufficient: $(0,0,0)$ and $(1,1,0)$ have the same $\\psi$ value and different $T$ values. No decoder can repair this collision. Conversely, retaining all three coordinates refines $T$ but retains more information than this task requires.\n\\end{example}","Suppose a family of tasks is given by $T_q:X\\to Y$, indexed by $q\\in Q$. Define the joint target\n\\[\n T_{\\mathrm{joint}}(x)=(T_q(x))_{q\\in Q}\\in Y^Q.\n\\]\nThen $\\phi$ is sufficient for the joint target if and only if it is sufficient for each individual task. The forward direction evaluates the decoded answer vector at $q$. For the reverse direction, choose a decoder $d_q$ for each query and define $d(z)(q)=d_q(z)$. The formal theorem permits an arbitrary query index type, and the simultaneous decoder selection uses classical choice.","\\begin{definition}[Deterministic experiment value]\nLet $\\Omega$ be finite, $A$ be a finite nonempty action set, $b:\\Omega\\to\\Q$ be a prior, and $u:\\Omega\\times A\\to\\Q$ be utility. For an observation $\\phi:\\Omega\\to Z$ and policy $\\pi:Z\\to A$, define\n\\[\n J_b(\\phi,\\pi)=\\sum_{\\omega\\in\\Omega}b(\\omega)u(\\omega,\\pi(\\phi(\\omega))).\n\\]\nWhen $Z$ is finite, write $J_b^*(\\phi)=\\max_{\\pi:Z\\to A}J_b(\\phi,\\pi)$.\n\\end{definition}","\\begin{theorem}[Deterministic information-value monotonicity]\nIf $\\phi\\succeq\\psi$, then $J_b^*(\\psi)\\le J_b^*(\\phi)$. If the acquisition costs satisfy $c_\\phi\\le c_\\psi$, then\n\\[\n J_b^*(\\psi)-c_\\psi\\le J_b^*(\\phi)-c_\\phi.\n\\]\n\\end{theorem}\n\\begin{proof}\nLet $d$ decode $\\psi$ from $\\phi$. Any coarse policy $\\pi$ lifts to $\\pi\\circ d$. For every world the lifted policy chooses exactly the same action, so its decision value is unchanged. A maximizing coarse policy exists because the policy space is finite and nonempty. Its lifted value is at most the best fine value. Subtracting the stated costs gives the second inequality.\n\\end{proof}","\\begin{definition}[Finite rational channel]\nA channel from $\\Omega$ to a finite observation alphabet $O$ consists of rational entries $E(o\\mid\\omega)\\ge0$ with $\\sum_o E(o\\mid\\omega)=1$ for every $\\omega$. Given a second channel $K(y\\mid o)$, their composite is\n\\[\n (EK)(y\\mid\\omega)=\\sum_o E(o\\mid\\omega)K(y\\mid o).\n\\]\nThe composite is a garbling of $E$.\n\\end{definition}","The unnormalized signal-action score and policy value are\n\\[\n S_E(o,a)=\\sum_\\omega b(\\omega)E(o\\mid\\omega)u(\\omega,a),\n \\qquad J_b(E,\\pi)=\\sum_o S_E(o,\\pi(o)).\n\\]\nThis form avoids dividing by the evidence mass. In particular, it remains well defined on observations whose predictive probability is zero.","\\begin{theorem}[Finite stochastic information-value monotonicity]\nEvery deterministic policy on $EK$ is weakly dominated in value by some deterministic policy on $E$. Consequently the best deterministic value of $EK$ is at most that of $E$.\n\\end{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields","\\begin{theorem}[Finite stochastic information-value monotonicity]\nEvery deterministic policy on $EK$ is weakly dominated in value by some deterministic policy on $E$. Consequently the best deterministic value of $EK$ is at most that of $E$.\n\\end{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","stochastic information-value monotonicity]\nEvery deterministic policy on $EK$ is weakly dominated in value by some deterministic policy on $E$. Consequently the best deterministic value of $EK$ is at most that of $E$.\n\\end{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","licy on $EK$ is weakly dominated in value by some deterministic policy on $E$. Consequently the best deterministic value of $EK$ is at most that of $E$.\n\\end{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","uently the best deterministic value of $EK$ is at most that of $E$.\n\\end{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","of $E$.\n\\end{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","nd{theorem}\n\\begin{proof}\nFor each fine observation $o$, select $a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","a_o\\in A$ maximizing $S_E(o,a)$. Such a selection exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","election exists because $A$ is finite and nonempty. For a coarse policy $\\pi$, expand the channel composition and rearrange finite sums:\n\\[\n J_b(EK,\\pi)=\\sum_o\\sum_y K(y\\mid o)S_E(o,\\pi(y)).\n\\]\nEvery inner score is at most $S_E(o,a_o)$. Nonnegativity and normalization of the row of $K$ give\n\\[\n \\sum_y K(y\\mid o)S_E(o,\\pi(y))\\le S_E(o,a_o).\n\\]\nSumming over $o$ proves domination by the fine policy $o\\mapsto a_o$. Taking a best coarse policy yields the optimal-value comparison.\n\\end{proof}","\\begin{example}[Information has a price]\nFor two equally likely worlds, let the action predict the world and let utility be $1$ for a correct prediction and $0$ otherwise. A constant observation gives value $1/2$. Revealing the world gives value $1$. If revelation costs $1/8$, its net value is $7/8$, exceeding the free constant experiment's $1/2$. If it costs $3/4$, its net value is $1/4$. The information order is unchanged across these two cases, while the preferred paid experiment changes.\n\\end{example}","A rational belief is a nonnegative vector $b$ on finite $\\Omega$ with $\\sum_\\omega b(\\omega)=1$. For a channel $E$, define the predictive mass\n\\[\n p(o)=\\sum_\\omega b(\\omega)E(o\\mid\\omega).\n\\]\nFor $p(o)>0$, the posterior is $b_o(\\omega)=b(\\omega)E(o\\mid\\omega)/p(o)$. The total updater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\","A rational belief is a nonnegative vector $b$ on finite $\\Omega$ with $\\sum_\\omega b(\\omega)=1$. For a channel $E$, define the predictive mass\n\\[\n p(o)=\\sum_\\omega b(\\omega)E(o\\mid\\omega).\n\\]\nFor $p(o)>0$, the posterior is $b_o(\\omega)=b(\\omega)E(o\\mid\\omega)/p(o)$. The total updater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\]\nPosterior-predict","A rational belief is a nonnegative vector $b$ on finite $\\Omega$ with $\\sum_\\omega b(\\omega)=1$. For a channel $E$, define the predictive mass\n\\[\n p(o)=\\sum_\\omega b(\\omega)E(o\\mid\\omega).\n\\]\nFor $p(o)>0$, the posterior is $b_o(\\omega)=b(\\omega)E(o\\mid\\omega)/p(o)$. The total updater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\]\nPosterior-predictive evaluation and the unnormal","A rational belief is a nonnegative vector $b$ on finite $\\Omega$ with $\\sum_\\omega b(\\omega)=1$. For a channel $E$, define the predictive mass\n\\[\n p(o)=\\sum_\\omega b(\\omega)E(o\\mid\\omega).\n\\]\nFor $p(o)>0$, the posterior is $b_o(\\omega)=b(\\omega)E(o\\mid\\omega)/p(o)$. The total updater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\]\nPosterior-predictive evaluation and the unnormalized decision formu","A rational belief is a nonnegative vector $b$ on finite $\\Omega$ with $\\sum_\\omega b(\\omega)=1$. For a channel $E$, define the predictive mass\n\\[\n p(o)=\\sum_\\omega b(\\omega)E(o\\mid\\omega).\n\\]\nFor $p(o)>0$, the posterior is $b_o(\\omega)=b(\\omega)E(o\\mid\\omega)/p(o)$. The total updater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\]\nPosterior-predictive evaluation and the unnormalized decision formula therefore agree exactly.","For $p(o)>0$, the posterior is $b_o(\\omega)=b(\\omega)E(o\\mid\\omega)/p(o)$. The total updater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\]\nPosterior-predictive evaluation and the unnormalized decision formula therefore agree exactly.","pdater in the source uses $b_o=b$ when $p(o)=0$. This fallback is normalized and has zero coefficient in predictive expectations. Nonnegativity of every summand implies that each Bayes numerator also vanishes on such an impossible observation. Hence, with the total-update convention, the division-free identity\n\\[\n p(o)b_o(\\omega)=b(\\omega)E(o\\mid\\omega)\n\\]\nholds on every branch. It follows that\n\\[\n \\sum_o p(o)\\sum_\\omega b_o(\\omega)u(\\omega,\\pi(o))=J_b(E,\\pi).\n\\]\nPosterior-predictive evaluation and the unnormalized decision formula therefore agree exactly.","For any nonnegative likelihood weight $L$, write\n\\[\n M_b(L)=\\sum_\\omega b(\\omega)L(\\omega),\\qquad\n C_L(b)(\\omega)=\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n\\]\nwhen $M_b(L)>0$. Given nonnegative $L_1,L_2$ and positive successive evidence masses, the source proves\n\\[\n M_b(L_1)M_{C_{L_1}(b)}(L_2)=M_b(L_1L_2),\\qquad\n C_{L_2}(C_{L_1}(b))=C_{L_1L_2}(b).\n\\]\nThese identities follow by expanding the finite sums and canceling the positive normalizers. Both conditioning orders agree when their respective intermediate masses are positive. Applying this algebra to physical observations requires the appropriate joint likelihood model. In particular, multiplying likelihoods is justified by a stated factorization of the evidence law, rather than merely b","For any nonnegative likelihood weight $L$, write\n\\[\n M_b(L)=\\sum_\\omega b(\\omega)L(\\omega),\\qquad\n C_L(b)(\\omega)=\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n\\]\nwhen $M_b(L)>0$. Given nonnegative $L_1,L_2$ and positive successive evidence masses, the source proves\n\\[\n M_b(L_1)M_{C_{L_1}(b)}(L_2)=M_b(L_1L_2),\\qquad\n C_{L_2}(C_{L_1}(b))=C_{L_1L_2}(b).\n\\]\nThese identities follow by expanding the finite sums and canceling the positive normalizers. Both conditioning orders agree when their respective intermediate masses are positive. Applying this algebra to physical observations requires the appropriate joint likelihood model. In particular, multiplying likelihoods is justified by a stated factorization of the evidence law, rather than merely by the existence of two reports.","\\begin{theorem}[Conditioning commutes with exact compression]\nLet $\\phi:\\Omega\\to Z$ map between finite spaces, and define the pushforward belief $\\bar b(z)=\\sum_{\\phi(\\omega)=z}b(\\omega)$. Suppose $L=\\bar L\\circ\\phi$ for a nonnegative likelihood $\\bar L:Z\\to\\Q$, and the evidence mass is positive. Then\n\\[\n \\phi_*C_L(b)=C_{\\bar L}(\\phi_*b).\n\\]\nSuch an exact $\\bar L$ exists if and only if $L$ is constant on every fiber of $\\phi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criter","\\begin{theorem}[Conditioning commutes with exact compression]\nLet $\\phi:\\Omega\\to Z$ map between finite spaces, and define the pushforward belief $\\bar b(z)=\\sum_{\\phi(\\omega)=z}b(\\omega)$. Suppose $L=\\bar L\\circ\\phi$ for a nonnegative likelihood $\\bar L:Z\\to\\Q$, and the evidence mass is positive. Then\n\\[\n \\phi_*C_L(b)=C_{\\bar L}(\\phi_*b).\n\\]\nSuch an exact $\\bar L$ exists if and only if $L$ is constant on every fiber of $\\phi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criterion} applied to the rational-valued target $L$.\n\\end{proof}","ing commutes with exact compression]\nLet $\\phi:\\Omega\\to Z$ map between finite spaces, and define the pushforward belief $\\bar b(z)=\\sum_{\\phi(\\omega)=z}b(\\omega)$. Suppose $L=\\bar L\\circ\\phi$ for a nonnegative likelihood $\\bar L:Z\\to\\Q$, and the evidence mass is positive. Then\n\\[\n \\phi_*C_L(b)=C_{\\bar L}(\\phi_*b).\n\\]\nSuch an exact $\\bar L$ exists if and only if $L$ is constant on every fiber of $\\phi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criterion} applied to the rational-valued target $L$.\n\\end{proof}","te spaces, and define the pushforward belief $\\bar b(z)=\\sum_{\\phi(\\omega)=z}b(\\omega)$. Suppose $L=\\bar L\\circ\\phi$ for a nonnegative likelihood $\\bar L:Z\\to\\Q$, and the evidence mass is positive. Then\n\\[\n \\phi_*C_L(b)=C_{\\bar L}(\\phi_*b).\n\\]\nSuch an exact $\\bar L$ exists if and only if $L$ is constant on every fiber of $\\phi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criterion} applied to the rational-valued target $L$.\n\\end{proof}","f $\\bar b(z)=\\sum_{\\phi(\\omega)=z}b(\\omega)$. Suppose $L=\\bar L\\circ\\phi$ for a nonnegative likelihood $\\bar L:Z\\to\\Q$, and the evidence mass is positive. Then\n\\[\n \\phi_*C_L(b)=C_{\\bar L}(\\phi_*b).\n\\]\nSuch an exact $\\bar L$ exists if and only if $L$ is constant on every fiber of $\\phi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criterion} applied to the rational-valued target $L$.\n\\end{proof}","hi$ for a nonnegative likelihood $\\bar L:Z\\to\\Q$, and the evidence mass is positive. Then\n\\[\n \\phi_*C_L(b)=C_{\\bar L}(\\phi_*b).\n\\]\nSuch an exact $\\bar L$ exists if and only if $L$ is constant on every fiber of $\\phi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criterion} applied to the rational-valued target $L$.\n\\end{proof}","hi$.