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longrightarrow","display":"\\longrightarrow","kind":"use","occurrences":[11,30,135,174,257]},{"id":"symbol-22449108dc31ebd8","tex":"\\longmapsto","aliases":"⟼ longmapsto","display":"\\longmapsto","kind":"use","occurrences":[135]}],"contexts":["\\begin{scopebox}[title=The central result]\nLet $D:U\\to V$ be linear over a field and let\n\\[\nq:V\\to V/\\im D,\\qquad O=\\Span\\{q(r_i):i\\in I\\}.\n\\]\nIf $V$ is finite dimensional, the least added dimension that repairs every $r_i$ is $\\dim O$. If an allowed subspace $A\\leq V$ satisfies $O\\subseteq q(A)$, the minimum can be attained with all added directions in $A$. Selecting entries from a fixed menu is a different optimization problem.\n\\end{scopebox}","Let $\\kk$ be a commutative ring. All modules are unital $\\kk$-modules and all maps in this chapter are $\\kk$-linear. Fix\n\\[\nD:U\\longrightarrow V,\n\\]\nand let $r:I\\to V$ be a family of residuals. The index set $I$ need not be finite for the algebraic statements. When finite enumeration is used later, finiteness will be imposed explicitly.","\\begin{definition}[Correction and repair]\\label{def:repair}\nA residual $r\\in V$ is \\emph{correctable} by $D$ when there is $u\\in U$ such that\n\\begin{equation}\\label{eq:correct}\nDu=-r.\n\\end{equation}\nAn added correction map $E:W\\to V$ \\emph{repairs} $r$ when there are $u\\in U$ and $w\\in W$ with\n\\begin{equation}\\label{eq:repair}\nDu+Ew=-r.\n\\end{equation}\nIt repairs the family when this condition holds for every $r_i$. The witnesses may depend on $i$.\n\\end{definition}","tion}[Correction and repair]\\label{def:repair}\nA residual $r\\in V$ is \\emph{correctable} by $D$ when there is $u\\in U$ such that\n\\begin{equation}\\label{eq:correct}\nDu=-r.\n\\end{equation}\nAn added correction map $E:W\\to V$ \\emph{repairs} $r$ when there are $u\\in U$ and $w\\in W$ with\n\\begin{equation}\\label{eq:repair}\nDu+Ew=-r.\n\\end{equation}\nIt repairs the family when this condition holds for every $r_i$. The witnesses may depend on $i$.\n\\end{definition}","def:repair}\nA residual $r\\in V$ is \\emph{correctable} by $D$ when there is $u\\in U$ such that\n\\begin{equation}\\label{eq:correct}\nDu=-r.\n\\end{equation}\nAn added correction map $E:W\\to V$ \\emph{repairs} $r$ when there are $u\\in U$ and $w\\in W$ with\n\\begin{equation}\\label{eq:repair}\nDu+Ew=-r.\n\\end{equation}\nIt repairs the family when this condition holds for every $r_i$. The witnesses may depend on $i$.\n\\end{definition}","The sign convention makes the repaired residual $r+Du+Ew$ equal to zero. In particular, the target of the equation remains $V$. A calculation in another space can establish \\eqref{eq:repair} only if a valid route back to $V$ is supplied.","The family condition does not ask for one common pair $(u,w)$ that corrects every residual at once. It asks for one common added map $E$ and separate coefficients for each residual. Nor does it require those witnesses to vary continuously, measurably, or linearly with the index. Such conditions would define different problems.","\\section{What the quotient remembers}\nWrite\n\\[\nQ_D=\\coker D=V/\\im D,\\qquad q_D:V\\longrightarrow Q_D.\n\\]\nThe class $q_D(r)$ records the part of $r$ that cannot be changed by the original correction map. Since $q_D(Du)=0$, all residuals that differ by an original correction have the same class.","\\begin{theorem}[Exact obstruction criterion]\\label{thm:correctable}\nFor every $r\\in V$,\n\\[\n\\exists u\\in U\\ (Du=-r)\\quad\\Longleftrightarrow\\quad q_D(r)=0.\n\\]\n\\end{theorem}\n\\begin{proof}\nApplying $q_D$ to $Du=-r$ gives $0=-q_D(r)$, hence $q_D(r)=0$. Conversely, $q_D(r)=0$ means $r\\in\\im D$. Choose $v\\in U$ with $Dv=r$. Then $D(-v)=-r$.\n\\end{proof}\n\\formal{correctable_iff_obstruction_zero}","This is an exact criterion for the declared linear correction equation. It is not a universal test for a nonlinear equation from which $D$ might have been obtained by differentiation. Chapter 7 gives counterexamples in both directions.","The quotient also has a universal property. If $L:V\\to Z$ annihilates $\\im D$, it descends to a unique linear map $\\overline L:Q_D\\to Z$ with $L=\\overline Lq_D$. Thus every linear detector that ignores correctable directions factors through the same quotient. This is the classical cokernel construction used in Paper III~\\cite{paperiii}. Canonicality is relative to $D$, not to an unspecified mathematical problem.","\\begin{example}[A visible residual]\\label{ex:plane}\nLet $D:\\Q\\to\\Q^2$ be $D(u)=(u,0)$ and let $r=(2,3)$. The quotient is identified with the second coordinate. The correction $u=-2$ removes the first coordinate but leaves $(0,3)$. The functional $L(x,y)=y$ satisfies $LD=0$ and $L(r)=3$, so no original correction eliminates $r$.\n\\end{example}","\\section{Repair as an image condition}\nAn added map $E:W\\to V$ contributes precisely the classes in $\\im(q_DE)$. This leads to the basic feasibility criterion.","\\begin{theorem}[A single residual]\\label{thm:single}\nFor every $E:W\\to V$ and $r\\in V$,\n\\begin{equation}\\label{eq:single}\n\\exists u,w\\ (Du+Ew=-r)\n\\quad\\Longleftrightarrow\\quad\nq_D(r)\\in\\im(q_DE).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nSuppose $Du+Ew=-r$. Applying the quotient gives $q_D(Ew)=-q_D(r)$, so $q_D(r)=q_D(E(-w))$ lies in the image.","Conversely, choose $w$ with $q_D(Ew)=q_D(r)$. Then $q_D(r-Ew)=0$, so there is $v\\in U$ with $Dv=r-Ew$. Taking $u=-v$ and $w'=-w$ gives\n\\[\nDu+Ew'=-(r-Ew)-Ew=-r.\n\\]\nThe witnesses solve the equation in $V$, not just in its quotient.\n\\end{proof}\n\\formal{repairable_iff_obstruction_mem_range}","\\Needspace{8\\baselineskip}\n\\begin{corollary}[Reconstruction from a quotient witness]\\label{cor:reconstruct}\nIf $q_D(Ew)=q_D(r)$, there are $u\\in U$ and $w'\\in W$ such that\n\\[\nDu+Ew'+r=0.\n\\]\n\\end{corollary}\n\\begin{proof}\nUse the witnesses constructed in the reverse direction of \\cref{thm:single} and move $r$ to the left.\n\\end{proof}\n\\formal{reconstruct_cancellation}","\\section{The obstruction span of a family}\n\\begin{definition}[Required obstruction span]\\label{def:span}\nFor a residual family $r:I\\to V$, put\n\\[\nO_D(r)=\\Span_{\\kk}\\{q_D(r_i):i\\in I\\}\\leq Q_D.\n\\]\nWe write $O$ when the family and correction map are understood.\n\\end{definition}","Every $q_D(r_i)$ lies in $O$. The use of a span matters because an added linear correction map supplies a submodule of quotient classes, not merely a list of unrelated vectors.","\\begin{theorem}[Family feasibility]\\label{thm:family}\nThe map $E:W\\to V$ repairs the family $r$ if and only if\n\\begin{equation}\\label{eq:family}\nO_D(r)\\subseteq\\im(q_DE).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nIf every residual is repairable, \\cref{thm:single} places every generator $q_D(r_i)$ in $\\im(q_DE)$. The image is a submodule, so it contains their span. Conversely, \\eqref{eq:family} places each generator in that image, and \\cref{thm:single} supplies its correction witnesses.\n\\end{proof}\n\\formal{family_repairable_iff_span_le}","For an empty family, $O=0$ and the family condition is vacuous. For a nonempty family whose residuals are already correctable, $O$ is still zero. Both cases should permit a repair of added dimension zero. The next chapter includes them without a special exception.","\\begin{scopebox}[title=What has been established]\nThe quotient determines feasibility only because the correction map and the original equation have been declared. A positive quotient witness can be lifted to actual cancellation in $V$. A disappearance caused by forgetting part of $V$ is a different operation and requires a preservation theorem.\n\\end{scopebox}","\\chapter{Minimum repair dimension}\\label{ch:minimal}\n\\section{A lower bound that every repair must satisfy}\nIn this chapter $\\kk$ is a field. The added space $W$ is finite dimensional when its numerical dimension is compared. The attainment result itself uses a linear section and does not require all ambient spaces to be finite dimensional.","\\begin{theorem}[Dimension lower bound]\\label{thm:lower}\nIf $E:W\\to V$ repairs $r$ and $W$ is finite dimensional, then $O=O_D(r)$ is finite dimensional and\n\\[\n\\dim_{\\kk}O\\leq\\dim_{\\kk}W.\n\\]\n\\end{theorem}\n\\begin{proof}\nBy \\cref{thm:family}, $O\\subseteq\\im(q_DE)$. The image of a linear map from a finite dimensional space is finite dimensional and has dimension at most that of its domain. Hence\n\\[\n\\dim O\\leq\\dim\\im(q_DE)\\leq\\dim W.\\qedhere\n\\]\n\\end{proof}\n\\formal{repair_dimension_lower_bound}","\\section{An attaining construction}\n\\begin{lemma}[A section of the quotient]\\label{lem:section}\nThere is a linear map $s:Q_D\\to V$ with\n\\[\nq_Ds=\\id_{Q_D}.\n\\]\n\\end{lemma}\n\\begin{proof}\nChoose a basis of the vector space $Q_D$. Since $q_D$ is surjective, choose a representative in $V$ of each basis vector. Extend this assignment linearly. Its composite with $q_D$ fixes the basis and therefore fixes all of $Q_D$.\n\\end{proof}\n\\formal{exists_obstruction_section}","{An attaining construction}\n\\begin{lemma}[A section of the quotient]\\label{lem:section}\nThere is a linear map $s:Q_D\\to V$ with\n\\[\nq_Ds=\\id_{Q_D}.\n\\]\n\\end{lemma}\n\\begin{proof}\nChoose a basis of the vector space $Q_D$. Since $q_D$ is surjective, choose a representative in $V$ of each basis vector. Extend this assignment linearly. Its composite with $q_D$ fixes the basis and therefore fixes all of $Q_D$.\n\\end{proof}\n\\formal{exists_obstruction_section}","This choice of section is generally nonunique. Changing a representative by a vector in $\\im D$ produces another valid section. The construction is an existence proof using classical choice, not an algorithm for selecting a basis.","\\begin{theorem}[Attainment on the obstruction span]\\label{thm:attain}\nLet $\\iota_O:O\\hookrightarrow Q_D$ be inclusion. There exists a linear map $E:O\\to V$ such that\n\\begin{equation}\\label{eq:section-on-O}\nq_DE=\\iota_O\n\\end{equation}\nand $E$ repairs the family.\n\\end{theorem}\n\\begin{proof}\nChoose the section $s$ from \\cref{lem:section} and set $E=s\\iota_O$. Then $q_DE=q_Ds\\iota_O=\\iota_O$. Every $z\\in O$ is therefore in $\\im(q_DE)$. Family feasibility follows from \\cref{thm:family}.\n\\end{proof}\n\\formal{exists_attaining_repair}","nt on the obstruction span]\\label{thm:attain}\nLet $\\iota_O:O\\hookrightarrow Q_D$ be inclusion. There exists a linear map $E:O\\to V$ such that\n\\begin{equation}\\label{eq:section-on-O}\nq_DE=\\iota_O\n\\end{equation}\nand $E$ repairs the family.\n\\end{theorem}\n\\begin{proof}\nChoose the section $s$ from \\cref{lem:section} and set $E=s\\iota_O$. Then $q_DE=q_Ds\\iota_O=\\iota_O$. Every $z\\in O$ is therefore in $\\im(q_DE)$. Family feasibility follows from \\cref{thm:family}.\n\\end{proof}\n\\formal{exists_attaining_repair}","\\[\n\\begin{tikzcd}[column sep=large,row sep=large]\nO \\arrow[r,\"E\"] \\arrow[dr,hook,\"\\iota_O\"'] & V \\arrow[d,\"q_D\"]\\\\\n& Q_D\n\\end{tikzcd}\n\\]\nThe triangle records exactly what is preserved: a required quotient class is lifted to a representative of that same class. The lift does not manufacture an unrelated object with a similar invariant.","\\begin{theorem}[The attained free minimum]\\label{thm:minimum}\nIf $V$ is finite dimensional, the minimum possible dimension of a finite dimensional added space that repairs the family is\n\\begin{equation}\\label{eq:minimum}\n\\min\\bigl\\{\\dim W: E:W\\to V\\text{ repairs }r\\bigr\\}=\\dim O_D(r).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nEvery admissible term on the left is at least $\\dim O$ by \\cref{thm:lower}. Because $V$ is finite dimensional, its quotient and the subspace $O$ are finite dimensional. The map in \\cref{thm:attain} is a repair with domain $O$, so the bound is attained.\n\\end{proof}\n\\formal{free_minimal_repair_rank}","}[The attained free minimum]\\label{thm:minimum}\nIf $V$ is finite dimensional, the minimum possible dimension of a finite dimensional added space that repairs the family is\n\\begin{equation}\\label{eq:minimum}\n\\min\\bigl\\{\\dim W: E:W\\to V\\text{ repairs }r\\bigr\\}=\\dim O_D(r).