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An example may explain why a definition was chosen. A failed route may isolate the exact boundary of a method. An alternative proof may carry a bridge to another field. A computation may reveal a stronger statement than the one originally sought. None of these objects is automatically part of the final implication\n\\[\nA\\vdash B.\n\\]\nThey may still be part of the mathematical event that produced it.","Proof Path Theory studies this larger object without changing the standard of proof. Its central separation is\n\\[\n\\boxed{\n\\text{certificate of validity}\n\\quad\\ne\\quad\n\\text{expanded discovery field}.\n}\n\\]\nThe certificate answers whether the conclusion follows. The discovery field records how the conclusion became reachable, what alternatives were considered, what was lost during compression, and which omitted structures may matter later.","\\begin{definition}[Evidence kind]\n\\label{res:2:1}\nAn evidence kind is a declared type attached to a record supporting a claim or justification. The core finite vocabulary used here is\n\\[\n\\begin{aligned}\n&\\text{formal proof},\\quad \\text{assumption},\\quad \\text{exact computation},\\\\\n&\\text{numerical evidence},\\quad \\text{conjecture},\\quad \\text{external result},\\\\\n&\\text{counterexample},\\quad \\text{interpretation}.\n\\end{aligned}\n\\]\nAn evidence record also states its source, scope, and whether it certifies the justification to which it is attached.\n\\end{definition}","\\begin{definition}[Typed proof field]\n\\label{res:3:1}\nA finite typed proof field is a tuple\n\\[\n\\Pcal=(V,\\Jcal,A,T,\\tau,\\Ecal,\\mathbf v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\op","\\begin{definition}[Typed proof field]\n\\label{res:3:1}\nA finite typed proof field is a tuple\n\\[\n\\Pcal=(V,\\Jcal,A,T,\\tau,\\Ecal,\\mathbf v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\opera","\\begin{definition}[Typed proof field]\n\\label{res:3:1}\nA finite typed proof field is a tuple\n\\[\n\\Pcal=(V,\\Jcal,A,T,\\tau,\\Ecal,\\mathbf v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. I","\\begin{definition}[Typed proof field]\n\\label{res:3:1}\nA finite typed proof field is a tuple\n\\[\n\\Pcal=(V,\\Jcal,A,T,\\tau,\\Ecal,\\mathbf v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","in{definition}[Typed proof field]\n\\label{res:3:1}\nA finite typed proof field is a tuple\n\\[\n\\Pcal=(V,\\Jcal,A,T,\\tau,\\Ecal,\\mathbf v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","a tuple\n\\[\n\\Pcal=(V,\\Jcal,A,T,\\tau,\\Ecal,\\mathbf v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","v,\\mathbf c),\n\\]\nwhere:\n\\begin{itemize}\n\\item $V$ is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","is a finite set of claim nodes.\n\\item $\\Jcal$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","al$ is a finite set of justifications.\n\\item Each $j\\in\\Jcal$ has a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","a finite premise set $\\operatorname{prem}(j)\\subseteq V$ and one conclusion $\\operatorname{con}(j)\\in V$.\n\\item $A\\subseteq V$ is the set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","he set of required premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","premises of the declared judgment.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","t.\n\\item $T\\subseteq V$ is the nonempty target set.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","t.\n\\item $\\tau$ assigns node and justification types.\n\\item $\\Ecal$ is a typed evidence ledger.\n\\item $\\mathbf v$ assigns nonnegative value coordinates to nodes.\n\\item $\\mathbf c$ assigns nonnegative storage, audit, recovery, and cognitive costs.\n\\end{itemize}\nThe dependency graph contains an edge $u\\to v$ whenever $u\\in\\operatorname{prem}(j)$ and $v=\\operatorname{con}(j)$ for some justification $j$. It is required to be acyclic.\n\\end{definition}","\\begin{definition}[Admissible leaf]\n\\label{res:3:2}\nA node is an admissible leaf when it is a required premise, a declared axiom, or an explicitly admitted external result. Write $L$ for this set, so $A\\subseteq L$. A certified justification with no premises can also derive a node, but it remains a justification and need not make that node a member of $L$. The distinction matters when extracting the smallest certificate.\n\\end{definition}","Required premises are not necessarily the entire assumption context. Admitted axioms and external results remain visible in the supporting ledger. The notation $A\\vdash B$ is relative to that declared background and does not erase those dependencies. A comparison between certificates must retain the assumptions under which each certificate is valid.","\\begin{definition}[Certified subfield]\n\\label{res:3:3}\nThe certified subfield $\\Pcal^{\\mathrm{cert}}$ contains every node of $\\Pcal$ and only those justifications accepted by the declared certification policy. Evidence and scope are part of that policy. A certification flag records acceptance: it is not itself a proof that the rule is sound under an intended mathematical interpretation. That separate soundness contract is made explicit in Section~\\ref{sec:old24}.\n\\end{definition}","\\begin{definition}[Derivability]\n\\label{res:3:4}\nLet $K\\subseteq V$ be a selected node set and let $H\\subseteq\\Jcal$ be a selected justification set. The derivable set $D(K,H)$ is the least subset of $K$ satisfying:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item every selected admissible leaf belongs to $D(K,H)$.\n\\item if $j\\in H$, $\\operatorname{prem}(j)\\subseteq D(K,H)$, and $\\operatorname{con}(j)\\in K$, then $\\operatorname{con}(j)\\in D(K,H)$.\n\\end{enumerate}\nWhen only certified justifications are allowed, write $D_{\\mathrm{cert}}(K,H)$.\n\\end{definition}","Derivability]\n\\label{res:3:4}\nLet $K\\subseteq V$ be a selected node set and let $H\\subseteq\\Jcal$ be a selected justification set. The derivable set $D(K,H)$ is the least subset of $K$ satisfying:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item every selected admissible leaf belongs to $D(K,H)$.\n\\item if $j\\in H$, $\\operatorname{prem}(j)\\subseteq D(K,H)$, and $\\operatorname{con}(j)\\in K$, then $\\operatorname{con}(j)\\in D(K,H)$.\n\\end{enumerate}\nWhen only certified justifications are allowed, write $D_{\\mathrm{cert}}(K,H)$.\n\\end{definition}","lected justification set. The derivable set $D(K,H)$ is the least subset of $K$ satisfying:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item every selected admissible leaf belongs to $D(K,H)$.\n\\item if $j\\in H$, $\\operatorname{prem}(j)\\subseteq D(K,H)$, and $\\operatorname{con}(j)\\in K$, then $\\operatorname{con}(j)\\in D(K,H)$.\n\\end{enumerate}\nWhen only certified justifications are allowed, write $D_{\\mathrm{cert}}(K,H)$.\n\\end{definition}","\\begin{proposition}[Finite derivability]\n\\label{res:3:5}\nFor a finite acyclic proof field, $D(K,H)$ is uniquely determined and can be computed in one topological pass through the dependency graph.\n\\end{proposition}","\\begin{definition}[Scoped branch assessment]\n\\label{res:22:1}\nLet $\\Pcal$ be a finite typed proof field with declared premises $A$ and target $t$. An exploratory branch is a finite sequence of candidate justifications whose certification and outstanding obligations are explicitly recorded. Its assessment is one of the following, relative to stated premises and rules:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item \\emph{Open}: obligations remain unresolved.\n\\item \\emph{Abandoned}: exploration stopped without a certificate of impossibility.\n\\item \\emph{Locally refuted}: a checked counterexample or obstruction excludes a stated intermediate step under specified hypotheses.\n\\item \\emph{Certified}: the selected justifications meet the declared proof certificate co","\\begin{definition}[Scoped branch assessment]\n\\label{res:22:1}\nLet $\\Pcal$ be a finite typed proof field with declared premises $A$ and target $t$. An exploratory branch is a finite sequence of candidate justifications whose certification and outstanding obligations are explicitly recorded. Its assessment is one of the following, relative to stated premises and rules:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item \\emph{Open}: obligations remain unresolved.\n\\item \\emph{Abandoned}: exploration stopped without a certificate of impossibility.\n\\item \\emph{Locally refuted}: a checked counterexample or obstruction excludes a stated intermediate step under specified hypotheses.\n\\item \\emph{Certified}: the selected justifications meet the declared proof certificate conditions.\n\\end{enumerate}\nAn assessment is a record of th","\\begin{definition}[Scoped branch assessment]\n\\label{res:22:1}\nLet $\\Pcal$ be a finite typed proof field with declared premises $A$ and target $t$. An exploratory branch is a finite sequence of candidate justifications whose certification and outstanding obligations are explicitly recorded. Its assessment is one of the following, relative to stated premises and rules:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item \\emph{Open}: obligations remain unresolved.\n\\item \\emph{Abandoned}: exploration stopped without a certificate of impossibility.\n\\item \\emph{Locally refuted}: a checked counterexample or obstruction excludes a stated intermediate step under specified hypotheses.\n\\item \\emph{Certified}: the selected justifications meet the declared proof certificate conditions.\n\\end{enumerate}\nAn assessment is a record of the branch, not a","ation stopped without a certificate of impossibility.\n\\item \\emph{Locally refuted}: a checked counterexample or obstruction excludes a stated intermediate step under specified hypotheses.\n\\item \\emph{Certified}: the selected justifications meet the declared proof certificate conditions.\n\\end{enumerate}\nAn assessment is a record of the branch, not a truth value assigned to every possible proof of $t$.\n\\end{definition}","\\begin{proposition}[Local failure does not imply global failure]\n\\label{res:22:2}\nA finite proof field may contain a locally refuted branch to target $t$ while also containing a certified derivation of $t$ from its declared premises.\n\\end{proposition}\n\\begin{proof}\nTake a premise $A$ and two alternative candidate derivations of $t$. Let the first use a proposed intermediate lemma $L$ that has a checked counterexample, so the corresponding justification is not certified. Let the second use a distinct certified justification from $A$ to $t$. The certified closure reaches $t$ by the second route. The failed first route remains a valid record of unsuccessful exploration, but cannot negate the existence of the second certificate. The same example can be represented by the alternative justification hypergraph developed earlier.\n\\end{proof}","anch to target $t$ while also containing a certified derivation of $t$ from its declared premises.\n\\end{proposition}\n\\begin{proof}\nTake a premise $A$ and two alternative candidate derivations of $t$. Let the first use a proposed intermediate lemma $L$ that has a checked counterexample, so the corresponding justification is not certified. Let the second use a distinct certified justification from $A$ to $t$. The certified closure reaches $t$ by the second route. The failed first route remains a valid record of unsuccessful exploration, but cannot negate the existence of the second certificate. The same example can be represented by the alternative justification hypergraph developed earlier.\n\\end{proof}","target $t$ while also containing a certified derivation of $t$ from its declared premises.\n\\end{proposition}\n\\begin{proof}\nTake a premise $A$ and two alternative candidate derivations of $t$. Let the first use a proposed intermediate lemma $L$ that has a checked counterexample, so the corresponding justification is not certified. Let the second use a distinct certified justification from $A$ to $t$. The certified closure reaches $t$ by the second route. The failed first route remains a valid record of unsuccessful exploration, but cannot negate the existence of the second certificate. The same example can be represented by the alternative justification hypergraph developed earlier.\n\\end{proof}","certified derivation of $t$ from its declared premises.\n\\end{proposition}\n\\begin{proof}\nTake a premise $A$ and two alternative candidate derivations of $t$. Let the first use a proposed intermediate lemma $L$ that has a checked counterexample, so the corresponding justification is not certified. Let the second use a distinct certified justification from $A$ to $t$. The certified closure reaches $t$ by the second route. The failed first route remains a valid record of unsuccessful exploration, but cannot negate the existence of the second certificate. The same example can be represented by the alternative justification hypergraph developed earlier.\n\\end{proof}","\\begin{proposition}[Scoped refutation survives unchanged context]\n\\label{res:22:3}\nSuppose a witness $w$ is verified by a sound checker to refute the predicate $L$ under assumptions $A$. In any later archive revision that retains the same interpretation of $L$, the same assumptions, and the checker's verification rules, $w$ remains a refutation of $L$. No conclusion follows about different predicates, altered assumptions, or alternative branches to a different goal.\n\\end{proposition}\n\\begin{proof}\nThe verification of $w$ depends only on the declared predicate, assumptions, and checking rules. An append-only collection of unrelated exploratory records does not alter those inputs. Therefore the same verification remains valid. When any of those inputs changes, the earlier check no longer certif","\\begin{proposition}[Scoped refutation survives unchanged context]\n\\label{res:22:3}\nSuppose a witness $w$ is verified by a sound checker to refute the predicate $L$ under assumptions $A$. In any later archive revision that retains the same interpretation of $L$, the same assumptions, and the checker's verification rules, $w$ remains a refutation of $L$. No conclusion follows about different predicates, altered assumptions, or alternative branches to a different goal.\n\\end{proposition}\n\\begin{proof}\nThe verification of $w$ depends only on the declared predicate, assumptions, and checking rules. An append-only collection of unrelated exploratory records does not alter those inputs. Therefore the same verification remains valid. When any of those inputs changes, the earlier check no longer certifies the revised claim without an explicit comparison theore","\\begin{proposition}[Scoped refutation survives unchanged context]\n\\label{res:22:3}\nSuppose a witness $w$ is verified by a sound checker to refute the predicate $L$ under assumptions $A$. In any later archive revision that retains the same interpretation of $L$, the same assumptions, and the checker's verification rules, $w$ remains a refutation of $L$. No conclusion follows about different predicates, altered assumptions, or alternative branches to a different goal.\n\\end{proposition}\n\\begin{proof}\nThe verification of $w$ depends only on the declared predicate, assumptions, and checking rules. An append-only collection of unrelated exploratory records does not alter those inputs. Therefore the same verification remains valid. When any of those inputs changes, the earlier check no longer certifies the revised claim without an explicit comparison theorem.\n\\end{proof}","und checker to refute the predicate $L$ under assumptions $A$. In any later archive revision that retains the same interpretation of $L$, the same assumptions, and the checker's verification rules, $w$ remains a refutation of $L$. No conclusion follows about different predicates, altered assumptions, or alternative branches to a different goal.\n\\end{proposition}\n\\begin{proof}\nThe verification of $w$ depends only on the declared predicate, assumptions, and checking rules. An append-only collection of unrelated exploratory records does not alter those inputs. Therefore the same verification remains valid. When any of those inputs changes, the earlier check no longer certifies the revised claim without an explicit comparison theorem.\n\\end{proof}","\\begin{definition}[Certificate]\n\\label{res:4:1}\nA certificate for a nonempty requested target set $T_0\\subseteq V$ is a pair\n\\[\n\\Ccal=(K,H),\n\\qquad K\\subseteq V,\n\\qquad H\\subseteq\\Jcal^{\\mathrm{cert}},\n\\]\nsuch that\n\\[\nA\\cup T_0\\subseteq K\n\\qquad\\text{and}\\qquad\nT_0\\subseteq D_{\\mathrm{cert}}(K,H).\n\\]\nOrder certificates componentwise: $(K,H)\\leq(K^{\\prime},H^{\\prime})$ means $K\\subseteq K^{\\prime}$ and $H\\subseteq H^{\\prime}$. A least certificate is contained in every certificate for the same request. Minimizing only the node set, only the justification set, or a numerical cost is a different objective.\n\\end{definition}","{definition}[Certificate]\n\\label{res:4:1}\nA certificate for a nonempty requested target set $T_0\\subseteq V$ is a pair\n\\[\n\\Ccal=(K,H),\n\\qquad K\\subseteq V,\n\\qquad H\\subseteq\\Jcal^{\\mathrm{cert}},\n\\]\nsuch that\n\\[\nA\\cup T_0\\subseteq K\n\\qquad\\text{and}\\qquad\nT_0\\subseteq D_{\\mathrm{cert}}(K,H).\n\\]\nOrder certificates componentwise: $(K,H)\\leq(K^{\\prime},H^{\\prime})$ means $K\\subseteq K^{\\prime}$ and $H\\subseteq H^{\\prime}$. A least certificate is contained in every certificate for the same request. Minimizing only the node set, only the justification set, or a numerical cost is a different objective.\n\\end{definition}","The usual target request is a subset of $T$, but an intermediate node may be requested for local analysis. Only a justification whose premises have been derived and whose conclusion is selected can contribute to the derivation. A record may retain other, inactive justifications without using them as proof steps. Extra records cannot lower a nonnegative additive cost, although zero costs can allow several minimizing records.","\\begin{definition}[AND-DAG]\n\\label{res:4:2}\nStarting at $T_0$, expand backward only at nodes outside $L$. The field is an AND-DAG relative to $T_0$ when every node reached outside $L$ has exactly one certified incoming justification. Every premise of that justification is then expanded in the same manner. The process may terminate at an admissible leaf or at a certified justification with no premises.