\n\\end{theorem}\n\\begin{proof}\nGrouping the normalizer by fibers gives $M_b(L)=M_{\\bar b}(\\bar L)$. On each fiber $\\phi^{-1}(z)$, the likelihood can be factored outside the sum. Thus\n\\[\n \\sum_{\\phi(\\omega)=z}\\frac{b(\\omega)L(\\omega)}{M_b(L)}\n =\\frac{\\bar b(z)\\bar L(z)}{M_{\\bar b}(\\bar L)}.\n\\]\nThe factorization criterion is Theorem~\\ref{thm:fiber-criterion} applied to the rational-valued target $L$.\n\\end{proof}","\\begin{example}[Rare evidence can reverse a posterior completely]\nUnder the uniform prior on $\\{0,1\\}$, take $L_0=(\\varepsilon,0)$ and $L_1=(0,\\varepsilon)$ with rational $\\varepsilon>0$. The two likelihoods differ pointwise by at most $\\varepsilon$, and both evidence masses equal $\\varepsilon/2$. Their posteriors are respectively $(1,0)$ and $(0,1)$, with full $L^1$ distance $2$. Arbitrarily small positive likelihood perturbations therefore do not imply uniform pointwise posterior stability without an additional evidence condition. The later model-error chapter uses weighted continuation comparisons rather than identifying nearby posterior points as the same controller state.\n\\end{example}","\\begin{definition}[Deterministic control sufficiency]\nLet $A_0$ be a finite nonempty set of admissible actions, let $f:S\\times A_0\\to S$ be transition, and let $r:S\\times A_0\\to\\Q$ be reward. A representation $\\phi:S\\to Z$ is control sufficient if, whenever $\\phi(s)=\\phi(t)$, every admissible action satisfies\n\\[\n r(s,a)=r(t,a),\\qquad \\phi(f(s,a))=\\phi(f(t,a)).\n\\]\nThe global action set is common to all states in this definition.\n\\end{definition}","Define the finite-horizon values by\n\\[\n V_0(s)=0,\\qquad Q_n(s,a)=r(s,a)+V_n(f(s,a)),\\qquad\n V_{n+1}(s)=\\max_{a\\in A_0}Q_n(s,a).\n\\]","\\begin{theorem}[Exact preservation of controlled value]\nIf $\\phi$ is control sufficient, then $V_n$ is constant on every fiber of $\\phi$ for every $n\\in\\N$. The same is true of $Q_n(\\cdot,a)$ for each admissible $a$, and Bellman-optimal action sets coincide within each fiber.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the values vanish. If $V_n$ is fiber constant and $\\phi(s)=\\phi(t)$, control sufficiency gives equal rewards and equal represented successors. Hence $Q_n(s,a)=Q_n(t,a)$ for every admissible $a$. Taking the maximum over the common action set proves equality of $V_{n+1}$. The action-value equality also makes all inequalities defining optimal action membership identical at $s$ and $t$.\n\\end{proof}","ntrol sufficient, then $V_n$ is constant on every fiber of $\\phi$ for every $n\\in\\N$. The same is true of $Q_n(\\cdot,a)$ for each admissible $a$, and Bellman-optimal action sets coincide within each fiber.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the values vanish. If $V_n$ is fiber constant and $\\phi(s)=\\phi(t)$, control sufficiency gives equal rewards and equal represented successors. Hence $Q_n(s,a)=Q_n(t,a)$ for every admissible $a$. Taking the maximum over the common action set proves equality of $V_{n+1}$. The action-value equality also makes all inequalities defining optimal action membership identical at $s$ and $t$.\n\\end{proof}","on every fiber of $\\phi$ for every $n\\in\\N$. The same is true of $Q_n(\\cdot,a)$ for each admissible $a$, and Bellman-optimal action sets coincide within each fiber.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the values vanish. If $V_n$ is fiber constant and $\\phi(s)=\\phi(t)$, control sufficiency gives equal rewards and equal represented successors. Hence $Q_n(s,a)=Q_n(t,a)$ for every admissible $a$. Taking the maximum over the common action set proves equality of $V_{n+1}$. The action-value equality also makes all inequalities defining optimal action membership identical at $s$ and $t$.\n\\end{proof}","\\cdot,a)$ for each admissible $a$, and Bellman-optimal action sets coincide within each fiber.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the values vanish. If $V_n$ is fiber constant and $\\phi(s)=\\phi(t)$, control sufficiency gives equal rewards and equal represented successors. Hence $Q_n(s,a)=Q_n(t,a)$ for every admissible $a$. Taking the maximum over the common action set proves equality of $V_{n+1}$. The action-value equality also makes all inequalities defining optimal action membership identical at $s$ and $t$.\n\\end{proof}","in{proof}\nAt horizon zero the values vanish. If $V_n$ is fiber constant and $\\phi(s)=\\phi(t)$, control sufficiency gives equal rewards and equal represented successors. Hence $Q_n(s,a)=Q_n(t,a)$ for every admissible $a$. Taking the maximum over the common action set proves equality of $V_{n+1}$. The action-value equality also makes all inequalities defining optimal action membership identical at $s$ and $t$.\n\\end{proof}","}\nAt horizon zero the values vanish. If $V_n$ is fiber constant and $\\phi(s)=\\phi(t)$, control sufficiency gives equal rewards and equal represented successors. Hence $Q_n(s,a)=Q_n(t,a)$ for every admissible $a$. Taking the maximum over the common action set proves equality of $V_{n+1}$. The action-value equality also makes all inequalities defining optimal action membership identical at $s$ and $t$.\n\\end{proof}","\\begin{example}[A finer representation can fail dynamic closure]\nConsider $S=\\{0,1,2\\}$, one action, zero reward, and transitions $f(0)=0$, $f(1)=2$, $f(2)=2$. The constant representation is control sufficient. The finer representation with fibers $\\{0,1\\}$ and $\\{2\\}$ refines that constant representation, but it merges states $0$ and $1$ whose successors fall into different fine fibers. It is therefore not control sufficient. This example is retained as an explicit formal counterexample to arbitrary refinement monotonicity.\n\\end{example}","Let $S$ be finite and replace deterministic transition by rational probability rows $P_a(s,s')\\ge0$ with $\\sum_{s'}P_a(s,s')=1$. The stochastic Bellman recursion is\n\\[\n V_0(s)=0,\\qquad Q_n(s,a)=r(s,a)+\\sum_{s'}P_a(s,s')V_n(s'),\\qquad\n V_{n+1}(s)=\\max_{a\\in A_0}Q_n(s,a).\n\\]\nEquality of entire raw transition rows is sufficient for compression, but can demand more than the abstract process needs. The appropriate condition compares their projected laws.","\\begin{definition}[Projected stochastic control sufficiency]\nA representation $\\phi:S\\to Z$ is projected stochastically sufficient if equal represented current states have equal immediate rewards and, for every admissible $a$ and every rational fiber-constant function $W:S\\to\\Q$,\n\\[\n \\sum_{s'}P_a(s,s')W(s')=\\sum_{s'}P_a(t,s')W(s')\n \\quad\\text{whenever }\\phi(s)=\\phi(t).\n\\]\n\\end{definition}","When $Z$ is finite, this expectation condition is equivalent to equality of the mass sent to every abstract fiber:\n\\[\n \\sum_{\\phi(s')=z'}P_a(s,s')=\\sum_{\\phi(s')=z'}P_a(t,s')\\qquad(z'\\in Z).\n\\]\nTo prove necessity, use the indicator of the fiber as $W$. To prove sufficiency, group the expectation by fibers and factor out the common value of $W$ on each nonempty fiber. Empty fibers contribute zero. These arguments explain why raw states may have distinct next-state probabilities while having identical controlled abstract behavior.","\\begin{theorem}[Exact stochastic quotient]\nSuppose $\\phi:S\\to Z$ is surjective and projected stochastically sufficient, with $S,Z$ finite. Choose a representative $s_z$ of every fiber, and define\n\\[\n \\bar r(z,a)=r(s_z,a),\\qquad\n \\bar P_a(z,z')=\\sum_{\\phi(s')=z'}P_a(s_z,s').\n\\]\nThese quantities give a normalized rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume t","\\begin{theorem}[Exact stochastic quotient]\nSuppose $\\phi:S\\to Z$ is surjective and projected stochastically sufficient, with $S,Z$ finite. Choose a representative $s_z$ of every fiber, and define\n\\[\n \\bar r(z,a)=r(s_z,a),\\qquad\n \\bar P_a(z,z')=\\sum_{\\phi(s')=z'}P_a(s_z,s').\n\\]\nThese quantities give a normalized rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(","\\begin{theorem}[Exact stochastic quotient]\nSuppose $\\phi:S\\to Z$ is surjective and projected stochastically sufficient, with $S,Z$ finite. Choose a representative $s_z$ of every fiber, and define\n\\[\n \\bar r(z,a)=r(s_z,a),\\qquad\n \\bar P_a(z,z')=\\sum_{\\phi(s')=z'}P_a(s_z,s').\n\\]\nThese quantities give a normalized rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s'","\\begin{theorem}[Exact stochastic quotient]\nSuppose $\\phi:S\\to Z$ is surjective and projected stochastically sufficient, with $S,Z$ finite. Choose a representative $s_z$ of every fiber, and define\n\\[\n \\bar r(z,a)=r(s_z,a),\\qquad\n \\bar P_a(z,z')=\\sum_{\\phi(s')=z'}P_a(s_z,s').\n\\]\nThese quantities give a normalized rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s')V_n(s'),\n\\]\nby projected equality and fiberwise grouping. Immediate rewards agree, so every action backup a","Suppose $\\phi:S\\to Z$ is surjective and projected stochastically sufficient, with $S,Z$ finite. Choose a representative $s_z$ of every fiber, and define\n\\[\n \\bar r(z,a)=r(s_z,a),\\qquad\n \\bar P_a(z,z')=\\sum_{\\phi(s')=z'}P_a(s_z,s').\n\\]\nThese quantities give a normalized rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s')V_n(s'),\n\\]\nby projected equality and fiberwise grouping. Immediate rewards agree, so every action backup agrees. Taking the maximum over the common action set proves the next horizon. The zero-horizon case is immediate.\n\\end{proof}","r, and define\n\\[\n \\bar r(z,a)=r(s_z,a),\\qquad\n \\bar P_a(z,z')=\\sum_{\\phi(s')=z'}P_a(s_z,s').\n\\]\nThese quantities give a normalized rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s')V_n(s'),\n\\]\nby projected equality and fiberwise grouping. Immediate rewards agree, so every action backup agrees. Taking the maximum over the common action set proves the next horizon. The zero-horizon case is immediate.\n\\end{proof}","d rational stochastic control system, independent of the representative choices for admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s')V_n(s'),\n\\]\nby projected equality and fiberwise grouping. Immediate rewards agree, so every action backup agrees. Taking the maximum over the common action set proves the next horizon. The zero-horizon case is immediate.\n\\end{proof}","r admissible actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s')V_n(s'),\n\\]\nby projected equality and fiberwise grouping. Immediate rewards agree, so every action backup agrees. Taking the maximum over the common action set proves the next horizon. The zero-horizon case is immediate.\n\\end{proof}","e actions. For every finite horizon,\n\\[\n \\bar V_n(\\phi(s))=V_n(s).\n\\]\n\\end{theorem}\n\\begin{proof}\nNonnegative fiber masses give nonnegative quotient probabilities; partitioning $S$ by the fibers shows that each quotient row sums to one. Sufficiency makes reward and projected mass unchanged when replacing $s_z$ by another representative. Inductively assume the value equality at horizon $n$. Then\n\\[\n \\sum_{z'}\\bar P_a(\\phi(s),z')\\bar V_n(z')\n =\\sum_{s'}P_a(s,s')V_n(s'),\n\\]\nby projected equality and fiberwise grouping. Immediate rewards agree, so every action backup agrees. Taking the maximum over the common action set proves the next horizon. The zero-horizon case is immediate.\n\\end{proof}","The concrete binding layer uses a small data-only propositional language. Formulas contain numbered atoms, truth, falsity, conjunction, disjunction, and implication. Derivations contain the corresponding introduction and elimination rules. Atoms are interpreted as arbitrary propositions under a valuation $v:\\N\\to\\operatorname{Prop}$. A recursive inference function either returns a formula or rejects. It accepts data representing a derivation, rather than executable tactics or imported source code.","\\begin{theorem}[Checker soundness]\nIf the inference checker accepts derivation $p$ for target $F$ under context $\\Gamma$, then every valuation satisfying $\\Gamma$ satisfies $F$.\n\\end{theorem}\n\\begin{proof}\nInduct on the derivation. A used assumption belongs to $\\Gamma$ and is true by the context hypothesis. Conjunction introduction combines the two inductive conclusions, and its projections select the corresponding conclusion. Implication introduction applies the induction hypothesis in the extended context; implication elimination applies a proved implication to a proved premise. The two disjunction injections construct the appropriate alternative. Case elimination applies the matching branch proof after extending the context with the observed alternative. Falsity elimi","\\begin{theorem}[Checker soundness]\nIf the inference checker accepts derivation $p$ for target $F$ under context $\\Gamma$, then every valuation satisfying $\\Gamma$ satisfies $F$.