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nEvery admissible term on the left is at least $\\dim O$ by \\cref{thm:lower}. Because $V$ is finite dimensional, its quotient and the subspace $O$ are finite dimensional. The map in \\cref{thm:attain} is a repair with domain $O$, so the bound is attained.\n\\end{proof}\n\\formal{free_minimal_repair_rank}","finite dimensional, the minimum possible dimension of a finite dimensional added space that repairs the family is\n\\begin{equation}\\label{eq:minimum}\n\\min\\bigl\\{\\dim W: E:W\\to V\\text{ repairs }r\\bigr\\}=\\dim O_D(r).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nEvery admissible term on the left is at least $\\dim O$ by \\cref{thm:lower}. Because $V$ is finite dimensional, its quotient and the subspace $O$ are finite dimensional. The map in \\cref{thm:attain} is a repair with domain $O$, so the bound is attained.\n\\end{proof}\n\\formal{free_minimal_repair_rank}","space that repairs the family is\n\\begin{equation}\\label{eq:minimum}\n\\min\\bigl\\{\\dim W: E:W\\to V\\text{ repairs }r\\bigr\\}=\\dim O_D(r).\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nEvery admissible term on the left is at least $\\dim O$ by \\cref{thm:lower}. Because $V$ is finite dimensional, its quotient and the subspace $O$ are finite dimensional. The map in \\cref{thm:attain} is a repair with domain $O$, so the bound is attained.\n\\end{proof}\n\\formal{free_minimal_repair_rank}","\\section{Directions restricted to a subspace}\nAn application usually prohibits some corrections. A first useful constraint is a subspace $A\\leq V$ containing every permitted added direction. Its quotient image is\n\\[\nq_D(A)=\\im\\bigl(q_D\\vert_A:A\\to Q_D\\bigr).\n\\]","\\begin{theorem}[Attainment inside a legal subspace]\\label{thm:legal}\nIf $O\\subseteq q_D(A)$, there exists a linear map $E:O\\to V$ satisfying\n\\[\n\\im E\\subseteq A,\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optim","\\begin{theorem}[Attainment inside a legal subspace]\\label{thm:legal}\nIf $O\\subseteq q_D(A)$, there exists a linear map $E:O\\to V$ satisfying\n\\[\n\\im E\\subseteq A,\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end","\\begin{theorem}[Attainment inside a legal subspace]\\label{thm:legal}\nIf $O\\subseteq q_D(A)$, there exists a linear map $E:O\\to V$ satisfying\n\\[\n\\im E\\subseteq A,\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","abel{thm:legal}\nIf $O\\subseteq q_D(A)$, there exists a linear map $E:O\\to V$ satisfying\n\\[\n\\im E\\subseteq A,\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","ubseteq q_D(A)$, there exists a linear map $E:O\\to V$ satisfying\n\\[\n\\im E\\subseteq A,\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","linear map $E:O\\to V$ satisfying\n\\[\n\\im E\\subseteq A,\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}",",\\qquad q_DE=\\iota_O,\n\\]\nand repairing the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","ng the entire family. If $V$ is finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","finite dimensional, its added dimension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","mension is optimal even among all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","ong all unrestricted repairs.\n\\end{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","d{theorem}\n\\begin{proof}\nConsider the surjection\n\\[\nf:A\\longrightarrow q_D(A),\\qquad a\\longmapsto q_D(a).\n\\]\nChoose a linear section $s_A:q_D(A)\\to A$. Let $j:O\\hookrightarrow q_D(A)$ and $\\iota_A:A\\hookrightarrow V$ be inclusions. Define\n\\[\nE=\\iota_As_Aj.\n\\]\nIts image is in $A$ by construction. Moreover, $q_D\\iota_As_Aj=j$ as a map into $Q_D$, which is precisely $\\iota_O$. By \\cref{thm:family}, $E$ repairs the family. The lower bound of \\cref{thm:lower} proves optimality when dimensions are finite.\n\\end{proof}\n\\formal{admissible_subspace_attainment}","\\begin{theorem}[Subspace impossibility]\\label{thm:legal-no-go}\nIf $O\\not\\subseteq q_D(A)$, no added map $E:W\\to V$ with image contained in $A$ repairs the family.\n\\end{theorem}\n\\begin{proof}\nSuch a map satisfies $\\im(q_DE)\\subseteq q_D(A)$. If it repaired the family, \\cref{thm:family} would imply $O\\subseteq\\im(q_DE)\\subseteq q_D(A)$, contrary to the hypothesis.\n\\end{proof}\n\\formal{admissible_subspace_no_go}","\\begin{corollary}[Rank difference in matrix coordinates]\\label{cor:rank-difference}\nLet the columns of a matrix $\\mathsf R$ be a finite residual family and let $\\mathsf D$ represent $D$. Then\n\\[\n\\dim O_D(r)=\\rank[\\mathsf D\\mid\\mathsf R]-\\rank\\mathsf D.\n\\]\n\\end{corollary}\n\\begin{proof}\nSet $M=\\im D+\\Span\\{r_i\\}$. The restriction $q_D\\vert_M$ has image $O$ and kernel $\\im D$. Rank and nullity give $\\dim M=\\dim\\im D+\\dim O$. The columns of $[\\mathsf D\\mid\\mathsf R]$ span $M$.\n\\end{proof}\n\\sourcescope{This coordinate corollary is an ordinary derivation included for the computational interpretation. The v0.2.0 Lean release proves the abstract attained dimension theorem, but does not yet identify this matrix formula with the Python rank implementation.}","x coordinates]\\label{cor:rank-difference}\nLet the columns of a matrix $\\mathsf R$ be a finite residual family and let $\\mathsf D$ represent $D$. Then\n\\[\n\\dim O_D(r)=\\rank[\\mathsf D\\mid\\mathsf R]-\\rank\\mathsf D.\n\\]\n\\end{corollary}\n\\begin{proof}\nSet $M=\\im D+\\Span\\{r_i\\}$. The restriction $q_D\\vert_M$ has image $O$ and kernel $\\im D$. Rank and nullity give $\\dim M=\\dim\\im D+\\dim O$. The columns of $[\\mathsf D\\mid\\mathsf R]$ span $M$.\n\\end{proof}\n\\sourcescope{This coordinate corollary is an ordinary derivation included for the computational interpretation. The v0.2.0 Lean release proves the abstract attained dimension theorem, but does not yet identify this matrix formula with the Python rank implementation.}","fference}\nLet the columns of a matrix $\\mathsf R$ be a finite residual family and let $\\mathsf D$ represent $D$. Then\n\\[\n\\dim O_D(r)=\\rank[\\mathsf D\\mid\\mathsf R]-\\rank\\mathsf D.\n\\]\n\\end{corollary}\n\\begin{proof}\nSet $M=\\im D+\\Span\\{r_i\\}$. The restriction $q_D\\vert_M$ has image $O$ and kernel $\\im D$. Rank and nullity give $\\dim M=\\dim\\im D+\\dim O$. The columns of $[\\mathsf D\\mid\\mathsf R]$ span $M$.\n\\end{proof}\n\\sourcescope{This coordinate corollary is an ordinary derivation included for the computational interpretation. The v0.2.0 Lean release proves the abstract attained dimension theorem, but does not yet identify this matrix formula with the Python rank implementation.}","The field hypothesis in the section argument is substantive. Over $\\Z$, take $D:\\Z\\to\\Z$ to be multiplication by two. The quotient map $q:\\Z\\to\\Z/2\\Z$ has no $\\Z$-linear section. A section would send the nonzero class to an integer $n$ with $2n=0$, forcing $n=0$ and contradicting $q(n)=1$. This elementary example explains why the module feasibility theorem does not automatically imply a splitting theorem over arbitrary rings.","\\chapter{Finite menus and conservation}\\label{ch:menu}\n\\section{A menu is an actual coefficient map}\nReturn to a commutative ring $\\kk$ and modules $U,V$. Let $e:J\\to V$ be a family of candidate directions and let $S\\subseteq J$ be finite. A coefficient vector is a function $a:S\\to\\kk$. Define\n\\begin{equation}\\label{eq:menu}\nE_S:\\kk^S\\longrightarrow V,\\qquad E_S(a)=\\sum_{j\\in S}a_je_j.\n\\end{equation}\nFinite addition and scalar multiplication show that $E_S$ is linear. The coefficient domain is the selected set itself, so no unselected direction appears in the equation.","ection{A menu is an actual coefficient map}\nReturn to a commutative ring $\\kk$ and modules $U,V$. Let $e:J\\to V$ be a family of candidate directions and let $S\\subseteq J$ be finite. A coefficient vector is a function $a:S\\to\\kk$. Define\n\\begin{equation}\\label{eq:menu}\nE_S:\\kk^S\\longrightarrow V,\\qquad E_S(a)=\\sum_{j\\in S}a_je_j.\n\\end{equation}\nFinite addition and scalar multiplication show that $E_S$ is linear. The coefficient domain is the selected set itself, so no unselected direction appears in the equation.","Write\n\\[\nO_S=\\Span_{\\kk}\\{q_D(e_j):j\\in S\\}.\n\\]","\\begin{lemma}[Quotient image of a menu]\\label{lem:menu-range}\nOne has $\\im(q_DE_S)=O_S$.\n\\end{lemma}\n\\begin{proof}\nFor every coefficient vector $a$,\n\\[\nq_D(E_Sa)=\\sum_{j\\in S}a_jq_D(e_j),\n\\]\nwhich lies in $O_S$. Conversely, every generator $q_D(e_j)$ is the image of the coefficient vector that equals one at $j$ and zero elsewhere. Since the image is a submodule, it contains their span.\n\\end{proof}\n\\formal{menu_quotient_range}","\\begin{theorem}[Exact finite menu feasibility]\\label{thm:menu}\nThe selected menu $S$ repairs the family exactly when\n\\begin{equation}\\label{eq:menu-criterion}\nO_D(r)\\subseteq O_S.\n\\end{equation}\nEquivalently,\n\\[\n\\forall i\\in I\\ \\exists u_i\\in U\\ \\exists a_i\\in\\kk^S,\n\\qquad Du_i+\\sum_{j\\in S}a_{ij}e_j=-r_i.\n\\]\n\\end{theorem}\n\\begin{proof}\nApply \\cref{thm:family} to the actual map $E_S$ and use \\cref{lem:menu-range}. Its reverse direction supplies the original correction witnesses and the selected coefficients, rather than only an abstract statement about dimensions.\n\\end{proof}\n\\formal{finite_menu_repair_iff_span}","\\Needspace{8\\baselineskip}\n\\begin{proposition}[Monotonicity of feasibility]\\label{prop:menu-mono}\nIf $S\\subseteq T$, then $O_S\\subseteq O_T$. Every family repairable with $S$ is therefore repairable with $T$.\n\\end{proposition}\n\\begin{proof}\nEach generator defining $O_S$ is also a generator defining $O_T$. Apply \\cref{thm:menu}. Equivalently, extend each coefficient vector by zero on $T\\setminus S$.\n\\end{proof}\n\\formal{repairable_menu_mono}","\\section{One obstruction, two selected entries}\n\\begin{example}[The difference between the two minima]\\label{ex:menu}\nOver $\\Q$, let\n\\[\nD(u)=(u,0,0),\\qquad r=(0,1,0),\n\\]\nand make only the following directions available:\n\\[\ne_0=(0,1,1),\\qquad e_1=(0,0,1).\n\\]\nThe quotient by $\\im D$ is the plane of the second and third coordinates. There $q_D(r)=(1,0)$, $q_D(e_0)=(1,1)$, and $q_D(e_1)=(0,1)$.\n\\end{example}","Using only $e_0$, the repair equation reads\n\\[\n(u,a,a)=(0,-1,0).\n\\]\nThe second coordinate forces $a=-1$ and the third forces $a=0$, a contradiction. Using only $e_1$, the second coordinate stays zero and cannot become $-1$. Both directions together give the exact witness\n\\begin{equation}\\label{eq:two-entry-witness}\nD(0)-e_0+e_1=(0,-1,0)=-r.\n\\end{equation}\nThe empty menu also fails. Its correction range is the first coordinate axis.","\\begin{center}\n\\begin{tabular}{@{}lll@{}}\n\\toprule\nSelected set & Quotient directions & Repairs $r$?","\\midrule\n$\\varnothing$ & $0$ & No","$\\{0\\}$ & $\\Span\\{(1,1)\\}$ & No","$\\{1\\}$ & $\\Span\\{(0,1)\\}$ & No","$\\{0,1\\}$ & $\\Q^2$ & Yes","The obstruction span has dimension one. A freely chosen direction $r$ repairs the residual with coefficient $-1$. The prescribed menu nevertheless requires two selected entries. Both statements are formalized in the examples module. The difference is not a contradiction: one cost counts independent added dimensions, the other counts named entries that must be purchased or enabled.","A weighting makes the distinction sharper. Assigning positive costs $c_0,c_1$ gives optimum $c_0+c_1$ for this menu. A subspace containing $e_0-e_1$ does not charge those two entries separately unless its cost model explicitly does so.","\\begin{proposition}[Invariant preservation]\\label{prop:invariant}\nLet $L:V\\to Z$ be linear. If $LD=0$ and $LE=0$, then\n\\[\nL(v+Du+Ew)=L(v)\n\\]\nfor all $v,u,w$.\n\\end{proposition}\n\\begin{proof}\nLinearity gives $L(v+Du+Ew)=L(v)+LD(u)+LE(w)$. The last two terms vanish.\n\\end{proof}\n\\formal{invariant_preserved}","\\begin{theorem}[Conservation obstruction]\\label{thm:conservation}\nIf $LD=0$, $LE=0$, and $L(r)\\ne0$, then $E$ does not repair $r$.\n\\end{theorem}\n\\begin{proof}\nA repaired residual would be $r+Du+Ew=0$. Applying \\cref{prop:invariant} would give $L(r)=L(0)=0$, contrary to the hypothesis.\n\\end{proof}\n\\formal{conservation_no_go}","\\begin{corollary}[Conservation for a selected menu]\\label{cor:menu-conservation}\nSuppose $LD=0$ and $L(e_j)=0$ for every $j\\in S$. If $L(r)\\ne0$, no coefficients from $S$ repair $r$.