\n\\end{definition}","Let $\\Anc_L(T_0)$ denote the premise nodes reached by this backward expansion, stopping at $L$. Let $H_{\\mathrm{core}}$ contain exactly the justifications selected at expanded nodes outside $L$.","\\begin{theorem}[Unique verification core]\n\\label{res:4:3}\n\\label{thm:and-core}\nLet $\\Pcal$ be a finite acyclic AND-DAG relative to $T_0$. Put\n\\[\nK_{\\mathrm{core}}=A\\cup T_0\\cup\\Anc_L(T_0).\n\\]\nThen $(K_{\\mathrm{core}},H_{\\mathrm{core}})$ is a certificate and is the least certificate under componentwise inclusion. In particular it is the unique inclusion-minimal certificate pair for this recorded inference problem.\n\\end{theorem}","\\begin{proof}\nBackward expansion terminates because the dependency graph is finite and acyclic. Every terminal node is either an admissible leaf or the conclusion of the selected justification with no premises. A forward topological pass therefore derives every expanded node. The required premises in $A$ are admissible and are retained even if the target does not use them. Thus the constructed pair is a certificate.","Now let $(K,H)$ be any certificate for the same request. It contains $A\\cup T_0$. If a requested or required predecessor node is outside $L$, its derivation must use its unique certified incoming justification. That justification belongs to $H$, and every one of its premises belongs to $K$. Repeating the argument forces all of $\\Anc_L(T_0)$ and $H_{\\mathrm{core}}$. At nodes in $L$ the argument stops. Hence $K_{\\mathrm{core}}\\subseteq K$ and $H_{\\mathrm{core}}\\subseteq H$, which proves the least-pair assertion.",",H)$ be any certificate for the same request. It contains $A\\cup T_0$. If a requested or required predecessor node is outside $L$, its derivation must use its unique certified incoming justification. That justification belongs to $H$, and every one of its premises belongs to $K$. Repeating the argument forces all of $\\Anc_L(T_0)$ and $H_{\\mathrm{core}}$. At nodes in $L$ the argument stops. Hence $K_{\\mathrm{core}}\\subseteq K$ and $H_{\\mathrm{core}}\\subseteq H$, which proves the least-pair assertion.","e request. It contains $A\\cup T_0$. If a requested or required predecessor node is outside $L$, its derivation must use its unique certified incoming justification. That justification belongs to $H$, and every one of its premises belongs to $K$. Repeating the argument forces all of $\\Anc_L(T_0)$ and $H_{\\mathrm{core}}$. At nodes in $L$ the argument stops. Hence $K_{\\mathrm{core}}\\subseteq K$ and $H_{\\mathrm{core}}\\subseteq H$, which proves the least-pair assertion.","\\begin{corollary}[Linear extraction]\n\\label{res:4:4}\nWith indexed certified adjacency and constant-time membership bookkeeping, the core can be extracted in time linear in $|A|$ plus the number of visited nodes, selected justifications, and premise incidences.\n\\end{corollary}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_02_and_dag_core.pdf}\n\\caption{The canonical \\protect\\path{and_dag.json} example. The AND gate requires both $A$ and $D$ for $L_1$. Its least certificate contains five nodes and four certified rules. The gold branch is certified but unnecessary for $B$. The dashed route to $F$ is uncertified and cannot enter the certificate.}\n\\label{fig:andcore}\n\\end{figure}","\\begin{problem}[Minimum Certificate Decision]\n\\label{res:5:1}\nGiven a finite acyclic certified proof field, a target set, nonnegative integer node and justification costs, and a budget $B$, determine whether a certificate of total cost at most $B$ exists.\n\\end{problem}","For NP-hardness, reduce Set Cover. Let $U=\\{u_1,\\ldots,u_m\\}$ be the universe, let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once","For NP-hardness, reduce Set Cover. Let $U=\\{u_1,\\ldots,u_m\\}$ be the universe, let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exi","For NP-hardness, reduce Set Cover. Let $U=\\{u_1,\\ldots,u_m\\}$ be the universe, let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set C","For NP-hardness, reduce Set Cover. Let $U=\\{u_1,\\ldots,u_m\\}$ be the universe, let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynom","For NP-hardness, reduce Set Cover. Let $U=\\{u_1,\\ldots,u_m\\}$ be the universe, let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","$U=\\{u_1,\\ldots,u_m\\}$ be the universe, let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}",", let $S_1,\\ldots,S_n\\subseteq U$ be the available sets, and let $k$ be the cover budget. Construct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","struct one optional admissible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","missible leaf node $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","de $s_i$ for each set $S_i$, with unit cost. Construct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","truct one zero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","ero-cost obligation node $e_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","_j$ for each element $u_j$. Whenever $u_j\\in S_i$, add a certified justification\n\\[\n\\{s_i\\}\\Longrightarrow e_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","_j.\n\\]\nAdd a zero-cost target node $t$ with the certified justification\n\\[\n\\{e_1,\\ldots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","ots,e_m\\}\\Longrightarrow t.\n\\]\nThe dependency graph is acyclic. A certificate for $t$ must derive every $e_j$. Deriving $e_j$ requires at least one selected source node $s_i$ with $u_j\\in S_i$. One selected $s_i$ may discharge several obligations, and its cost is paid once. Therefore a certificate of cost at most $k$ exists exactly when the original Set Cover instance has a cover of size at most $k$. The reduction is polynomial.\n\\end{proof}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_04_certificate_complexity.pdf}\n\\caption{The Set Cover example as a membership matrix. Filled marks belong to the selected rows $S_1$ and $S_3$; open marks show unselected memberships. Every obligation is covered. The target requires all four obligations through one AND rule. The exact minimum is two, although the minimizing pair need not be unique.}\n\\label{fig:setcover}\n\\end{figure}","\\begin{definition}[Certified proof category]\n\\label{res:6:1}\nFix a directed graph of typed mathematical states and certified primitive transformations. Its path category $\\mathsf{Cert}$ has these states as objects and finite composable paths as morphisms. The empty path is an identity, and composition is concatenation. One may further identify paths by a declared congruence compatible with concatenation. Associativity and the identity laws then hold by construction.\n\\end{definition}","A state may be a proposition, a diagram, an exact finite computation, a coordinate representation, or a structured collection of claims. Candidate analogies may live in a larger discovery graph, but they do not enter $\\mathsf{Cert}$ until their logical force and domain have been certified.","\\begin{proposition}[Functorial transport]\n\\label{res:6:2}\nLet $F:\\mathsf{Cert}_1\\to\\mathsf{Cert}_2$ be a functor. If\n\\[\nA\\xrightarrow{f_1}S_1\\xrightarrow{f_2}\\cdots\\xrightarrow{f_r}B\n\\]\nis a certified path in $\\mathsf{Cert}_1$, then\n\\[\nF(A)\\xrightarrow{F(f_1)}F(S_1)\\xrightarrow{F(f_2)}\\cdots\\xrightarrow{F(f_r)}F(B)\n\\]\nis a certified path in $\\mathsf{Cert}_2$.\n\\end{proposition}","\\begin{definition}[Compositional certificate map]\n\\label{res:6:3}\nA compositional certificate map is a functor\n\\[\n\\Ccal:\\mathsf{Expanded}\\longrightarrow\\mathsf{Certificate}\n\\]\nthat sends expanded certified paths to checkable certificate paths and preserves identity and composition.\n\\end{definition}","\\begin{proof}\nProceed by induction on path length. The length-zero case is an identity. Assume the first $r$ morphisms have an auditable composite certificate and append one certified morphism. Functoriality identifies the new certificate with the composition of the earlier certificate and the primitive certificate of the appended morphism. The declared composition rule completes the audit.\n\\end{proof}","\\begin{definition}[Coherent views of a proof field]\n\\label{res:23:1}\nChoose a finite collection of views $V$, fibers $F_v$ of local proof states or annotations, and for each declared directed edge $e:v\\to w$ a restriction map $e^*:F_w\\to F_v$. The \\emph{strict realization set} is\n\\[\n\\Gamma(F)=\\big\\{(x_v)\\in\\prod_{v\\in V}F_v : e^*(x_w)=x_v\\text{ for every declared edge }e\\big\\}.\n\\]\nThe \\emph{extension fiber} of prescribed information $s$ on $U\\subseteq V$ consists of those global coherent families whose restrictions equal $s$. If the edges form a category, the usual functorial composition equations are additional explicit obligations.\n\\end{definition}","proof field]\n\\label{res:23:1}\nChoose a finite collection of views $V$, fibers $F_v$ of local proof states or annotations, and for each declared directed edge $e:v\\to w$ a restriction map $e^*:F_w\\to F_v$. The \\emph{strict realization set} is\n\\[\n\\Gamma(F)=\\big\\{(x_v)\\in\\prod_{v\\in V}F_v : e^*(x_w)=x_v\\text{ for every declared edge }e\\big\\}.\n\\]\nThe \\emph{extension fiber} of prescribed information $s$ on $U\\subseteq V$ consists of those global coherent families whose restrictions equal $s$. If the edges form a category, the usual functorial composition equations are additional explicit obligations.\n\\end{definition}","field]\n\\label{res:23:1}\nChoose a finite collection of views $V$, fibers $F_v$ of local proof states or annotations, and for each declared directed edge $e:v\\to w$ a restriction map $e^*:F_w\\to F_v$. The \\emph{strict realization set} is\n\\[\n\\Gamma(F)=\\big\\{(x_v)\\in\\prod_{v\\in V}F_v : e^*(x_w)=x_v\\text{ for every declared edge }e\\big\\}.\n\\]\nThe \\emph{extension fiber} of prescribed information $s$ on $U\\subseteq V$ consists of those global coherent families whose restrictions equal $s$. If the edges form a category, the usual functorial composition equations are additional explicit obligations.\n\\end{definition}","cal proof states or annotations, and for each declared directed edge $e:v\\to w$ a restriction map $e^*:F_w\\to F_v$. The \\emph{strict realization set} is\n\\[\n\\Gamma(F)=\\big\\{(x_v)\\in\\prod_{v\\in V}F_v : e^*(x_w)=x_v\\text{ for every declared edge }e\\big\\}.\n\\]\nThe \\emph{extension fiber} of prescribed information $s$ on $U\\subseteq V$ consists of those global coherent families whose restrictions equal $s$. If the edges form a category, the usual functorial composition equations are additional explicit obligations.\n\\end{definition}","\\begin{theorem}[Finite arrow-cover gluing]\n\\label{res:23:2}\nSuppose patches $U_i\\subseteq V$ cover every view and every declared edge has both endpoints in some $U_i$. Let $s_i$ be coherent assignments on $U_i$ that agree on all overlaps. There is a unique global coherent assignment $s\\in\\Gamma(F)$ restricting to every $s_i$.\n\\end{theorem}\n\\begin{proof}\nFor each view $v$, choose any patch containing it and define $s(v)$ to be the value supplied by that patch. Agreement on overlaps makes this independent of the choice. For a declared edge $e:v\\to w$, choose a patch containing both endpoints. The local assignment on that patch obeys $e^*s(w)=s(v)$, so the global assignment is coherent. It is the only possible glue because every coordinate was prescribed by at least one patch.\n\\end{p","\\begin{theorem}[Finite arrow-cover gluing]\n\\label{res:23:2}\nSuppose patches $U_i\\subseteq V$ cover every view and every declared edge has both endpoints in some $U_i$. Let $s_i$ be coherent assignments on $U_i$ that agree on all overlaps. There is a unique global coherent assignment $s\\in\\Gamma(F)$ restricting to every $s_i$.\n\\end{theorem}\n\\begin{proof}\nFor each view $v$, choose any patch containing it and define $s(v)$ to be the value supplied by that patch. Agreement on overlaps makes this independent of the choice. For a declared edge $e:v\\to w$, choose a patch containing both endpoints. The local assignment on that patch obeys $e^*s(w)=s(v)$, so the global assignment is coherent. It is the only possible glue because every coordinate was prescribed by at least one patch.\n\\end{proof}","nite arrow-cover gluing]\n\\label{res:23:2}\nSuppose patches $U_i\\subseteq V$ cover every view and every declared edge has both endpoints in some $U_i$. Let $s_i$ be coherent assignments on $U_i$ that agree on all overlaps. There is a unique global coherent assignment $s\\in\\Gamma(F)$ restricting to every $s_i$.\n\\end{theorem}\n\\begin{proof}\nFor each view $v$, choose any patch containing it and define $s(v)$ to be the value supplied by that patch. Agreement on overlaps makes this independent of the choice. For a declared edge $e:v\\to w$, choose a patch containing both endpoints. The local assignment on that patch obeys $e^*s(w)=s(v)$, so the global assignment is coherent. It is the only possible glue because every coordinate was prescribed by at least one patch.\n\\end{proof}","dpoints in some $U_i$. Let $s_i$ be coherent assignments on $U_i$ that agree on all overlaps. There is a unique global coherent assignment $s\\in\\Gamma(F)$ restricting to every $s_i$.\n\\end{theorem}\n\\begin{proof}\nFor each view $v$, choose any patch containing it and define $s(v)$ to be the value supplied by that patch. Agreement on overlaps makes this independent of the choice. For a declared edge $e:v\\to w$, choose a patch containing both endpoints. The local assignment on that patch obeys $e^*s(w)=s(v)$, so the global assignment is coherent. It is the only possible glue because every coordinate was prescribed by at least one patch.\n\\end{proof}","here is a unique global coherent assignment $s\\in\\Gamma(F)$ restricting to every $s_i$.\n\\end{theorem}\n\\begin{proof}\nFor each view $v$, choose any patch containing it and define $s(v)$ to be the value supplied by that patch. Agreement on overlaps makes this independent of the choice. For a declared edge $e:v\\to w$, choose a patch containing both endpoints. The local assignment on that patch obeys $e^*s(w)=s(v)$, so the global assignment is coherent. It is the only possible glue because every coordinate was prescribed by at least one patch.\n\\end{proof}","\\begin{example}[Object coverage cannot replace arrow coverage]\n\\label{res:23:3}\nLet $V=\\{v,w\\}$ with an edge $v\\to w$, both fibers $\\{0,1\\}$, and identity restriction. Singleton patches prescribe $s(v)=0$ and $s(w)=1$. These assignments cover all views and agree on their empty overlap. There is no global coherent assignment with those coordinates. The missing ingredient is the edge condition, not the availability of the two individual values.\n\\end{example}","A second connection is quantitative. Work over $\\mathbb F_2$ with $E=\\mathbb F_2^n$. Let $A:E\\to\\mathbb F_2^m$ impose coherence, $P:E\\to\\mathbb F_2^p$ retain prescribed information, and $x\\in E$ be a raw candidate. Define\n\\[\n\\mathfrak K_P(x)=\\{y\\in E:Ay=0,\\ Py=Px\\},\\qquad\nM=\\begin{bmatrix}A","A second connection is quantitative. Work over $\\mathbb F_2$ with $E=\\mathbb F_2^n$. Let $A:E\\to\\mathbb F_2^m$ impose coherence, $P:E\\to\\mathbb F_2^p$ retain prescribed information, and $x\\in E$ be a raw candidate. Define\n\\[\n\\mathfrak K_P(x)=\\{y\\in E:Ay=0,\\ Py=Px\\},\\qquad\nM=\\begin{bmatrix}A\\\\P\\end{bmatrix},\\quad\nb=\\begin{bmatrix}0\\\\Px\\end{bmatrix}.\n\\]\nThis is a finite linear realization fiber. The equation $My=b$ does not by itself produce a solution. Compatibility of the prescribed information must first be established.","Px\\end{bmatrix}.\n\\]\nThis is a finite linear realization fiber. The equation $My=b$ does not by itself produce a solution. Compatibility of the prescribed information must first be established.","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row v","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradict","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive a","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field","\\begin{theorem}[Finite realization alternative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","ernative and ambiguity]\n\\label{res:23:4}\nExactly one of the following holds:\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","egin{enumerate}[label=\\textup{(\\roman*)}]\n\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","\\item $My=b$ has a solution $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","tion $y_0$. Its complete solution set is $y_0+H$, where $H=\\ker A\\cap\\ker P$, and\n\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","\\[\n|\\mathfrak K_P(x)|=2^{n-\\operatorname{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","ame{rank}(M)}.\n\\]\n\\item A row vector $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","or $\\lambda$ exists with $\\lambda M=0$ and $\\lambda b=1$. It is a directly checkable certificate that $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","t $\\mathfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","hfrak K_P(x)=\\varnothing$.\n\\end{enumerate}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","}\n\\end{theorem}\n\\begin{proof}\nIf $My_0=b$, any other solution differs from $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","rom $y_0$ by an element of $\\ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","ker M=H$, and every such difference gives another solution. Rank nullity gives $\\dim H=n-\\operatorname{rank}(M)$. If $b$ is not in the column space of $M$, linear algebra supplies a linear functional that vanishes on that column space and takes value $1$ on $b$. Expressing it as a row vector gives $\\lambda$. Conversely such a witness contradicts any proposed equation $My=b$, since it would imply $0=\\lambda My=\\lambda b=1$. These cases are mutually exclusive and exhaustive over a field.