\n\\end{theorem}\n\\begin{proof}\nInduct on the derivation. A used assumption belongs to $\\Gamma$ and is true by the context hypothesis. Conjunction introduction combines the two inductive conclusions, and its projections select the corresponding conclusion. Implication introduction applies the induction hypothesis in the extended context; implication elimination applies a proved implication to a proved premise. The two disjunction injections construct the appropriate alternative. Case elimination applies the matching branch proof after extending the context with the observed alternative. Falsity elimination follows","\\begin{theorem}[Checker soundness]\nIf the inference checker accepts derivation $p$ for target $F$ under context $\\Gamma$, then every valuation satisfying $\\Gamma$ satisfies $F$.\n\\end{theorem}\n\\begin{proof}\nInduct on the derivation. A used assumption belongs to $\\Gamma$ and is true by the context hypothesis. Conjunction introduction combines the two inductive conclusions, and its projections select the corresponding conclusion. Implication introduction applies the induction hypothesis in the extended context; implication elimination applies a proved implication to a proved premise. The two disjunction injections construct the appropriate alternative. Case elimination applies the matching branch proof after extending the context with the observed alternative. Falsity elimination follows from the impossible pro","\\begin{theorem}[Checker soundness]\nIf the inference checker accepts derivation $p$ for target $F$ under context $\\Gamma$, then every valuation satisfying $\\Gamma$ satisfies $F$.\n\\end{theorem}\n\\begin{proof}\nInduct on the derivation. A used assumption belongs to $\\Gamma$ and is true by the context hypothesis. Conjunction introduction combines the two inductive conclusions, and its projections select the corresponding conclusion. Implication introduction applies the induction hypothesis in the extended context; implication elimination applies a proved implication to a proved premise. The two disjunction injections construct the appropriate alternative. Case elimination applies the matching branch proof after extending the context with the observed alternative. Falsity elimination follows from the impossible proposition. The checker's shape and equality","\\begin{theorem}[Checker soundness]\nIf the inference checker accepts derivation $p$ for target $F$ under context $\\Gamma$, then every valuation satisfying $\\Gamma$ satisfies $F$.\n\\end{theorem}\n\\begin{proof}\nInduct on the derivation. A used assumption belongs to $\\Gamma$ and is true by the context hypothesis. Conjunction introduction combines the two inductive conclusions, and its projections select the corresponding conclusion. Implication introduction applies the induction hypothesis in the extended context; implication elimination applies a proved implication to a proved premise. The two disjunction injections construct the appropriate alternative. Case elimination applies the matching branch proof after extending the context with the observed alternative. Falsity elimination follows from the impossible proposition. The checker's shape and equality tests ensure","\\begin{theorem}[Checker soundness]\nIf the inference checker accepts derivation $p$ for target $F$ under context $\\Gamma$, then every valuation satisfying $\\Gamma$ satisfies $F$.\n\\end{theorem}\n\\begin{proof}\nInduct on the derivation. A used assumption belongs to $\\Gamma$ and is true by the context hypothesis. Conjunction introduction combines the two inductive conclusions, and its projections select the corresponding conclusion. Implication introduction applies the induction hypothesis in the extended context; implication elimination applies a proved implication to a proved premise. The two disjunction injections construct the appropriate alternative. Case elimination applies the matching branch proof after extending the context with the observed alternative. Falsity elimination follows from the impossible proposition. The checker's shape and equality tests ensure that the formulas in these rules match exactly.\n\\end{proof}","Programs are finite abstract syntax trees built from Boolean literals, three input variables, negation, conjunction, disjunction, and exclusive-or. The input domain is the eight triples in $\\{0,1\\}^3$. Let $N(p)$ be the tree-node count: each literal or variable contributes one, negation adds one to its child, and a binary constructor adds one to the sum of its child counts.","The instrumented evaluator evaluates both children of a binary operation. A structural induction proves that its output equals the ordinary evaluator and that its visit count is exactly $N(p)$ for every input. Thus node count is a valid eager-traversal meter for this interpreter. It is not a short-circuit meter, hardware runtime, memory-allocation count, or Boolean circuit size with shared subexpressions.","\\begin{theorem}[Complete Boolean equivalence check]\nLet $\\sigma(p)$ be the eight output bits of $p$, in the fixed lexicographic input order. Then\n\\[\n \\sigma(p)=\\sigma(q)\\quad\\Longleftrightarrow\\quad\n \\forall x\\in\\{0,1\\}^3,\\ p(x)=q(x).\n\\]\nThe executable checker comparing all eight outputs is sound and complete.\n\\end{theorem}\n\\begin{proof}\nEquality of signatures gives equality at every entry. Every possible input occurs in the enumerated list, so every input is covered. Conversely, pointwise equality makes the two mapped output lists equal. This is exhaustive finite checking, not a sampled approximation.\n\\end{proof}","The initial grammar layer has five atoms: two literals and three variables. If a layer is $L$, expansion retains $L$, adds all negations, and adds all ordered pairs under each of the three binary operations. Let $M_d$ count emitted list entries at depth $d$. The source proves\n\\[\n M_0=5,\\qquad M_{d+1}=2M_d+3M_d^2.\n\\]\nThe entries can contain duplicates. The node bound is $B_0=1$, $B_{d+1}=2B_d+1$, and every expression of height at most $d$ occurs in the depth-$d$ list. In particular $M_1=85$ and $M_2=21845$.","s: two literals and three variables. If a layer is $L$, expansion retains $L$, adds all negations, and adds all ordered pairs under each of the three binary operations. Let $M_d$ count emitted list entries at depth $d$. The source proves\n\\[\n M_0=5,\\qquad M_{d+1}=2M_d+3M_d^2.\n\\]\nThe entries can contain duplicates. The node bound is $B_0=1$, $B_{d+1}=2B_d+1$, and every expression of height at most $d$ occurs in the depth-$d$ list. In particular $M_1=85$ and $M_2=21845$.","e variables. If a layer is $L$, expansion retains $L$, adds all negations, and adds all ordered pairs under each of the three binary operations. Let $M_d$ count emitted list entries at depth $d$. The source proves\n\\[\n M_0=5,\\qquad M_{d+1}=2M_d+3M_d^2.\n\\]\nThe entries can contain duplicates. The node bound is $B_0=1$, $B_{d+1}=2B_d+1$, and every expression of height at most $d$ occurs in the depth-$d$ list. In particular $M_1=85$ and $M_2=21845$.","is $L$, expansion retains $L$, adds all negations, and adds all ordered pairs under each of the three binary operations. Let $M_d$ count emitted list entries at depth $d$. The source proves\n\\[\n M_0=5,\\qquad M_{d+1}=2M_d+3M_d^2.\n\\]\nThe entries can contain duplicates. The node bound is $B_0=1$, $B_{d+1}=2B_d+1$, and every expression of height at most $d$ occurs in the depth-$d$ list. In particular $M_1=85$ and $M_2=21845$.","sion retains $L$, adds all negations, and adds all ordered pairs under each of the three binary operations. Let $M_d$ count emitted list entries at depth $d$. The source proves\n\\[\n M_0=5,\\qquad M_{d+1}=2M_d+3M_d^2.\n\\]\nThe entries can contain duplicates. The node bound is $B_0=1$, $B_{d+1}=2B_d+1$, and every expression of height at most $d$ occurs in the depth-$d$ list. In particular $M_1=85$ and $M_2=21845$.","Generation is reserved before a layer is constructed. The declared generation charge is the sum of the layer-entry counts through the chosen depth. A candidate comparison is conservatively charged\n\\[\n T(r,d)=8\\bigl(N(r)+B_d+1\\bigr).\n\\]\nThe examined prefix is limited by both a user trial cap and the remaining budget divided by $T(r,d)$. Its generation plus examination charge is proved at most the supplied budget. If the generation charge does not fit, search returns the original reference and incurs zero modeled search charge. If the entire layer fits both limits, the result is optimal among equivalent expressions of height at most $d$. A partial prefix provides optimality only over that prefix.","examined prefix is limited by both a user trial cap and the remaining budget divided by $T(r,d)$. Its generation plus examination charge is proved at most the supplied budget. If the generation charge does not fit, search returns the original reference and incurs zero modeled search charge. If the entire layer fits both limits, the result is optimal among equivalent expressions of height at most $d$. A partial prefix provides optimality only over that prefix.","\\begin{example}[A smaller artifact need not repay its discovery]\nThe reference $\\neg(\\neg x_1\\vee\\neg x_2)$ has six nodes. The equivalent conjunction $x_1\\wedge x_2$ has three. The archived depth-one search examines the complete 85-entry layer under its stated limits and returns the conjunction, with modeled search charge $6890$. Each eager evaluation subsequently saves three visits. A large deployment can repay the exploration cost; a single evaluation cannot. The synthesis theorem supplies semantic safety and a budget bound even in the unprofitable case, and the promotion layer separately checks the campaign economics.\n\\end{example}","Semantic entries carry an expression, its complete eight-bit signature, its node count, and equations certifying the two cached fields. The index retains at most one representative of each signature and keeps a smallest representative among inserted expressions. There are at most $2^8=256$ entries, regardless of the size of the submitted list.","\\begin{proposition}[Recursive semantic coverage]\nFor every depth $d$, recursively expanding the retained representatives and compacting again preserves the smallest equivalent node count obtainable from the full uncompressed grammar layer at that depth, with the same reference fallback.\n\\end{proposition}\n\\begin{proof}\nCompaction gives every input expression an equivalent retained representative of no greater node count. Equivalence is a congruence for each grammar constructor. Replacing either child by a no-larger equivalent representative therefore preserves the result and does not increase its node count. Negation has the same property. Induction on depth supplies semantic and cost coverage of the full grammar by recursive representatives. Conversely, eve","A complete finite semantic algebra supports a stronger conclusion than searching a deeper finite layer. Let $D$ be a semantic domain with interpretations of the Boolean constructors. A cost function $c:D\\to\\N$ is a lower certificate if it satisfies the constructor inequalities\n\\begin{align*}\n c(\\text{literal }b)&\\le1,& c(\\text{variable }j)&\\le1,","A complete finite semantic algebra supports a stronger conclusion than searching a deeper finite layer. Let $D$ be a semantic domain with interpretations of the Boolean constructors. A cost function $c:D\\to\\N$ is a lower certificate if it satisfies the constructor inequalities\n\\begin{align*}\n c(\\text{literal }b)&\\le1,& c(\\text{variable }j)&\\le1,\\\\\n c(\\neg s)&\\le c(s)+1,&\n c(s\\mathbin{\\circ}t)&\\le c(s)+c(t)+1\n \\quad(\\circ\\in\\{\\wedge,\\vee,\\oplus\\}).\n\\end{align*}","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bou","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bound to the other's","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bound to the other's witness, obtaining bot","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bound to the other's witness, obtaining both inequalities between their co","\\begin{theorem}[Attained global semantic minimum]\n\\label{thm:semantic-minimum}\nSuppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bound to the other's witness, obtaining both inequalities between their costs.\n\\end{proof}","uppose $c$ satisfies those inequalities and, for every semantic state $s$, a witness expression $w_s$ has semantics $s$ and node count $c(s)$. Then for every expression $p$ with semantics $s$,\n\\[\n N(w_s)=c(s)\\le N(p).