\n\\end{corollary}\n\\begin{proof}\nFor every $a:S\\to\\kk$,\n\\[\nL(E_Sa)=\\sum_{j\\in S}a_jL(e_j)=0.\n\\]\nThus $LE_S=0$, and \\cref{thm:conservation} applies.\n\\end{proof}\n\\formal{conservation_menu_no_go}","For instance, let $D(u)=(u,0,0)$ and permit additional directions only on the third coordinate axis. The invariant $L(x,y,z)=y$ annihilates every legal correction. It detects $r=(0,1,0)$, so no legal repair exists. Enlarging the number of copies of the third coordinate direction cannot help.","The theorem is only as meaningful as the invariant it uses. In a domain application one must establish that $L$ is the intended native invariant and that every permitted transformation really preserves it. An abstract functional with a suggestive name is not such a proof.","\\chapter{Faithful transport of obstructions}\\label{ch:transport}\n\\section{The induced quotient map}\nSuppose two correction systems are related by a commuting square\n\\begin{equation}\\label{eq:square}\n\\begin{tikzcd}[column sep=large,row sep=large]\nU \\arrow[r,\"D\"] \\arrow[d,\"a\"'] & V \\arrow[d,\"b\"]\\\\\nU' \\arrow[r,\"D'\"'] & V'.\n\\end{tikzcd}\n\\qquad bD=D'a.\n\\end{equation}\nAll four spaces are modules over the same commutative ring. Denote the quotient maps by $q_D$ and $q_{D'}$.","U' \\arrow[r,\"D'\"'] & V'.\n\\end{tikzcd}\n\\qquad bD=D'a.\n\\end{equation}\nAll four spaces are modules over the same commutative ring. Denote the quotient maps by $q_D$ and $q_{D'}$.","\\begin{lemma}[The correction image descends]\\label{lem:range-maps}\nThe map $b$ sends $\\im D$ into $\\im D'$.\n\\end{lemma}\n\\begin{proof}\nIf $v=Du$, then $b(v)=bD(u)=D'a(u)$, which lies in $\\im D'$.\n\\end{proof}\n\\formal{correction_range_maps}","It follows that there is a well defined linear map\n\\[\nB:Q_D\\longrightarrow Q_{D'},\\qquad B(q_D(v))=q_{D'}(b(v)).\n\\]\nIndeed, changing $v$ by an element of $\\im D$ changes $b(v)$ by an element of $\\im D'$. This is the quotient map constructed in the formal source, not an assumed function on unidentified obstruction spaces.","\\begin{proposition}[Naturality]\\label{prop:naturality}\nFor every residual $r\\in V$,\n\\begin{equation}\\label{eq:naturality}\nB(q_D(r))=q_{D'}(b(r)).\n\\end{equation}\n\\end{proposition}\n\\begin{proof}\nThis is the defining equation of the induced quotient map.\n\\end{proof}\n\\formal{transport_obstruction_naturality}","Naturality transports a correctable residual to a correctable one. The converse can fail because $B$ may annihilate a nonzero class. A second condition is needed to use target vanishing as a source certificate.","\\section{Faithfulness on precisely the required data}\n\\begin{definition}[Faithfulness on a submodule]\\label{def:faithful}\nFor $O\\leq Q_D$, the map $B$ is \\emph{faithful on $O$} when\n\\[\n\\forall z\\in O,\\qquad Bz=0\\Longrightarrow z=0.\n\\]\n\\end{definition}","This condition is equivalent to $\\ker B\\cap O=0$. It does not require injectivity on all of $Q_D$ when only a smaller family of obstruction classes matters.","\\begin{proposition}[Equivalent injectivity criterion]\\label{prop:faithful-inj}\nThe map $B$ is faithful on $O$ if and only if its restriction $B\\vert_O:O\\to Q_{D'}$ is injective.\n\\end{proposition}\n\\begin{proof}\nIf $B$ is faithful and $Bx=By$ for $x,y\\in O$, then $x-y\\in O$ and $B(x-y)=0$. Faithfulness gives $x=y$. Conversely, if $B\\vert_O$ is injective and $Bz=0=B0$ for $z\\in O$, injectivity gives $z=0$.\n\\end{proof}\n\\formal{faithfulOn_iff_injective}","begin{proposition}[Equivalent injectivity criterion]\\label{prop:faithful-inj}\nThe map $B$ is faithful on $O$ if and only if its restriction $B\\vert_O:O\\to Q_{D'}$ is injective.\n\\end{proposition}\n\\begin{proof}\nIf $B$ is faithful and $Bx=By$ for $x,y\\in O$, then $x-y\\in O$ and $B(x-y)=0$. Faithfulness gives $x=y$. Conversely, if $B\\vert_O$ is injective and $Bz=0=B0$ for $z\\in O$, injectivity gives $z=0$.\n\\end{proof}\n\\formal{faithfulOn_iff_injective}","\\begin{theorem}[Reflection of correctability]\\label{thm:faithful}\nLet $O=O_D(r)$. If $B$ is faithful on $O$, then for every $i\\in I$,\n\\[\nb(r_i)\\text{ is correctable by }D'\n\\quad\\Longleftrightarrow\\quad\nr_i\\text{ is correctable by }D.\n\\]\n\\end{theorem}\n\\begin{proof}\nBy \\cref{thm:correctable}, target correctability means $q_{D'}(b(r_i))=0$. Naturality identifies this class with $B(q_D(r_i))$. Because $q_D(r_i)\\in O$, faithfulness then yields $q_D(r_i)=0$, which is source correctability. Conversely, if $q_D(r_i)=0$, naturality and linearity give $q_{D'}(b(r_i))=B0=0$.\n\\end{proof}\n\\formal{transport_faithfulness_on_obstruction_span}","egin{theorem}[Reflection of correctability]\\label{thm:faithful}\nLet $O=O_D(r)$. If $B$ is faithful on $O$, then for every $i\\in I$,\n\\[\nb(r_i)\\text{ is correctable by }D'\n\\quad\\Longleftrightarrow\\quad\nr_i\\text{ is correctable by }D.\n\\]\n\\end{theorem}\n\\begin{proof}\nBy \\cref{thm:correctable}, target correctability means $q_{D'}(b(r_i))=0$. Naturality identifies this class with $B(q_D(r_i))$. Because $q_D(r_i)\\in O$, faithfulness then yields $q_D(r_i)=0$, which is source correctability. Conversely, if $q_D(r_i)=0$, naturality and linearity give $q_{D'}(b(r_i))=B0=0$.\n\\end{proof}\n\\formal{transport_faithfulness_on_obstruction_span}","]\\label{thm:faithful}\nLet $O=O_D(r)$. If $B$ is faithful on $O$, then for every $i\\in I$,\n\\[\nb(r_i)\\text{ is correctable by }D'\n\\quad\\Longleftrightarrow\\quad\nr_i\\text{ is correctable by }D.\n\\]\n\\end{theorem}\n\\begin{proof}\nBy \\cref{thm:correctable}, target correctability means $q_{D'}(b(r_i))=0$. Naturality identifies this class with $B(q_D(r_i))$. Because $q_D(r_i)\\in O$, faithfulness then yields $q_D(r_i)=0$, which is source correctability. Conversely, if $q_D(r_i)=0$, naturality and linearity give $q_{D'}(b(r_i))=B0=0$.\n\\end{proof}\n\\formal{transport_faithfulness_on_obstruction_span}","O$, then for every $i\\in I$,\n\\[\nb(r_i)\\text{ is correctable by }D'\n\\quad\\Longleftrightarrow\\quad\nr_i\\text{ is correctable by }D.\n\\]\n\\end{theorem}\n\\begin{proof}\nBy \\cref{thm:correctable}, target correctability means $q_{D'}(b(r_i))=0$. Naturality identifies this class with $B(q_D(r_i))$. Because $q_D(r_i)\\in O$, faithfulness then yields $q_D(r_i)=0$, which is source correctability. Conversely, if $q_D(r_i)=0$, naturality and linearity give $q_{D'}(b(r_i))=B0=0$.\n\\end{proof}\n\\formal{transport_faithfulness_on_obstruction_span}","is correctable by }D'\n\\quad\\Longleftrightarrow\\quad\nr_i\\text{ is correctable by }D.\n\\]\n\\end{theorem}\n\\begin{proof}\nBy \\cref{thm:correctable}, target correctability means $q_{D'}(b(r_i))=0$. Naturality identifies this class with $B(q_D(r_i))$. Because $q_D(r_i)\\in O$, faithfulness then yields $q_D(r_i)=0$, which is source correctability. Conversely, if $q_D(r_i)=0$, naturality and linearity give $q_{D'}(b(r_i))=B0=0$.\n\\end{proof}\n\\formal{transport_faithfulness_on_obstruction_span}","\\begin{proposition}[A witness to information loss]\\label{prop:loss-witness}\nIf $B$ is not faithful on $O$, there is $z\\in O$ such that\n\\[\nz\\ne0,\\qquad Bz=0.\n\\]\n\\end{proposition}\n\\begin{proof}\nNegating the quantified implication in \\cref{def:faithful} gives an element of $O$ whose image is zero and which is itself nonzero.\n\\end{proof}\n\\formal{nonfaithful_witness}","\\section{A transport that conceals failure}\n\\begin{example}[Correct in the target, false in the source]\\label{ex:loss}\nLet\n\\[\nD:\\Q\\to\\Q^2,\\quad D(u)=(u,0),\\qquad\nD'=\\id_{\\Q},\\quad a=\\id_{\\Q},\\quad b(x,y)=x.\n\\]\nThen $bD=D'a$. For $r=(0,1)$ there is no $u$ with $(u,0)=(0,-1)$, so the source residual is not correctable. Its image $b(r)=0$ is correctable by $D'$.\n\\end{example}","Here $Q_D$ is one dimensional and $Q_{D'}=0$. The induced map $B$ annihilates every source obstruction. It is therefore not faithful on the span of $q_D(r)$. The formal example proves both the false positive produced by projection and the failure of the faithfulness hypothesis.","A second elementary calculation clarifies why faithfulness is imposed on a span. The map $B:\\Q^2\\to\\Q$, $B(x,y)=x+y$, sends each standard basis vector to the nonzero number one. Nevertheless, it kills their difference $(1,-1)$. Survival of generators is not injectivity on their span. This calculation is explanatory and has no separate declaration in the present formal source.","Assume $\\kk$, $U$, $I$, and $J$ have specified finite enumerations. The ring $\\kk$ is commutative, $U,V$ are $\\kk$-modules, and equality in $V$ is decidable. The target module $V$ need not itself have a finite enumeration. Fix a linear order on $J$ so that the candidate universe can be enumerated in a definite order.","For each finite subset $S\\subseteq J$, define the Boolean test\n\\begin{equation}\\label{eq:boolean-check}\n\\operatorname{check}(S)=\\operatorname{decide}\\left(\n\\forall i\\in I\\ \\exists u\\in U\\ \\exists a\\in\\kk^S,\\quad Du+E_Sa=-r_i\n\\right).\n\\end{equation}\nAll quantified domains are finite. Decidable equality in $V$ determines each candidate equation.","\\begin{proposition}[The Boolean test has the intended meaning]\\label{prop:check}\nThe equality $\\operatorname{check}(S)=\\mathrm{true}$ holds if and only if $E_S$ repairs the residual family.\n\\end{proposition}\n\\begin{proof}\nThe Boolean is defined by deciding precisely the quantified equations in \\cref{def:repair} for the map $E_S$. No intermediate rank guess or unproved feasibility oracle replaces those equations.\n\\end{proof}\n\\formal{repairMenuCheck_spec}","Additional admissibility is a supplied function $\\operatorname{allowed}:\\mathcal P(J)\\to\\{\\mathrm{false},\\mathrm{true}\\}$. The cost is any supplied function $c:\\mathcal P(J)\\to\\N$. The formal theorem does not require $c$ to be additive in the selected entries.","\\section{Enumerating menus and selecting an optimum}\nEnumerate $J$ as a sorted list. Take every sublist and convert it to a finite set. The resulting list, denoted $\\mathcal L_J$, contains every subset of $J$.","\\begin{lemma}[Coverage of the menu universe]\\label{lem:allmenus}\nEvery finite subset $S\\subseteq J$ occurs in $\\mathcal L_J$.\n\\end{lemma}\n\\begin{proof}\nFilter the sorted full list by membership in $S$. The result is a sublist of the full list and therefore appears among its sublists. Converting this filtered list to a finite set gives exactly $S$.\n\\end{proof}\n\\formal{mem_allMenus}","For a finite list $\\mathcal L$ of candidates, a Boolean predicate $h$, and a natural cost $c$, define $\\operatorname{bestByCheck}(h,c,\\mathcal L)$ to filter by $h$ and then return an argument of minimum cost. It returns \\code{none} exactly when the filtered list is empty. Ties need not define a mathematically unique optimum.","\\begin{lemma}[Finite selection]\\label{lem:best}\nIf this selector returns $x$, then $x\\in\\mathcal L$, $h(x)=\\mathrm{true}$, and $c(x)\\leq c(y)$ for every $y\\in\\mathcal L$ with $h(y)=\\mathrm{true}$. It returns \\code{none} if and only if no listed candidate passes $h$.\n\\end{lemma}\n\\begin{proof}\nEvery element of the filtered list is an original list member satisfying $h$. A minimum cost element of that finite list has cost no larger than that of any other member. The selector returns no element exactly when the filtered list is empty, which is exactly the stated rejection condition.\n\\end{proof}\n\\formal{bestByCheck_spec}","The repair optimizer is\n\\[\n\\operatorname{bestRepairMenu}\n=\\operatorname{bestByCheck}\\bigl(\nS\\mapsto\\operatorname{allowed}(S)\\wedge\\operatorname{check}(S),\\ c,\\ \\mathcal L_J\\bigr).\n\\]","\\begin{theorem}[Soundness and minimum cost]\\label{thm:optimizer}\nIf the optimizer returns \\code{some S}, then $S$ is allowed, $E_S$ repairs the family, and\n\\[\nc(S)\\leq c(T)\n\\]\nfor every allowed subset $T\\subseteq J$ whose menu map repairs the family.\n\\end{theorem}\n\\begin{proof}\nApply \\cref{lem:best}. Passing the conjunction gives both admissibility and $\\operatorname{check}(S)=\\mathrm{true}$. Proposition~\\ref{prop:check} converts the latter into actual repair feasibility. Every competing $T$ occurs in $\\mathcal L_J$ by \\cref{lem:allmenus}, so the same finite selection lemma compares its cost to $c(S)$.