\n\\end{proof}","This negative certificate is a concrete instance of the protected obstruction principle from Komorebi Space. Its scope is exactly the finite linear system $My=b$. It is not a certificate that some broad mathematical conjecture is unprovable.","Suppose the realization fiber is nonempty and a new linear observation $Q:E\\to\\mathbb F_2^q$ becomes available. The right question is not how many bits appear in the output of $Q$ but how many remaining degrees of freedom in the fiber it distinguishes.","\\begin{theorem}[Exact observation gain]\n\\label{res:23:5}\nGive the nonempty finite affine fiber $y_0+H$ the uniform distribution. Then the Shannon entropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induce","\\begin{theorem}[Exact observation gain]\n\\label{res:23:5}\nGive the nonempty finite affine fiber $y_0+H$ the uniform distribution. Then the Shannon entropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$,","\\begin{theorem}[Exact observation gain]\n\\label{res:23:5}\nGive the nonempty finite affine fiber $y_0+H$ the uniform distribution. Then the Shannon entropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}","\\begin{theorem}[Exact observation gain]\n\\label{res:23:5}\nGive the nonempty finite affine fiber $y_0+H$ the uniform distribution. Then the Shannon entropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","gain]\n\\label{res:23:5}\nGive the nonempty finite affine fiber $y_0+H$ the uniform distribution. Then the Shannon entropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","uniform distribution. Then the Shannon entropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","ntropy of the observed value $Qy$ is\n\\[\nH(Qy)=\\operatorname{rank}(Q|_H)\\quad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","ad\\text{bits}.\n\\]\nFor every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","r every observation value that actually occurs, the corresponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","ponding conditional fiber contains\n\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","\\[\n2^{\\dim H-\\operatorname{rank}(Q|_H)}\n\\]\nrealizations. Thus the uncertainty reduction is exactly $\\operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","operatorname{rank}(Q|_H)$ bits.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","s.\n\\end{theorem}\n\\begin{proof}\nThe image $Q(H)$ is a vector subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","subspace of dimension $r=\\operatorname{rank}(Q|_H)$, so the observed values form the affine set $Qy_0+Q(H)$ of size $2^r$. Every such value has a preimage that is a coset of $H\\cap\\ker Q$, whose dimension is $\\dim H-r$ by rank nullity. Equal fiber cardinalities make the induced output distribution uniform on $2^r$ values. Its entropy is $\\log_2(2^r)=r$, and each conditional fiber has cardinality $2^{\\dim H-r}$.\n\\end{proof}","\\begin{example}[One informative question]\n\\label{res:23:6}\nLet $y=(a,b,c)\\in\\mathbb F_2^3$ with $a+b+c=0$ and protected value $a=0$. The realization fiber is $\\{(0,0,0),(0,1,1)\\}$. It has one bit of ambiguity. Observing $b$ resolves that bit exactly. Observing $a$ supplies no new information, despite being a perfectly valid measurement. This difference is measured by the rank of the observation restricted to $H$, rather than by the output width of the observation.\n\\end{example}","e}[One informative question]\n\\label{res:23:6}\nLet $y=(a,b,c)\\in\\mathbb F_2^3$ with $a+b+c=0$ and protected value $a=0$. The realization fiber is $\\{(0,0,0),(0,1,1)\\}$. It has one bit of ambiguity. Observing $b$ resolves that bit exactly. Observing $a$ supplies no new information, despite being a perfectly valid measurement. This difference is measured by the rank of the observation restricted to $H$, rather than by the output width of the observation.\n\\end{example}","Fix a node set $V$, a set $L\\subseteq V$ of admissible leaves, and a set $J$ of justifications. A justification $j$ has a premise set $A_j$, a conclusion $c_j$, and a certification predicate. Define its support by\n\\[\nS_j=A_j\\cup\\{c_j\\}.\n\\]\nA set $X\\subseteq V$ is a \\emph{model of the declared field} when $L\\subseteq X$ and, for every certified $j$,\n\\[\nA_j\\subseteq X\\quad\\Longrightarrow\\quad c_j\\in X.\n\\]\nEach justification is one implication with conjunctive premises. Alternative justifications remain separate implications. Uncertified attempts and failed-route annotations do not contribute a deduction rule merely by appearing in the archive.","Let $\\operatorname{Der}_F$ be the inductively generated set obtained from the leaves by certified rules. For a finite field, repeated forward closure computes this set. It is not necessary to enumerate all valuations to perform that computation.","Every declared field of this positive-implication form has a model: the full set $V$. It contains every leaf and every rule conclusion. Nonexistence in the following extension problems therefore concerns prescribed local values or additional preservation requirements, not this unrestricted model space. This does not make every node derivable. Derivability is membership in the least model, not membership in some model.","\\begin{theorem}[Soundness and relative completeness]\n\\label{res:24:1}\n\\label{thm:relative-completeness}\nFor every node $v$,\n\\[\nv\\in\\operatorname{Der}_F\n\\quad\\Longleftrightarrow\\quad\n\\text{$v\\in X$ for every model $X$ of the declared field}.\n\\]\nIf $v$ is not derivable, $\\operatorname{Der}_F$ itself is a model omitting $v$.\n\\end{theorem}\n\\begin{proof}\nEvery model contains the leaves. Induction on a derivation shows that it contains the conclusion of every certified application whose premises have already been derived. Thus every model contains $\\operatorname{Der}_F$. Conversely, the inductive constructors show that $\\operatorname{Der}_F$ contains the leaves and is closed under every certified rule. It is therefore a model. Applying the universal condition to this particular model proves the reverse implication,","\\begin{theorem}[Soundness and relative completeness]\n\\label{res:24:1}\n\\label{thm:relative-completeness}\nFor every node $v$,\n\\[\nv\\in\\operatorname{Der}_F\n\\quad\\Longleftrightarrow\\quad\n\\text{$v\\in X$ for every model $X$ of the declared field}.\n\\]\nIf $v$ is not derivable, $\\operatorname{Der}_F$ itself is a model omitting $v$.\n\\end{theorem}\n\\begin{proof}\nEvery model contains the leaves. Induction on a derivation shows that it contains the conclusion of every certified application whose premises have already been derived. Thus every model contains $\\operatorname{Der}_F$. Conversely, the inductive constructors show that $\\operatorname{Der}_F$ contains the leaves and is closed under every certified rule. It is therefore a model. Applying the universal condition to this particular model proves the reverse implication, and the same model gives the stated countermodel.\n\\end{proof}","_F\n\\quad\\Longleftrightarrow\\quad\n\\text{$v\\in X$ for every model $X$ of the declared field}.\n\\]\nIf $v$ is not derivable, $\\operatorname{Der}_F$ itself is a model omitting $v$.\n\\end{theorem}\n\\begin{proof}\nEvery model contains the leaves. Induction on a derivation shows that it contains the conclusion of every certified application whose premises have already been derived. Thus every model contains $\\operatorname{Der}_F$. Conversely, the inductive constructors show that $\\operatorname{Der}_F$ contains the leaves and is closed under every certified rule. It is therefore a model. Applying the universal condition to this particular model proves the reverse implication, and the same model gives the stated countermodel.\n\\end{proof}","e declared field}.\n\\]\nIf $v$ is not derivable, $\\operatorname{Der}_F$ itself is a model omitting $v$.\n\\end{theorem}\n\\begin{proof}\nEvery model contains the leaves. Induction on a derivation shows that it contains the conclusion of every certified application whose premises have already been derived. Thus every model contains $\\operatorname{Der}_F$. Conversely, the inductive constructors show that $\\operatorname{Der}_F$ contains the leaves and is closed under every certified rule. It is therefore a model. Applying the universal condition to this particular model proves the reverse implication, and the same model gives the stated countermodel.\n\\end{proof}","For an intended interpretation $\\mu:V\\to\\mathrm{Prop}$, the transfer to mathematical truth has explicit hypotheses: every leaf must satisfy $\\mu$, and every certified rule must preserve $\\mu$ from all its premises to its conclusion. Under those hypotheses, induction gives $v\\in\\operatorname{Der}_F\\Rightarrow\\mu(v)$. Lean checks this statement as \\path{ProofPath.Field.semantic_soundness}. It does not infer those hypotheses from a JSON certification flag.","For a patch $U\\subseteq V$, let $\\mathcal M_F(U)$ consist of subsets $X_U\\subseteq U$ which contain $L\\cap U$ and obey every certified rule whose entire support lies in $U$. If $U\\subseteq W$, define\n\\[\n\\rho_{U,W}:\\mathcal M_F(W)\\longrightarrow\\mathcal M_F(U),\n\\qquad \\rho_{U,W}(X_W)=X_W\\cap U.\n\\]\nThis map is well defined. A rule supported in $U$ is also supported in $W$, and its conclusion remains in $U$. The restriction therefore preserves both the local leaves and the local implication conditions.","patch $U\\subseteq V$, let $\\mathcal M_F(U)$ consist of subsets $X_U\\subseteq U$ which contain $L\\cap U$ and obey every certified rule whose entire support lies in $U$. If $U\\subseteq W$, define\n\\[\n\\rho_{U,W}:\\mathcal M_F(W)\\longrightarrow\\mathcal M_F(U),\n\\qquad \\rho_{U,W}(X_W)=X_W\\cap U.\n\\]\nThis map is well defined. A rule supported in $U$ is also supported in $W$, and its conclusion remains in $U$. The restriction therefore preserves both the local leaves and the local implication conditions.","\\begin{proposition}[The proof-field view laws]\n\\label{res:24:2}\nThe restriction maps satisfy\n\\[\n\\rho_{U,U}=\\mathrm{id},\\qquad\n\\rho_{U,W}\\rho_{W,Z}=\\rho_{U,Z}\\quad(U\\subseteq W\\subseteq Z).\n\\]\nConsequently $U\\mapsto\\mathcal M_F(U)$ is a contravariant set-valued view system on the inclusion order of patches.\n\\end{proposition}\n\\begin{proof}\nBoth laws are the corresponding intersection identities. For the second, $(X\\cap W)\\cap U=X\\cap U$ because $U\\subseteq W$. The values outside a patch are absent, or equivalently normalized to false in the Lean representation. There are no additional hidden coordinates.\n\\end{proof}","tion}[The proof-field view laws]\n\\label{res:24:2}\nThe restriction maps satisfy\n\\[\n\\rho_{U,U}=\\mathrm{id},\\qquad\n\\rho_{U,W}\\rho_{W,Z}=\\rho_{U,Z}\\quad(U\\subseteq W\\subseteq Z).\n\\]\nConsequently $U\\mapsto\\mathcal M_F(U)$ is a contravariant set-valued view system on the inclusion order of patches.\n\\end{proposition}\n\\begin{proof}\nBoth laws are the corresponding intersection identities. For the second, $(X\\cap W)\\cap U=X\\cap U$ because $U\\subseteq W$. The values outside a patch are absent, or equivalently normalized to false in the Lean representation. There are no additional hidden coordinates.\n\\end{proof}","label{res:24:2}\nThe restriction maps satisfy\n\\[\n\\rho_{U,U}=\\mathrm{id},\\qquad\n\\rho_{U,W}\\rho_{W,Z}=\\rho_{U,Z}\\quad(U\\subseteq W\\subseteq Z).\n\\]\nConsequently $U\\mapsto\\mathcal M_F(U)$ is a contravariant set-valued view system on the inclusion order of patches.\n\\end{proposition}\n\\begin{proof}\nBoth laws are the corresponding intersection identities. For the second, $(X\\cap W)\\cap U=X\\cap U$ because $U\\subseteq W$. The values outside a patch are absent, or equivalently normalized to false in the Lean representation. There are no additional hidden coordinates.\n\\end{proof}","The strict Komorebi realization is now an actual set of coherent local models,\n\\[\n\\Gamma(\\mathcal M_F)=\n\\left\\{(X_U)_U:\\rho_{U,W}(X_W)=X_U\\text{ whenever }U\\subseteq W\\right\\}.\n\\]\nThere is an explicit bijection\n\\[\n\\operatorname{Models}(F)\\simeq\\Gamma(\\mathcal M_F),\n\\qquad X\\longmapsto(X\\cap U)_U.\n\\]\nIts inverse evaluates a coherent family at the full patch $V$. Coherence shows that every other component is precisely the restriction of that value. The Lean declaration \\path{ProofPath.Field.modelEquivSections} constructs this equivalence, using \\path{ViewDiagram} and the actual local restriction maps.","\\studyhead{01}{The bridge is a restriction map}\nFor one certified rule $\\{a,b\\}\\to c$ and no admissible leaves ($L=\\varnothing$), there are seven global Boolean rule models. Only $(1,1,0)$ violates the implication. On either two-node patch below, the complete support is absent, so all four local valuations are admissible.","\\begin{figure}[H]\\centering\n\\includegraphics[width=\\linewidth]{fig_16_restriction_bridge.pdf}\n\\caption{The local model sets and their actual restrictions. Each arrow is $\\rho_{U,W}(X)=X\\cap U$. The two ways of restricting a global model to $\\{c\\}$ agree. The diagram is counted directly from the finite rule model, with seven global models, four models on each two-node patch, and two overlap values.}\n\\label{fig:restriction-models}\n\\end{figure}","Let $(U_i)_{i\\in I}$ cover the nodes. A family $X_i\\in\\mathcal M_F(U_i)$ is matching when\n\\[\nX_i\\cap U_j=X_j\\cap U_i\\qquad\\text{for every }i,j.\n\\]\nThis equality includes agreement about false values, not merely common true values.","\\begin{theorem}[Whole-support gluing]\n\\label{res:24:3}\n\\label{thm:support-gluing}\nSuppose every certified justification has its whole support in some patch:\n\\[\n\\forall j\\text{ certified},\\quad\\exists i\\quad S_j\\subseteq U_i.\n\\]\nThen every matching family $(X_i)$ has a unique global field model $X$ satisfying $X\\cap U_i=X_i$ for all $i$.\n\\end{theorem}\n\\begin{proof}\nSet $X=\\bigcup_iX_i$. Matching and node coverage imply $X\\cap U_i=X_i$. A leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{p","\\begin{theorem}[Whole-support gluing]\n\\label{res:24:3}\n\\label{thm:support-gluing}\nSuppose every certified justification has its whole support in some patch:\n\\[\n\\forall j\\text{ certified},\\quad\\exists i\\quad S_j\\subseteq U_i.\n\\]\nThen every matching family $(X_i)$ has a unique global field model $X$ satisfying $X\\cap U_i=X_i$ for all $i$.\n\\end{theorem}\n\\begin{proof}\nSet $X=\\bigcup_iX_i$. Matching and node coverage imply $X\\cap U_i=X_i$. A leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{proof}","upport gluing]\n\\label{res:24:3}\n\\label{thm:support-gluing}\nSuppose every certified justification has its whole support in some patch:\n\\[\n\\forall j\\text{ certified},\\quad\\exists i\\quad S_j\\subseteq U_i.\n\\]\nThen every matching family $(X_i)$ has a unique global field model $X$ satisfying $X\\cap U_i=X_i$ for all $i$.\n\\end{theorem}\n\\begin{proof}\nSet $X=\\bigcup_iX_i$. Matching and node coverage imply $X\\cap U_i=X_i$. A leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{proof}","ied justification has its whole support in some patch:\n\\[\n\\forall j\\text{ certified},\\quad\\exists i\\quad S_j\\subseteq U_i.\n\\]\nThen every matching family $(X_i)$ has a unique global field model $X$ satisfying $X\\cap U_i=X_i$ for all $i$.\n\\end{theorem}\n\\begin{proof}\nSet $X=\\bigcup_iX_i$. Matching and node coverage imply $X\\cap U_i=X_i$. A leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{proof}","\\text{ certified},\\quad\\exists i\\quad S_j\\subseteq U_i.\n\\]\nThen every matching family $(X_i)$ has a unique global field model $X$ satisfying $X\\cap U_i=X_i$ for all $i$.\n\\end{theorem}\n\\begin{proof}\nSet $X=\\bigcup_iX_i$. Matching and node coverage imply $X\\cap U_i=X_i$. A leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{proof}","{theorem}\n\\begin{proof}\nSet $X=\\bigcup_iX_i$. Matching and node coverage imply $X\\cap U_i=X_i$. A leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{proof}","leaf belongs to some patch and hence to its local model, so $L\\subseteq X$. Consider a certified rule whose premises belong to $X$. Choose a patch containing its entire support. Matching implies that every premise is true in that patch's local model, whose rule condition forces the conclusion. Hence $X$ obeys all certified rules. Any other global model with the specified restrictions agrees with $X$ at every node because the patches cover the nodes. This proves uniqueness.\n\\end{proof}","\\begin{example}[Pairwise dependencies do not suffice]\n\\label{res:24:4}\nTake nodes $a,b,c$, no admissible leaves ($L=\\varnothing$), and one certified rule\n\\[\n\\{a,b\\}\\longrightarrow c.\n\\]\nUse patches $U_0=\\{a,c\\}$ and $U_1=\\{b,c\\}$. The local valuations $X_0=\\{a\\}$ and $X_1=\\{b\\}$ agree on their overlap: both assign false to $c$. Both are local models, since neither patch contains the full rule support. They cover every node and both pairwise premise-to-conclusion dependencies. Nevertheless, a global model realizing them would contain $a,b$ and omit $c$, contradicting the rule. Thus replacing a conjunctive hyperedge by unrelated pairwise checks can lose precisely the constraint that matters.\n\\end{example}","ertified rule\n\\[\n\\{a,b\\}\\longrightarrow c.\n\\]\nUse patches $U_0=\\{a,c\\}$ and $U_1=\\{b,c\\}$. The local valuations $X_0=\\{a\\}$ and $X_1=\\{b\\}$ agree on their overlap: both assign false to $c$. Both are local models, since neither patch contains the full rule support. They cover every node and both pairwise premise-to-conclusion dependencies. Nevertheless, a global model realizing them would contain $a,b$ and omit $c$, contradicting the rule. Thus replacing a conjunctive hyperedge by unrelated pairwise checks can lose precisely the constraint that matters.\n\\end{example}","[\n\\{a,b\\}\\longrightarrow c.\n\\]\nUse patches $U_0=\\{a,c\\}$ and $U_1=\\{b,c\\}$. The local valuations $X_0=\\{a\\}$ and $X_1=\\{b\\}$ agree on their overlap: both assign false to $c$. Both are local models, since neither patch contains the full rule support. They cover every node and both pairwise premise-to-conclusion dependencies. Nevertheless, a global model realizing them would contain $a,b$ and omit $c$, contradicting the rule. Thus replacing a conjunctive hyperedge by unrelated pairwise checks can lose precisely the constraint that matters.