\n\\]\nNo expression-height bound appears in this conclusion. Two such attained cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bound to the other's witness, obtaining both inequalities between their costs.\n\\end{proof}","d cost certificates have identical cost functions.\n\\end{theorem}\n\\begin{proof}\nStructural induction proves $c(\\operatorname{sem}(p))\\le N(p)$. The leaf inequalities give the base cases. For a unary constructor, use its inequality and the child's induction hypothesis. For a binary constructor, use its inequality and both child hypotheses. Applying this lower bound to any expression with semantics $s$ and using the attained witness proves global minimality. For two certificates, apply each one's lower bound to the other's witness, obtaining both inequalities between their costs.\n\\end{proof}","For the three-input Boolean backend, $D$ consists of all 256 truth signatures. A proposed bank must have exactly 256 rows. Each row's stored expression is checked for its exact signature and claimed node count, and the bank is checked against every unary and binary constructor inequality. An accepted bank therefore carries the hypotheses of Theorem~\\ref{thm:semantic-minimum}. Its query returns a globally node-minimal expression within this entire Boolean tree grammar, including all finite heights.","\\begin{definition}[Dimension-indexed affine signature]\nOver a ring $R$, a signature from dimension $n$ to dimension $m$ is a pair $(A,b)$ with $A\\in R^{m\\times n}$ and $b\\in R^m$, denoting $x\\mapsto Ax+b$. Dimensions are part of the type. Composition and pointwise addition are\n\\[\n (B,c)\\circ(A,b)=(BA,Bb+c),\\qquad\n (A,b)+(A',b')=(A+A',b+b').\n\\]\n\\end{definition}","\\begin{theorem}[Complete zero-and-basis probes]\nTwo affine signatures $f,g:R^n\\to R^m$ are equal if and only if they agree at $0$ and at every standard basis vector $e_j$. Thus $n+1$ vector probes establish equality on all inputs. A mismatch in signatures has a mismatch among these probes.\n\\end{theorem}\n\\begin{proof}\nThe zero probe identifies the offsets because $f(0)=b_f$. At $e_j$, the output is the $j$th matrix column plus the offset. After canceling the now-equal offsets, equality of the basis probes identifies every column. Equal coefficients give equal evaluations for all inputs. The mismatch assertion is the contrapositive.\n\\end{proof}","{theorem}[Complete zero-and-basis probes]\nTwo affine signatures $f,g:R^n\\to R^m$ are equal if and only if they agree at $0$ and at every standard basis vector $e_j$. Thus $n+1$ vector probes establish equality on all inputs. A mismatch in signatures has a mismatch among these probes.\n\\end{theorem}\n\\begin{proof}\nThe zero probe identifies the offsets because $f(0)=b_f$. At $e_j$, the output is the $j$th matrix column plus the offset. After canceling the now-equal offsets, equality of the basis probes identifies every column. Equal coefficients give equal evaluations for all inputs. The mismatch assertion is the contrapositive.\n\\end{proof}","\\begin{example}[The zero probe is necessary]\nOver $\\Q$, consider scalar maps $f(x)=x$ and $g(x)=2x-1$. They agree at the single basis vector $x=1$, but $f(0)=0$ and $g(0)=-1$. Omitting the zero probe would accept a false affine equivalence. Over a field, equality of the linear matrices alone also leaves arbitrary affine offsets unconstrained.\n\\end{example}","Typed programs combine affine atoms, identities, sequential composition, and pointwise addition. Composition connects only matching intermediate dimensions. An independent recursive execution semantics and coefficient-normalization semantics are defined. Structural induction proves\n\\[\n \\operatorname{eval}(\\operatorname{normalize}(p),x)=\\operatorname{run}(p,x)\n \\qquad\\text{for all }x.\n\\]\nHence two programs are equivalent if and only if their normal signatures coincide. Decidable coefficient comparison is a sound and complete checker when equality in $R$ is decidable, as it is for the rational and finite-field executable examples.","An independent recursive execution semantics and coefficient-normalization semantics are defined. Structural induction proves\n\\[\n \\operatorname{eval}(\\operatorname{normalize}(p),x)=\\operatorname{run}(p,x)\n \\qquad\\text{for all }x.\n\\]\nHence two programs are equivalent if and only if their normal signatures coincide. Decidable coefficient comparison is a sound and complete checker when equality in $R$ is decidable, as it is for the rational and finite-field executable examples.","One-hole typed contexts permit replacement at arbitrary nesting depth. Equivalence is preserved under composition and pointwise addition, so plugging equivalent programs into such a context preserves its semantics. A declared additive nonnegative primitive tariff, with $m$ charged for adding two $m$-dimensional outputs, also makes contextual cost monotone: a no-more-expensive replacement cannot increase the whole context's declared cost. This is a compositional guarantee for the explicit syntax and tariff.","An exact normal form can improve verification without improving evaluation. The affine backend therefore checks semantics and cost separately. Its dense arithmetic tariff is\n\\[\n D(n,m)=m(2n+1),\n\\]\ncharging $n$ multiplications, $n$ accumulations, and one bias addition per output row. The coefficient-composition tariff is\n\\[\n C(n,m,k)=2knm+D(m,k).\n\\]\nThese are declared scalar-operation tariffs. They exclude bit complexity, memory traffic, allocation, and physical timing.","Replacing a composed program by one dense atom with its normalized coefficients preserves every output. Its per-call execution tariff becomes $D(n,m)$. The source's no-slower selector retains the original program whenever that dense tariff is greater than its current execution cost. Normalization work is charged separately for campaign comparisons.","\\begin{example}[Beneficial and harmful fusion]\nA dense $2\\to3\\to2$ chain has per-call tariff $D(2,3)+D(3,2)=29$. Its fused $2\\to2$ map costs $10$ per call, and coefficient composition costs $38$. Thus $38+10q\\le29q$ exactly when $q\\ge2$, with strict savings exactly when $q\\ge3$.","In contrast, a $4\\to1\\to4$ bottleneck costs $D(4,1)+D(1,4)=21$ per call, while its dense fused $4\\to4$ form costs $36$. Fusion remains semantically correct but loses its cheap intermediate bottleneck. The cost-aware selector therefore keeps the original chain in that comparison.\n\\end{example}","Work over a field $R$. For reference $f=(A,b):R^n\\to R^m$, a candidate supplies an intermediate width $k$, first and second affine signatures\n\\[\n a(x)=Ux+u_0,\\qquad d(y)=Vy+v_0,\n\\]\nand matrices $L\\in R^{k\\times m}$, $H\\in R^{n\\times k}$. The checker verifies\n\\[\n VU=A,\\qquad Vu_0+v_0=b,\\qquad LAH=I_k.\n\\]\nThe first two identities give the exact upper implementation. The third is an independently checkable lower witness. Validation uses polynomial matrix identities; an untrusted rank routine is not part of the logical justification.","\\begin{theorem}[Certified minimum affine width]\n\\label{thm:affine-width}\nAn accepted candidate has $k=\\operatorname{rank}(A)$. Every exact two-stage affine factorization of $f$ through width $\\ell$ satisfies $k\\le\\ell$. Under the dense two-stage tariff,\n\\[\n F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff","\\begin{theorem}[Certified minimum affine width]\n\\label{thm:affine-width}\nAn accepted candidate has $k=\\operatorname{rank}(A)$. Every exact two-stage affine factorization of $f$ through width $\\ell$ satisfies $k\\le\\ell$. Under the dense two-stage tariff,\n\\[\n F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","inimum affine width]\n\\label{thm:affine-width}\nAn accepted candidate has $k=\\operatorname{rank}(A)$. Every exact two-stage affine factorization of $f$ through width $\\ell$ satisfies $k\\le\\ell$. Under the dense two-stage tariff,\n\\[\n F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","atorname{rank}(A)$. Every exact two-stage affine factorization of $f$ through width $\\ell$ satisfies $k\\le\\ell$. Under the dense two-stage tariff,\n\\[\n F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","$k\\le\\ell$. Under the dense two-stage tariff,\n\\[\n F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","tariff,\n\\[\n F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","F(n,m,k)=D(n,k)+D(k,m),\n\\]\nthe accepted candidate is no more expensive than any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","any such two-stage factorization.\n\\end{theorem}\n\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","\\begin{proof}\nThe upper factorization gives $\\operatorname{rank}(A)=\\operatorname{rank}(VU)\\le k$. The lower witness gives\n\\[\n k=\\operatorname{rank}(I_k)=\\operatorname{rank}(LAH)\\le\\operatorname{rank}(A).\n\\]\nThus rank equals $k$. Any exact factorization through $\\ell$ has $\\operatorname{rank}(A)\\le\\ell$, by the same upper rank inequality. Finally,\n\\[\n F(n,m,k)=k(2n+1)+m(2k+1)\n\\]\nis increasing in $k$, so minimum width gives minimum tariff in this class.\n\\end{proof}","Exact factorization alone does not prove minimum width. A redundant factorization can be correct while failing to supply a valid $LAH=I_k$. Conversely, a lower rank witness alone does not prove that the proposed affine implementation has the correct offset. Both directions are needed.","\\begin{example}[A checked rank-one bottleneck]\nIn the archived finite-field example over $\\mathbf F_5$, $A$ is the $4\\times4$ all-ones matrix and $b=(1,2,3,4)$. The first stage computes $y=x_1+x_2+x_3+x_4$. The second emits $(y+1,y+2,y+3,y+4)$. Taking $L$ to select the first output coordinate and $H$ to select the first input coordinate gives $LAH=(1)$. The width-one factorization is therefore exact and width minimal.","Its tariff is $9+12=21$, compared with dense cost $36$, saving $15$ per call. The declared factor-checking charge is $105$. For this already atomic reference, the campaign inequality is exactly\n\\[\n 105+g\\le15q,\n\\]\nwhere $g$ is the declared generation work and $q$ the deployment count. With $g=0$, seven calls break even. Generation, validation, and deployment are all visible in this comparison.\n\\end{example}","The menu selector compares three equivalent implementations: the current program, its dense fused atom, and the accepted minimum-width two-stage plan. Its selected tariff is exactly\n\\[\n \\min\\{\\operatorname{cost}(p),\\ D(n,m),\\ F(n,m,k)\\}.\n\\]\nThis proves a minimum over the explicit menu, and also comparison with every exact two-dense-stage factorization. It does not prove a globally cheapest affine program in the larger language. Sparse schedules, sharing, special primitives, and hardware-specific algorithms have different admissible implementations or tariffs. An identity program already costs zero in this syntax and remains preferable to a paid dense factorization.","Suppose a stored candidate has been validated for $(A,b)$ and a later target is $(A,b')$ with exactly the same linear matrix. Keep $U,u_0,V,L,H$ and replace only the second-stage offset by\n\\[\n v_0'=b'-Vu_0.\n\\]\nThen $VU=A$ is unchanged, $Vu_0+v_0'=b'$, and $LAH=I_k$ remains true. The same width and two-stage tariff are retained. A second retargeting overwrites the offset rather than accumulating prior offset changes.","Writing $I$ for issuance and $C$ for the specified episode certificate, the resulting conclusion is\n\\[\n \\Prob(I\\cap C^{\\mathsf c})\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\nThis is an unconditional probability over the original outcome space. It allows arbitrary dependence between scheduled checkpoints. It retains failed and unfinished runs in that space, with no plan issued on those runs. It does not assert a nominal bound conditional on issuance or completion, eventual issuance, or a productivity lower bound. The final recorder theorem derives the execution-to-prefix correspondence inside its generative model. The model's primitive source law remains an explicit premise.","\\subsection{A running computational example}\nConsider an affine map over $\\mathbb F_5$ with four inputs and four outputs, all-ones linear coefficient matrix, and offset $(1,2,3,4)$. The direct dense evaluation tariff is $4(2\\cdot4+1)=36$. Its rank-one factorization has a $4\\to1$ stage followed by a $1\\to4$ stage, with tariff $9+12=21$. Thus each declared call saves $15$ arithmetic cost units. For $1{,}000$ calls the gross saving is $15{,}000$.","subsection{A running computational example}\nConsider an affine map over $\\mathbb F_5$ with four inputs and four outputs, all-ones linear coefficient matrix, and offset $(1,2,3,4)$. The direct dense evaluation tariff is $4(2\\cdot4+1)=36$. Its rank-one factorization has a $4\\to1$ stage followed by a $1\\to4$ stage, with tariff $9+12=21$. Thus each declared call saves $15$ arithmetic cost units. For $1{,}000$ calls the gross saving is $15{,}000$.","xample}\nConsider an affine map over $\\mathbb F_5$ with four inputs and four outputs, all-ones linear coefficient matrix, and offset $(1,2,3,4)$. The direct dense evaluation tariff is $4(2\\cdot4+1)=36$. Its rank-one factorization has a $4\\to1$ stage followed by a $1\\to4$ stage, with tariff $9+12=21$. Thus each declared call saves $15$ arithmetic cost units. For $1{,}000$ calls the gross saving is $15{,}000$.","The delivered exact-quote example gives warm inspection setup $133$ and cold generation setup $196$, both with success probability $1$. Their net values are $14{,}867$ and $14{,}804$. These numbers are exact within the declared model. They are not wall-clock measurements, energy savings, or an empirical result about an external generation service. Later examples introduce uncertainty and show how certification can change an action, or lead a conservative planner to stop.","\\begin{definition}[Guarded rational system]\nLet $S$ be a state space, $A(s)$ a finite nonempty set of actions, and $O$ a\nfinite observation space. A guarded system consists of a rational reward\n$r(s,a)$, a transition $T(s,a,o)$, and rational probabilities $P(o\\mid s,a)$\nsatisfying\n\\[\n P(o\\mid s,a)\\geq0,\\qquad \\sum_{o\\in O}P(o\\mid s,a)=1.