\n\\end{proof}\n\\formal{menu_optimality_certificate}","e{some S}, then $S$ is allowed, $E_S$ repairs the family, and\n\\[\nc(S)\\leq c(T)\n\\]\nfor every allowed subset $T\\subseteq J$ whose menu map repairs the family.\n\\end{theorem}\n\\begin{proof}\nApply \\cref{lem:best}. Passing the conjunction gives both admissibility and $\\operatorname{check}(S)=\\mathrm{true}$. Proposition~\\ref{prop:check} converts the latter into actual repair feasibility. Every competing $T$ occurs in $\\mathcal L_J$ by \\cref{lem:allmenus}, so the same finite selection lemma compares its cost to $c(S)$.\n\\end{proof}\n\\formal{menu_optimality_certificate}","n $S$ is allowed, $E_S$ repairs the family, and\n\\[\nc(S)\\leq c(T)\n\\]\nfor every allowed subset $T\\subseteq J$ whose menu map repairs the family.\n\\end{theorem}\n\\begin{proof}\nApply \\cref{lem:best}. Passing the conjunction gives both admissibility and $\\operatorname{check}(S)=\\mathrm{true}$. Proposition~\\ref{prop:check} converts the latter into actual repair feasibility. Every competing $T$ occurs in $\\mathcal L_J$ by \\cref{lem:allmenus}, so the same finite selection lemma compares its cost to $c(S)$.\n\\end{proof}\n\\formal{menu_optimality_certificate}","\\subseteq J$ whose menu map repairs the family.\n\\end{theorem}\n\\begin{proof}\nApply \\cref{lem:best}. Passing the conjunction gives both admissibility and $\\operatorname{check}(S)=\\mathrm{true}$. Proposition~\\ref{prop:check} converts the latter into actual repair feasibility. Every competing $T$ occurs in $\\mathcal L_J$ by \\cref{lem:allmenus}, so the same finite selection lemma compares its cost to $c(S)$.\n\\end{proof}\n\\formal{menu_optimality_certificate}","\\begin{theorem}[Complete rejection within the declared universe]\\label{thm:none}\nThe optimizer returns \\code{none} if and only if no allowed subset of $J$ repairs the family.\n\\end{theorem}\n\\begin{proof}\nIf it returns \\code{none}, \\cref{lem:best} excludes every passing member of $\\mathcal L_J$. Any allowed feasible subset would occur there by \\cref{lem:allmenus} and pass the Boolean test by \\cref{prop:check}, a contradiction. Conversely, if no allowed subset repairs the family, no listed subset satisfies the filter, so the selector returns \\code{none}.\n\\end{proof}\n\\formal{menu_search_none_iff}","\\section{A fully evaluated example over two elements}\\label{sec:f2}\nTake the data of \\cref{ex:menu} over $\\F_2$. Negation is now the identity. The equation is\n\\[\n(u,a,a+b)=(0,1,0).\n\\]\nThus the full menu succeeds with $u=0$, $a=b=1$. The two singleton menus still fail.","\\midrule\n$\\varnothing$ & 0 & Infeasible & Infeasible","$\\{0\\}$ & 1 & Infeasible & Infeasible","$\\{1\\}$ & 1 & Infeasible & Infeasible","$\\{0,1\\}$ & 2 & Feasible & Not allowed","\\section{What the algorithm does not promise}\nThe reference solver is deliberately finite and exhaustive. With $t=|J|$, there are $2^t$ candidate subsets. For a subset of size $s$, each residual has at most $|U|\\,|\\kk|^s$ candidate coefficient and correction assignments to examine. Summing this upper bound over all subsets gives\n\\[\n|I|\\,|U|\\sum_{s=0}^t\\binom ts|\\kk|^s\n=|I|\\,|U|(1+|\\kk|)^t\n\\]\npossible equation tests before accounting for short circuiting and inadmissible subsets. This is an elementary enumeration bound, not a formal runtime theorem about the compiled Lean implementation.","The results in this chapter are first stated over $\\Z$. An integer $p$ need not be prime for the finite divisibility implications. The interpretation as a $p$-adic lift uses a positive prime $p$.","\\begin{lemma}[A linear digit from B\\'ezout data]\\label{lem:digit}\nLet $p,A,c\\in\\Z$. If there are $s,b\\in\\Z$ with $sA+bp=1$, then there is $t\\in\\Z$ such that\n\\[\np\\mid c+At.\n\\]\n\\end{lemma}\n\\begin{proof}\nSet $t=-sc$. Then\n\\[\nc+At=c(1-As)=pbc.\\qedhere\n\\]\n\\end{proof}\n\\formal{Arithmetic.linear_digit_exists}","\\begin{lemma}[Quadratic correction identity]\\label{lem:quad}\nLet $p,m,x,a,c,t\\in\\Z$ satisfy\n\\[\nx^2-a=mc,\\qquad p\\mid c+2xt,\\qquad p\\mid m.\n\\]\nThen\n\\[\nmp\\mid (x+mt)^2-a.\n\\]\n\\end{lemma}\n\\begin{proof}\nWrite $c+2xt=pd$ and $m=ps$. Direct expansion gives\n\\begin{align*}\n(x+mt)^2-a\n&=(x^2-a)+2mxt+m^2t^2","\\begin{lemma}[Quadratic correction identity]\\label{lem:quad}\nLet $p,m,x,a,c,t\\in\\Z$ satisfy\n\\[\nx^2-a=mc,\\qquad p\\mid c+2xt,\\qquad p\\mid m.\n\\]\nThen\n\\[\nmp\\mid (x+mt)^2-a.\n\\]\n\\end{lemma}\n\\begin{proof}\nWrite $c+2xt=pd$ and $m=ps$. Direct expansion gives\n\\begin{align*}\n(x+mt)^2-a\n&=(x^2-a)+2mxt+m^2t^2\\\\\n&=m(c+2xt)+m^2t^2\\\\\n&=mp(d+st^2).\\qedhere\n\\end{align*}\nThe last expression displays the required quotient explicitly.\n\\end{proof}\n\\formal{Arithmetic.quadratic_digit_correction}","The term $m^2t^2$ is not discarded as an approximation. Its divisibility follows from $p\\mid m$. This is what makes a linear digit equation sufficient for the finite quadratic step.","\\section{The quadratic lifting theorem}\n\\begin{theorem}[One higher precision]\\label{thm:hensel}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","tion{The quadratic lifting theorem}\n\\begin{theorem}[One higher precision]\\label{thm:hensel}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","quadratic lifting theorem}\n\\begin{theorem}[One higher precision]\\label{thm:hensel}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","in{theorem}[One higher precision]\\label{thm:hensel}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","m}[One higher precision]\\label{thm:hensel}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","on]\\label{thm:hensel}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","el}\nLet $p,x,a\\in\\Z$ and $k\\in\\N$ with $k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","k\\geq1$. Assume\n\\[\np^k\\mid x^2-a,\\qquad \\gcd(2x,p)=1.\n\\]\nThen there exists $t\\in\\Z$ such that\n\\begin{equation}\\label{eq:hensel}\np^{k+1}\\mid (x+p^kt)^2-a.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nChoose $c\\in\\Z$ with $x^2-a=p^kc$. Coprimality supplies B\\'ezout data for $2x$ and $p$, so \\cref{lem:digit} gives $t$ with $p\\mid c+2xt$. Since $k\\geq1$, one has $p\\mid p^k$. Apply \\cref{lem:quad} with $m=p^k$.\n\\end{proof}\n\\formal{Arithmetic.quadratic_hensel_step}","The formal hypothesis is \\code{IsCoprime (2 * x) p}, expressed by B\\'ezout data. The theorem constructs an integer representative of a correction class. It does not state that this representative is already the unique digit between zero and $p-1$, nor that an infinite succession of such choices has been constructed in a complete field.","\\section{An exact seven-adic trace}\\label{sec:trace}\nFor $x^2=2$ at $p=7$, start with $x_1=3$. At each step, let\n\\[\nc_k=\\frac{x_k^2-2}{7^k},\\qquad\nc_k+2x_kt_k\\equiv0\\pmod7,\\qquad\nx_{k+1}=x_k+7^kt_k.\n\\]\nThe following table records normalized choices for this example. Every displayed digit equation is an exact integer calculation.","\\begin{center}\n\\begin{tabular}{@{}rrrrr@{}}\n\\toprule\n$k$ & $x_k$ & $c_k\\bmod7$ & $t_k$ & $x_{k+1}$","All representatives are congruent to three modulo seven, so $2x_k\\equiv6\\pmod7$ remains a unit. The formal statement \\decl{Arithmetic.hensel_seven_trace} verifies the seven congruences\n\\[\n7^k\\mid x_k^2-2\\qquad (1\\leq k\\leq7)\n\\]\nfor the displayed representatives. It is a finite conjunction of exact divisibility facts. The general one step theorem is proved separately.","For example,\n\\[\n10^2-2=98=7^2\\cdot2,\n\\]\nand the next digit satisfies $2+2\\cdot10\\cdot2=42$. Hence\n\\[\n108=10+7^2\\cdot2,\n\\qquad 108^2-2=7^3\\cdot34.\n\\]\nThe new digit corrects the residual quotient at precisely the next required modulus.","\\section{Why a singular derivative is inconclusive}\n\\begin{proposition}[A modular solution that does not lift]\\label{prop:mod4}\nThe equation $x^2=2$ has no solution in $\\Z/4\\Z$.\n\\end{proposition}\n\\begin{proof}\nThe four residue classes have squares $0,1,0,1$, respectively. None has square two.\n\\end{proof}\n\\formal{Arithmetic.singular_mod_four_impossible}","Nevertheless, $x=0$ solves the equation modulo two. Its two possible lifts modulo four are zero and two, and both fail. Thus existence at one precision is not itself an existence theorem at the next.","\\begin{proposition}[Singularity does not prohibit lifting]\\label{prop:singular-success}\nFor every $k\\in\\N$,\n\\[\n2^k\\mid 1^2-1.\n\\]\n\\end{proposition}\n\\begin{proof}\nThe residual is zero, which is divisible by every integer.\n\\end{proof}\n\\formal{Arithmetic.singular_derivative_can_still_lift}","For the equation $x^2-1=0$, the derivative at $x=1$ is two and is not a unit modulo two. The exact root still persists at every dyadic precision. A unit derivative is a sufficient hypothesis for \\cref{thm:hensel}, not a necessary condition for every possible lift.","\\chapter{Nonlinear limits and native realizations}\\label{ch:limits}\n\\section{A linearized problem is a different equation}\nSuppose a nonlinear map $F$ is differentiable at an approximation $x_0$. Its linearized correction equation is\n\\begin{equation}\\label{eq:linearization}\nDF(x_0)u=-F(x_0).\n\\end{equation}\nThe cokernel criterion completely decides \\eqref{eq:linearization} once the derivative is fixed. It does not, by itself, decide whether there is $x$ with $F(x)=0$. Higher order terms are absent from the linearized equation, and they may either create a solution or prevent the linear prediction from reaching one.","el{ch:limits}\n\\section{A linearized problem is a different equation}\nSuppose a nonlinear map $F$ is differentiable at an approximation $x_0$. Its linearized correction equation is\n\\begin{equation}\\label{eq:linearization}\nDF(x_0)u=-F(x_0).\n\\end{equation}\nThe cokernel criterion completely decides \\eqref{eq:linearization} once the derivative is fixed. It does not, by itself, decide whether there is $x$ with $F(x)=0$. Higher order terms are absent from the linearized equation, and they may either create a solution or prevent the linear prediction from reaching one.","its}\n\\section{A linearized problem is a different equation}\nSuppose a nonlinear map $F$ is differentiable at an approximation $x_0$. Its linearized correction equation is\n\\begin{equation}\\label{eq:linearization}\nDF(x_0)u=-F(x_0).\n\\end{equation}\nThe cokernel criterion completely decides \\eqref{eq:linearization} once the derivative is fixed. It does not, by itself, decide whether there is $x$ with $F(x)=0$. Higher order terms are absent from the linearized equation, and they may either create a solution or prevent the linear prediction from reaching one.","\\begin{proposition}[A nonlinear root with a nonzero first order obstruction]\\label{prop:nonlinear-root}\nFor $a\\in\\R\\setminus\\{0\\}$, define $F_a(x)=x^2-a^2$. At zero,\n\\[\nF_a'(0)=0,\n\\]\nand the correction equation $0\\cdot u=a^2$ has no solution. Nevertheless, $F_a(a)=0$.\n\\end{proposition}\n\\begin{proof}\nDifferentiation gives $F_a'(x)=2x$, so the derivative at zero is zero. Since $a\\ne0$, its square is nonzero, while $0\\cdot u=0$ for every $u$. Finally, $F_a(a)=a^2-a^2=0$.\n\\end{proof}\n\\formal{Nonlinear.nonlinear_root_despite_first_order_obstruction}","n}[A nonlinear root with a nonzero first order obstruction]\\label{prop:nonlinear-root}\nFor $a\\in\\R\\setminus\\{0\\}$, define $F_a(x)=x^2-a^2$. At zero,\n\\[\nF_a'(0)=0,\n\\]\nand the correction equation $0\\cdot u=a^2$ has no solution. Nevertheless, $F_a(a)=0$.\n\\end{proposition}\n\\begin{proof}\nDifferentiation gives $F_a'(x)=2x$, so the derivative at zero is zero. Since $a\\ne0$, its square is nonzero, while $0\\cdot u=0$ for every $u$. Finally, $F_a(a)=a^2-a^2=0$.\n\\end{proof}\n\\formal{Nonlinear.nonlinear_root_despite_first_order_obstruction}","h a nonzero first order obstruction]\\label{prop:nonlinear-root}\nFor $a\\in\\R\\setminus\\{0\\}$, define $F_a(x)=x^2-a^2$. At zero,\n\\[\nF_a'(0)=0,\n\\]\nand the correction equation $0\\cdot u=a^2$ has no solution. Nevertheless, $F_a(a)=0$.\n\\end{proposition}\n\\begin{proof}\nDifferentiation gives $F_a'(x)=2x$, so the derivative at zero is zero. Since $a\\ne0$, its square is nonzero, while $0\\cdot u=0$ for every $u$. Finally, $F_a(a)=a^2-a^2=0$.