\n\\end{example}","\\studyhead{02}{Agreement can hide a missing rule}\nKeep the same rule and choose the local data $X_0=\\{a\\}$ on $U_0=\\{a,c\\}$ and $X_1=\\{b\\}$ on $U_1=\\{b,c\\}$. Both local models assign false to their shared coordinate $c$. They match exactly, yet their only assembled valuation violates the global rule.","\\begin{figure}[H]\\centering\n\\includegraphics[width=\\linewidth]{fig_15_whole_support.pdf}\n\\caption{The explicit counterexample to replacing whole-support coverage with pairwise dependency coverage. The local data force $(a,b,c)=(1,1,0)$, so both premises hold and the conclusion fails. Green nodes carry value one and open nodes carry value zero.}\n\\label{fig:whole-support-example}\n\\end{figure}","{\\sffamily\\bfseries\\color{KomorebiTeal}The successful gluing argument}\\par\\smallskip\nA patch containing the entire support $\\{a,b,c\\}$ must enforce the conjunctive rule. If all local models match and every certified support occurs in a patch, their union obeys every rule. Node coverage then gives uniqueness. This is the content of Theorem~\\ref{thm:support-gluing}, not an inference from pairwise agreement alone.","The exact binary model admits a general algebraic formulation. Let $K$ be a field, let $E,R,S$ be $K$-vector spaces, and let $D:E\\to R$ and $p:E\\to S$ be linear. The prescribed problem is to repair $x$ while retaining $p(x)$:\n\\[\nDy=0,\\qquad p(y)=p(x).\n\\]\nThe admissible correction image is $B=D(\\ker p)$. The elementary correction equivalence gives feasibility exactly when $Dx\\in B$.","\\begin{theorem}[A linear certificate alternative]\n\\label{res:24:5}\n\\label{thm:linear-alternative}\nThe protected problem is infeasible if and only if there is a linear functional $\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy","\\begin{theorem}[A linear certificate alternative]\n\\label{res:24:5}\n\\label{thm:linear-alternative}\nThe protected problem is infeasible if and only if there is a linear functional $\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a m","\\begin{theorem}[A linear certificate alternative]\n\\label{res:24:5}\n\\label{thm:linear-alternative}\nThe protected problem is infeasible if and only if there is a linear functional $\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its","\\begin{theorem}[A linear certificate alternative]\n\\label{res:24:5}\n\\label{thm:linear-alternative}\nThe protected problem is infeasible if and only if there is a linear functional $\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In f","\\begin{theorem}[A linear certificate alternative]\n\\label{res:24:5}\n\\label{thm:linear-alternative}\nThe protected problem is infeasible if and only if there is a linear functional $\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alte…","ternative}\nThe protected problem is infeasible if and only if there is a linear functional $\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","\\lambda:R\\to K$ satisfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","isfying\n\\[\n\\lambda(Dh)=0\\quad\\text{whenever }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","r }p(h)=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","=0,\n\\qquad \\lambda(Dx)=1.\n\\]\nFor a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","For a finite matrix $M$ and right side $b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","$b$, the corresponding alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","ng alternative is\n\\[\n\\nexists y\\ (My=b)\n\\quad\\Longleftrightarrow\\quad\n\\exists\\ell\\ (\\ell M=0\\ \\text{and}\\ \\ell b=1).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}",").\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","in{proof}\nIf $Dx\\notin B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","B$, linear separation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","eparation of a vector from a subspace gives a functional vanishing on $B$ and nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","nonzero at $Dx$. One construction extends a basis of $B$ by $Dx$ and then to a basis of $R$. Assign value zero to the basis of $B$ and value one to $Dx$, and extend linearly. Equivalently, normalize any separating functional by its nonzero value. Conversely, if $y$ were feasible, then $h=x-y$ would satisfy $p(h)=0$ and $Dh=Dx$, contradicting the displayed functional equations. For a matrix, take $B$ to be its column image. In finite coordinates the separating functional is a row vector, giving the stated alternative.\n\\end{proof}","\\studyhead{03}{An obstruction is a checkable equation}\nIn the binary example from \\path{data/bridge_examples.json}, compatibility demands $a+b=0$ while the protected observations require $a=0$ and $b=1$. The variable $c$ is unconstrained. Its freedom cannot remove the incompatible requirement on the first two coordinates.","\\begin{figure}[H]\\centering\n\\includegraphics[width=\\linewidth]{fig_19_obstruction_witness.pdf}\n\\caption{A normalized negative witness in $\\mathbb F_2$. Adding the three rows gives zero on the left and one on the right. The witness is $\\ell=(1,1,1)$, which satisfies $\\ell M=0$ and $\\ell b=1$ for the original stacked system.}\n\\label{fig:obstruction-witness}\n\\end{figure}","The exact system and witness can be inspected without trusting a search procedure:\n\\[\n M=\\begin{pmatrix}1&1&0\\\\1&0&0\\\\0&1&0\\end{pmatrix},\\qquad\n b=\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix},\\qquad\n \\ell M=(0,0,0),\\quad \\ell b=1.\n\\]\nIf $My=b$ had a solution, multiplying by $\\ell$ would give $0=1$. This is a certificate for the declared protected problem. A search that merely stops supplies no comparable implication.\n\\clearpage","1\\end{pmatrix},\\qquad\n \\ell M=(0,0,0),\\quad \\ell b=1.\n\\]\nIf $My=b$ had a solution, multiplying by $\\ell$ would give $0=1$. This is a certificate for the declared protected problem. A search that merely stops supplies no comparable implication.\n\\clearpage","Let $T:E\\to R$ and $Q:E\\to W$ be linear, and fix a known point $x_0$. Put\n\\[\n\\mathcal A=\\{y:Ty=Tx_0\\},\\quad H=\\ker T,\\quad\nQ_H=Q|_H.\n\\]\nTranslation $h\\mapsto x_0+h$ identifies $H$ with $\\mathcal A$. The observation image is the translate $Qx_0+Q(H)$, and the fiber above the attained value $Qx_0$ is a translate of $\\ker Q_H$.","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts.","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An atta","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. E","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on","\\begin{theorem}[Exact observation counts]\n\\label{res:24:6}\n\\label{thm:checked-counts}\nSuppose $K$ is a finite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\l…","nite field of cardinality $q$ and $E$ is a finite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","ite-dimensional $K$-vector space. Set\n$k=\\dim_K H$ and $r=\\dim_K Q(H)$. Then\n\\[\n|\\mathcal A|=q^k,\\qquad\n|Q(\\mathcal A)|=q^r,\\qquad\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|=q^{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","{k-r}.\n\\]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","]\nIn particular the exact count identity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","dentity is\n\\[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","[\n|\\mathcal A|=\n|\\{y\\in\\mathcal A:Qy=Qx_0\\}|\\,q^r.\n\\]\nEvery attained observation fiber has the same size. For the uniform distribution on $\\mathcal A$, the observation removes exactly $r\\log_2q$ bits of uncertainty.\n\\end{theorem}\n\\begin{proof}\nThe two translation equivalences reduce the counts to finite vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","e vector spaces. Such a space of dimension $d$ has $q^d$ elements. Rank-nullity for $Q_H$ gives $\\dim\\ker Q_H+r=k$, proving the product identity and all the counts. An attained value can be treated by choosing a point attaining it as the base point. Equal fiber sizes imply a uniform observation distribution and a uniform conditional distribution on each fiber. The entropy difference is therefore $\\log_2(q^k)-\\log_2(q^{k-r})=r\\log_2q$.\n\\end{proof}","For the protected problem, set $T=(D,p)$ and choose an actual feasible $y_0$, so $Ty_0=(0,px)$. This reduces its solution fiber to the theorem without changing the protected values. A concrete binary example has $a+b+c=0$ and $a=0$. Its two realizations are $(0,0,0)$ and $(0,1,1)$. Observing $b$ has rank one on the ambiguity space and determines the remaining coordinate. Repeating the same observation in two output rows does not double the information.","\\studyhead{04}{Ask what an observation distinguishes}\nReturn to the two vertices $000$ and $011$ from the opening cube. The ambiguity space is the span of $011$. Reading the protected coordinate $a$ repeats known information. Reading $b$ separates the two points. Reading $b$ twice creates two output rows but no additional distinction.","\\begin{figure}[H]\\centering\n\\includegraphics[width=\\linewidth]{fig_18_observation_partitions.pdf}\n\\caption{Three observation maps on the same feasible binary fiber. The last column is the size of each attained conditional fiber. Gains of zero, one, and one bit follow from the rank of the map restricted to $H$, under the uniform distribution. The duplicated output is not twice as informative.}\n\\label{fig:observation-partitions}\n\\end{figure}","For example, the repeated observation has matrix\n\\[\n Q=\\begin{pmatrix}0&1&0\\\\0&1&0\\end{pmatrix}.\n\\]\nIts two rows are identical. On $H=\\langle011\\rangle$ it has rank one, so $|\\mathcal A|=2$, $|Q(\\mathcal A)|=2$, and every attained observation fiber has size one. These are count identities before they are interpreted probabilistically.","0&1&0\\end{pmatrix}.\n\\]\nIts two rows are identical. On $H=\\langle011\\rangle$ it has rank one, so $|\\mathcal A|=2$, $|Q(\\mathcal A)|=2$, and every attained observation fiber has size one. These are count identities before they are interpreted probabilistically.","\\medskip\nThe general finite-field formula in Theorem~\\ref{thm:checked-counts} retains the same separation: $q^k$ initial possibilities, $q^r$ attained outcomes, and $q^{k-r}$ realizations per outcome.\n\\clearpage","Fix the field $F$, including its admitted leaves and certified rules. For a positive observation set $S\\subseteq V$, let\n\\[\n  \\operatorname{cl}_F(S)=\\operatorname{Der}_{F[S]},\n\\]\nwhere $F[S]$ is obtained by admitting $L\\cup S$ as leaves and changing nothing else. For a negative observation set $N\\subseteq V$, define\n\\[\n \\operatorname{Ext}_F(S,N)=\n \\{X\\subseteq V:X\\text{ is an }F\\text{-model},\\ S\\subseteq X,\\ X\\cap N=\\varnothing\\}.\n\\]\nPositive and negative observations are both part of the preservation contract. Dropping the negative observations can turn an impossible request into a different, feasible problem.","\\begin{theorem}[Least protected extension and a complete obstruction]\n\\label{res:new:extension}\nThe extension fiber is nonempty exactly when\n\\[\n \\operatorname{cl}_F(S)\\cap N=\\varnothing.\n\\]\nWhen this condition holds, $\\operatorname{cl}_F(S)$ is its least member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ con","\\begin{theorem}[Least protected extension and a complete obstruction]\n\\label{res:new:extension}\nThe extension fiber is nonempty exactly when\n\\[\n \\operatorname{cl}_F(S)\\cap N=\\varnothing.\n\\]\nWhen this condition holds, $\\operatorname{cl}_F(S)$ is its least member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible.","\\begin{theorem}[Least protected extension and a complete obstruction]\n\\label{res:new:extension}\nThe extension fiber is nonempty exactly when\n\\[\n \\operatorname{cl}_F(S)\\cap N=\\varnothing.\n\\]\nWhen this condition holds, $\\operatorname{cl}_F(S)$ is its least member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node deriv…","label{res:new:extension}\nThe extension fiber is nonempty exactly when\n\\[\n \\operatorname{cl}_F(S)\\cap N=\\varnothing.\n\\]\nWhen this condition holds, $\\operatorname{cl}_F(S)$ is its least member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","tension fiber is nonempty exactly when\n\\[\n \\operatorname{cl}_F(S)\\cap N=\\varnothing.\n\\]\nWhen this condition holds, $\\operatorname{cl}_F(S)$ is its least member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","is its least member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","member. More generally,\n\\[\n X\\in\\operatorname{Ext}_F(S,N)\n \\quad\\Longleftrightarrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","arrow\\quad\n X\\text{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","{ is an }F\\text{-model and }\n \\operatorname{cl}_F(S)\\subseteq X\\subseteq V\\setminus N.\n\\]\nIf no extension exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","ion exists, some $n\\in N$ has a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","s a derivation from $L\\cup S$ using only the declared certified rules.\n\\end{theorem}\n\\begin{proof}\nInduction on a derivation shows that every model containing $S$ contains $\\operatorname{cl}_F(S)$. The inductive closure is itself a model and contains $S$. If it avoids $N$, it is therefore a permitted extension, and the preceding inclusion proves leastness. If it meets $N$, every model containing $S$ contains a forbidden node, so no extension is possible. The displayed classification follows from the same inclusions. A point of the nonempty intersection is exactly the conclusion of the stated forbidden-node derivation.\n\\end{proof}","This describes an interval \\emph{inside the family of rule models}, not an arbitrary interval of the power set. Closure under the rules remains essential. Nor is the fiber generally an affine space. For the rule $\\{a,b\\}\\to c$, with no leaves, $S=\\varnothing$ and $N=\\{c\\}$, the allowed models are precisely $\\varnothing$, $\\{a\\}$, and $\\{b\\}$. There are three, not a power of two. The rank and entropy formulas for vector spaces cannot be imported into this model without additional structure.","\\begin{corollary}[Exact extension of a partial valuation]\n\\label{res:new:partial}\nLet $S\\subseteq U\\subseteq V$. A global model whose true nodes on $U$ are exactly $S$ exists if and only if\n\\[\n \\operatorname{cl}_F(S)\\cap U=S.\n\\]\n\\end{corollary}\n\\begin{proof}\nSet $N=U\\setminus S$. Closure always contains $S$, so avoiding $N$ is equivalent to the displayed equality. Apply Theorem~\\ref{res:new:extension}.\n\\end{proof}","\\begin{proposition}[Staged closure and persistent obstructions]\n\\label{res:new:stages}\nFor positive sets $S,T$,\n\\[\n \\operatorname{cl}_F\\bigl(\\operatorname{cl}_F(S)\\cup T\\bigr)\n   =\\operatorname{cl}_F(S\\cup T).\n\\]\nIf $S\\subseteq S'$ and $N\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the o","\\begin{proposition}[Staged closure and persistent obstructions]\n\\label{res:new:stages}\nFor positive sets $S,T$,\n\\[\n \\operatorname{cl}_F\\bigl(\\operatorname{cl}_F(S)\\cup T\\bigr)\n   =\\operatorname{cl}_F(S\\cup T).\n\\]\nIf $S\\subseteq S'$ and $N\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller on","\\begin{proposition}[Staged closure and persistent obstructions]\n\\label{res:new:stages}\nFor positive sets $S,T$,\n\\[\n \\operatorname{cl}_F\\bigl(\\operatorname{cl}_F(S)\\cup T\\bigr)\n   =\\operatorname{cl}_F(S\\cup T).\n\\]\nIf $S\\subseteq S'$ and $N\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclus","\\begin{proposition}[Staged closure and persistent obstructions]\n\\label{res:new:stages}\nFor positive sets $S,T$,\n\\[\n \\operatorname{cl}_F\\bigl(\\operatorname{cl}_F(S)\\cup T\\bigr)\n   =\\operatorname{cl}_F(S\\cup T).\n\\]\nIf $S\\subseteq S'$ and $N\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclusion follows by inclu","\\begin{proposition}[Staged closure and persistent obstructions]\n\\label{res:new:stages}\nFor positive sets $S,T$,\n\\[\n \\operatorname{cl}_F\\bigl(\\operatorname{cl}_F(S)\\cup T\\bigr)\n   =\\operatorname{cl}_F(S\\cup T).\n\\]\nIf $S\\subseteq S'$ and $N\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclusion follows by inclusion of the extension fibers.\n\\end{proof}","F(S\\cup T).\n\\]\nIf $S\\subseteq S'$ and $N\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclusion follows by inclusion of the extension fibers.\n\\end{proof}","\\subseteq N'$, then\n$\\operatorname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclusion follows by inclusion of the extension fibers.\n\\end{proof}","rname{Ext}_F(S',N')\\subseteq\\operatorname{Ext}_F(S,N)$.\nConsequently an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclusion follows by inclusion of the extension fibers.\n\\end{proof}","y an infeasible request stays infeasible under those additions, while the field and the meaning of its rules remain fixed.\n\\end{proposition}\n\\begin{proof}\nClosure is extensive, monotone, and idempotent because it is the least model containing its seed. The model $\\operatorname{cl}_F(S\\cup T)$ contains $\\operatorname{cl}_F(S)$ and $T$, giving one inclusion in the first equation. Extensivity gives $S\\cup T\\subseteq\\operatorname{cl}_F(S)\\cup T$, and monotonicity gives the other. A model satisfying the larger positive and negative prescriptions also satisfies the smaller ones. The last conclusion follows by inclusion of the extension fibers.\n\\end{proof}","The module \\path{ProofPath.Extension} checks these statements on the existing \\path{Field} type. The Python module \\path{extension.py} uses the same \\path{ProofFieldViews} snapshot as the local-model construction. It computes closure by forward rule scans and records each newly derived node. At most $|V|$ successful insertions occur. If a forbidden node is reached, a causal subtrace is extracted. It need not be the shortest or cheapest refutation.","\\begin{figure}[htbp]\\centering\n\\includegraphics[width=\\linewidth]{fig_23_protected_closure.pdf}\n\\caption{Two protected requests for the same rule field. With only $a$ prescribed true and $c$ prescribed false, the least model is $\\{a\\}$. Adding $b$ to the positive data derives the forbidden node $c$. The obstruction is the certified rule application, not the act of abandoning a search.