\n\\]\nIts finite-horizon Bellman values are\n\\begin{align}\n V_0(s)&=0,\\label{eq:bellmanzero}","\\begin{definition}[Guarded rational system]\nLet $S$ be a state space, $A(s)$ a finite nonempty set of actions, and $O$ a\nfinite observation space. A guarded system consists of a rational reward\n$r(s,a)$, a transition $T(s,a,o)$, and rational probabilities $P(o\\mid s,a)$\nsatisfying\n\\[\n P(o\\mid s,a)\\geq0,\\qquad \\sum_{o\\in O}P(o\\mid s,a)=1.\n\\]\nIts finite-horizon Bellman values are\n\\begin{align}\n V_0(s)&=0,\\label{eq:bellmanzero}\\\\\n Q_h(s,a)&=r(s,a)+\\sum_oP(o\\mid s,a)V_h(T(s,a,o)),\\label{eq:qvalue}\\\\\n V_{h+1}(s)&=\\max_{a\\in A(s)}Q_h(s,a).\\label{eq:bellman}\n\\end{align}\n\\end{definition}","Here $h$ in $Q_h$ counts \\emph{continuation} steps. Choosing an acquisition\naction and then deploying a successful artifact requires two total actions,\nso its first-action score is $Q_1$, not $Q_0$. This indexing also applies to\n\\texttt{robustChoice md h s} in the formal source.","\\begin{theorem}[Optimality over complete adaptive plans]\nEvery horizon $H$ has a legal deterministic observation tree attaining\n$V_H(s)$. Every other legal deterministic observation tree has value at most\n$V_H(s)$. If each admissible transition preserves an invariant, the attaining\ntree preserves that invariant on every branch.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, the empty tree has value zero. At horizon $h+1$, choose an\naction attaining the finite maximum in \\eqref{eq:bellman}; after each observation,\nattach an optimal horizon-$h$ tree at the successor state. Induction proves\nequality of the resulting expected value and the Bellman value. For an arbitrary\nlegal tree, induction bounds each child value by the corresponding optimal\ncontinuation. Multiplica","\\begin{theorem}[Optimality over complete adaptive plans]\nEvery horizon $H$ has a legal deterministic observation tree attaining\n$V_H(s)$. Every other legal deterministic observation tree has value at most\n$V_H(s)$. If each admissible transition preserves an invariant, the attaining\ntree preserves that invariant on every branch.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, the empty tree has value zero. At horizon $h+1$, choose an\naction attaining the finite maximum in \\eqref{eq:bellman}; after each observation,\nattach an optimal horizon-$h$ tree at the successor state. Induction proves\nequality of the resulting expected value and the Bellman value. For an arbitrary\nlegal tree, induction bounds each child value by the corresponding optimal\ncontinuation. Multiplication by nonnegative probabilities, summation, and the\nmaximum","\\begin{theorem}[Optimality over complete adaptive plans]\nEvery horizon $H$ has a legal deterministic observation tree attaining\n$V_H(s)$. Every other legal deterministic observation tree has value at most\n$V_H(s)$. If each admissible transition preserves an invariant, the attaining\ntree preserves that invariant on every branch.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, the empty tree has value zero. At horizon $h+1$, choose an\naction attaining the finite maximum in \\eqref{eq:bellman}; after each observation,\nattach an optimal horizon-$h$ tree at the successor state. Induction proves\nequality of the resulting expected value and the Bellman value. For an arbitrary\nlegal tree, induction bounds each child value by the corresponding optimal\ncontinuation. Multiplication by nonnegative probabilities, summation, and the\nmaximum over admissible first actions give the upper bound. Invariant\npreservation fo","timality over complete adaptive plans]\nEvery horizon $H$ has a legal deterministic observation tree attaining\n$V_H(s)$. Every other legal deterministic observation tree has value at most\n$V_H(s)$. If each admissible transition preserves an invariant, the attaining\ntree preserves that invariant on every branch.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, the empty tree has value zero. At horizon $h+1$, choose an\naction attaining the finite maximum in \\eqref{eq:bellman}; after each observation,\nattach an optimal horizon-$h$ tree at the successor state. Induction proves\nequality of the resulting expected value and the Bellman value. For an arbitrary\nlegal tree, induction bounds each child value by the corresponding optimal\ncontinuation. Multiplication by nonnegative probabilities, summation, and the\nmaximum over admissible first actions give the upper bound. Invariant\npreservation follows by the same tree induction. The invariant argument applies\nto every observation branch, including a branch of probability zero.\n\\end{proof}","other legal deterministic observation tree has value at most\n$V_H(s)$. If each admissible transition preserves an invariant, the attaining\ntree preserves that invariant on every branch.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, the empty tree has value zero. At horizon $h+1$, choose an\naction attaining the finite maximum in \\eqref{eq:bellman}; after each observation,\nattach an optimal horizon-$h$ tree at the successor state. Induction proves\nequality of the resulting expected value and the Bellman value. For an arbitrary\nlegal tree, induction bounds each child value by the corresponding optimal\ncontinuation. Multiplication by nonnegative probabilities, summation, and the\nmaximum over admissible first actions give the upper bound. Invariant\npreservation follows by the same tree induction. The invariant argument applies\nto every observation branch, including a branch of probability zero.\n\\end{proof}","\\begin{example}[The artifact-backed rank-one comparison]\nThe exact quote model retained in \\path{AVOC/RobustQuoteExamples.lean} has\nwarm setup $133$, cold setup $196$, success probability $1$ for both, and gross\ndeployment value $15000$ for each over $1000$ calls. With initial credit $196$\nand two total actions available,\n\\[\n Q_1(\\mathrm{inspectWarm})=15000-133=14867,\n \\qquad\n Q_1(\\mathrm{generateCold})=15000-196=14804.\n\\]\nWarm inspection is better by $63$ in this exact model. At one total action,\nacquisition has only its negative immediate reward, and stopping is preferable\nwhen neither artifact is ready. The deployment opportunity creates the change\nbetween these two horizons.\n\\end{example}","The exact quote model retained in \\path{AVOC/RobustQuoteExamples.lean} has\nwarm setup $133$, cold setup $196$, success probability $1$ for both, and gross\ndeployment value $15000$ for each over $1000$ calls. With initial credit $196$\nand two total actions available,\n\\[\n Q_1(\\mathrm{inspectWarm})=15000-133=14867,\n \\qquad\n Q_1(\\mathrm{generateCold})=15000-196=14804.\n\\]\nWarm inspection is better by $63$ in this exact model. At one total action,\nacquisition has only its negative immediate reward, and stopping is preferable\nwhen neither artifact is ready. The deployment opportunity creates the change\nbetween these two horizons.\n\\end{example}","Quotes carry more than a success estimate. A protected identity records the\nartifact kind, exact source signature, workload call count, cost model, provider,\nchecker, environment, and schema. A setup band has natural endpoints. Gross\nand probability bands have rational endpoints, with probability endpoints in\n$[0,1]$ and nonnegative gross lower endpoints. A submitted quote validates only\nwhen it equals the protected approved quote. Changing either identity or any\nband changes the evidence.","For kind $k\\in\\{w,c\\}$, write\n\\[\n c_k\\in[c_k^-,c_k^+],\\qquad\n g_k\\in[g_k^-,g_k^+],\\qquad\n p_k\\in[p_k^-,p_k^+].\n\\]\nThe formal nominal model reserves $c_k^+$, credits $g_k^-$, and uses $p_k^-$.\nA compatible realization supplies exact values inside these bands. In the true\ncomparison system, rewards and report probabilities use those exact values,\nbut the state transitions and action guards remain those of the nominal model.\nIn particular, setup reservations continue to use the upper setup bound.","This common guard matters. If the true setup is $1$ and the upper quoted setup\nis $3$, a budget of $2$ still makes the acquisition inadmissible for both systems\nin the theorem. The comparator does not obtain an extra feasible action by\nknowing the smaller true fee. A theorem against that larger feasible set would\nrequire a different comparison argument.","Let $w_c(k)=c_k^+-c_k^-$, $w_g(k)=g_k^+-g_k^-$, and\n$w_p(k)=p_k^+-p_k^-$. The source uses the uniform quantities\n\\begin{align}\n \\varepsilon&=w_c(w)+w_c(c)+w_g(w)+w_g(c),","Let $w_c(k)=c_k^+-c_k^-$, $w_g(k)=g_k^+-g_k^-$, and\n$w_p(k)=p_k^+-p_k^-$. The source uses the uniform quantities\n\\begin{align}\n \\varepsilon&=w_c(w)+w_c(c)+w_g(w)+w_g(c),\\\\\n \\delta&=2\\bigl(w_p(w)+w_p(c)\\bigr),\\\\\n M&=c_w^++c_c^++g_w^++g_c^+.\n\\end{align}\nThey bound reward error, Boolean-row $L^1$ error, and nominal one-step reward\nmagnitude respectively. For an acquisition of kind $k$, the sharper local\nreward error is $w_c(k)$ and its row error is $2w_p(k)$. For deployment of kind\n$k$, the local reward error is $w_g(k)$ and its row error is zero. Stop has zero\nprimitive errors.","M&=c_w^++c_c^++g_w^++g_c^+.\n\\end{align}\nThey bound reward error, Boolean-row $L^1$ error, and nominal one-step reward\nmagnitude respectively. For an acquisition of kind $k$, the sharper local\nreward error is $w_c(k)$ and its row error is $2w_p(k)$. For deployment of kind\n$k$, the local reward error is $w_g(k)$ and its row error is zero. Stop has zero\nprimitive errors.","\\begin{lemma}[Bernoulli row error and expectation transport]\nFor Bernoulli parameters $p,q$,\n\\[\n \\sum_{o\\in\\{0,1\\}}|P_p(o)-P_q(o)|=2|p-q|.\n\\]\nIf $|W(t)|\\leq B$, then replacing an observation row $P$ by $\\widehat P$\nchanges its expected continuation by at most\n\\[\n \\left|\\sum_o(P(o)-\\widehat P(o))W(T(s,a,o))\\right|\n \\leq B\\sum_o|P(o)-\\widehat P(o)|.\n\\]\n\\end{lemma}\n\\begin{proof}\nThe two Bernoulli row differences are $p-q$ and $q-p$, which have equal\nabsolute values. For the second statement, apply the triangle inequality to the\nfinite sum and use the uniform bound on each continuation value.\n\\end{proof}","oulli row error and expectation transport]\nFor Bernoulli parameters $p,q$,\n\\[\n \\sum_{o\\in\\{0,1\\}}|P_p(o)-P_q(o)|=2|p-q|.\n\\]\nIf $|W(t)|\\leq B$, then replacing an observation row $P$ by $\\widehat P$\nchanges its expected continuation by at most\n\\[\n \\left|\\sum_o(P(o)-\\widehat P(o))W(T(s,a,o))\\right|\n \\leq B\\sum_o|P(o)-\\widehat P(o)|.\n\\]\n\\end{lemma}\n\\begin{proof}\nThe two Bernoulli row differences are $p-q$ and $q-p$, which have equal\nabsolute values. For the second statement, apply the triangle inequality to the\nfinite sum and use the uniform bound on each continuation value.\n\\end{proof}","error and expectation transport]\nFor Bernoulli parameters $p,q$,\n\\[\n \\sum_{o\\in\\{0,1\\}}|P_p(o)-P_q(o)|=2|p-q|.\n\\]\nIf $|W(t)|\\leq B$, then replacing an observation row $P$ by $\\widehat P$\nchanges its expected continuation by at most\n\\[\n \\left|\\sum_o(P(o)-\\widehat P(o))W(T(s,a,o))\\right|\n \\leq B\\sum_o|P(o)-\\widehat P(o)|.