\n\\end{proof}\n\\formal{Nonlinear.nonlinear_root_despite_first_order_obstruction}","rst order obstruction]\\label{prop:nonlinear-root}\nFor $a\\in\\R\\setminus\\{0\\}$, define $F_a(x)=x^2-a^2$. At zero,\n\\[\nF_a'(0)=0,\n\\]\nand the correction equation $0\\cdot u=a^2$ has no solution. Nevertheless, $F_a(a)=0$.\n\\end{proposition}\n\\begin{proof}\nDifferentiation gives $F_a'(x)=2x$, so the derivative at zero is zero. Since $a\\ne0$, its square is nonzero, while $0\\cdot u=0$ for every $u$. Finally, $F_a(a)=a^2-a^2=0$.\n\\end{proof}\n\\formal{Nonlinear.nonlinear_root_despite_first_order_obstruction}","The root can lie arbitrarily close to zero by choosing a small nonzero $a$. This produces a family of nearby counterexamples. It does not assert that one fixed equation with a nonzero residual at zero has roots converging to zero.","\\begin{proposition}[A solvable linearization without a real root]\\label{prop:no-real-root}\nLet $G(x)=x^2+x+1$. Then\n\\[\nG'(0)=1,\\qquad \\exists u\\in\\R\\ (u=-1),\n\\]\nso the linearized correction at zero is solvable. However, $G$ has no real root.\n\\end{proposition}\n\\begin{proof}\nThe derivative is $2x+1$, which equals one at zero, and $G(0)=1$. Thus $u=-1$ solves the linearized equation. But\n\\[\nG(x)=\\left(x+\\frac12\\right)^2+\\frac34>0\n\\]\nfor every real $x$.\n\\end{proof}\n\\formal{Nonlinear.linearized_solution_without_real_root}","ut a real root]\\label{prop:no-real-root}\nLet $G(x)=x^2+x+1$. Then\n\\[\nG'(0)=1,\\qquad \\exists u\\in\\R\\ (u=-1),\n\\]\nso the linearized correction at zero is solvable. However, $G$ has no real root.\n\\end{proposition}\n\\begin{proof}\nThe derivative is $2x+1$, which equals one at zero, and $G(0)=1$. Thus $u=-1$ solves the linearized equation. But\n\\[\nG(x)=\\left(x+\\frac12\\right)^2+\\frac34>0\n\\]\nfor every real $x$.\n\\end{proof}\n\\formal{Nonlinear.linearized_solution_without_real_root}","\\textbf{An exact linear repair} provides witnesses for $Du+Ew=-r$ in the original target. Its validity is governed by quotient image membership.","Constructive repair fits that criterion in a direct way. The native equation supplies $V$ and its residuals. The allowed transformations supply $D$ and the permissible class of added maps. The quotient organizes which residual classes those maps can remove. A transport that is used to decide the problem must preserve the necessary obstruction data.","Fix matrices\n\\[\n D\\in\\Q^{m\\times n},\\qquad R\\in\\Q^{m\\times s},\\qquad E\\in\\Q^{m\\times q}.\n\\]\nThe residuals are the columns of $R$. For a selected menu $S$, let $\\widetilde E_S$ retain the original candidate indexing and replace unselected columns by zero. The supplied coefficient matrices have shapes $U\\in\\Q^{n\\times s}$ and $W\\in\\Q^{q\\times s}$. Their correction equation is\n\\begin{equation}\\label{eq:rational-original}\n DU+\\widetilde E_SW=-R.\n\\end{equation}\nKeeping the original indexing avoids any ambiguity about a permutation of candidate columns. Coefficients on a zeroed column have no effect. The admissibility predicate and cost function belong to the original problem, not to the certificate producer.","\\begin{proposition}[Successful certificate]\\label{prop:rational-success}\nIf an admissible $S$ and rational matrices $U,W$ satisfy \\eqref{eq:rational-original}, they repair every original residual.\n\\end{proposition}\n\\begin{proof}\nTake the $i$th column of the displayed identity. It is the original equation $Du_i+\\sum_{j\\in S}w_{ji}e_j=-r_i$.\n\\end{proof}\n\\formal{RationalCertificates.checkRepair_sound}","\\section{Separating obstructions}\n\\begin{definition}[Separating row]\nA separating certificate for $S$ is a row $Y\\in\\Q^{1\\times m}$ satisfying\n\\begin{equation}\\label{eq:rational-separator}\n YD=0,\\qquad Y\\widetilde E_S=0,\\qquad YR\\ne0.\n\\end{equation}\nThe last condition asks that at least one original residual is detected. No residual index needs to be selected when the family is empty.\n\\end{definition}","\\begin{theorem}[Sound separation]\\label{thm:rational-separation}\nA separating certificate rules out a repair by the selected menu.\n\\end{theorem}\n\\begin{proof}\nMultiplying a supposed correction equation on the left by $Y$ gives $0=-YR$, contradicting \\eqref{eq:rational-separator}.\n\\end{proof}\n\\formal{RationalCertificates.annihilator_kills_residual}\n\\formal{RationalCertificates.checkSeparator_sound}","\\begin{proposition}[Linear alternative]\\label{prop:rational-alternative}\nFor a fixed rational menu, either the correction matrices exist or a separating row exists, but not both.\n\\end{proposition}\n\\begin{proof}\nExclusivity follows from the preceding theorem. If no correction matrices exist, some residual $r_i$ lies outside the column space of $[D\\ \\widetilde E_S]$. Choose a basis for that space, adjoin $r_i$, and extend to a basis of $\\Q^m$. The functional taking value one on $r_i$ and zero on all other chosen basis vectors is a rational separating row.\n\\end{proof}\n\\sourcescope{This completeness argument is an ordinary proof. The new Lean module proves certificate soundness, not a general algorithm or formal theorem producing a separator for every infeasible rational menu.}","proposition}[Linear alternative]\\label{prop:rational-alternative}\nFor a fixed rational menu, either the correction matrices exist or a separating row exists, but not both.\n\\end{proposition}\n\\begin{proof}\nExclusivity follows from the preceding theorem. If no correction matrices exist, some residual $r_i$ lies outside the column space of $[D\\ \\widetilde E_S]$. Choose a basis for that space, adjoin $r_i$, and extend to a basis of $\\Q^m$. The functional taking value one on $r_i$ and zero on all other chosen basis vectors is a rational separating row.\n\\end{proof}\n\\sourcescope{This completeness argument is an ordinary proof. The new Lean module proves certificate soundness, not a general algorithm or formal theorem producing a separator for every infeasible rational menu.}","\\label{prop:rational-alternative}\nFor a fixed rational menu, either the correction matrices exist or a separating row exists, but not both.\n\\end{proposition}\n\\begin{proof}\nExclusivity follows from the preceding theorem. If no correction matrices exist, some residual $r_i$ lies outside the column space of $[D\\ \\widetilde E_S]$. Choose a basis for that space, adjoin $r_i$, and extend to a basis of $\\Q^m$. The functional taking value one on $r_i$ and zero on all other chosen basis vectors is a rational separating row.\n\\end{proof}\n\\sourcescope{This completeness argument is an ordinary proof. The new Lean module proves certificate soundness, not a general algorithm or formal theorem producing a separator for every infeasible rational menu.}","ed rational menu, either the correction matrices exist or a separating row exists, but not both.\n\\end{proposition}\n\\begin{proof}\nExclusivity follows from the preceding theorem. If no correction matrices exist, some residual $r_i$ lies outside the column space of $[D\\ \\widetilde E_S]$. Choose a basis for that space, adjoin $r_i$, and extend to a basis of $\\Q^m$. The functional taking value one on $r_i$ and zero on all other chosen basis vectors is a rational separating row.\n\\end{proof}\n\\sourcescope{This completeness argument is an ordinary proof. The new Lean module proves certificate soundness, not a general algorithm or formal theorem producing a separator for every infeasible rational menu.}","\\section{Dimension certificates without a reported rank}\nSuppose an augmentation with $t$ columns succeeds. Its optimality can be certified by two additional matrices\n\\[\n H\\in\\Q^{t\\times m},\\qquad Q\\in\\Q^{s\\times t}\n\\]\nwith\n\\begin{equation}\\label{eq:rational-dimension}\n HD=0,\\qquad HRQ=I_t.\n\\end{equation}\nThese identities make a lower bound on the required dimension explicit.","\\begin{theorem}[Dimension lower certificate]\\label{thm:rational-lower}\nAssume \\eqref{eq:rational-dimension}. If an arbitrary competing augmentation $E'\\in\\Q^{m\\times w}$ has coefficients satisfying $DU'+E'W'=-R$, then $t\\leq w$.\n\\end{theorem}\n\\begin{proof}\nMultiplication by $H$ and $Q$ gives $HE'W'Q=-I_t$, hence\n\\[\n (HE')(-W'Q)=I_t.\n\\]\nThus $HE':\\Q^w\\to\\Q^t$ has a right inverse. It is surjective, and consequently $t\\leq w$.\n\\end{proof}\n\\formal{RationalCertificates.dimension_certificate_right_inverse}\n\\formal{RationalCertificates.dimension_certificate_lower_bound}","mension lower certificate]\\label{thm:rational-lower}\nAssume \\eqref{eq:rational-dimension}. If an arbitrary competing augmentation $E'\\in\\Q^{m\\times w}$ has coefficients satisfying $DU'+E'W'=-R$, then $t\\leq w$.\n\\end{theorem}\n\\begin{proof}\nMultiplication by $H$ and $Q$ gives $HE'W'Q=-I_t$, hence\n\\[\n (HE')(-W'Q)=I_t.\n\\]\nThus $HE':\\Q^w\\to\\Q^t$ has a right inverse. It is surjective, and consequently $t\\leq w$.\n\\end{proof}\n\\formal{RationalCertificates.dimension_certificate_right_inverse}\n\\formal{RationalCertificates.dimension_certificate_lower_bound}","A successful augmentation with $t$ columns attains this bound. The verifier checks the original repair identity and \\eqref{eq:rational-dimension}. It does not trust a numerical rank supplied by the search program.\n\\formal{RationalCertificates.checkDimension_sound}","\\begin{proposition}[Availability of lower certificates]\nLet $O$ be the span of the residual classes in $\\Q^m/\\im D$, and let $t=\\dim O$. Then matrices $H,Q$ satisfying \\eqref{eq:rational-dimension} exist.\n\\end{proposition}\n\\begin{proof}\nChoose coordinates $O\\cong\\Q^t$, extend that map to the whole quotient, and compose with the quotient map to obtain $H$. Then $HD=0$. The columns of $HR$ span $\\Q^t$, so $HR$ has a linear right inverse $Q$.\n\\end{proof}\n\\sourcescope{The availability argument is an ordinary finite-dimensional proof. The concrete producer in the next chapter finds and exports such data. Its search implementation is not itself proved correct in Lean.}","roposition}[Availability of lower certificates]\nLet $O$ be the span of the residual classes in $\\Q^m/\\im D$, and let $t=\\dim O$. Then matrices $H,Q$ satisfying \\eqref{eq:rational-dimension} exist.\n\\end{proposition}\n\\begin{proof}\nChoose coordinates $O\\cong\\Q^t$, extend that map to the whole quotient, and compose with the quotient map to obtain $H$. Then $HD=0$. The columns of $HR$ span $\\Q^t$, so $HR$ has a linear right inverse $Q$.\n\\end{proof}\n\\sourcescope{The availability argument is an ordinary finite-dimensional proof. The concrete producer in the next chapter finds and exports such data. Its search implementation is not itself proved correct in Lean.}","r certificates]\nLet $O$ be the span of the residual classes in $\\Q^m/\\im D$, and let $t=\\dim O$. Then matrices $H,Q$ satisfying \\eqref{eq:rational-dimension} exist.\n\\end{proposition}\n\\begin{proof}\nChoose coordinates $O\\cong\\Q^t$, extend that map to the whole quotient, and compose with the quotient map to obtain $H$. Then $HD=0$. The columns of $HR$ span $\\Q^t$, so $HR$ has a linear right inverse $Q$.\n\\end{proof}\n\\sourcescope{The availability argument is an ordinary finite-dimensional proof. The concrete producer in the next chapter finds and exports such data. Its search implementation is not itself proved correct in Lean.}","\\section{Exact rank and menu cost}\nFor $A\\in\\Q^{a\\times b}$, a certificate of rank $d$ consists of appropriately shaped rational matrices with\n\\begin{equation}\\label{eq:rational-rank}\n A=FG,\\qquad PAQ=I_d,\n\\end{equation}\nwhere the inner dimension of $FG$ is $d$.","\\begin{proposition}[Rank certificate]\nThe identities \\eqref{eq:rational-rank} imply $\\rank A=d$.\n\\end{proposition}\n\\begin{proof}\nThe factorization gives $\\rank A\\leq d$. The identity factorization through $A$ gives $d\\leq\\rank A$.\n\\end{proof}\n\\formal{RationalCertificates.checkRank_sound}","For comparison with the original quotient statement, ordinary rank-nullity gives\n\\begin{equation}\\label{eq:rank-difference-new}\n \\dim O=\\rank[D\\ R]-\\rank D.