}\n\\label{fig:protected-closure}\n\\end{figure}","Whole-support coverage is easy to inspect and is sufficient for gluing. It can nevertheless be unnecessarily strong. Suppose $F$ contains $a\\to b$, $b\\to c$, and $a\\to c$, with patches $\\{a,b\\}$ and $\\{b,c\\}$. The last support $\\{a,c\\}$ is absent from both patches. Yet the first two visible rules already imply the last rule. Requiring another patch would duplicate information rather than restore a missing constraint.","Let $\\mathcal U=(U_i)_i$ cover every node. Define $F_{\\mathcal U}$ to have the same nodes and leaves as $F$ but only those certified rules whose entire support lies in some $U_i$. The distinction between the two fields is explicit. We call the cover \\emph{semantically sufficient for the declared rules} when\n\\[\n \\forall j\\text{ certified in }F,\\qquad\n c_j\\in\\operatorname{cl}_{F_{\\mathcal U}}(A_j).\n \\tag{SC}\\label{eq:semantic-cover}\n\\]\nThe word semantic here refers to the Boolean rule-model semantics of Section~\\ref{sec:old24}, not to an automatic interpretation of node labels as arbitrary mathematical truths.","\\begin{theorem}[Exact criterion for universal gluing]\n\\label{res:new:semantic-cover}\nUnder node coverage, the following conditions are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item Every matching family of local $F$-models on the patches has a unique global $F$-model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: ev","\\begin{theorem}[Exact criterion for universal gluing]\n\\label{res:new:semantic-cover}\nUnder node coverage, the following conditions are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item Every matching family of local $F$-models on the patches has a unique global $F$-model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is a","\\begin{theorem}[Exact criterion for universal gluing]\n\\label{res:new:semantic-cover}\nUnder node coverage, the following conditions are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item Every matching family of local $F$-models on the patches has a unique global $F$-model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i","\\begin{theorem}[Exact criterion for universal gluing]\n\\label{res:new:semantic-cover}\nUnder node coverage, the following conditions are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item Every matching family of local $F$-models on the patches has a unique global $F$-model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equ","erion for universal gluing]\n\\label{res:new:semantic-cover}\nUnder node coverage, the following conditions are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item Every matching family of local $F$-models on the patches has a unique global $F$-model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","for universal gluing]\n\\label{res:new:semantic-cover}\nUnder node coverage, the following conditions are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item Every matching family of local $F$-models on the patches has a unique global $F$-model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","model extending it.\n\\item Every model of $F_{\\mathcal U}$ is a model of $F$.\n\\item Condition~\\eqref{eq:semantic-cover} holds.\n\\item For every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","or every positive seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","seed $S$, $\\operatorname{cl}_{F_{\\mathcal U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","U}}(S)=\\operatorname{cl}_F(S)$.\n\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","\\end{enumerate}\nWhole-support coverage implies these conditions, but the converse need not hold.\n\\end{theorem}\n\\begin{proof}\nA matching family assembles uniquely to a subset $X\\subseteq V$ because the patches cover all nodes and agree on their common coordinates. Its local rule conditions say exactly that $X$ is a model of $F_{\\mathcal U}$. Conversely, a model of $F_{\\mathcal U}$ restricts to an $F$-model on every patch: every original rule supported in that patch is among the retained rules. Those restrictions match. Thus (i) and (ii) are equivalent, and uniqueness follows from node coverage.","Assume (ii) and fix a certified rule $j$. The set $\\operatorname{cl}_{F_{\\mathcal U}}(A_j)$ is a model of $F_{\\mathcal U}$ containing all of $A_j$. By (ii) it obeys the original rule and contains $c_j$, proving (iii). Conversely, if $X$ is a model of $F_{\\mathcal U}$ containing $A_j$, leastness gives $\\operatorname{cl}_{F_{\\mathcal U}}(A_j)\\subseteq X$. Condition (iii) therefore forces $c_j\\in X$. Every original rule is satisfied, which proves (ii).","Under (ii) the two fields have exactly the same models, since every $F$-model already obeys the retained rules. Their least models containing any seed are equal, giving (iv). If (iv) holds, apply it to $A_j$. The conclusion $c_j$ lies in $\\operatorname{cl}_F(A_j)$ by the original rule, so it lies in the retained closure as required by (iii). Finally, whole-support coverage retains every certified rule. The three-rule example above proves that this condition is not necessary.\n\\end{proof}","There is a constructive negative result as well. If (SC) fails for $j$, let\n\\[\n X=\\operatorname{cl}_{F_{\\mathcal U}}(A_j).\n\\]\nThen $A_j\\subseteq X$ and $c_j\\notin X$. The restrictions $X\\cap U_i$ are matching local $F$-models, but their only assembled subset violates the original rule. They are a checkable obstruction to \\emph{universal gluing for this cover}. They do not show that every matching family on the cover fails.","\\studyhead{05}{A missing support can be redundant}\nThe patches $U_0=\\{a,b\\}$ and $U_1=\\{b,c\\}$ do not contain the support of $a\\to c$. Their visible rules $a\\to b$ and $b\\to c$ nevertheless provide a certified derivation of that implication. Compare this with the missing conjunctive support in Figure~\\ref{fig:whole-support-example}: neither visible patch there provides a substitute for $\\{a,b\\}\\to c$.","A change of proof-field representation needs more than a map of node names. Let $F$ and $G$ be declared fields. A rule transport consists of a node map $f:V_F\\to V_G$ and a rule map $h:J_F\\to J_G$ satisfying\n\\[\n\\begin{aligned}\n f(L_F)&\\subseteq L_G,","A change of proof-field representation needs more than a map of node names. Let $F$ and $G$ be declared fields. A rule transport consists of a node map $f:V_F\\to V_G$ and a rule map $h:J_F\\to J_G$ satisfying\n\\[\n\\begin{aligned}\n f(L_F)&\\subseteq L_G,\\\\\n j\\text{ certified in }F&\\Longrightarrow h(j)\\text{ certified in }G,\\\\\n A^G_{h(j)}&\\subseteq f(A^F_j),\\\\\n c^G_{h(j)}&=f(c^F_j).\n\\end{aligned}\n\\]\nThe premise inclusion points in the direction needed for executing the translated step. The mapped target rule may need fewer premises, but it must not introduce a new premise unsupported by the source step. This is a sufficient, deliberately concrete transport interface. A more general translation could map one rule to an entire derivation.","\\begin{theorem}[Compositional rule transport]\n\\label{res:new:transport}\nRule transports admit identities and associative composition. A transport $(f,h):F\\to G$ sends every derivation of $v$ in $F$ to a derivation of $f(v)$ in $G$. Every $G$-model pulls back along $f$ to an $F$-model. In particular,\n\\[\n Y\\in\\operatorname{Ext}_G(f(S),f(N))\n   \\Longrightarrow f^{-1}(Y)\\in\\operatorname{Ext}_F(S,N).\n\\]\nAn obstruction in the source extension problem therefore implies an obstruction in the target extension problem with the mapped observations.\n\\end{theorem}\n\\begin{proof}\nIdentity maps meet each displayed condition. For two composable transports, compose the node and rule maps. Applying the second map to the first premise inclusion and then using the second inclusion proves the composite premise condition. The conclusion and leaf conditions also compose.","\\begin{theorem}[Compositional rule transport]\n\\label{res:new:transport}\nRule transports admit identities and associative composition. A transport $(f,h):F\\to G$ sends every derivation of $v$ in $F$ to a derivation of $f(v)$ in $G$. Every $G$-model pulls back along $f$ to an $F$-model. In particular,\n\\[\n Y\\in\\operatorname{Ext}_G(f(S),f(N))\n   \\Longrightarrow f^{-1}(Y)\\in\\operatorname{Ext}_F(S,N).\n\\]\nAn obstruction in the source extension problem therefore implies an obstruction in the target extension problem with the mapped observations.\n\\end{theorem}\n\\begin{proof}\nIdentity maps meet each displayed condition. For two composable transports, compose the node and rule maps. Applying the second map to the first premise inclusion and then using the second inclusion proves the composite premise condition. The conclusion and leaf conditions also compose. Function composition gives th","\\begin{theorem}[Compositional rule transport]\n\\label{res:new:transport}\nRule transports admit identities and associative composition. A transport $(f,h):F\\to G$ sends every derivation of $v$ in $F$ to a derivation of $f(v)$ in $G$. Every $G$-model pulls back along $f$ to an $F$-model. In particular,\n\\[\n Y\\in\\operatorname{Ext}_G(f(S),f(N))\n   \\Longrightarrow f^{-1}(Y)\\in\\operatorname{Ext}_F(S,N).\n\\]\nAn obstruction in the source extension problem therefore implies an obstruction in the target extension problem with the mapped observations.\n\\end{theorem}\n\\begin{proof}\nIdentity maps meet each displayed condition. For two composable transports, compose the node and rule maps. Applying the second map to the first premise inclusion and then using the second inclusion proves the composite premise condition. The conclusion and leaf conditions also compose. Function composition gives the ident","\\begin{theorem}[Compositional rule transport]\n\\label{res:new:transport}\nRule transports admit identities and associative composition. A transport $(f,h):F\\to G$ sends every derivation of $v$ in $F$ to a derivation of $f(v)$ in $G$. Every $G$-model pulls back along $f$ to an $F$-model. In particular,\n\\[\n Y\\in\\operatorname{Ext}_G(f(S),f(N))\n   \\Longrightarrow f^{-1}(Y)\\in\\operatorname{Ext}_F(S,N).\n\\]\nAn obstruction in the source extension problem therefore implies an obstruction in the target extension problem with the mapped observations.\n\\end{theorem}\n\\begin{proof}\nIdentity maps meet each displayed condition. For two composable transports, compose the node and rule maps. Applying the second map to the first premise inclusion and then using the second inclusion proves the composite premise condition. The conclusion and leaf conditions also compose. Function composition gives the identity and associativity laws","\\begin{theorem}[Compositional rule transport]\n\\label{res:new:transport}\nRule transports admit identities and associative composition. A transport $(f,h):F\\to G$ sends every derivation of $v$ in $F$ to a derivation of $f(v)$ in $G$. Every $G$-model pulls back along $f$ to an $F$-model. In particular,\n\\[\n Y\\in\\operatorname{Ext}_G(f(S),f(N))\n   \\Longrightarrow f^{-1}(Y)\\in\\operatorname{Ext}_F(S,N).\n\\]\nAn obstruction in the source extension problem therefore implies an obstruction in the target extension problem with the mapped observations.\n\\end{theorem}\n\\begin{proof}\nIdentity maps meet each displayed condition. For two composable transports, compose the node and rule maps. Applying the second map to the first premise inclusion and then using the second inclusion proves the composite premise condition. The conclusion and leaf conditions also compose. Function composition gives the identity and associativity laws.","If $Y$ is a target model, $f^{-1}(Y)$ contains all source leaves. When it contains every premise of a source rule, $Y$ contains every premise of the mapped rule and hence its conclusion. This proves the model pullback. Positive and negative observations pull back by their image definitions. If a target extension existed while the source problem were infeasible, its pullback would contradict that infeasibility.\n\\end{proof}","Positive seeds can be added before transport. The same rule transport sends $F[S]$ to $G[f(S)]$, and hence\n\\[\n f\\bigl(\\operatorname{cl}_F(S)\\bigr)\n      \\subseteq\\operatorname{cl}_G(f(S)).\n\\]\nA forbidden source consequence $n\\in N\\cap\\operatorname{cl}_F(S)$ maps to the explicit forbidden consequence $f(n)\\in f(N)\\cap\\operatorname{cl}_G(f(S))$. The formal development checks this witness map, not only the existence statement obtained by contraposition.","\\begin{example}[Why transport is not an equivalence]\n\\label{res:new:collapse}\nLet the source have nodes $a,b$, no leaves, and no rules. The request $a$ true and $b$ false is feasible. Map both nodes to one target node $u$, again with no leaves or rules. This is a rule transport, but the image request requires $u$ both true and false and is infeasible. Thus source feasibility need not imply target feasibility, and target infeasibility need not imply source infeasibility.\n\\end{example}","\\begin{definition}[Finite indexed problem and trace]\n\\label{res:trace:problem}\nAn indexed problem $P$ specifies a finite address space $V_n=\\{0,\\ldots,n-1\\}$, admitted leaves $L$, a list of certified rules, positive observations $S$, and forbidden nodes $N$. A rule at address $j$ has a finite premise list $A_j$ and conclusion $b_j$. Two identical rows may have different rule addresses. Every listed node must belong to $V_n$. Node lists are individually duplicate free, but $L$, $S$, and $N$ need not be disjoint.","te indexed problem and trace]\n\\label{res:trace:problem}\nAn indexed problem $P$ specifies a finite address space $V_n=\\{0,\\ldots,n-1\\}$, admitted leaves $L$, a list of certified rules, positive observations $S$, and forbidden nodes $N$. A rule at address $j$ has a finite premise list $A_j$ and conclusion $b_j$. Two identical rows may have different rule addresses. Every listed node must belong to $V_n$. Node lists are individually duplicate free, but $L$, $S$, and $N$ need not be disjoint.","An indexed problem $P$ specifies a finite address space $V_n=\\{0,\\ldots,n-1\\}$, admitted leaves $L$, a list of certified rules, positive observations $S$, and forbidden nodes $N$. A rule at address $j$ has a finite premise list $A_j$ and conclusion $b_j$. Two identical rows may have different rule addresses. Every listed node must belong to $V_n$. Node lists are individually duplicate free, but $L$, $S$, and $N$ need not be disjoint.","ndexed problem $P$ specifies a finite address space $V_n=\\{0,\\ldots,n-1\\}$, admitted leaves $L$, a list of certified rules, positive observations $S$, and forbidden nodes $N$. A rule at address $j$ has a finite premise list $A_j$ and conclusion $b_j$. Two identical rows may have different rule addresses. Every listed node must belong to $V_n$. Node lists are individually duplicate free, but $L$, $S$, and $N$ need not be disjoint.","oblem $P$ specifies a finite address space $V_n=\\{0,\\ldots,n-1\\}$, admitted leaves $L$, a list of certified rules, positive observations $S$, and forbidden nodes $N$. A rule at address $j$ has a finite premise list $A_j$ and conclusion $b_j$. Two identical rows may have different rule addresses. Every listed node must belong to $V_n$. Node lists are individually duplicate free, but $L$, $S$, and $N$ need not be disjoint.","A trace $\\sigma=(j_1,\\ldots,j_k)$ is a list of rule addresses. Replay begins with the list $K_0=L\\mathbin{+\\!+}S$. It rejects an unknown rule address or a step whose premises are not all known. Otherwise it adds the conclusion. Strict replay additionally requires the conclusion to be new. Causal replay permits a conclusion already known. Lists are compared by membership when they represent sets.\n\\end{definition}","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known no","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis ap","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis applies to the remaining trace. The stricter novelty test c","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis applies to the remaining trace. The stricter novelty test can reject a step","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis applies to the remaining trace. The stricter novelty test can reject a step but cannot invalidate this argument for an accepted one. T","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis applies to the remaining trace. The stricter novelty test can reject a step but cannot invalidate this argument for an accepted one. The composition identity follows by the same induction, with the","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis applies to the remaining trace. The stricter novelty test can reject a step but cannot invalidate this argument for an accepted one. The composition identity follows by the same induction, with the lookup and enabling tests taking","\\begin{theorem}[Replay preserves derivability and composes]\n\\label{res:trace:replay}\nSuppose every member of $K$ belongs to $\\operatorname{cl}_F(S)$. If either replay mode accepts a trace starting from $K$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq\\operatorname{cl}_F(S).\n\\]\nFor a fixed replay mode, replaying $\\sigma\\mathbin{+\\!+}\\tau$ is the same as replaying $\\sigma$ and then replaying $\\tau$ from the returned state. Failure at either stage propagates to the concatenated execution.\n\\end{theorem}\n\\begin{proof}\nInduct on the trace. The empty trace returns the initial state. For a nonempty trace, a successful first step identifies a certified rule whose premises are already derivable. Closure under that rule proves its conclusion derivable. Adding the conclusion preserves all previously known nodes, and the induction hypothesis applies to the remaining trace. The stricter novelty test can reject a step but cannot invalidate this argument for an accepted one. The composition identity follows by the same induction, with the lookup and enabling tests taking the same branch on both …","\\begin{theorem}[Acceptance identifies the least seeded model]\n\\label{res:trace:least}\nA complete-closure certificate consists of a strict trace $\\sigma$ and a reported node list $K$. The checker accepts precisely when replay succeeds, its output has the same members as $K$, and $K$ contains the admitted leaves and satisfies every declared rule. Acceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: re","\\begin{theorem}[Acceptance identifies the least seeded model]\n\\label{res:trace:least}\nA complete-closure certificate consists of a strict trace $\\sigma$ and a reported node list $K$. The checker accepts precisely when replay succeeds, its output has the same members as $K$, and $K$ contains the admitted leaves and satisfies every declared rule. Acceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, w","\\begin{theorem}[Acceptance identifies the least seeded model]\n\\label{res:trace:least}\nA complete-closure certificate consists of a strict trace $\\sigma$ and a reported node list $K$. The checker accepts precisely when replay succeeds, its output has the same members as $K$, and $K$ contains the admitted leaves and satisfies every declared rule. Acceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, while the model test provides saturation under rules not used in the trace.