\n\\]\n\\end{lemma}\n\\begin{proof}\nThe two Bernoulli row differences are $p-q$ and $q-p$, which have equal\nabsolute values. For the second statement, apply the triangle inequality to the\nfinite sum and use the uniform bound on each continuation value.\n\\end{proof}","Induction on the Bellman recursion gives $|\\widehat V_h(s)|\\leq hM$ for the\nnominal value. The source's accumulated value radius and local Q radius are\n\\begin{align}\n D_h&=\\sum_{j=0}^{h-1}(\\varepsilon+\\delta jM),\\label{eq:value-radius}","Induction on the Bellman recursion gives $|\\widehat V_h(s)|\\leq hM$ for the\nnominal value. The source's accumulated value radius and local Q radius are\n\\begin{align}\n D_h&=\\sum_{j=0}^{h-1}(\\varepsilon+\\delta jM),\\label{eq:value-radius}\\\\\n \\rho_h(a)&=D_h+\\varepsilon_a+\\delta_a hM,\\label{eq:local-radius}\\\\\n \\Gamma_h&=D_h+\\varepsilon+\\delta hM.\\label{eq:global-radius}\n\\end{align}\nThe Bellman residual is zero here because the nominal tables are exact Bellman\nvalues. The broader model-error framework also allows a supplied table with a\nproved nonzero residual, but no residual term is silently introduced in these\nquote results.","\\begin{theorem}[Value and Q enclosures]\nFor every compatible realization, every state, and every continuation horizon,\n\\begin{align}\n |V_h(s)-\\widehat V_h(s)|&\\leq D_h,\\\\\n |Q_h(s,a)-\\widehat Q_h(s,a)|&\\leq\\rho_h(a)\\leq\\Gamma_h.\n\\end{align}\nThus $L_h(s,a)=\\widehat Q_h(s,a)-\\rho_h(a)$ and\n$U_h(s,a)=\\widehat Q_h(s,a)+\\rho_h(a)$ enclose the true Q value.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the value error is zero. In a single backup, insert and subtract\nthe expectation of the nominal continuation under the true row. The reward\nerror contributes at most $\\varepsilon_a$, the continuation-value error at most\n$D_h$, and row transport at most $\\delta_a hM$. Taking a maximum over the same\nfeasible action set is nonexpansive in the uniform score error. Therefore\n$D_{h+1}\\leq D_h+\\varepsilon+\\delta hM$, whose accumulated upper bound is\n\\eqref{eq:value-radius}. The local Q statement follows from the same inserted\nexpect","|Q_h(s,a)-\\widehat Q_h(s,a)|&\\leq\\rho_h(a)\\leq\\Gamma_h.\n\\end{align}\nThus $L_h(s,a)=\\widehat Q_h(s,a)-\\rho_h(a)$ and\n$U_h(s,a)=\\widehat Q_h(s,a)+\\rho_h(a)$ enclose the true Q value.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the value error is zero. In a single backup, insert and subtract\nthe expectation of the nominal continuation under the true row. The reward\nerror contributes at most $\\varepsilon_a$, the continuation-value error at most\n$D_h$, and row transport at most $\\delta_a hM$. Taking a maximum over the same\nfeasible action set is nonexpansive in the uniform score error. Therefore\n$D_{h+1}\\leq D_h+\\varepsilon+\\delta hM$, whose accumulated upper bound is\n\\eqref{eq:value-radius}. The local Q statement follows from the same inserted\nexpectation before taking a maximum.\n\\end{proof}","\\Gamma_h.\n\\end{align}\nThus $L_h(s,a)=\\widehat Q_h(s,a)-\\rho_h(a)$ and\n$U_h(s,a)=\\widehat Q_h(s,a)+\\rho_h(a)$ enclose the true Q value.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the value error is zero. In a single backup, insert and subtract\nthe expectation of the nominal continuation under the true row. The reward\nerror contributes at most $\\varepsilon_a$, the continuation-value error at most\n$D_h$, and row transport at most $\\delta_a hM$. Taking a maximum over the same\nfeasible action set is nonexpansive in the uniform score error. Therefore\n$D_{h+1}\\leq D_h+\\varepsilon+\\delta hM$, whose accumulated upper bound is\n\\eqref{eq:value-radius}. The local Q statement follows from the same inserted\nexpectation before taking a maximum.\n\\end{proof}",",a)=\\widehat Q_h(s,a)-\\rho_h(a)$ and\n$U_h(s,a)=\\widehat Q_h(s,a)+\\rho_h(a)$ enclose the true Q value.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero the value error is zero. In a single backup, insert and subtract\nthe expectation of the nominal continuation under the true row. The reward\nerror contributes at most $\\varepsilon_a$, the continuation-value error at most\n$D_h$, and row transport at most $\\delta_a hM$. Taking a maximum over the same\nfeasible action set is nonexpansive in the uniform score error. Therefore\n$D_{h+1}\\leq D_h+\\varepsilon+\\delta hM$, whose accumulated upper bound is\n\\eqref{eq:value-radius}. The local Q statement follows from the same inserted\nexpectation before taking a maximum.\n\\end{proof}","in{proof}\nAt horizon zero the value error is zero. In a single backup, insert and subtract\nthe expectation of the nominal continuation under the true row. The reward\nerror contributes at most $\\varepsilon_a$, the continuation-value error at most\n$D_h$, and row transport at most $\\delta_a hM$. Taking a maximum over the same\nfeasible action set is nonexpansive in the uniform score error. Therefore\n$D_{h+1}\\leq D_h+\\varepsilon+\\delta hM$, whose accumulated upper bound is\n\\eqref{eq:value-radius}. The local Q statement follows from the same inserted\nexpectation before taking a maximum.\n\\end{proof}","The executable robust choice maximizes $L_h(s,a)$ over feasible actions. It is\na maximizer of a \\emph{certified surrogate lower bound}. It is not asserted to\nsolve the exact minimax problem over all compatible realizations. A bound can\nbe valid and substantially below the actual worst-case score.","\\begin{proposition}[Distinct-competitor dominance]\nIf $a$ is feasible and\n\\[\n U_h(s,b)\\leq L_h(s,a)\\quad\n \\text{for every feasible }b\\ne a,\n\\]\nthen $a$ is optimal in every compatible realization.\n\\end{proposition}\n\\begin{proof}\nFor a distinct competitor, chain the three inequalities\n$Q_h(s,b)\\leq U_h(s,b)\\leq L_h(s,a)\\leq Q_h(s,a)$.\nThe comparison with $a$ itself is reflexive.\n\\end{proof}","\\begin{example}[A deliberately conservative abstract quote fixture]\nThe retained uncertainty example assigns zero setup to both kinds, warm gross\n$200$, cold gross $60$, warm success band $[3/5,1]$, and exact cold success $1$.\nStart with both statuses untried and use $h=1$, hence two total actions. The\nnominal scores are warm $120$, cold $60$, and stop $0$. Here\n$\\varepsilon=0$, $\\delta=4/5$, $M=260$, and $D_1=0$. The local warm radius\nis $208$, while the cold and stop radii are zero:\n\\[\n\\begin{array}{c|r|r|r}\n \\text{action}&\\widehat Q_1&\\rho_1&L_1","mple}[A deliberately conservative abstract quote fixture]\nThe retained uncertainty example assigns zero setup to both kinds, warm gross\n$200$, cold gross $60$, warm success band $[3/5,1]$, and exact cold success $1$.\nStart with both statuses untried and use $h=1$, hence two total actions. The\nnominal scores are warm $120$, cold $60$, and stop $0$. Here\n$\\varepsilon=0$, $\\delta=4/5$, $M=260$, and $D_1=0$. The local warm radius\nis $208$, while the cold and stop radii are zero:\n\\[\n\\begin{array}{c|r|r|r}\n \\text{action}&\\widehat Q_1&\\rho_1&L_1","abstract quote fixture]\nThe retained uncertainty example assigns zero setup to both kinds, warm gross\n$200$, cold gross $60$, warm success band $[3/5,1]$, and exact cold success $1$.\nStart with both statuses untried and use $h=1$, hence two total actions. The\nnominal scores are warm $120$, cold $60$, and stop $0$. Here\n$\\varepsilon=0$, $\\delta=4/5$, $M=260$, and $D_1=0$. The local warm radius\nis $208$, while the cold and stop radii are zero:\n\\[\n\\begin{array}{c|r|r|r}\n \\text{action}&\\widehat Q_1&\\rho_1&L_1","example assigns zero setup to both kinds, warm gross\n$200$, cold gross $60$, warm success band $[3/5,1]$, and exact cold success $1$.\nStart with both statuses untried and use $h=1$, hence two total actions. The\nnominal scores are warm $120$, cold $60$, and stop $0$. Here\n$\\varepsilon=0$, $\\delta=4/5$, $M=260$, and $D_1=0$. The local warm radius\nis $208$, while the cold and stop radii are zero:\n\\[\n\\begin{array}{c|r|r|r}\n \\text{action}&\\widehat Q_1&\\rho_1&L_1\\\\\\hline\n \\mathrm{inspectWarm}&120&208&-88\\\\\n \\mathrm{generateCold}&60&0&60\\\\\n \\mathrm{stop}&0&0&0\n\\end{array}\n\\]\nThe lower-Q rule chooses cold although warm's exact compatible Q range is\n$[120,200]$. This is a clear example of conservatism in the certified envelope.\nIt does not show that cold is the exact maximin choice.\n\\end{example}","\\mathrm{stop}&0&0&0\n\\end{array}\n\\]\nThe lower-Q rule chooses cold although warm's exact compatible Q range is\n$[120,200]$. This is a clear example of conservatism in the certified envelope.\nIt does not show that cold is the exact maximin choice.\n\\end{example}","s, local radii, and numerical order through\n\\path{uncertain_warm_local_radius},\n\\path{uncertain_cold_local_radius}, and\n\\path{uncertainty_changes_numerical_order}. The displayed first-action\nranking is a direct reconstruction at the stated initial state and horizon.\nThis fixture is \\emph{abstract quote planning}, not an artifact-backed saving\nexample. Its protected rank-one source has dense cost $36$ per call and the\nfixture has one call, so gross values $200$ and $60$ cannot come from bounded\nexecution savings for that source. The separate $1000$-call example above is\nthe artifact-backed comparison. These two roles must remain distinct.","warm_local_radius},\n\\path{uncertain_cold_local_radius}, and\n\\path{uncertainty_changes_numerical_order}. The displayed first-action\nranking is a direct reconstruction at the stated initial state and horizon.\nThis fixture is \\emph{abstract quote planning}, not an artifact-backed saving\nexample. Its protected rank-one source has dense cost $36$ per call and the\nfixture has one call, so gross values $200$ and $60$ cannot come from bounded\nexecution savings for that source. The separate $1000$-call example above is\nthe artifact-backed comparison. These two roles must remain distinct.","_radius},\n\\path{uncertain_cold_local_radius}, and\n\\path{uncertainty_changes_numerical_order}. The displayed first-action\nranking is a direct reconstruction at the stated initial state and horizon.\nThis fixture is \\emph{abstract quote planning}, not an artifact-backed saving\nexample. Its protected rank-one source has dense cost $36$ per call and the\nfixture has one call, so gross values $200$ and $60$ cannot come from bounded\nexecution savings for that source. The separate $1000$-call example above is\nthe artifact-backed comparison. These two roles must remain distinct.","merical_order}. The displayed first-action\nranking is a direct reconstruction at the stated initial state and horizon.\nThis fixture is \\emph{abstract quote planning}, not an artifact-backed saving\nexample. Its protected rank-one source has dense cost $36$ per call and the\nfixture has one call, so gross values $200$ and $60$ cannot come from bounded\nexecution savings for that source. The separate $1000$-call example above is\nthe artifact-backed comparison. These two roles must remain distinct.","Let $a_h(s)$ maximize $L_h(s,a)$ and let $a^*$ maximize the true $Q_h$.\nThen\n\\begin{align*}\n V_{h+1}(s)-Q_h(s,a_h(s))\n &\\leq U_h(s,a^*)-L_h(s,a_h(s))","Let $a_h(s)$ maximize $L_h(s,a)$ and let $a^*$ maximize the true $Q_h$.\nThen\n\\begin{align*}\n V_{h+1}(s)-Q_h(s,a_h(s))\n &\\leq U_h(s,a^*)-L_h(s,a_h(s))\\\\\n &\\leq U_h(s,a^*)-L_h(s,a^*)\\\\\n &\\leq 2\\Gamma_h.\n\\end{align*}\nThe comparison controls the Bellman loss of the actual choice at each state.\nThe policy is rebuilt at every remaining horizon with the same frozen evidence.","\\begin{theorem}[Expected frozen-policy regret]\nFor a horizon $H$, let $J_H^{\\rm lower}(s)$ be the expected value of the complete\nadaptive policy induced by robust lower-Q selection. Under any compatible\nrealization,\n\\[\n 0\\leq V_H(s)-J_H^{\\rm lower}(s)\n \\leq\\sum_{h=0}^{H-1}2\\Gamma_h.\n\\]\nThe comparator is the true optimal legal adaptive plan with the same\nconservative reservation guards.\n\\end{theorem}\n\\begin{proof}\nWrite the regret at a node with $h+1$ remaining steps as its local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one.","\\begin{theorem}[Expected frozen-policy regret]\nFor a horizon $H$, let $J_H^{\\rm lower}(s)$ be the expected value of the complete\nadaptive policy induced by robust lower-Q selection. Under any compatible\nrealization,\n\\[\n 0\\leq V_H(s)-J_H^{\\rm lower}(s)\n \\leq\\sum_{h=0}^{H-1}2\\Gamma_h.\n\\]\nThe comparator is the true optimal legal adaptive plan with the same\nconservative reservation guards.