\n\\end{equation}\nIndeed, the quotient map restricted to $\\im D+\\im R$ has kernel $\\im D$ and image $O$. The current matrix verifier does not depend on a formal identification of this formula with the earlier abstract Lean quotient API. It obtains the lower bound directly from a right inverse.","Minimum menu cost is a separate optimization problem. To certify a successful menu $S$, the verifier requires a separating row for every admissible menu $T$ with strictly smaller declared cost. It quantifies over all subsets of the original finite candidate universe. It does not trust an incomplete producer-supplied list.","The rational example $D(u)=(u,0,0)$, $r=(0,1,0)$, $e_0=(0,1,1)$, $e_1=(0,0,1)$ is checked in the new matrix layer. Both entries repair the residual with coefficients $(-1,1)$. Rows $(0,1,0)$, $(0,1,-1)$, and $(0,1,0)$ separate the empty menu, the first singleton, and the second singleton. Thus the minimum cardinality is two. Restricting admissibility to at most one entry gives a checked complete rejection.\n\\formal{RationalMenus.menu_certificate}\n\\formal{RationalMenus.two_entries_minimal}\n\\formal{RationalMenus.restricted_certificate}\n\\formal{RationalMenus.no_small_menu}","\\chapter{Constructive matrix attainment}\\label{ch:matrix-attainment}\n\\section{A complete detector}\nFix $D\\in\\Q^{m\\times n}$ and $R\\in\\Q^{m\\times s}$. A split correction certificate consists of\n\\[\n P\\in\\Q^{n\\times m},\\quad K\\in\\Q^{m\\times d},\\quad L\\in\\Q^{d\\times m}\n\\]\nwith\n\\begin{equation}\\label{eq:split-certified}\n LD=0,\\qquad DP+KL=I_m.\n\\end{equation}\nThe first identity makes $L$ a detector that ignores original corrections. The second proves that no further information has been discarded: every target vector has the decomposition $v=DPv+KLv$. There is no assumption that $LK=I_d$, and redundant detector coordinates are permitted.","imes s}$. A split correction certificate consists of\n\\[\n P\\in\\Q^{n\\times m},\\quad K\\in\\Q^{m\\times d},\\quad L\\in\\Q^{d\\times m}\n\\]\nwith\n\\begin{equation}\\label{eq:split-certified}\n LD=0,\\qquad DP+KL=I_m.\n\\end{equation}\nThe first identity makes $L$ a detector that ignores original corrections. The second proves that no further information has been discarded: every target vector has the decomposition $v=DPv+KLv$. There is no assumption that $LK=I_d$, and redundant detector coordinates are permitted.","consists of\n\\[\n P\\in\\Q^{n\\times m},\\quad K\\in\\Q^{m\\times d},\\quad L\\in\\Q^{d\\times m}\n\\]\nwith\n\\begin{equation}\\label{eq:split-certified}\n LD=0,\\qquad DP+KL=I_m.\n\\end{equation}\nThe first identity makes $L$ a detector that ignores original corrections. The second proves that no further information has been discarded: every target vector has the decomposition $v=DPv+KLv$. There is no assumption that $LK=I_d$, and redundant detector coordinates are permitted.","\\begin{theorem}[Exact detection]\\label{thm:split-exact}\nUnder \\eqref{eq:split-certified},\n\\[\n (\\exists X,\\ DX=-R)\\quad\\Longleftrightarrow\\quad LR=0.\n\\]\nWhen $LR=0$, the explicit witness is $X=-PR$.\n\\end{theorem}\n\\begin{proof}\nA correction gives $0=LDX=-LR$. Conversely, $LR=0$ gives $R=(DP+KL)R=DPR$, so $D(-PR)=-R$.\n\\end{proof}\n\\formal{RationalCertificates.split_correctable_iff}","At the quotient level, $L$ induces an injection $\\Q^m/\\im D\\to\\Q^d$. If $Lv=0$, the split identity gives $v=DPv$, proving that its class is zero. The image is $\\im L$. Hence the required residual obstruction span is isomorphic to the column space of $C=LR$. This quotient interpretation is an ordinary consequence of the displayed identities.","\\section{The attaining construction}\nSuppose the detected residual matrix $C=LR$ has certificate data\n\\[\n F\\in\\Q^{d\\times t},\\quad G\\in\\Q^{t\\times s},\\quad\n J\\in\\Q^{t\\times d},\\quad Q\\in\\Q^{s\\times t}\n\\]\nsatisfying\n\\begin{equation}\\label{eq:factor-certified}\n LR=FG,\\qquad JLRQ=I_t.\n\\end{equation}\nThe new augmentation and its correction coefficients are not unspecified witnesses. They are\n\\begin{equation}\\label{eq:attaining-formula}\n \\boxed{E=KF,\\qquad U=-PR,\\qquad W=-G.}\n\\end{equation}","n{The attaining construction}\nSuppose the detected residual matrix $C=LR$ has certificate data\n\\[\n F\\in\\Q^{d\\times t},\\quad G\\in\\Q^{t\\times s},\\quad\n J\\in\\Q^{t\\times d},\\quad Q\\in\\Q^{s\\times t}\n\\]\nsatisfying\n\\begin{equation}\\label{eq:factor-certified}\n LR=FG,\\qquad JLRQ=I_t.\n\\end{equation}\nThe new augmentation and its correction coefficients are not unspecified witnesses. They are\n\\begin{equation}\\label{eq:attaining-formula}\n \\boxed{E=KF,\\qquad U=-PR,\\qquad W=-G.}\n\\end{equation}","\\begin{theorem}[Constructive minimum augmentation]\\label{thm:matrix-attainment}\nAssume \\eqref{eq:split-certified} and \\eqref{eq:factor-certified}. The matrices in \\eqref{eq:attaining-formula} satisfy $DU+EW=-R$. Every competing rational augmentation repairing the same residual family has at least $t$ added dimensions.\n\\end{theorem}\n\\begin{proof}\nThe constructed matrices satisfy\n\\[\n DU+EW=-DPR-KFG=-DPR-KLR=-(DP+KL)R=-R.\n\\]\nFor the lower bound set $H=JL$. Then $HD=0$ and $HRQ=I_t$. Apply Theorem~\\ref{thm:rational-lower} to every competing repair.\n\\end{proof}\n\\formal{RationalCertificates.split_constructs_repair}\n\\formal{RationalCertificates.checkOptimalConstruction_sound}","label{thm:matrix-attainment}\nAssume \\eqref{eq:split-certified} and \\eqref{eq:factor-certified}. The matrices in \\eqref{eq:attaining-formula} satisfy $DU+EW=-R$. Every competing rational augmentation repairing the same residual family has at least $t$ added dimensions.\n\\end{theorem}\n\\begin{proof}\nThe constructed matrices satisfy\n\\[\n DU+EW=-DPR-KFG=-DPR-KLR=-(DP+KL)R=-R.\n\\]\nFor the lower bound set $H=JL$. Then $HD=0$ and $HRQ=I_t$. Apply Theorem~\\ref{thm:rational-lower} to every competing repair.\n\\end{proof}\n\\formal{RationalCertificates.split_constructs_repair}\n\\formal{RationalCertificates.checkOptimalConstruction_sound}","rix-attainment}\nAssume \\eqref{eq:split-certified} and \\eqref{eq:factor-certified}. The matrices in \\eqref{eq:attaining-formula} satisfy $DU+EW=-R$. Every competing rational augmentation repairing the same residual family has at least $t$ added dimensions.\n\\end{theorem}\n\\begin{proof}\nThe constructed matrices satisfy\n\\[\n DU+EW=-DPR-KFG=-DPR-KLR=-(DP+KL)R=-R.\n\\]\nFor the lower bound set $H=JL$. Then $HD=0$ and $HRQ=I_t$. Apply Theorem~\\ref{thm:rational-lower} to every competing repair.\n\\end{proof}\n\\formal{RationalCertificates.split_constructs_repair}\n\\formal{RationalCertificates.checkOptimalConstruction_sound}","ent}\nAssume \\eqref{eq:split-certified} and \\eqref{eq:factor-certified}. The matrices in \\eqref{eq:attaining-formula} satisfy $DU+EW=-R$. Every competing rational augmentation repairing the same residual family has at least $t$ added dimensions.\n\\end{theorem}\n\\begin{proof}\nThe constructed matrices satisfy\n\\[\n DU+EW=-DPR-KFG=-DPR-KLR=-(DP+KL)R=-R.\n\\]\nFor the lower bound set $H=JL$. Then $HD=0$ and $HRQ=I_t$. Apply Theorem~\\ref{thm:rational-lower} to every competing repair.\n\\end{proof}\n\\formal{RationalCertificates.split_constructs_repair}\n\\formal{RationalCertificates.checkOptimalConstruction_sound}","\\section{How the Python producer obtains the matrices}\nThe executable producer uses exact rational row reduction. It selects independent columns of $D$ and extends them by standard basis vectors to a basis $B$ of $\\Q^m$. The inverse $B^{-1}$ supplies the image coordinates and the complementary detector coordinates. The selected image coordinates are placed in the original domain to obtain $P$, while the complementary basis vectors form $K$ and the corresponding rows of $B^{-1}$ form $L$.","the matrices}\nThe executable producer uses exact rational row reduction. It selects independent columns of $D$ and extends them by standard basis vectors to a basis $B$ of $\\Q^m$. The inverse $B^{-1}$ supplies the image coordinates and the complementary detector coordinates. The selected image coordinates are placed in the original domain to obtain $P$, while the complementary basis vectors form $K$ and the corresponding rows of $B^{-1}$ form $L$.","cer uses exact rational row reduction. It selects independent columns of $D$ and extends them by standard basis vectors to a basis $B$ of $\\Q^m$. The inverse $B^{-1}$ supplies the image coordinates and the complementary detector coordinates. The selected image coordinates are placed in the original domain to obtain $P$, while the complementary basis vectors form $K$ and the corresponding rows of $B^{-1}$ form $L$.","rational row reduction. It selects independent columns of $D$ and extends them by standard basis vectors to a basis $B$ of $\\Q^m$. The inverse $B^{-1}$ supplies the image coordinates and the complementary detector coordinates. The selected image coordinates are placed in the original domain to obtain $P$, while the complementary basis vectors form $K$ and the corresponding rows of $B^{-1}$ form $L$.","For $C=LR$, the producer selects independent residual columns to form $F$. A selection matrix $Q$ records those columns. A left inverse $J$ of $F$ gives $G=JC$. The producer then forms $E,U,W$ using \\eqref{eq:attaining-formula}, and checks every identity again with exact rational arithmetic before export.","Zero-dimensional targets, original domains, obstruction spaces, and residual families are retained. For an already correctable residual family the construction has $t=0$, so it adds no directions. The input limit of the supplied prototype is an implementation resource bound, not a restriction on Theorem~\\ref{thm:matrix-attainment}.","\\section{A full intermediate trace}\\label{sec:matrix-worked}\nTake\n\\[\n D=\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix},\\qquad\n R=\\begin{pmatrix}2&4\\\\1&2\\\\1&2\\end{pmatrix}.\n\\]\nThe producer selects the first column of $D$ and the complementary second and third coordinate axes. It obtains\n\\[\n P=\\begin{pmatrix}1&0&0\\end{pmatrix},\\quad\n K=\\begin{pmatrix}0&0","1&2\\end{pmatrix}.\n\\]\nThe producer selects the first column of $D$ and the complementary second and third coordinate axes. It obtains\n\\[\n P=\\begin{pmatrix}1&0&0\\end{pmatrix},\\quad\n K=\\begin{pmatrix}0&0","1&2\\end{pmatrix}.\n\\]\nThe producer selects the first column of $D$ and the complementary second and third coordinate axes. It obtains\n\\[\n P=\\begin{pmatrix}1&0&0\\end{pmatrix},\\quad\n K=\\begin{pmatrix}0&0\\\\1&0\\\\0&1\\end{pmatrix},\\quad\n L=\\begin{pmatrix}0&1&0\\\\0&0&1\\end{pmatrix}.\n\\]\nThus $LD=0$, $DP=\\operatorname{diag}(1,0,0)$, and $KL=\\operatorname{diag}(0,1,1)$. The residual coordinates are\n\\[\n C=LR=\\begin{pmatrix}1&2","0&0&1\\end{pmatrix}.\n\\]\nThus $LD=0$, $DP=\\operatorname{diag}(1,0,0)$, and $KL=\\operatorname{diag}(0,1,1)$. The residual coordinates are\n\\[\n C=LR=\\begin{pmatrix}1&2","0&0&1\\end{pmatrix}.\n\\]\nThus $LD=0$, $DP=\\operatorname{diag}(1,0,0)$, and $KL=\\operatorname{diag}(0,1,1)$. The residual coordinates are\n\\[\n C=LR=\\begin{pmatrix}1&2\\\\1&2\\end{pmatrix}.\n\\]\nA rank-one certificate is\n\\[\n F=\\begin{pmatrix}1","1&2\\end{pmatrix}.\n\\]\nA rank-one certificate is\n\\[\n F=\\begin{pmatrix}1\\\\1\\end{pmatrix},\\quad\n G=\\begin{pmatrix}1&2\\end{pmatrix},\\quad\n J=\\begin{pmatrix}1&0\\end{pmatrix},\\quad\n Q=\\begin{pmatrix}1\\\\0\\end{pmatrix}.\n\\]\nThe result is\n\\[\n E=\\begin{pmatrix}0","0\\end{pmatrix}.\n\\]\nThe result is\n\\[\n E=\\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix},\\qquad\n U=\\begin{pmatrix}-2&-4\\end{pmatrix},\\qquad\n W=\\begin{pmatrix}-1&-2\\end{pmatrix}.\n\\]\nThe original equations become\n\\[\n DU+EW=\n \\begin{pmatrix}-2&-4","1\\end{pmatrix},\\qquad\n U=\\begin{pmatrix}-2&-4\\end{pmatrix},\\qquad\n W=\\begin{pmatrix}-1&-2\\end{pmatrix}.\n\\]\nThe original equations become\n\\[\n DU+EW=\n \\begin{pmatrix}-2&-4\\\\0&0\\\\0&0\\end{pmatrix}+\n \\begin{pmatrix}0&0\\\\-1&-2\\\\-1&-2\\end{pmatrix}=-R.\n\\]\nThe lower witness is $H=JL=(0,1,0)$, with $HD=0$ and $HRQ=1$. The ambient cokernel has dimension two, but the residual classes span only one direction. The construction therefore adds one direction and proves that zero added directions are insufficient.\n\\formal{Case000.certificate_checked}\n\\formal{Case000.original_equations_checked}\n\\formal{Case000.optimal}","-1&-2\\end{pmatrix}=-R.\n\\]\nThe lower witness is $H=JL=(0,1,0)$, with $HD=0$ and $HRQ=1$. The ambient cokernel has dimension two, but the residual classes span only one direction. The construction therefore adds one direction and proves that zero added directions are insufficient.