\n\\end{proof}","d a reported node list $K$. The checker accepts precisely when replay succeeds, its output has the same members as $K$, and $K$ contains the admitted leaves and satisfies every declared rule. Acceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, while the model test provides saturation under rules not used in the trace.\n\\end{proof}","itted leaves and satisfies every declared rule. Acceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, while the model test provides saturation under rules not used in the trace.\n\\end{proof}","and satisfies every declared rule. Acceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, while the model test provides saturation under rules not used in the trace.\n\\end{proof}","ceptance implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, while the model test provides saturation under rules not used in the trace.\n\\end{proof}","e implies\n\\[\n \\boxed{K=\\operatorname{cl}_F(S).}\n\\]\nHere equality means equality of the represented sets, not equality of list order.\n\\end{theorem}\n\\begin{proof}\nTheorem~\\ref{res:trace:replay} gives $K\\subseteq\\operatorname{cl}_F(S)$ because every initial node is an admitted leaf or a positive observation. Extensivity of replay ensures that $K$ contains $S$. The final closedness test makes $K$ an $F$-model, so leastness of seeded closure gives the reverse inclusion. Both tests are needed: replay provides derivability, while the model test provides saturation under rules not used in the trace.\n\\end{proof}","The connection to the prior theory is exact. In Lean, \\path{closed_iff_model} identifies the Boolean rule check with the existing \\path{Field.Model} predicate, and \\path{checkClosure_exact} proves the displayed equality with \\path{Field.closure}. For a valid indexed problem, \\path{valid_closure_bounded} proves that every member of that closure has an address below $n$. Thus totalizing rule lookup in the abstract \\path{Field} cannot introduce a conclusion at an unnamed address: only successful lookups are certified.","Logical meaning remains a separate, explicit hypothesis. If an interpretation $\\mu:V_n\\to\\mathrm{Prop}$ satisfies the admitted leaves and rules, and every observation in $S$ is true under $\\mu$, then every member of an accepted closure is true under $\\mu$. This follows by applying the prior model-soundness theorem to the derived nodes. Acceptance does not establish the truth of arbitrarily attached sentence labels.","\\begin{definition}[Checked forbidden-node witness]\n\\label{res:trace:witness}\nA witness is a pair $(v,\\tau)$, consisting of a target address and a causal trace. Its checker accepts when replay from $L\\cup S$ succeeds, the returned state contains $v$, and $v\\in N$.\n\\end{definition}","\\begin{theorem}[Sound and complete obstruction witnesses]\n\\label{res:trace:complete}\nFor the finite-premise indexed rule field,\n\\[\n \\boxed{\n \\text{an accepted forbidden-node witness exists}\n \\quad\\Longleftrightarrow\\quad\n \\operatorname{Ext}_F(S,N)=\\varnothing.\n }\n\\]\nThe statement does not require the dependency graph to be acyclic.\n\\end{theorem}\n\\begin{proof}\nAn accepted witness derives a node in $N$. Every model containing $L\\cup S$ contains that node, so no protected extension exists.","begin{theorem}[Sound and complete obstruction witnesses]\n\\label{res:trace:complete}\nFor the finite-premise indexed rule field,\n\\[\n \\boxed{\n \\text{an accepted forbidden-node witness exists}\n \\quad\\Longleftrightarrow\\quad\n \\operatorname{Ext}_F(S,N)=\\varnothing.\n }\n\\]\nThe statement does not require the dependency graph to be acyclic.\n\\end{theorem}\n\\begin{proof}\nAn accepted witness derives a node in $N$. Every model containing $L\\cup S$ contains that node, so no protected extension exists.","plete obstruction witnesses]\n\\label{res:trace:complete}\nFor the finite-premise indexed rule field,\n\\[\n \\boxed{\n \\text{an accepted forbidden-node witness exists}\n \\quad\\Longleftrightarrow\\quad\n \\operatorname{Ext}_F(S,N)=\\varnothing.\n }\n\\]\nThe statement does not require the dependency graph to be acyclic.\n\\end{theorem}\n\\begin{proof}\nAn accepted witness derives a node in $N$. Every model containing $L\\cup S$ contains that node, so no protected extension exists.","Take $V_4=\\{0,1,2,3\\}$, $L=\\{0\\}$, $S=\\varnothing$, and the two rules $r_0:0\\to1$ and $r_1:1\\to2$. The least closure is $\\{0,1,2\\}$. Node $3$ has no derivation from these inputs.","The trace $[0]$ derives exactly $\\{0,1\\}$ but leaves $r_1$ enabled with its conclusion absent. It is a valid partial derivation, not a complete closure. Conversely, $\\{0,1,2,3\\}$ is closed under both rules, but replay cannot establish its unsupported node $3$. A model need not be the least model.","With $N=\\{3\\}$, the exact closure gives a positive result. With $N=\\{2\\}$, the same closure supports a negative result with witness $(2,[0,1])$. The second problem is not a different interpretation of the same request: its protected observations have changed. A checker must receive the fixed request separately from the proposed answer.","race:verdict}\nThe complete result format contains a Boolean feasibility flag, a reported closure, a strict closure trace, and an optional forbidden-node witness. Both branches require the complete-closure check. A positive result additionally requires no obstruction witness and no forbidden node in the closure. A negative result requires an accepted obstruction witness. For every accepted result $C$,\n\\[\n C.K=\\operatorname{cl}_F(S),\\qquad\n \\bigl(C.\\mathrm{feasible}=\\mathrm{true}\\bigr)\n \\Longleftrightarrow\n \\operatorname{Ext}_F(S,N)\\ne\\varnothing.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe complete-closure check proves the first equality. In the positive branch, the closure is a model containing $S$ and avoiding $N$, hence is an extension. In the negative branch, the witness excludes every extension by Theorem~\\ref{res:trace:complete}. Case analysis on the Boolean flag gives the equivalence.\n\\end{proof}","erdict}\nThe complete result format contains a Boolean feasibility flag, a reported closure, a strict closure trace, and an optional forbidden-node witness. Both branches require the complete-closure check. A positive result additionally requires no obstruction witness and no forbidden node in the closure. A negative result requires an accepted obstruction witness. For every accepted result $C$,\n\\[\n C.K=\\operatorname{cl}_F(S),\\qquad\n \\bigl(C.\\mathrm{feasible}=\\mathrm{true}\\bigr)\n \\Longleftrightarrow\n \\operatorname{Ext}_F(S,N)\\ne\\varnothing.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe complete-closure check proves the first equality. In the positive branch, the closure is a model containing $S$ and avoiding $N$, hence is an extension. In the negative branch, the witness excludes every extension by Theorem~\\ref{res:trace:complete}. Case analysis on the Boolean flag gives the equivalence.\n\\end{proof}","n the closure. A negative result requires an accepted obstruction witness. For every accepted result $C$,\n\\[\n C.K=\\operatorname{cl}_F(S),\\qquad\n \\bigl(C.\\mathrm{feasible}=\\mathrm{true}\\bigr)\n \\Longleftrightarrow\n \\operatorname{Ext}_F(S,N)\\ne\\varnothing.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe complete-closure check proves the first equality. In the positive branch, the closure is a model containing $S$ and avoiding $N$, hence is an extension. In the negative branch, the witness excludes every extension by Theorem~\\ref{res:trace:complete}. Case analysis on the Boolean flag gives the equivalence.\n\\end{proof}","negative result requires an accepted obstruction witness. For every accepted result $C$,\n\\[\n C.K=\\operatorname{cl}_F(S),\\qquad\n \\bigl(C.\\mathrm{feasible}=\\mathrm{true}\\bigr)\n \\Longleftrightarrow\n \\operatorname{Ext}_F(S,N)\\ne\\varnothing.\n\\]\n\\end{corollary}\n\\begin{proof}\nThe complete-closure check proves the first equality. In the positive branch, the closure is a model containing $S$ and avoiding $N$, hence is an extension. In the negative branch, the witness excludes every extension by Theorem~\\ref{res:trace:complete}. Case analysis on the Boolean flag gives the equivalence.\n\\end{proof}","\\begin{theorem}[Lossless encoding at the typed syntax boundary]\n\\label{res:trace:codec}\nLet $e_P,d_P$ and $e_C,d_C$ be the stated Lean encoders and partial decoders for problem and certificate records. For every typed problem $P$ and certificate $C$,\n\\[\n d_P(e_P(P))=\\operatorname{some}(P),\\qquad\n d_C(e_C(C))=\\operatorname{some}(C).\n\\]\nConsequently both encoders are injective on their typed domains. These identities do not require the encoded record to pass the separate validity checks.\n\\end{theorem}\n\\begin{proof}\nThe natural-number and Boolean constructors decode to their original values. Induction on a list proves the corresponding array round trip. The rule, witness, and optional-witness decoders then recover their components. Unfolding the exact-length record encoders and decoders gives","\\begin{theorem}[Lossless encoding at the typed syntax boundary]\n\\label{res:trace:codec}\nLet $e_P,d_P$ and $e_C,d_C$ be the stated Lean encoders and partial decoders for problem and certificate records. For every typed problem $P$ and certificate $C$,\n\\[\n d_P(e_P(P))=\\operatorname{some}(P),\\qquad\n d_C(e_C(C))=\\operatorname{some}(C).\n\\]\nConsequently both encoders are injective on their typed domains. These identities do not require the encoded record to pass the separate validity checks.\n\\end{theorem}\n\\begin{proof}\nThe natural-number and Boolean constructors decode to their original values. Induction on a list proves the corresponding array round trip. The rule, witness, and optional-witness decoders then recover their components. Unfolding the exact-length record encoders and decoders gives the two ident","\\begin{theorem}[Lossless encoding at the typed syntax boundary]\n\\label{res:trace:codec}\nLet $e_P,d_P$ and $e_C,d_C$ be the stated Lean encoders and partial decoders for problem and certificate records. For every typed problem $P$ and certificate $C$,\n\\[\n d_P(e_P(P))=\\operatorname{some}(P),\\qquad\n d_C(e_C(C))=\\operatorname{some}(C).\n\\]\nConsequently both encoders are injective on their typed domains. These identities do not require the encoded record to pass the separate validity checks.\n\\end{theorem}\n\\begin{proof}\nThe natural-number and Boolean constructors decode to their original values. Induction on a list proves the corresponding array round trip. The rule, witness, and optional-witness decoders then recover their components. Unfolding the exact-length record encoders and decoders gives the two identities. If two encoded records are equal, applying the decoder gives equal \\texttt{some} values, hence equal origin","\\begin{theorem}[Lossless encoding at the typed syntax boundary]\n\\label{res:trace:codec}\nLet $e_P,d_P$ and $e_C,d_C$ be the stated Lean encoders and partial decoders for problem and certificate records. For every typed problem $P$ and certificate $C$,\n\\[\n d_P(e_P(P))=\\operatorname{some}(P),\\qquad\n d_C(e_C(C))=\\operatorname{some}(C).\n\\]\nConsequently both encoders are injective on their typed domains. These identities do not require the encoded record to pass the separate validity checks.\n\\end{theorem}\n\\begin{proof}\nThe natural-number and Boolean constructors decode to their original values. Induction on a list proves the corresponding array round trip. The rule, witness, and optional-witness decoders then recover their components. Unfolding the exact-length record encoders and decoders gives the two identities. If two encoded records are equal, applying the decoder gives equal \\texttt{some} values, hence equal original records.\n\\end{pro","\\begin{theorem}[Lossless encoding at the typed syntax boundary]\n\\label{res:trace:codec}\nLet $e_P,d_P$ and $e_C,d_C$ be the stated Lean encoders and partial decoders for problem and certificate records. For every typed problem $P$ and certificate $C$,\n\\[\n d_P(e_P(P))=\\operatorname{some}(P),\\qquad\n d_C(e_C(C))=\\operatorname{some}(C).\n\\]\nConsequently both encoders are injective on their typed domains. These identities do not require the encoded record to pass the separate validity checks.\n\\end{theorem}\n\\begin{proof}\nThe natural-number and Boolean constructors decode to their original values. Induction on a list proves the corresponding array round trip. The rule, witness, and optional-witness decoders then recover their components. Unfolding the exact-length record encoders and decoders gives the two identities. If two encoded records are equal, applying the decoder gives equal \\texttt{some} values, hence equal original records.\n\\end{proof}","\\begin{theorem}[Accepted JSON records certify their decoded answer]\n\\label{res:trace:json}\nSuppose the JSON checker accepts the separately supplied problem and certificate syntax trees. Then both decode to typed records $P,C$, both pass their index and shape checks, and the decoded certificate satisfies the closure equality and feasibility equivalence of Corollary~\\ref{res:trace:verdict} for that decoded problem.\n\\end{theorem}\n\\begin{proof}\nThe executable checker rejects any failed decoding or validity check. In the remaining branch it calls the same complete result checker used in Corollary~\\ref{res:trace:verdict}. Its acceptance therefore supplies that corollary's hypothesis. The round-trip identities additionally show that checking encoded typed records agrees with checking those records directly.\n\\end{proof}","A representation is sufficient only relative to specified questions and a specified class of possible archives. Fix a set $\\mathscr A$ of admissible proof archives. A representation scheme is a map $C:\\mathscr A\\to\\mathscr R$. Its output for an archive $\\Pcal$ is $C(\\Pcal)$. Distinguishing the scheme from one output is essential: a decoder chosen only after seeing one fixed archive could simply store the answer.","\\begin{definition}[Task]\n\\label{res:7:1}\nA task is a map $q:\\mathscr A\\to Y_q$ to a declared answer set. A task family $\\Qcal$ is a set of such maps. Examples include asking for the target certificate, a dependency list, a specified boundary example, or the claims affected by a recorded revision. A teaching or discovery objective becomes an exact task only after its required output or acceptance criterion has been specified.\n\\end{definition}","\\begin{definition}[Task-sufficient representation]\n\\label{res:7:2}\nThe scheme $C:\\mathscr A\\to\\mathscr R$ is sufficient for $q$ when there is a single decoder $d_q:C(\\mathscr A)\\to Y_q$ such that\n\\[\nq(\\Pcal)=d_q(C(\\Pcal))\\qquad\\text{for every }\\Pcal\\in\\mathscr A.\n\\]\nIt is $\\Qcal$-sufficient when this holds for every $q\\in\\Qcal$. The decoder may depend on the task and the scheme, but not on hidden knowledge of the particular archive being decoded.\n\\end{definition}","\\begin{theorem}[Task information loss criterion]\n\\label{res:7:3}\nLet $C$ be a compression map and let $q$ be a task. If there exist proof fields $\\Pcal_1$ and $\\Pcal_2$ such that\n\\[\nC(\\Pcal_1)=C(\\Pcal_2)\n\\qquad\\text{but}\\qquad\nq(\\Pcal_1)\\ne q(\\Pcal_2),\n\\]\nthen the compressed representation is not sufficient for $q$.\n\\end{theorem}","\\begin{proof}\nSuppose a decoder $d_q$ existed. Applying it to the common compressed representation would have to produce both $q(\\Pcal_1)$ and $q(\\Pcal_2)$. Since these answers differ, no such decoder exists.\n\\end{proof}","\\begin{definition}[Recoverability preorder]\n\\label{res:7:4}\nFor a fixed task family $\\Qcal$, write\n\\[\nR_1\\preceq_{\\Qcal} R_2\n\\]\nwhen every task in $\\Qcal$ that factors through the representation scheme $R_1$ also factors through $R_2$. Both schemes have domain $\\mathscr A$, and each factorization uses one decoder valid on that whole domain.\n\\end{definition}","\\begin{proposition}[Recoverability is a preorder]\n\\label{res:7:5}\nThe relation $\\preceq_{\\Qcal}$ is reflexive and transitive.\n\\end{proposition}","\\begin{proof}\nEvery representation supports every task it supports, so the relation is reflexive. If every task supported by $R_1$ is supported by $R_2$, and every task supported by $R_2$ is supported by $R_3$, then every task supported by $R_1$ is supported by $R_3$. Thus the relation is transitive.\n\\end{proof}","Define $R_1\\sim_{\\Qcal}R_2$ when both $R_1\\preceq_{\\Qcal}R_2$ and $R_2\\preceq_{\\Qcal}R_1$.","\\begin{corollary}[Quotient poset]\n\\label{res:7:6}\nThe recoverability preorder descends to a partial order on the equivalence classes under $\\sim_{\\Qcal}$.\n\\end{corollary}","\\begin{definition}[Retention projection]\n\\label{res:7:7}\nFix a finite node universe $V$. For a mask $K\\subseteq V$, define\n\\[\nP_K(S)=S\\cap K,\n\\qquad S\\subseteq V.\n\\]\n\\end{definition}","\\begin{theorem}[Retention projection algebra]\n\\label{res:7:8}\nFor masks $K,L\\subseteq V$,\n\\[\nP_K\\circ P_L=P_{K\\cap L}=P_L\\circ P_K.\n\\]\nConsequently every retention projection is idempotent, and the masks form a Boolean lattice with meet $K\\cap L$ and join $K\\cup L$.\n\\end{theorem}","\\begin{proof}\nFor every $S\\subseteq V$,\n\\[\nP_K(P_L(S))=(S\\cap L)\\cap K=S\\cap(K\\cap L).\n\\]\nIntersection is associative and commutative, which gives the composition identity and commutativity. Taking $L=K$ gives idempotence. The power set of $V$ with intersection, union, complement, bottom $\\varnothing$, and top $V$ is a Boolean lattice.\n\\end{proof}","\\begin{corollary}[Task monotonicity under retention]\n\\label{res:7:9}\nIf $K\\subseteq L$, every task whose required nodes and metadata are retained by $P_K$ is also supported by $P_L$.