\n\\end{theorem}\n\\begin{proof}\nWrite the regret at a node with $h+1$ remaining steps as its local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one. This yields the cumulativ","\\begin{theorem}[Expected frozen-policy regret]\nFor a horizon $H$, let $J_H^{\\rm lower}(s)$ be the expected value of the complete\nadaptive policy induced by robust lower-Q selection. Under any compatible\nrealization,\n\\[\n 0\\leq V_H(s)-J_H^{\\rm lower}(s)\n \\leq\\sum_{h=0}^{H-1}2\\Gamma_h.\n\\]\nThe comparator is the true optimal legal adaptive plan with the same\nconservative reservation guards.\n\\end{theorem}\n\\begin{proof}\nWrite the regret at a node with $h+1$ remaining steps as its local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one. This yields the cumulative upper bound.\nNonnegativity follows because the selected complete tree is legal and $V_H$\ndominates the value of every legal tree.\n\\end{proof}","a horizon $H$, let $J_H^{\\rm lower}(s)$ be the expected value of the complete\nadaptive policy induced by robust lower-Q selection. Under any compatible\nrealization,\n\\[\n 0\\leq V_H(s)-J_H^{\\rm lower}(s)\n \\leq\\sum_{h=0}^{H-1}2\\Gamma_h.\n\\]\nThe comparator is the true optimal legal adaptive plan with the same\nconservative reservation guards.\n\\end{theorem}\n\\begin{proof}\nWrite the regret at a node with $h+1$ remaining steps as its local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one. This yields the cumulative upper bound.\nNonnegativity follows because the selected complete tree is legal and $V_H$\ndominates the value of every legal tree.\n\\end{proof}","n. Under any compatible\nrealization,\n\\[\n 0\\leq V_H(s)-J_H^{\\rm lower}(s)\n \\leq\\sum_{h=0}^{H-1}2\\Gamma_h.\n\\]\nThe comparator is the true optimal legal adaptive plan with the same\nconservative reservation guards.\n\\end{theorem}\n\\begin{proof}\nWrite the regret at a node with $h+1$ remaining steps as its local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one. This yields the cumulative upper bound.\nNonnegativity follows because the selected complete tree is legal and $V_H$\ndominates the value of every legal tree.\n\\end{proof}","leq\\sum_{h=0}^{H-1}2\\Gamma_h.\n\\]\nThe comparator is the true optimal legal adaptive plan with the same\nconservative reservation guards.\n\\end{theorem}\n\\begin{proof}\nWrite the regret at a node with $h+1$ remaining steps as its local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one. This yields the cumulative upper bound.\nNonnegativity follows because the selected complete tree is legal and $V_H$\ndominates the value of every legal tree.\n\\end{proof}","ts local Bellman loss\nplus the expected regret of the selected continuation. The local term is at\nmost $2\\Gamma_h$. By induction, every child regret is bounded by the common\nsum $\\sum_{j<h}2\\Gamma_j$. Taking the expectation leaves that constant unchanged\nbecause probabilities sum to one. This yields the cumulative upper bound.\nNonnegativity follows because the selected complete tree is legal and $V_H$\ndominates the value of every legal tree.\n\\end{proof}","This is the argument implemented by \\path{plan_regret_le} in\n\\path{AVOC/GuardedPolicyRegret.lean} and\n\\path{robust_policy_regret} in \\path{AVOC/RobustQuotePlanning.lean}.\nOrdinary finite sums simplify the displayed bound to\n\\[\n \\varepsilon H(H+1)+\\frac{\\delta M}{3}H(H-1)(H+1).\n\\]\nThis last closed form is a prose algebraic consequence of the source's sums,\nnot a newly compiled Lean declaration. The envelope is deliberately broad and\nmay be vacuous for a particular application. It remains a valid certificate.\nThe regret is an expected policy-value difference. It does not bound every\nrealized sequence of successes and failures.","Let $K$ be the number of successes in $n$ Bernoulli trials with rational\nparameter $p\\in[0,1]$. Define\n\\[\n b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe co","Let $K$ be the number of successes in $n$ Bernoulli trials with rational\nparameter $p\\in[0,1]$. Define\n\\[\n b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational","Let $K$ be the number of successes in $n$ Bernoulli trials with rational\nparameter $p\\in[0,1]$. Define\n\\[\n b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cov","Let $K$ be the number of successes in $n$ Bernoulli trials with rational\nparameter $p\\in[0,1]$. Define\n\\[\n b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","ber of successes in $n$ Bernoulli trials with rational\nparameter $p\\in[0,1]$. Define\n\\[\n b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","uccesses in $n$ Bernoulli trials with rational\nparameter $p\\in[0,1]$. Define\n\\[\n b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","b_{n,k}(p)=\\binom nk p^k(1-p)^{n-k},\\qquad 0\\leq k\\leq n.\n\\]\nChoose a positive integer grid size $m$ and nonnegative rational allowance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","owance\n$\\alpha$. The grid cells have endpoints\n\\[\n \\ell_i=\\frac{i}{m},\\qquad u_i=\\frac{i+1}{m},\\qquad 0\\leq i<m.\n\\]\nFor observed count $k$, the envelope of its point mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","oint mass over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","s over cell $i$ is\n\\[\n B_{n,k,i}=\\binom nk u_i^k(1-\\ell_i)^{n-k}.\n\\]\nRetain the cell exactly when\n\\[\n B_{n,k,i}>\\frac{\\alpha}{n+1}.\n\\]\nReturn the closed interval spanning all retained cells. If there are no retained\ncells, return $[0,1]$. The returned interval is therefore always a valid band.\nThe construction performs exact rational arithmetic and strict comparison.\nBoundary cells cover $0$ and $1$ as well as interior parameters.","\\begin{theorem}[Finite-sample grid coverage]\nIf $K$ has the binomial count law $b_{n,k}(p)$, the returned interval $I(K)$ obeys\n\\[\n \\Prob\\{p\\notin I(K)\\}\\leq\\alpha.\n\\]\nThis holds for every rational $p\\in[0,1]$, without assuming that $p$ is a grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the","\\begin{theorem}[Finite-sample grid coverage]\nIf $K$ has the binomial count law $b_{n,k}(p)$, the returned interval $I(K)$ obeys\n\\[\n \\Prob\\{p\\notin I(K)\\}\\leq\\alpha.\n\\]\nThis holds for every rational $p\\in[0,1]$, without assuming that $p$ is a grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ a","\\begin{theorem}[Finite-sample grid coverage]\nIf $K$ has the binomial count law $b_{n,k}(p)$, the returned interval $I(K)$ obeys\n\\[\n \\Prob\\{p\\notin I(K)\\}\\leq\\alpha.\n\\]\nThis holds for every rational $p\\in[0,1]$, without assuming that $p$ is a grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate","\\begin{theorem}[Finite-sample grid coverage]\nIf $K$ has the binomial count law $b_{n,k}(p)$, the returned interval $I(K)$ obeys\n\\[\n \\Prob\\{p\\notin I(K)\\}\\leq\\alpha.\n\\]\nThis holds for every rational $p\\in[0,1]$, without assuming that $p$ is a grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate.\n\\end{proof}","b\\{p\\notin I(K)\\}\\leq\\alpha.\n\\]\nThis holds for every rational $p\\in[0,1]$, without assuming that $p$ is a grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate.\n\\end{proof}","in[0,1]$, without assuming that $p$ is a grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate.\n\\end{proof}","grid\npoint.\n\\end{theorem}\n\\begin{proof}\nSome grid cell contains $p$. In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate.\n\\end{proof}","In that cell, $p\\leq u_i$ and\n$1-p\\leq1-\\ell_i$, hence $b_{n,k}(p)\\leq B_{n,k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate.\n\\end{proof}",",k,i}$. If $p$ is outside the\nreturned interval, the containing cell could not have been retained: any\nretained cell lies inside the hull. Therefore\n$B_{n,k,i}\\leq\\alpha/(n+1)$, and the mass of this excluded count $k$ is at most\nthat threshold. There are $n+1$ possible counts, so summing their failure masses\ngives at most $\\alpha$. If no cells were retained, the returned full band cannot\nexclude $p$ and the conclusion is immediate.\n\\end{proof}","\\begin{example}[Ten successes are evidence, not certainty]\nFor $n=10$, $m=10$, and $\\alpha=1/10$, ten successes retain cells\n$\\{6,7,8,9\\}$ and return $[3/5,1]$. Ten failures retain\n$\\{0,1,2,3\\}$ and return $[0,2/5]$. These endpoint calculations are explicit\nLean theorems. By contrast, the empirical singleton $\\{K/n\\}$ has no such\ncoverage property. At $n=1$ and $p=1/2$, both possible singletons exclude the\ntrue parameter, so its miss probability is $1$.\n\\end{example}","nty]\nFor $n=10$, $m=10$, and $\\alpha=1/10$, ten successes retain cells\n$\\{6,7,8,9\\}$ and return $[3/5,1]$. Ten failures retain\n$\\{0,1,2,3\\}$ and return $[0,2/5]$. These endpoint calculations are explicit\nLean theorems. By contrast, the empirical singleton $\\{K/n\\}$ has no such\ncoverage property. At $n=1$ and $p=1/2$, both possible singletons exclude the\ntrue parameter, so its miss probability is $1$.\n\\end{example}","The relevant declarations include \\path{ten_successes_lower},\n\\path{ten_successes_upper}, \\path{ten_failures_lower},\n\\path{ten_failures_upper}, and\n\\path{empirical_singleton_counterexample}. Nothing here says the true\nparameter has posterior probability $9/10$ of lying inside an observed band.\nThe guarantee concerns repeated-data failure probability under the specified\nlaw.","A schedule is a finite set $J$ of checkpoints. Each checkpoint fixes an\nidentity stratum, sample size $n_j$, grid $m_j>0$, true parameter $p_j$, and\nnonnegative error allocation $\\alpha_j$. Its count statistic has the stated\nbinomial marginal. Counts can be dependent, including different prefixes of\none source and checkpoints that share other underlying randomness.","\\begin{theorem}[Scheduled and selected failure bounds]\nLet $B_j$ be the event that checkpoint $j$'s interval excludes $p_j$. Then\n\\[\n \\Prob\\Bigl(\\bigcup_{j\\in J}B_j\\Bigr)\n \\leq\\sum_{j\\in J}\\alpha_j.\n\\]\nFor any data-dependent selector $j(\\omega)\\in J$,\n\\[\n \\Prob\\{\\omega\\in B_{j(\\omega)}\\}\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor each outcome $\\omega$, the indicator of a union is at most the sum of the\nindividual indicators. Multiply by the nonnegative rational weight of\n$\\omega$ and sum over the finite probability space. Interchange the two finite\nsums to obtain the union bound. Each checkpoint has failure mass at most its\nallocated $\\alpha_j$ by finite-grid coverage. Finally, selected failure implies\nfailure of at least one scheduled","\\begin{theorem}[Scheduled and selected failure bounds]\nLet $B_j$ be the event that checkpoint $j$'s interval excludes $p_j$. Then\n\\[\n \\Prob\\Bigl(\\bigcup_{j\\in J}B_j\\Bigr)\n \\leq\\sum_{j\\in J}\\alpha_j.\n\\]\nFor any data-dependent selector $j(\\omega)\\in J$,\n\\[\n \\Prob\\{\\omega\\in B_{j(\\omega)}\\}\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor each outcome $\\omega$, the indicator of a union is at most the sum of the\nindividual indicators. Multiply by the nonnegative rational weight of\n$\\omega$ and sum over the finite probability space. Interchange the two finite\nsums to obtain the union bound. Each checkpoint has failure mass at most its\nallocated $\\alpha_j$ by finite-grid coverage. Finally, selected failure implies\nfailure of at least one scheduled checkpoint, so it is a subset of","\\begin{theorem}[Scheduled and selected failure bounds]\nLet $B_j$ be the event that checkpoint $j$'s interval excludes $p_j$. Then\n\\[\n \\Prob\\Bigl(\\bigcup_{j\\in J}B_j\\Bigr)\n \\leq\\sum_{j\\in J}\\alpha_j.\n\\]\nFor any data-dependent selector $j(\\omega)\\in J$,\n\\[\n \\Prob\\{\\omega\\in B_{j(\\omega)}\\}\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor each outcome $\\omega$, the indicator of a union is at most the sum of the\nindividual indicators. Multiply by the nonnegative rational weight of\n$\\omega$ and sum over the finite probability space. Interchange the two finite\nsums to obtain the union bound. Each checkpoint has failure mass at most its\nallocated $\\alpha_j$ by finite-grid coverage. Finally, selected failure implies\nfailure of at least one scheduled checkpoint, so it is a subset of the union.\nThis last incl","\\begin{theorem}[Scheduled and selected failure bounds]\nLet $B_j$ be the event that checkpoint $j$'s interval excludes $p_j$. Then\n\\[\n \\Prob\\Bigl(\\bigcup_{j\\in J}B_j\\Bigr)\n \\leq\\sum_{j\\in J}\\alpha_j.\n\\]\nFor any data-dependent selector $j(\\omega)\\in J$,\n\\[\n \\Prob\\{\\omega\\in B_{j(\\omega)}\\}\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor each outcome $\\omega$, the indicator of a union is at most the sum of the\nindividual indicators. Multiply by the nonnegative rational weight of\n$\\omega$ and sum over the finite probability space. Interchange the two finite\nsums to obtain the union bound. Each checkpoint has failure mass at most its\nallocated $\\alpha_j$ by finite-grid coverage. Finally, selected failure implies\nfailure of at least one scheduled checkpoint, so it is a subset of the union.