\n\\formal{Case000.certificate_checked}\n\\formal{Case000.original_equations_checked}\n\\formal{Case000.optimal}","The fractional case uses $D=(2/3,4/3)^{\\mathsf T}$ and $R=(1/2,5/6)^{\\mathsf T}$. It exercises nonintegral splitting and correction coefficients while retaining exact arithmetic. Empty and rank-deficient cases are included in the same exported corpus. Every case has its own original input file, certificate file, and formal optimality statement.","The generated Lean data use typed pairs in $\\Z\\times\\N$. Their decoder returns the intended rational when the denominator is nonzero and explicitly rejects a zero denominator at the cell level. Each generated input and certificate additionally proves denominator validity for all its cells.\n\\formal{RationalCertificates.decodeCell_zero}\n\\formal{RationalCertificates.decodeCell_nonzero}\n\\formal{RationalCertificates.decodeCell_sound}\n\\formal{RationalCertificates.checkWireMatrix_sound}\n\\formal{RationalCertificates.decodeWireMatrix_cell}","The original matrices $D,R$ are emitted separately from certificate data. The certificate module imports them and cannot replace them through a certificate field. A canonical problem hash binds the JSON records. A diagnostic Lean echo of the decoded input matrices is compared with the original JSON values for every exported case. That echo is an additional execution check, never a theorem premise.","\\section{Legality and coefficient rings}\nFree minimum dimension is not minimum menu cost. The constructed direction $KF$ need not occur among supplied menu entries. When legal directions form $\\im A$, an additional identity $KF=AZ$ certifies that the augmentation lies in that legal subspace. This identity also rewrites every added correction as $A(ZW)$.\n\\formal{RationalCertificates.constructed_repair_in_legal_image}","The availability of splitting also depends on the coefficient ring. For $D=(2)$ over $\\Z$, the condition $LD=0$ forces $L=0$. The split identity would then require $2P=1$, which is impossible over $\\Z$. Over $\\Q$, $P=1/2$ is permitted. The current producer and checker are rational, not a certificate of integral realizability.","Fix rational matrices\n\\[\n D\\in\\Q^{m\\times n},\\qquad E\\in\\Q^{m\\times q},\\qquad R\\in\\Q^{m\\times s}.\n\\]\nThe columns $e_j$ of $E$ are the available new directions. The finite family\n\\[\n \\mathcal A\\subseteq\\mathcal P(\\{0,\\ldots,q-1\\})\n\\]\ncontains exactly the admissible selections. It need not be downward closed. Nonnegative integer weights $c_j$ define the cost\n\\[\n c(S)=\\sum_{j\\in S}c_j.\n\\]\nThese matrices, selections, and costs are fixed original input. A certificate cannot replace them with another problem.","Let $\\widetilde E_S$ retain each column of $E$ selected by $S$ and replace every other column by zero. This convention preserves the original candidate indexing. A coefficient table $W\\in\\Q^{q\\times s}$ may have arbitrary values in unselected rows, but those rows contribute nothing. A proposed repair consists of matrices $U,W$ with\n\\begin{equation}\n DU+\\widetilde E_SW=-R.\n \\label{eq:menu-cover-primal}\n\\end{equation}\nEvery selected direction is charged, even if its coefficient row happens to be zero. Thus cost measures the selected menu, not merely the support of a particular coefficient table.","\\begin{definition}[Separating block]\nA separating block is a pair $(B,Y)$, where $B\\subseteq\\{0,\\ldots,q-1\\}$ and $Y\\in\\Q^{1\\times m}$ satisfy\n\\begin{equation}\n YD=0,\\qquad Y\\widetilde E_B=0,\\qquad YR\\ne0.\n \\label{eq:menu-cover-dual}\n\\end{equation}\nThe set $B$ need not itself belong to $\\mathcal A$.\n\\end{definition}","\\begin{lemma}[Column-mask identity]\nFor every row $Y$ and column $j$,\n\\[\n (Y\\widetilde E_S)_{0j}=\n \\begin{cases}\n (YE)_{0j},&j\\in S,","\\begin{lemma}[Column-mask identity]\nFor every row $Y$ and column $j$,\n\\[\n (Y\\widetilde E_S)_{0j}=\n \\begin{cases}\n (YE)_{0j},&j\\in S,\\\\\n 0,&j\\notin S.\n \\end{cases}\n\\]\n\\end{lemma}\n\\begin{proof}\nThe mask either retains the entire $j$th column or replaces it by zero. Expanding the finite matrix product gives the two cases.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.mul_masked_entry}","0,&j\\notin S.\n \\end{cases}\n\\]\n\\end{lemma}\n\\begin{proof}\nThe mask either retains the entire $j$th column or replaces it by zero. Expanding the finite matrix product gives the two cases.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.mul_masked_entry}","\\begin{theorem}[Downward validity of a separator]\nIf $(B,Y)$ is a separating block and $T\\subseteq B$, then $Y$ also separates $T$. In particular, $T$ cannot satisfy \\eqref{eq:menu-cover-primal}.\n\\end{theorem}\n\\begin{proof}\nFor a selected column $j\\in T$, one also has $j\\in B$, so $(YE)_{0j}=0$ by the block identity. Every unselected column is zero after masking. Thus $Y\\widetilde E_T=0$. The other two equations in \\eqref{eq:menu-cover-dual} are unchanged.","If a repair for $T$ existed, multiplication by $Y$ would give\n\\[\n 0=YDU+Y\\widetilde E_TW=-YR,\n\\]\ncontradicting the detection condition.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.separates_subset}","Let $\\mathcal B$ be a finite list of separating blocks. Say that $\\mathcal B$ covers $T$ when some $(B,Y)\\in\\mathcal B$ satisfies $T\\subseteq B$. Validity of the block list and coverage of the relevant selections are separate obligations.","\\begin{theorem}[Certified minimum menu cost]\nSuppose the following data have been checked:\n\\begin{enumerate}\n \\item $S\\in\\mathcal A$ and the original repair equation holds for $S,U,W$.\n \\item Every member of $\\mathcal B$ is a valid separating block.\n \\item Every $T\\in\\mathcal A$ with $c(T)<c(S)$ is covered by $\\mathcal B$.\n\\end{enumerate}\nThen $S$ has minimum cost among all admissible feasible selections.\n\\end{theorem}\n\\begin{proof}\nThe first condition supplies an admissible feasible selection. If a cheaper admissible feasible $T$ existed, the third condition would place it inside a separating block. Downward validity would make $T$ infeasible, a contradiction.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.checkCoveredOptimal_sound}","nd the original repair equation holds for $S,U,W$.\n \\item Every member of $\\mathcal B$ is a valid separating block.\n \\item Every $T\\in\\mathcal A$ with $c(T)<c(S)$ is covered by $\\mathcal B$.\n\\end{enumerate}\nThen $S$ has minimum cost among all admissible feasible selections.\n\\end{theorem}\n\\begin{proof}\nThe first condition supplies an admissible feasible selection. If a cheaper admissible feasible $T$ existed, the third condition would place it inside a separating block. Downward validity would make $T$ infeasible, a contradiction.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.checkCoveredOptimal_sound}","ting block.\n \\item Every $T\\in\\mathcal A$ with $c(T)<c(S)$ is covered by $\\mathcal B$.\n\\end{enumerate}\nThen $S$ has minimum cost among all admissible feasible selections.\n\\end{theorem}\n\\begin{proof}\nThe first condition supplies an admissible feasible selection. If a cheaper admissible feasible $T$ existed, the third condition would place it inside a separating block. Downward validity would make $T$ infeasible, a contradiction.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.checkCoveredOptimal_sound}","\\begin{theorem}[Certified rejection of the admissible family]\nIf every block is valid and every member of $\\mathcal A$ is covered, then no admissible menu repairs the residual family.\n\\end{theorem}\n\\begin{proof}\nEvery admitted selection lies inside a valid separating block and is therefore infeasible.\n\\end{proof}\n\\formal{KomorebiRepair.RationalCertificates.checkCoveredNoRepair_sound}","When $\\mathcal A$ is empty, this conclusion is vacuous. The implementation reports that fact explicitly. It does not manufacture a nonzero obstruction or claim that the original equations have no solution without the admissibility restriction.","The producer evaluates all $2^q$ selections in this bounded implementation. It chooses a feasible admitted selection of minimum declared cost, with a deterministic tie break used only for reproducibility. Alternatively, it records that none of the admitted selections passed the exact linear alternative.","Take\n\\[\n D=\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix},\\qquad\n E=\\begin{pmatrix}0&0\\\\1&0\\\\1&1\\end{pmatrix},\\qquad\n R=\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix},\n\\]\nwith both candidate weights equal to one and every subset admitted. The chosen menu is $S=\\{0,1\\}$, with\n\\[\n U=(0),\\qquad W=\\begin{pmatrix}-1","0\\end{pmatrix},\n\\]\nwith both candidate weights equal to one and every subset admitted. The chosen menu is $S=\\{0,1\\}$, with\n\\[\n U=(0),\\qquad W=\\begin{pmatrix}-1","0\\end{pmatrix},\n\\]\nwith both candidate weights equal to one and every subset admitted. The chosen menu is $S=\\{0,1\\}$, with\n\\[\n U=(0),\\qquad W=\\begin{pmatrix}-1\\\\1\\end{pmatrix}.\n\\]\nIts exact repair is\n\\[\n DU+EW=\\begin{pmatrix}0","1\\end{pmatrix}.\n\\]\nIts exact repair is\n\\[\n DU+EW=\\begin{pmatrix}0\\\\-1\\\\0\\end{pmatrix}=-R.\n\\]\nTwo separating blocks suffice:\n\\[\n B_0=\\{0\\},\\quad Y_0=(0,1,-1),\\qquad\n B_1=\\{1\\},\\quad Y_1=(0,1,0).\n\\]\nBoth satisfy $Y_iD=0$, annihilate their masked candidate matrices, and give $Y_iR=1$.","0\\end{pmatrix}=-R.\n\\]\nTwo separating blocks suffice:\n\\[\n B_0=\\{0\\},\\quad Y_0=(0,1,-1),\\qquad\n B_1=\\{1\\},\\quad Y_1=(0,1,0).\n\\]\nBoth satisfy $Y_iD=0$, annihilate their masked candidate matrices, and give $Y_iR=1$.","midrule\n$\\varnothing$ & 0 & $B_0$ or $B_1$","$\\{0\\}$ & 1 & $B_0$","$\\{1\\}$ & 1 & $B_1$","bottomrule\n\\end{tabular}\n\\end{center}\nThus cost two is minimal. Restricting admissibility to the three cheaper menus turns the same two blocks into a complete no-admissible-repair certificate. The unrestricted one-dimensional repair direction $R$ still exists, but it is not one of the two listed candidate directions.","A third candidate can make the cost distinction sharper. Add the direct direction $e_0=(0,1,0)$ at cost five, and price $e_1=(0,1,1)$ and $e_2=(0,0,1)$ at one each. One menu entry can now repair the residual, but the cheapest repair selects two entries, $\\{1,2\\}$, at total cost two. Minimum free dimension, minimum menu cardinality, and minimum menu cost are three different quantities.","A separate example shows why rejection covers help. Let\n\\[\n D=\\begin{pmatrix}1\\\\0\\end{pmatrix},\\qquad\n E=\\begin{pmatrix}2&3&4&5\\\\0&0&0&0\\end{pmatrix},\\qquad\n R=\\begin{pmatrix}0\\\\1\\end{pmatrix}.\n\\]\nThe row $Y=(0,1)$ separates the full four-entry menu. One block therefore excludes all sixteen subsets. Adding $(0,1)$ as a new candidate would change the problem and allow a repair. The certified rejection of the original menu would remain correct.","1\\end{pmatrix}.\n\\]\nThe row $Y=(0,1)$ separates the full four-entry menu. One block therefore excludes all sixteen subsets. Adding $(0,1)$ as a new candidate would change the problem and allow a repair. The certified rejection of the original menu would remain correct.","The solver is not restricted to coefficients in a finite field. For\n\\[\n D=\\begin{pmatrix}1\\\\0\\end{pmatrix},\\quad\n E=\\begin{pmatrix}0\\\\2/3\\end{pmatrix},\\quad\n R=\\begin{pmatrix}1/2\\\\5/6\\end{pmatrix},\n\\]\nthe selected direction at cost seven has witnesses\n\\[\n U=-1/2,\\qquad W=-5/4.\n\\]\nThe checker verifies the exact rational equation, with no tolerance or floating-point rank test.","5/6\\end{pmatrix},\n\\]\nthe selected direction at cost seven has witnesses\n\\[\n U=-1/2,\\qquad W=-5/4.\n\\]\nThe checker verifies the exact rational equation, with no tolerance or floating-point rank test.","The generated input module contains $D,E,R$, the candidate weights, and the literal admissibility table. The certificate module imports those inputs and supplies only coefficient matrices and separating blocks. Typed rational decoding and nonzero denominator proofs are retained from the preceding bridge.","Let $J=\\{0,\\ldots,q-1\\}$ be the original candidate index set. The matrices $D,E,R$ and the original correction equation remain fixed:\n\\[\n DU+\\widetilde E_S W=-R.