\n\\end{corollary}","\\begin{proof}\nEvery item retained by $K$ is retained by $L$. Hence every task requirement contained in $K$ is contained in $L$.\n\\end{proof}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_05_recoverability_preorder.pdf}\n\\caption{A concrete retention diamond for the four archives $(C_0,x,y)$ with $x,y\\in\\{0,1\\}$. Arrows discard coordinates. The two middle schemes are incomparable for the task family $\\{q_x,q_y\\}$. This is not a universal ranking of teaching, discovery, and formal presentations.}\n\\label{fig:preorder}\n\\end{figure}","\\begin{proof}\nLet the archives be $\\Pcal_{xy}=(C_0,x,y)$ for $(x,y)\\in\\{0,1\\}^2$, where $C_0$ is a fixed certificate retained by every allowed mask. Let $q_x(\\Pcal_{xy})=x$ and $q_y(\\Pcal_{xy})=y$. Retaining $x$ is sufficient for $q_x$. Any mask omitting $x$ identifies $\\Pcal_{0y}$ with $\\Pcal_{1y}$, so it is not sufficient for that task. Thus the least mask for $\\{q_x\\}$ retains exactly $C_0$ and $x$. The same argument gives the least mask $C_0$ and $y$ for $\\{q_y\\}$. Their joint task family requires both records. These masks are also the unique minima for unit cost per retained side record.\n\\end{proof}","egin{proof}\nLet the archives be $\\Pcal_{xy}=(C_0,x,y)$ for $(x,y)\\in\\{0,1\\}^2$, where $C_0$ is a fixed certificate retained by every allowed mask. Let $q_x(\\Pcal_{xy})=x$ and $q_y(\\Pcal_{xy})=y$. Retaining $x$ is sufficient for $q_x$. Any mask omitting $x$ identifies $\\Pcal_{0y}$ with $\\Pcal_{1y}$, so it is not sufficient for that task. Thus the least mask for $\\{q_x\\}$ retains exactly $C_0$ and $x$. The same argument gives the least mask $C_0$ and $y$ for $\\{q_y\\}$. Their joint task family requires both records. These masks are also the unique minima for unit cost per retained side record.\n\\end{proof}","(C_0,x,y)$ for $(x,y)\\in\\{0,1\\}^2$, where $C_0$ is a fixed certificate retained by every allowed mask. Let $q_x(\\Pcal_{xy})=x$ and $q_y(\\Pcal_{xy})=y$. Retaining $x$ is sufficient for $q_x$. Any mask omitting $x$ identifies $\\Pcal_{0y}$ with $\\Pcal_{1y}$, so it is not sufficient for that task. Thus the least mask for $\\{q_x\\}$ retains exactly $C_0$ and $x$. The same argument gives the least mask $C_0$ and $y$ for $\\{q_y\\}$. Their joint task family requires both records. These masks are also the unique minima for unit cost per retained side record.\n\\end{proof}","for $(x,y)\\in\\{0,1\\}^2$, where $C_0$ is a fixed certificate retained by every allowed mask. Let $q_x(\\Pcal_{xy})=x$ and $q_y(\\Pcal_{xy})=y$. Retaining $x$ is sufficient for $q_x$. Any mask omitting $x$ identifies $\\Pcal_{0y}$ with $\\Pcal_{1y}$, so it is not sufficient for that task. Thus the least mask for $\\{q_x\\}$ retains exactly $C_0$ and $x$. The same argument gives the least mask $C_0$ and $y$ for $\\{q_y\\}$. Their joint task family requires both records. These masks are also the unique minima for unit cost per retained side record.\n\\end{proof}",",y)\\in\\{0,1\\}^2$, where $C_0$ is a fixed certificate retained by every allowed mask. Let $q_x(\\Pcal_{xy})=x$ and $q_y(\\Pcal_{xy})=y$. Retaining $x$ is sufficient for $q_x$. Any mask omitting $x$ identifies $\\Pcal_{0y}$ with $\\Pcal_{1y}$, so it is not sufficient for that task. Thus the least mask for $\\{q_x\\}$ retains exactly $C_0$ and $x$. The same argument gives the least mask $C_0$ and $y$ for $\\{q_y\\}$. Their joint task family requires both records. These masks are also the unique minima for unit cost per retained side record.\n\\end{proof}","\\begin{theorem}[All predicates distinguish exactly injective encodings]\n\\label{res:7:11}\n\\label{thm:injective-compression}\nLet $C:A\\to B$. Every predicate $q:A\\to\\mathrm{Prop}$ factors through $C$ if and only if $C$ is injective.\n\\end{theorem}\n\\begin{proof}\nIf $C$ is injective, define the decoder of $q$ at $b$ by the predicate $\\exists a\\ (C(a)=b\\ \\text{and}\\ q(a))$. At $b=C(a_0)$, injectivity makes this equivalent to $q(a_0)$. Conversely, if $C(a)=C(a')$, use the predicate $q(z)\\equiv(z=a)$. Its decoded values at the equal encodings must agree, giving $a'=a$. This is the general collision criterion specialized to all predicates.\n\\end{proof}","dicates distinguish exactly injective encodings]\n\\label{res:7:11}\n\\label{thm:injective-compression}\nLet $C:A\\to B$. Every predicate $q:A\\to\\mathrm{Prop}$ factors through $C$ if and only if $C$ is injective.\n\\end{theorem}\n\\begin{proof}\nIf $C$ is injective, define the decoder of $q$ at $b$ by the predicate $\\exists a\\ (C(a)=b\\ \\text{and}\\ q(a))$. At $b=C(a_0)$, injectivity makes this equivalent to $q(a_0)$. Conversely, if $C(a)=C(a')$, use the predicate $q(z)\\equiv(z=a)$. Its decoded values at the equal encodings must agree, giving $a'=a$. This is the general collision criterion specialized to all predicates.\n\\end{proof}","ly injective encodings]\n\\label{res:7:11}\n\\label{thm:injective-compression}\nLet $C:A\\to B$. Every predicate $q:A\\to\\mathrm{Prop}$ factors through $C$ if and only if $C$ is injective.\n\\end{theorem}\n\\begin{proof}\nIf $C$ is injective, define the decoder of $q$ at $b$ by the predicate $\\exists a\\ (C(a)=b\\ \\text{and}\\ q(a))$. At $b=C(a_0)$, injectivity makes this equivalent to $q(a_0)$. Conversely, if $C(a)=C(a')$, use the predicate $q(z)\\equiv(z=a)$. Its decoded values at the equal encodings must agree, giving $a'=a$. This is the general collision criterion specialized to all predicates.\n\\end{proof}","es:7:11}\n\\label{thm:injective-compression}\nLet $C:A\\to B$. Every predicate $q:A\\to\\mathrm{Prop}$ factors through $C$ if and only if $C$ is injective.\n\\end{theorem}\n\\begin{proof}\nIf $C$ is injective, define the decoder of $q$ at $b$ by the predicate $\\exists a\\ (C(a)=b\\ \\text{and}\\ q(a))$. At $b=C(a_0)$, injectivity makes this equivalent to $q(a_0)$. Conversely, if $C(a)=C(a')$, use the predicate $q(z)\\equiv(z=a)$. Its decoded values at the equal encodings must agree, giving $a'=a$. This is the general collision criterion specialized to all predicates.\n\\end{proof}","to\\mathrm{Prop}$ factors through $C$ if and only if $C$ is injective.\n\\end{theorem}\n\\begin{proof}\nIf $C$ is injective, define the decoder of $q$ at $b$ by the predicate $\\exists a\\ (C(a)=b\\ \\text{and}\\ q(a))$. At $b=C(a_0)$, injectivity makes this equivalent to $q(a_0)$. Conversely, if $C(a)=C(a')$, use the predicate $q(z)\\equiv(z=a)$. Its decoded values at the equal encodings must agree, giving $a'=a$. This is the general collision criterion specialized to all predicates.\n\\end{proof}","The collision obstruction of Theorem~\\ref{res:7:3} has a converse. It gives a complete test for exact task recovery at the level of sets. Write $\\operatorname{im}C=C(\\mathscr A)$ for the attained output set. A decoder is required on this set and nowhere else.","\\begin{theorem}[Attained factorization and unique decoding]\n\\label{res:new:factorization}\nFor $C:\\mathscr A\\to\\mathcal R$ and $q:\\mathscr A\\to Y$, the following are equivalent:\n\\[\n\\begin{aligned}\n &\\exists d:\\operatorname{im}C\\to Y\\quad\n       \\forall a\\in\\mathscr A,\\quad q(a)=d(C(a)),","\\begin{theorem}[Attained factorization and unique decoding]\n\\label{res:new:factorization}\nFor $C:\\mathscr A\\to\\mathcal R$ and $q:\\mathscr A\\to Y$, the following are equivalent:\n\\[\n\\begin{aligned}\n &\\exists d:\\operatorname{im}C\\to Y\\quad\n       \\forall a\\in\\mathscr A,\\quad q(a)=d(C(a)),\\\\\n &\\forall a,b\\in\\mathscr A,\\quad C(a)=C(b)\\Longrightarrow q(a)=q(b).\n\\end{aligned}\n\\]\nWhen a decoder exists it is unique on $\\operatorname{im}C$. Otherwise there are two archives with the same encoding and different task answers.\n\\end{theorem}\n\\begin{proof}\nA decoder applied to equal encodings gives equal answers. Conversely, for an attained value $r$, choose $a$ with $C(a)=r$ and define $d(r)=q(a)$. Constancy on each fiber makes the value independent of the choice. At $r=C(a)$ it gives the required answer. Any other decoder must give that same answer at every attained value, proving uniqueness. Negating fiber constancy supplies the pair in the last statement. The construction is set theoretic and uses classical choice in the Lean implementation. It does not provide an eff","&\\forall a,b\\in\\mathscr A,\\quad C(a)=C(b)\\Longrightarrow q(a)=q(b).\n\\end{aligned}\n\\]\nWhen a decoder exists it is unique on $\\operatorname{im}C$. Otherwise there are two archives with the same encoding and different task answers.\n\\end{theorem}\n\\begin{proof}\nA decoder applied to equal encodings gives equal answers. Conversely, for an attained value $r$, choose $a$ with $C(a)=r$ and define $d(r)=q(a)$. Constancy on each fiber makes the value independent of the choice. At $r=C(a)$ it gives the required answer. Any other decoder must give that same answer at every attained value, proving uniqueness. Negating fiber constancy supplies the pair in the last statement. The construction is set theoretic and uses classical choice in the Lean implementation. It does not provide an efficient decoder for arbitrary infinite archive classes.\n\\end{","&\\forall a,b\\in\\mathscr A,\\quad C(a)=C(b)\\Longrightarrow q(a)=q(b).\n\\end{aligned}\n\\]\nWhen a decoder exists it is unique on $\\operatorname{im}C$. Otherwise there are two archives with the same encoding and different task answers.\n\\end{theorem}\n\\begin{proof}\nA decoder applied to equal encodings gives equal answers. Conversely, for an attained value $r$, choose $a$ with $C(a)=r$ and define $d(r)=q(a)$. Constancy on each fiber makes the value independent of the choice. At $r=C(a)$ it gives the required answer. Any other decoder must give that same answer at every attained value, proving uniqueness. Negating fiber constancy supplies the pair in the last statement. The construction is set theoretic and uses classical choice in the Lean implementation. It does not provide an efficient decoder for arbitrary infinite archive classes.\n\\end{proof}","ligned}\n\\]\nWhen a decoder exists it is unique on $\\operatorname{im}C$. Otherwise there are two archives with the same encoding and different task answers.\n\\end{theorem}\n\\begin{proof}\nA decoder applied to equal encodings gives equal answers. Conversely, for an attained value $r$, choose $a$ with $C(a)=r$ and define $d(r)=q(a)$. Constancy on each fiber makes the value independent of the choice. At $r=C(a)$ it gives the required answer. Any other decoder must give that same answer at every attained value, proving uniqueness. Negating fiber constancy supplies the pair in the last statement. The construction is set theoretic and uses classical choice in the Lean implementation. It does not provide an efficient decoder for arbitrary infinite archive classes.\n\\end{proof}","\\begin{example}[The empty-domain boundary]\n\\label{res:new:empty-decoder}\nTake $\\mathscr A=\\varnothing$, $\\mathcal R=\\{*\\}$, and $Y=\\varnothing$. There is one encoding and one task on the empty archive class. The attained set is empty, so its empty decoder exists. There is no total decoder $\\{*\\}\\to\\varnothing$. Thus attained decoding and decoding on the entire output type are not equivalent without an additional hypothesis.\n\\end{example}","Suppose tasks $q_i:\\mathscr A\\to Y_i$ are indexed by $i\\in I$. Their joint representation is\n\\[\n Q(a)=(q_i(a))_{i\\in I}.\n\\]\nIt records every distinction asked for by the task family and makes no demand to recover the entire source archive.","\\begin{theorem}[Joint task representation and compositional recovery]\n\\label{res:new:joint}\nAn encoding $C$ is sufficient for every $q_i$ if and only if it is sufficient for $Q$. In particular every sufficient encoding recovers the joint representation on attained values, while $Q$ itself recovers each task by projection.","If a representation $R$ is recoverable from $C$, then every task recoverable from $R$ is recoverable from $C$. Recovery of representations is reflexive and transitive.\n\\end{theorem}\n\\begin{proof}\nIf $Q$ factors through $C$, each coordinate factors by projection. Conversely, equal values of $C$ give equal values of every $q_i$. Function extensionality gives equal joint vectors, and Theorem~\\ref{res:new:factorization} supplies the joint decoder. This argument allows a different answer type for each task and also covers an empty task family.","For the second assertion, write $R=\\pi\\circ C$ and $q=d\\circ R$ on their attained sets. Then $q=(d\\circ\\pi)\\circ C$. The identity decoder proves reflexivity, and composition of two recovery maps proves transitivity. All maps are applied only to outputs in their declared domains.\n\\end{proof}","The word least concerns retained distinctions, not bytes, running time, or the cognitive cost of reading an archive. Two archives are identified by $Q$ precisely when every declared task gives the same answer on them. Any sufficient representation may retain more, but it cannot identify archives that $Q$ distinguishes. For a finite archive family, the reference implementation labels these joint-answer classes and checks that every task is constant within a class. Costs remain a separate optimization problem.","Let a node $v$ carry finite nonnegative numerical coordinates\n\\[\n\\mathbf v(v)=\n\\bigl(V(v),D(v),M(v),F(v),A(v)\\bigr),\n\\]\nrepresenting declared verification, discovery, meaning, future bridge, and auditability weights. Summation requires an additive numerical model. The weights may be supplied or estimated, but an ordinal ranking alone does not justify adding them. Changing a numerical scale can change the resulting loss and retention problem.","\\begin{definition}[Additive loss vector]\n\\label{res:9:1}\nFor a retained node set $K\\subseteq V$, define\n\\[\n\\Loss(\\Pcal,K)\n=\n\\sum_{v\\notin K}\\mathbf v(v).\n\\]\nNormalize a coordinate by its total only when that total is positive. If the total is zero, nonnegativity makes every contribution in that coordinate zero. The implementation records normalized loss zero there as a convention, rather than dividing by zero.\n\\end{definition}","\\begin{proposition}[Zero verification loss can coexist with positive discovery loss]\n\\label{res:9:2}\nThere exist proof fields and certificates $K$ for which the verification coordinate of $\\Loss(\\Pcal,K)$ is zero while the discovery and bridge coordinates are positive.\n\\end{proposition}","\\begin{proof}\nTake an AND-DAG certificate from $A$ to $B$ and add an optional example node $E$ that is not an ancestor of $B$. Assign positive discovery and bridge values to $E$ and zero verification value. The unique core omits $E$ while preserving the entire verification ancestry. The verification loss is zero and the other declared losses are positive.\n\\end{proof}","\\begin{definition}[Future task model]\n\\label{res:10:1}\nLet $\\Tcal$ be a finite task set with probabilities $p_t\\ge0$ satisfying $\\sum_t p_t=1$. Let $R_t\\subseteq V$ be the node set required by task $t$, and let $c(v)\\ge0$ be the recovery cost of node $v$ when it is absent.\n\\end{definition}","\\begin{definition}[Expected compression debt]\n\\label{res:10:2}\nThe expected compression debt of retained set $K$ is\n\\[\n\\Debt(K)\n=\n\\sum_{t\\in\\Tcal}p_t\n\\sum_{v\\in R_t\\setminus K}c(v).\n\\]\n\\end{definition}","\\begin{proposition}[Debt is monotone under retention]\n\\label{res:10:3}\nIf $K_1\\subseteq K_2$, then\n\\[\n\\Debt(K_2)\\le\\Debt(K_1).\n\\]\n\\end{proposition}","\\begin{proof}\nFor every task $t$, the missing set $R_t\\setminus K_2$ is contained in $R_t\\setminus K_1$. Recovery costs are nonnegative. The taskwise cost cannot increase, and neither can its probability-weighted sum.\n\\end{proof}","\\begin{corollary}[Irreversible compression accumulates debt]\n\\label{res:10:4}\nFor a descending sequence of retained sets\n\\[\nK_0\\supseteq K_1\\supseteq\\cdots\\supseteq K_r,\n\\]\nthe expected debt is nondecreasing.\n\\end{corollary}","The same theorem can support several legitimate representations:\n\\[\n\\Rcal(B)=\n\\{R_{\\mathrm{cert}},R_{\\mathrm{journal}},R_{\\mathrm{teach}},\nR_{\\mathrm{formal}},R_{\\mathrm{discovery}},R_{\\mathrm{archive}}\\}.\n\\]\nThe certificate is optimized for checking. The journal form balances rigor and readable length. The teaching form expands conceptual bottlenecks. The formal form exposes primitive dependencies. The discovery form retains alternatives and failed routes. The archive preserves reconstruction information.","\\begin{proposition}[Portfolio robustness]\n\\label{res:11:2}\nFor representation schemes $C_i:\\mathscr A\\to R_i$, retain the tagged portfolio $C(\\Pcal)=(C_i(\\Pcal))_i$. Every task recoverable from any one $C_i$ is recoverable from $C$.\n\\end{proposition}","\\begin{proof}\nIf $q=d_i\\circ C_i$, compose $d_i$ with the $i$th coordinate projection from the portfolio. This gives a decoder of $q$ from $C$. Retaining the coordinate labels is part of the construction. An untagged union that loses source identities need not permit the same recovery.\n\\end{proof}","\\begin{definition}[Proof compiler]\n\\label{res:12:1}\nLet $g$ be a presentation goal and let $\\Pcal$ be a versioned proof field. A proof compiler is a map on the declared archive class\n\\[\n\\Gamma_g:\\mathscr A\\longrightarrow R_g\n\\]\nthat selects, orders, and renders the material required by $g$ while preserving links to canonical claim and justification identifiers.\n\\end{definition}","\\begin{definition}[Latent bridge graph]\n\\label{res:13:1}\nA latent bridge graph is a finite directed graph\n\\[\n\\Bcal=(W,E_B)\n\\]\nwhose nodes include retained proof objects, omitted proof objects, and anticipated future targets. An edge $x\\to y$ means that $x$ has been declared a possible transfer route toward $y$. The edge may be certified, conjectural, or exploratory, but its status must be recorded.\n\\end{definition}","Let $S\\subseteq W$ be the retained starting set. Let $F\\subseteq W$ be a future target set with weights $w_f\\ge0$.","\\begin{definition}[Future reach utility]\n\\label{res:13:2}\nThe weighted future reach of $S$ is\n\\[\nU(S)=\\sum_{f\\in F}w_f\\,\n\\mathbf 1\\{f\\in\\Reach_{\\Bcal}(S)\\}.\n\\]\n\\end{definition}","\\begin{definition}[Bridge-critical node]\n\\label{res:13:3}\nA node $v$ is bridge critical for target $f$ relative to starts $S$ when $f$ is reachable from $S$ in $\\Bcal$ and becomes unreachable after $v$ is removed.