\nThis last inclusion holds pointwise even if selection sees all outcomes.\n\\end{proof}","j$'s interval excludes $p_j$. Then\n\\[\n \\Prob\\Bigl(\\bigcup_{j\\in J}B_j\\Bigr)\n \\leq\\sum_{j\\in J}\\alpha_j.\n\\]\nFor any data-dependent selector $j(\\omega)\\in J$,\n\\[\n \\Prob\\{\\omega\\in B_{j(\\omega)}\\}\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor each outcome $\\omega$, the indicator of a union is at most the sum of the\nindividual indicators. Multiply by the nonnegative rational weight of\n$\\omega$ and sum over the finite probability space. Interchange the two finite\nsums to obtain the union bound. Each checkpoint has failure mass at most its\nallocated $\\alpha_j$ by finite-grid coverage. Finally, selected failure implies\nfailure of at least one scheduled checkpoint, so it is a subset of the union.\nThis last inclusion holds pointwise even if selection sees all outcomes.\n\\end{proof}","\\{\\omega\\in B_{j(\\omega)}\\}\\leq\\sum_{j\\in J}\\alpha_j.\n\\]\n\\end{theorem}\n\\begin{proof}\nFor each outcome $\\omega$, the indicator of a union is at most the sum of the\nindividual indicators. Multiply by the nonnegative rational weight of\n$\\omega$ and sum over the finite probability space. Interchange the two finite\nsums to obtain the union bound. Each checkpoint has failure mass at most its\nallocated $\\alpha_j$ by finite-grid coverage. Finally, selected failure implies\nfailure of at least one scheduled checkpoint, so it is a subset of the union.\nThis last inclusion holds pointwise even if selection sees all outcomes.\n\\end{proof}","\\begin{example}[Why the schedule's total allowance is necessary]\nTake a fair Boolean outcome $\\omega$ and two checkpoint bad events\n$B_0=\\{\\omega=0\\}$ and $B_1=\\{\\omega=1\\}$. Each has probability $1/2$.\nSelecting $j(\\omega)=\\omega$ makes selected failure certain. The total allowance\n$1/2+1/2=1$ remains valid, while either single checkpoint's allowance fails.\nThis is the exact obstruction proved by\n\\path{fairBool_adaptive_exceeds_single_budget}.\n\\end{example}","The count-law obligation is substantial. Fair marginals do not imply a binomial\nsum. Two identical fair bits each have parameter $1/2$, but their two-bit count\nis either zero or two and never one. The source contains this obstruction as\n\\path{correlatedPair_not_product}.","The primitive model uses a finite rational probability space and, for each\nidentity $k$, a finite latent Boolean vector of length $N_k$. Its success\nparameter $p_k$ is fixed. The fresh-outcome contract states that after agreement\nwith any specified pattern on a previously revealed set of positions, a fresh\nposition's joint success mass is $p_k$ times that history mass. In notation,\n\\[\n \\Prob\\{H\\text{ matches and }X_i=1\\}=p_k\\Prob\\{H\\text{ matches}\\},\n \\qquad i\\text{ fresh}.\n\\]\nThe multiplication-only form is meaningful for histories of mass zero and does\nnot divide by them. Induction determines the mass of every full pattern, then\nsumming patterns by success count gives its binomial law. Truncation yields the\nsame law for each fixed prefix.","Every accepted generative step emits an ordinary recorder command and an\neffective reveal marker. Whole-run induction proves that the emitted commands\nreproduce the actual raw recorder state and that the reveal state equals the\ninherited reveal machine on its effective trace. In particular, for every\nidentity,\n\\begin{equation}\n \\mathrm{samples}_k(\\mathrm{terminalLog})\n =\\mathrm{source}_k.\\mathrm{take}\n     \\bigl(|\\mathrm{samples}_k(\\mathrm{terminalLog})|\\bigr).\n \\label{eq:recorder-prefix}\n\\end{equation}","\\begin{example}[Replay and faults do not create new evidence]\nLet an identity's source be $[\\mathrm{true},\\mathrm{false}]$. Begin ID $7$, report\nID $7$, then report ID $7$ again. The sample list is $[\\mathrm{true}]$ and the\nremaining source begins with false. Begin ID $8$ and finish it as a fault. The\nsample list stays unchanged. Begin and report ID $9$. It appends false, yielding\n$[\\mathrm{true},\\mathrm{false}]$. A further fresh report rejects after source\nexhaustion, while a replay of ID $7$ remains possible. The replay supplies no\nextra success and the fault supplies no concealed binary observation.\n\\end{example}","hrm{true},\\mathrm{false}]$. Begin ID $7$, report\nID $7$, then report ID $7$ again. The sample list is $[\\mathrm{true}]$ and the\nremaining source begins with false. Begin ID $8$ and finish it as a fault. The\nsample list stays unchanged. Begin and report ID $9$. It appends false, yielding\n$[\\mathrm{true},\\mathrm{false}]$. A further fresh report rejects after source\nexhaustion, while a replay of ID $7$ remains possible. The replay supplies no\nextra success and the fault supplies no concealed binary observation.\n\\end{example}","der-to-policy endpoint reads counts from the actual terminal log.\nRejected whole runs supply no counts. Successful runs with \\emph{any} pending\noperation also supply no counts, even when the pending identity is unrelated to\nthe selected checkpoint. A fully resolved run still needs enough success or\nrejection reports for both selected checkpoints. A checkpoint always reads the\nfirst predetermined $n_j$ reports, so later reports cannot change that count.","\\begin{theorem}[Recorder-to-policy bad-issuance bound]\nAssume finite rational latent sources satisfy the primitive conditional laws;\nthe finite schedule, grids, and nonnegative allowances are predetermined;\nscheduled prefix lengths fit within their sources; selected strata match the\nspecified true probabilities; and deterministic setup/gross memberships hold.\nThen\n\\begin{equation}\n \\Prob\\{\\text{a plan is issued and lacks its episode certificate}\\}\n \\leq \\sum_{j\\in J}\\alpha_j.\n \\label{eq:issuance}\n\\end{equation}\nThe episode certificate supplies a compatible realization with the stated exact\nparameters, value and Q enclosures, and expected whole-policy regret for every\nfinite horizon under frozen quotes.\n\\end{theorem}\n\\begin{proof}\nThe primitive conditional law yields binomial prefix count laws at all scheduled\ncheckpoints. The generative run refinement gives \\eqref{eq:recorder-prefix}.\nWhenever gated counts are available, taking the fixed first $n_j$ reports\ntherefore gives the corresponding latent prefix count. On the event where all\nscheduled bands cover their true parameters, eithe…","ble realization with the stated exact\nparameters, value and Q enclosures, and expected whole-policy regret for every\nfinite horizon under frozen quotes.\n\\end{theorem}\n\\begin{proof}\nThe primitive conditional law yields binomial prefix count laws at all scheduled\ncheckpoints. The generative run refinement gives \\eqref{eq:recorder-prefix}.\nWhenever gated counts are available, taking the fixed first $n_j$ reports\ntherefore gives the corresponding latent prefix count. On the event where all\nscheduled bands cover their true parameters, either the gates suppress issuance\nor both selected learned bands contain the correct parameters. Combine those\nmemberships with the supplied deterministic memberships to construct a\ncompatible realization. The deterministic frozen-episode theorem then supplies\nthe policy certificate. Thus bad issuance is a subset of the scheduled band\nfailure union, whose probability is bounded by the total allowance.\n\\end{proof}","The probability in \\eqref{eq:issuance} ranges over the original outcome space,\nincluding rejected, unresolved, and insufficient-data runs. There is no claim\nthat a plan will be issued, that any operation finishes, or that collection is\ncheap. If the issuance probability is $q>0$, the generic conditional implication\nis only\n\\[\n \\Prob\\{\\text{uncertified}\\mid\\text{issued}\\}\n \\leq \\min\\left(1,\\frac{\\sum_j\\alpha_j}{q}\\right).\n\\]\nThe same nominal allowance need not hold after conditioning on issuance.","\\begin{example}[Predictable availability still biases completed runs]\nLet $(X_1,X_2)$ be independent fair bits. Reveal $X_1$, and reveal $X_2$ by the\ndeadline only if $X_1=1$. Availability of the second report depends on the past,\nnot on $X_2$. Completion probability is $1/2$. Conditional on completion,\n$X_1=1$ with probability one, and the two-bit count has probability zero of\nbeing zero. A genuine two-trial binomial count with parameter $1/2$ has zero\ncount probability $1/4$.\n\\end{example}","ll biases completed runs]\nLet $(X_1,X_2)$ be independent fair bits. Reveal $X_1$, and reveal $X_2$ by the\ndeadline only if $X_1=1$. Availability of the second report depends on the past,\nnot on $X_2$. Completion probability is $1/2$. Conditional on completion,\n$X_1=1$ with probability one, and the two-bit count has probability zero of\nbeing zero. A genuine two-trial binomial count with parameter $1/2$ has zero\ncount probability $1/4$.\n\\end{example}","_1,X_2)$ be independent fair bits. Reveal $X_1$, and reveal $X_2$ by the\ndeadline only if $X_1=1$. Availability of the second report depends on the past,\nnot on $X_2$. Completion probability is $1/2$. Conditional on completion,\n$X_1=1$ with probability one, and the two-bit count has probability zero of\nbeing zero. A genuine two-trial binomial count with parameter $1/2$ has zero\ncount probability $1/4$.\n\\end{example}","A direct interval illustration uses ten independent fair bits and issues an\ninterval only when all ten are true. Its issuance probability is $2^{-10}$.\nThe implemented $n=m=10$, $\\alpha=1/10$ interval is then $[3/5,1]$, which excludes\n$1/2$. Conditional on issuance, failure is certain. Unconditional bad issuance\nis only $2^{-10}$ and satisfies the allowance. This finite calculation is an\nillustrative consequence of the checked interval endpoints and the elementary\nproduct law, not a new Lean declaration.","Conditioning and ordering are different obstructions. Suppose two independent\nfair attempts start together. Success takes one unit of time and rejection\ntakes two; resolve ties by the initiation ID. Record every final outcome and\nwait until all intents have resolved. The completion order is\n\\[\n\\begin{array}{c|c|c}\n\\text{initiation outcomes}&\\text{completion-order outcomes}&\\text{mass}\\\\\\hline\n(0,0)&(0,0)&1/4\\\\\n(0,1)&(1,0)&1/4\\\\\n(1,0)&(1,0)&1/4\\\\\n(1,1)&(1,1)&1/4\n\\end{array}\n\\]\nThe first completed bit succeeds with probability $3/4$. No report was missing,\nno terminal event was forged, and no operation remains pending. Nevertheless,\nthe first-completed prefix is not a fair source prefix.","(1,1)&(1,1)&1/4\n\\end{array}\n\\]\nThe first completed bit succeeds with probability $3/4$. No report was missing,\nno terminal event was forged, and no operation remains pending. Nevertheless,\nthe first-completed prefix is not a fair source prefix.","For the implemented interval, the stronger retained finite experiment starts\ntwenty independent fair attempts and lets all successes complete before all\nrejections. If there are at least ten successes, the first ten completed reports\nare all true. Their interval is $[3/5,1]$ and excludes $1/2$. That event has mass\n\\[\n \\frac{1}{2^{20}}\\sum_{k=10}^{20}\\binom{20}{k}\n =\\frac{308333}{524288}>\\frac{1}{10}.\n\\]\nThus actual interval failure exceeds its nominal allowance despite complete,\nfully resolved lifecycle records. The exact Python evidence is preserved under\n\\path{results/v0.41.0/review_completion_order.py} and its JSON receipt. It is\na finite computational obstruction with an explicit rational derivation here,\nnot a newly kernel-checked theorem.","For a successful paid trace with total evaluation debit $e$ and switching debit\n$u$, the formal accounting identities are\n\\begin{align}\n \\mathrm{evaluationSpent}_{\\rm final}\n &=\\mathrm{evaluationSpent}_{\\rm initial}+e,","For a successful paid trace with total evaluation debit $e$ and switching debit\n$u$, the formal accounting identities are\n\\begin{align}\n \\mathrm{evaluationSpent}_{\\rm final}\n &=\\mathrm{evaluationSpent}_{\\rm initial}+e,\\\\\n \\mathrm{switchSpent}_{\\rm final}\n &=\\mathrm{switchSpent}_{\\rm initial}+u.\n\\end{align}\nFrom a safe initial journal, their sum stays within protected credits.\nSuccessful replay also cannot decrease the protected utility of the active\nversion. These are \\path{paid_trace_exact_accounting},\n\\path{paid_trace_no_overspend}, and\n\\path{replay_utility_nondecreasing} in\n\\path{AVOC/DurableUpgradeLaws.lean}."]}