\n\\]\nHere $\\widetilde E_S$ zeros unselected columns of $E$ while retaining its original indexing. Costs are nonnegative integers attached to those same columns:\n\\[\n c(S)=\\sum_{j\\in S}w_j,\\qquad w_j\\in\\N.\n\\]\nThe admissibility rule either permits every subset or supplies a fixed finite table of permitted subsets. An empty table permits none. It is different from the all-subsets rule.","\\begin{definition}[Selection interval]\nFor $A\\subseteq B\\subseteq J$, define\n\\[\n \\mathcal I(A,B)=\\{T\\subseteq J:A\\subseteq T\\subseteq B\\}.\n\\]\nThe set $A$ contains candidates already required, and $B$ contains all candidates still permitted by the branch decisions. The unresolved indices form $B\\setminus A$.\n\\end{definition}","\\begin{lemma}[Exact branch partition]\nIf $j\\in B\\setminus A$, then\n\\[\n \\mathcal I(A,B)=\n \\mathcal I(A,B\\setminus\\{j\\})\\;\\sqcup\\;\n \\mathcal I(A\\cup\\{j\\},B).\n\\]\n\\end{lemma}\n\\begin{proof}\nEvery $T\\in\\mathcal I(A,B)$ either excludes $j$ or includes $j$. In the first case it belongs to the left interval, and in the second it belongs to the right interval. Conversely, each child interval is contained in the parent. Their membership conditions on $j$ are incompatible, so the union is disjoint.\n\\end{proof}","n B\\setminus A$, then\n\\[\n \\mathcal I(A,B)=\n \\mathcal I(A,B\\setminus\\{j\\})\\;\\sqcup\\;\n \\mathcal I(A\\cup\\{j\\},B).\n\\]\n\\end{lemma}\n\\begin{proof}\nEvery $T\\in\\mathcal I(A,B)$ either excludes $j$ or includes $j$. In the first case it belongs to the left interval, and in the second it belongs to the right interval. Conversely, each child interval is contained in the parent. Their membership conditions on $j$ are incompatible, so the union is disjoint.\n\\end{proof}","Nonnegative additive costs are monotone under inclusion:\n\\[\n A\\subseteq T\\quad\\Longrightarrow\\quad c(A)\\leq c(T).\n\\]\nTherefore, when an incumbent repair has cost $c_\\ast$, the finite check\n\\[\n c_\\ast\\leq c(A)\n\\]\nexcludes every strictly cheaper repair in $\\mathcal I(A,B)$. It does not say the interval is infeasible. It only says that it cannot improve the incumbent.\n\\formal{KomorebiRepair.RationalCertificates.additiveCost_mono}","\\begin{proof}\nThe sum for $c(T)$ contains all terms in the sum for $c(A)$ and only additional nonnegative terms. Apply transitivity to $c_\\ast\\leq c(A)\\leq c(T)$.\n\\end{proof}","A row $Y$ is checked against the largest selection $B$ in the interval:\n\\[\n YD=0,\\qquad Y\\widetilde E_B=0,\\qquad YR\\ne0.\n\\]\nEvery $T\\subseteq B$ uses only columns annihilated by $Y$. Consequently, the same row proves that no $T\\in\\mathcal I(A,B)$ repairs the family. This reuses the downward separator theorem from the previous chapter.","The residual matrix in the last inequality is the original $R$. Projecting it to zero in some unrelated representation is not accepted as a repair.","If admissibility is specified by a list $\\mathcal A$, a terminal check establishes\n\\[\n \\nexists T\\in\\mathcal A:\\quad A\\subseteq T\\subseteq B.\n\\]\nIt examines the original list, not the powerset of $J$. For the all-subsets rule this terminal certificate is unavailable.\n\\formal{KomorebiRepair.RationalCertificates.intervalExcluded_sound}","It is not enough that $A$ itself be inadmissible. A larger selection in the interval might be permitted. For example, a rule that permits only the pair $\\{0,1\\}$ excludes the empty selection but not the interval from the empty selection to the full pair. The verifier retains this distinction and does not assume that admissibility is downward closed.","\\begin{definition}[Accepted branch tree]\nA tree for $\\mathcal I(A,B)$ consists of one of the three terminal certificates above, or a split at $j\\in B\\setminus A$ with accepted trees for both child intervals. In lower-bound mode a cost threshold is fixed throughout the tree. In infeasibility mode cost leaves are rejected.\n\\end{definition}","\\begin{theorem}[Structural certificate soundness]\nAn accepted lower-bound tree with threshold $c_\\ast$ proves\n\\[\n T\\in\\mathcal I(A,B),\\quad T\\text{ admissible and feasible}\n \\quad\\Longrightarrow\\quad c_\\ast\\leq c(T).\n\\]\nAn accepted infeasibility tree proves that no admissible selection in $\\mathcal I(A,B)$ is feasible.\n\\end{theorem}\n\\formal{KomorebiRepair.RationalCertificates.checkBranch_sound}","\\begin{corollary}[Minimum cost certificate]\nSuppose an admissible selection $S$ has supplied coefficients $U,W$ satisfying the original repair equation. An accepted lower-bound tree of threshold $c(S)$ on $\\mathcal I(\\varnothing,J)$ proves that $S$ has minimum cost among all admissible feasible selections.\n\\end{corollary}\n\\formal{KomorebiRepair.RationalCertificates.checkTreeOptimal_sound}\n\\begin{proof}\nThe supplied coefficient equation proves feasibility of $S$. Every selection lies in $\\mathcal I(\\varnothing,J)$. Structural soundness therefore gives $c(S)\\leq c(T)$ for every admissible feasible competitor $T$.\n\\end{proof}","le selection $S$ has supplied coefficients $U,W$ satisfying the original repair equation. An accepted lower-bound tree of threshold $c(S)$ on $\\mathcal I(\\varnothing,J)$ proves that $S$ has minimum cost among all admissible feasible selections.\n\\end{corollary}\n\\formal{KomorebiRepair.RationalCertificates.checkTreeOptimal_sound}\n\\begin{proof}\nThe supplied coefficient equation proves feasibility of $S$. Every selection lies in $\\mathcal I(\\varnothing,J)$. Structural soundness therefore gives $c(S)\\leq c(T)$ for every admissible feasible competitor $T$.\n\\end{proof}","coefficients $U,W$ satisfying the original repair equation. An accepted lower-bound tree of threshold $c(S)$ on $\\mathcal I(\\varnothing,J)$ proves that $S$ has minimum cost among all admissible feasible selections.\n\\end{corollary}\n\\formal{KomorebiRepair.RationalCertificates.checkTreeOptimal_sound}\n\\begin{proof}\nThe supplied coefficient equation proves feasibility of $S$. Every selection lies in $\\mathcal I(\\varnothing,J)$. Structural soundness therefore gives $c(S)\\leq c(T)$ for every admissible feasible competitor $T$.\n\\end{proof}","cepted lower-bound tree of threshold $c(S)$ on $\\mathcal I(\\varnothing,J)$ proves that $S$ has minimum cost among all admissible feasible selections.\n\\end{corollary}\n\\formal{KomorebiRepair.RationalCertificates.checkTreeOptimal_sound}\n\\begin{proof}\nThe supplied coefficient equation proves feasibility of $S$. Every selection lies in $\\mathcal I(\\varnothing,J)$. Structural soundness therefore gives $c(S)\\leq c(T)$ for every admissible feasible competitor $T$.\n\\end{proof}","I(\\varnothing,J)$ proves that $S$ has minimum cost among all admissible feasible selections.\n\\end{corollary}\n\\formal{KomorebiRepair.RationalCertificates.checkTreeOptimal_sound}\n\\begin{proof}\nThe supplied coefficient equation proves feasibility of $S$. Every selection lies in $\\mathcal I(\\varnothing,J)$. Structural soundness therefore gives $c(S)\\leq c(T)$ for every admissible feasible competitor $T$.\n\\end{proof}","\\begin{corollary}[Scoped complete rejection]\nAn accepted infeasibility tree on $\\mathcal I(\\varnothing,J)$ proves that the declared admissible family contains no repair.\n\\end{corollary}\n\\formal{KomorebiRepair.RationalCertificates.checkTreeNoRepair_sound}","There is no claim that directions outside $J$ cannot repair the residual. Nor does a rejected certificate establish infeasibility. A missing child, a bad separator, or an exhausted resource budget leaves the outcome unresolved.","The Python producer traverses intervals and retains exact row-reduction witnesses. It first checks whether the original admissibility table excludes the interval. It then uses the incumbent cost to test a bound leaf. If necessary, it solves the rational linear problem on the upper selection $B$. An infeasible upper selection supplies a separator that ends the whole branch.","If the lower selection $A$ is admitted and feasible, its coefficient matrix supplies a new incumbent and the interval ends at a cost leaf. Otherwise, the producer selects an unresolved index and explores both decisions. The current heuristic visits low-cost candidates first. Correctness of this heuristic is not an assumption of any theorem.","Let $b_1,b_2,b_3$ be the standard basis of $\\Q^3$. Take $D(u)=ub_1$ and residual $r=b_2+b_3$. There are 24 candidate entries:\n\\[\n a_0=b_2,\\qquad a_1=b_3,\\qquad a_j=b_1\\quad(2\\leq j<24).\n\\]\nTheir costs are two, three, and ten for each remaining entry. All subsets are admissible.","The selected repair is $S=\\{0,1\\}$, with cost five, original coefficient zero, and coefficient $-1$ on each selected menu direction. Its full coefficient matrix satisfies the original equation exactly.","The proof tree has five nodes. At the root it splits on candidate zero. The excluded child lacks the only second-coordinate correction and has separator $(0,1,0)$. The included child splits on candidate one. Excluding candidate one has separator $(0,0,1)$. Including both reaches lower selection $\\{0,1\\}$ with cost five and terminates at a bound leaf.","There are $2^{24}=16{,}777{,}216$ possible selections, but the producer used six distinct linear solves and submitted five tree nodes. Lean checks that tree and proves the minimum-cost conclusion for every selection. This is not an experiment that visited sixteen million cases. It is one structured problem with a short exhaustive mathematical argument.","A second 24-candidate example puts every available correction on the first coordinate axis of $\\Q^2$ and sets the residual to $(0,1)$. The row $(0,1)$ supplies a one-node certificate rejecting all selections. This illustrates compressed reasoning, not a claim that arbitrary difficult search problems have short certificates.","\\appendix\n\\chapter{Formal correspondence}\\label{app:correspondence}\n\\section{How the prose matches the source}\nThe original formal source is the supplied \\emph{Constructive Repair Lean v0.2.0} release~\\cite{formal}. The namespace is \\code{KomorebiRepair}. The notation $Q_D$, $q_D$, $O_D(r)$, and $E_S$ corresponds to \\code{ObstructionSpace}, \\code{obstructionMap}, \\code{obstructionSpan}, and \\code{menuMap}, respectively.","The mathematical statements here use $\\exists u,w$ with the same sign convention as the source. Families quantify $\\forall i\\,\\exists u_i,w_i$. They do not assume one shared correction vector. The algebraic base is a commutative ring. The dimension and section results specialize to a field, and finiteness is stated when numerical dimensions or terminating enumeration are used.","\\record{Arithmetic.singular_derivative_can_still_lift}{54}{The exact root one of the equation $x^2=1$ persists at every dyadic depth. See \\cref{prop:singular-success}.}","\\chapter{Verification and provenance}\\label{app:evidence}\n\\section{The evidence attached to the formal release}\nThe v0.2.0 verification record reports successful compilation of ten proof modules, an axiom audit of all 49 named theorem declarations, and individual replay of those modules through Lean's kernel. The allowed transitive axiom set is\n\\[\n\\{\\code{propext},\\ \\code{Classical.choice},\\ \\code{Quot.sound}\\}.\n\\]\nNo additional assumption axiom, placeholder axiom, or native decision axiom is allowed in the accepted theorem library.","\\toprule Case & $m$ & $n$ & $s$ & $d$ & $t$ & Example class","This exact workflow was executed on the fractional example. It ended with \\nolinkurl{KERNEL_CHECKED_OPTIMAL_REPAIR} and $t=1$. It requires no network access or publication service. An existing output directory is refused so that previous evidence is not silently overwritten. The large compiler and dependency kit are not duplicated in the small source handoff.","The single-command workflow was executed on the fractional problem and on the four-candidate globally infeasible menu. The former returns \\nolinkurl{KERNEL_CHECKED_OPTIMAL_MENU}, cost seven, with coefficient $-5/4$. The latter returns \\nolinkurl{KERNEL_CHECKED_NO_ADMISSIBLE_MENU}, using one separator for sixteen subsets. Both commands compile the sources, audit all relevant theorem dependencies, replay the modules, and match the Lean input echo to the original matrices and constraints."]}