\n\\end{definition}","In graph language, $v$ dominates an initially reachable $f$ relative to $S$ when every directed path from $S$ to $f$ passes through $v$. Paths of length zero are allowed in the reachability convention. Standard graph terminology is used throughout \\cite{diestel}.","\\begin{theorem}[Criticality equals domination]\n\\label{res:13:4}\nAssume $f$ is reachable from $S$. A node $v\\notin S$ is bridge critical for $f$ if and only if $v$ dominates $f$ relative to $S$.\n\\end{theorem}","\\begin{proof}\nIf $v$ dominates $f$, every path from $S$ to $f$ uses $v$. Removing $v$ removes every such path, so $f$ becomes unreachable. Conversely, if removing $v$ makes $f$ unreachable, then there could not have been a path from $S$ to $f$ avoiding $v$. Therefore every path used $v$, which is exactly domination.\n\\end{proof}","\\begin{corollary}[Weighted criticality]\n\\label{res:13:5}\nFor $v\\notin S$, the utility lost by deleting $v$ is the total weight of initially reachable future targets dominated by $v$:\n\\[\nU(S)-U_{\\Bcal\\setminus\\{v\\}}(S)\n=\\sum_{\\substack{f\\in F\\cap\\Reach_{\\Bcal}(S)","\\begin{corollary}[Weighted criticality]\n\\label{res:13:5}\nFor $v\\notin S$, the utility lost by deleting $v$ is the total weight of initially reachable future targets dominated by $v$:\n\\[\nU(S)-U_{\\Bcal\\setminus\\{v\\}}(S)\n=\\sum_{\\substack{f\\in F\\cap\\Reach_{\\Bcal}(S)\\\\v\\text{ dominates }f}}w_f.\n\\]\nDeleted target vertices are counted as unreachable.\n\\end{corollary}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_07_bridge_dominators.pdf}\n\\caption{A declared bridge graph. The boundary theorem $C$ dominates future targets $A^\\prime$ and $B^\\prime$ but not $C^\\prime$, which has a bypass from $E$. These arrows describe future reach in the supplied model, not proof certification. Deleting a bridge vertex is different from merely retaining a different source set.}\n\\label{fig:dominator}\n\\end{figure}","Let $C\\subseteq W$ be the candidate nodes that may be retained and let $M$ be a mandatory base set. For $S\\subseteq C$, define\n\\[\nU_M(S)=U(M\\cup S).\n\\]\nThe bridge graph remains fixed: retention changes the starting set, not the available intermediate vertices or edges. Deleting unretained intermediate vertices would define a different objective, for which the coverage proof below need not apply.","\\begin{theorem}[Bridge coverage is monotone and submodular]\n\\label{res:14:1}\n\\label{thm:bridge-submodular}\nThe future reach utility $U_M$ is a nonnegative monotone submodular set function on $C$.\n\\end{theorem}","\\begin{proof}\nFor each candidate $x\\in C$, let $R_x\\subseteq F$ be the set of future targets reachable from $M\\cup\\{x\\}$. Let $R_M$ be the targets already reachable from $M$. Reachability from a union of starting nodes is the union of their individual reach sets, so\n\\[\nU_M(S)=\n\\sum_{f\\in R_M\\cup\\bigcup_{x\\in S}R_x}w_f.\n\\]\nAdding a candidate cannot remove a reachable target, so $U_M$ is monotone.","Now let $S\\subseteq T\\subseteq C$ and let $x\\in C\\setminus T$. The marginal gain of adding $x$ is the total weight of targets in $R_x$ that are not already covered. Since $T$ covers at least every target covered by $S$, the set newly covered from $T$ is contained in the set newly covered from $S$. Therefore\n\\[\nU_M(S\\cup\\{x\\})-U_M(S)\n\\ge\nU_M(T\\cup\\{x\\})-U_M(T),\n\\]\nwhich is submodularity.\n\\end{proof}","\\begin{theorem}[Greedy cardinality guarantee]\n\\label{res:14:2}\n\\label{thm:bridge-greedy}\nLet $k$ be an integer with $1\\leq k\\leq|C|$. Starting from the empty supplement, let $S_k$ result from repeatedly adding an unselected candidate with largest marginal future reach. Let $O_k$ be an optimal supplement of size at most $k$. Then\n\\[\nU_M(S_k)-U_M(\\varnothing)\n\\ge\n\\left(1-\\left(1-\\frac1k\\right)^k\\right)\n\\bigl(U_M(O_k)-U_M(\\varnothing)\\bigr),\n\\]\nand therefore the greedy gain is at least $(1-1/e)$ times the optimal gain.\n\\end{theorem}","Let $k$ be an integer with $1\\leq k\\leq|C|$. Starting from the empty supplement, let $S_k$ result from repeatedly adding an unselected candidate with largest marginal future reach. Let $O_k$ be an optimal supplement of size at most $k$. Then\n\\[\nU_M(S_k)-U_M(\\varnothing)\n\\ge\n\\left(1-\\left(1-\\frac1k\\right)^k\\right)\n\\bigl(U_M(O_k)-U_M(\\varnothing)\\bigr),\n\\]\nand therefore the greedy gain is at least $(1-1/e)$ times the optimal gain.\n\\end{theorem}","\\begin{proof}\nWrite $G_i=U_M(S_i)-U_M(\\varnothing)$ and $G^*=U_M(O_k)-U_M(\\varnothing)$. By monotonicity and submodularity, the total marginal contribution of the at most $k$ elements of $O_k$ beyond $S_i$ is at least $G^*-G_i$. Hence at least one candidate has marginal gain at least $(G^*-G_i)/k$. Greedy chooses a candidate with at least that gain, so\n\\[\nG^*-G_{i+1}\n\\le\n\\left(1-\\frac1k\\right)(G^*-G_i).\n\\]\nIterating from $G_0=0$ gives\n\\[\nG^*-G_k\n\\le\n\\left(1-\\frac1k\\right)^kG^*.\n\\]\nRearranging gives the first bound. The inequality $(1-1/k)^k\\le e^{-1}$ gives the second.\n\\end{proof}","_M(S_i)-U_M(\\varnothing)$ and $G^*=U_M(O_k)-U_M(\\varnothing)$. By monotonicity and submodularity, the total marginal contribution of the at most $k$ elements of $O_k$ beyond $S_i$ is at least $G^*-G_i$. Hence at least one candidate has marginal gain at least $(G^*-G_i)/k$. Greedy chooses a candidate with at least that gain, so\n\\[\nG^*-G_{i+1}\n\\le\n\\left(1-\\frac1k\\right)(G^*-G_i).\n\\]\nIterating from $G_0=0$ gives\n\\[\nG^*-G_k\n\\le\n\\left(1-\\frac1k\\right)^kG^*.\n\\]\nRearranging gives the first bound. The inequality $(1-1/k)^k\\le e^{-1}$ gives the second.\n\\end{proof}","rnothing)$ and $G^*=U_M(O_k)-U_M(\\varnothing)$. By monotonicity and submodularity, the total marginal contribution of the at most $k$ elements of $O_k$ beyond $S_i$ is at least $G^*-G_i$. Hence at least one candidate has marginal gain at least $(G^*-G_i)/k$. Greedy chooses a candidate with at least that gain, so\n\\[\nG^*-G_{i+1}\n\\le\n\\left(1-\\frac1k\\right)(G^*-G_i).\n\\]\nIterating from $G_0=0$ gives\n\\[\nG^*-G_k\n\\le\n\\left(1-\\frac1k\\right)^kG^*.\n\\]\nRearranging gives the first bound. The inequality $(1-1/k)^k\\le e^{-1}$ gives the second.\n\\end{proof}","ginal contribution of the at most $k$ elements of $O_k$ beyond $S_i$ is at least $G^*-G_i$. Hence at least one candidate has marginal gain at least $(G^*-G_i)/k$. Greedy chooses a candidate with at least that gain, so\n\\[\nG^*-G_{i+1}\n\\le\n\\left(1-\\frac1k\\right)(G^*-G_i).\n\\]\nIterating from $G_0=0$ gives\n\\[\nG^*-G_k\n\\le\n\\left(1-\\frac1k\\right)^kG^*.\n\\]\nRearranging gives the first bound. The inequality $(1-1/k)^k\\le e^{-1}$ gives the second.\n\\end{proof}","\\begin{warning}[Arbitrary costs]\n\\label{res:14:3}\nFor $k=0$, both gains are zero without using the displayed division by $k$. For a budget at least $|C|$, retaining all candidates is optimal. The displayed theorem is the cardinality form of the standard submodular greedy bound \\cite{nemhauser}. A gain-per-cost rule for unequal costs is a different algorithm and does not inherit this guarantee. Zero-cost candidates also need an explicit convention.\n\\end{warning}","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_08_submodular_greedy.pdf}\n\\caption{Greedy and exact retention coincide on the committed finite instance. The square and circle marks are deliberately overlaid. The general cardinality guarantee is $1-1/e$, not equality with the optimum on every graph.}\n\\label{fig:greedy}\n\\end{figure}","\\begin{definition}[Versioned proof object]\n\\label{res:15:1}\nA versioned proof object is a sequence\n\\[\n\\Pcal_0\\longrightarrow\\Pcal_1\\longrightarrow\\cdots\n\\]\nwhose nodes and justifications carry stable identifiers. Each transition records additions, removals, altered scopes, evidence changes, and a digest of its parent version.\n\\end{definition}","\\begin{definition}[Append-only certified update]\n\\label{res:15:2}\nAn update $\\Pcal_i\\to\\Pcal_{i+1}$ is append-only certified when every existing node, admissible-leaf status, evidence record, interpretation, and certified justification remains unchanged. The later field may add records and certified rules but may not withdraw an earlier admitted leaf or change the scope of an earlier inference.\n\\end{definition}","\\begin{theorem}[Derivability is monotone under append-only updates]\n\\label{res:15:3}\nEvery target derivable in $\\Pcal_i$ remains derivable in $\\Pcal_{i+1}$ under an append-only certified update.\n\\end{theorem}","\\begin{proof}\nA certificate in $\\Pcal_i$ uses only nodes and certified justifications that remain unchanged in $\\Pcal_{i+1}$. The same selected subfield therefore satisfies the same derivability closure in the later version.\n\\end{proof}","\\begin{definition}[Syntactic affected set]\n\\label{res:15:4}\nFor a changed node set $X\\subseteq V$, define\n\\[\n\\operatorname{Aff}(X)=X\\cup\\Desc(X),\n\\]\nwhere descendants are computed in the declared dependency graph.\n\\end{definition}","\\begin{theorem}[Minimal closed syntactic review set]\n\\label{res:15:5}\nThe set $\\operatorname{Aff}(X)$ is the unique smallest node set $Y$ such that $X\\subseteq Y$ and $Y$ is closed under outgoing dependency edges: whenever $u\\in Y$ and $u\\to v$ is a declared dependency edge, then $v\\in Y$.\n\\end{theorem}","\\begin{proof}\nBy construction, every node reachable from $X$ lies in $\\operatorname{Aff}(X)$, so the set contains $X$ and is closed under outgoing edges. Let $Y$ be any other set with those two properties. Every directed path beginning in $X$ remains inside $Y$ by induction on path length. Hence every descendant of $X$ lies in $Y$, and $\\operatorname{Aff}(X)\\subseteq Y$.\n\\end{proof}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_09_version_invalidation.pdf}\n\\caption{Changing $L_1$ marks $L_1$, $L_2$, and $B$ for syntactic review. The lower branch has no forward dependency path from $L_1$. Membership in the review set is not a claim that a theorem has become false.}\n\\label{fig:invalidation}\n\\end{figure}","\\begin{proposition}[Digest stability]\n\\label{res:16:3}\nFix a deterministic serialization $s$ and a deterministic hash function $h$. Identical serialized byte sequences have identical digests under $h$. Distinct serialized byte sequences have distinct digests unless they form a collision under $h$. Detecting a record change additionally requires that the change alter its serialized bytes.\n\\end{proposition}","\\begin{proof}\nIf $s(r)=s(r')$, applying the function $h$ gives equal digests. If $s(r)\\ne s(r')$ but $h(s(r))=h(s(r'))$, the two serialized strings form a hash collision by definition. Thus change detection at the record level requires both preservation of the audited distinction by $s$ and exclusion of a collision for the resulting bytes. No injectivity of a fixed-length hash on arbitrary byte strings is asserted.\n\\end{proof}","For example, joining string parts with a separator is ambiguous if the same separator may occur inside a part. The tuples $(\\texttt{a|b},\\texttt{c})$ and $(\\texttt{a},\\texttt{b|c})$ both become \\texttt{a|b|c}. No hash function can recover the lost tuple boundary. The earlier \\path{provenance_digest} helper had this defect with a control-character separator. Python release 1.4.0 replaces that encoding with a versioned, domain-separated JSON array of strings. Historical digests must retain their earlier format and version. The snapshot digest has its own unchanged serialization contract.","\\begin{proposition}[Hashing cannot add recoverable tasks]\n\\label{res:new:hash}\nLet $s:\\mathscr A\\to B$ be serialization and $h:B\\to D$ be any deterministic map. Every task recoverable from $h\\circ s$ is recoverable from $s$. If $h$ is injective on the attained set $s(\\mathscr A)$, the two encodings support exactly the same tasks.\n\\end{proposition}\n\\begin{proof}\nIf $s(a)=s(b)$, then $h(s(a))=h(s(b))$. Any task constant on the digest fibers is therefore constant on the serialization fibers, and Theorem~\\ref{res:new:factorization} supplies its decoder. If $h$ is injective on attained serializations, equality of digests conversely implies equality of serializations, so the two fiber partitions coincide and the same criterion applies in both directions.\n\\end{proof}","Every task recoverable from $h\\circ s$ is recoverable from $s$. If $h$ is injective on the attained set $s(\\mathscr A)$, the two encodings support exactly the same tasks.\n\\end{proposition}\n\\begin{proof}\nIf $s(a)=s(b)$, then $h(s(a))=h(s(b))$. Any task constant on the digest fibers is therefore constant on the serialization fibers, and Theorem~\\ref{res:new:factorization} supplies its decoder. If $h$ is injective on attained serializations, equality of digests conversely implies equality of serializations, so the two fiber partitions coincide and the same criterion applies in both directions.\n\\end{proof}","Let $X_i=\\theta+\\varepsilon_i$ be numerical estimates of a fixed quantity $\\theta$. Assume $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2>0$, and common pairwise correlation $\\rho$. For $n>1$, assume $-1/(n-1)<\\rho\\leq1$, so the mean has positive variance. These variables model estimates, not proof truth values. The mean has variance\n\\[\n\\operatorname{Var}(\\bar X)\n=\n\\sigma^2\\left(\\frac{1-\\rho}{n}+\\rho\\right).\n\\]","\\begin{theorem}[Effective reviewer count]\n\\label{res:17:1}\n\\label{thm:review-count}\nUnder the equicorrelated model, define the variance-equivalent reviewer count by\n\\[\n\\boxed{\n n_{\\mathrm{eff}}\n =\n \\frac{n}{1+(n-1)\\rho}\n}\n\\]\nso that $\\operatorname{Var}(\\bar X)=\\sigma^2/n_{\\mathrm{eff}}$. This effective count is a positive real number and need not be an integer.\n\\end{theorem}","\\begin{proof}\nThe mean of $m$ independent reviewers of variance $\\sigma^2$ has variance $\\sigma^2/m$. Equating this with the correlated mean variance gives\n\\[\n\\frac{\\sigma^2}{m}\n=\n\\sigma^2\\left(\\frac{1-\\rho}{n}+\\rho\\right)\n=\n\\sigma^2\\frac{1+(n-1)\\rho}{n}.\n\\]\nSolving for $m$ gives the formula.\n\\end{proof}","For $n=1$, the effective count is one. At the excluded boundary $\\rho=-1/(n-1)$, the mean has zero variance and there is no finite variance-equivalent count. For fixed $0<\\rho\\leq1$, $n_{\\mathrm{eff}}\\to1/\\rho$ as $n\\to\\infty$. Negative correlations can give $n_{\\mathrm{eff}}>n$, but a fixed negative correlation is not admissible for arbitrarily large $n$.","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=\\linewidth]{fig_10_effective_reviewers.pdf}\n\\caption{Effective reviewer count under the specified equicorrelation model. The logarithmic vertical scale keeps all four curves visible. At $n=100$ and $\\rho=0.1$, the effective count is about $9.17$. This formula models variance reduction, not the probability that a mathematical claim is true.}\n\\label{fig:reviewers}\n\\end{figure}","For a general positive definite covariance matrix $\\Sigma$, consider unbiased linear aggregates $w^\\top X$ with $\\mathbf 1^\\top w=1$.","\\begin{theorem}[Minimum-variance covariance-aware weights]\n\\label{res:17:2}\nThe unique minimum-variance unbiased linear aggregate has weights\n\\[\nw^*\n=\n\\frac{\\Sigma^{-1}\\mathbf 1}\n{\\mathbf 1^\\top\\Sigma^{-1}\\mathbf 1}.\n\\]\nIts variance is\n\\[\n\\frac{1}{\\mathbf 1^\\top\\Sigma^{-1}\\mathbf 1}.\n\\]\n\\end{theorem}","\\begin{proof}\nMinimize $w^\\top\\Sigma w$ subject to $\\mathbf 1^\\top w=1$. The Lagrangian is\n\\[\nL(w,\\lambda)=w^\\top\\Sigma w-2\\lambda(\\mathbf 1^\\top w-1).\n\\]\nStationarity gives $\\Sigma w=\\lambda\\mathbf 1$, hence $w=\\lambda\\Sigma^{-1}\\mathbf 1$. The constraint gives\n\\[\n\\lambda=(\\mathbf 1^\\top\\Sigma^{-1}\\mathbf 1)^{-1}.\n\\]\nPositive definiteness makes the objective strictly convex, so the solution is unique. Substitution gives the variance formula.\n\\end{proof}","\\begin{warning}[Bias is not covariance]\n\\label{res:17:3}\nCovariance-aware weighting does not remove a common bias that is absent from the covariance model. If every reviewer shares bias $b$, the mean squared error contains the irreducible term $b^2$.\n\\end{warning}","\\begin{proposition}[MSE with a shared interpretive offset]\n\\label{res:25:1}\nSuppose reviewers report $J_i=\\theta+b+\\varepsilon_i$, where $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance en","\\begin{proposition}[MSE with a shared interpretive offset]\n\\label{res:25:1}\nSuppose reviewers report $J_i=\\theta+b+\\varepsilon_i$, where $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squa","\\begin{proposition}[MSE with a shared interpretive offset]\n\\label{res:25:1}\nSuppose reviewers report $J_i=\\theta+b+\\varepsilon_i$, where $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus s","\\begin{proposition}[MSE with a shared interpretive offset]\n\\label{res:25:1}\nSuppose reviewers report $J_i=\\theta+b+\\varepsilon_i$, where $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\e","\\begin{proposition}[MSE with a shared interpretive offset]\n\\label{res:25:1}\nSuppose reviewers report $J_i=\\theta+b+\\varepsilon_i$, where $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","shared interpretive offset]\n\\label{res:25:1}\nSuppose reviewers report $J_i=\\theta+b+\\varepsilon_i$, where $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","e $\\theta$ is the target quantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","uantity, $b$ is a fixed shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","d shared offset, $\\mathbb E\\varepsilon_i=0$, $\\operatorname{Var}(\\varepsilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","psilon_i)=\\sigma^2$, and $\\operatorname{Cov}(\\varepsilon_i,\\varepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","arepsilon_j)=\\rho\\sigma^2$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","$ for $i\\ne j$. For admissible $\\rho$, the mean squared error of the reviewer average is\n\\[\n\\boxed{\\mathbb E(\\overline J_n-\\theta)^2=b^2+\\sigma^2\\left(\\rho+\\frac{1-\\rho}{n}\\right).}\n\\]\nFor fixed $\\rho\\ge0$, it approaches $b^2+\\rho\\sigma^2$ as $n\\to\\infty$.\n\\end{proposition}\n\\begin{proof}\nThe estimation error is $\\overline J_n-\\theta=b+\\overline\\varepsilon_n$. Its mean is $b$, and its variance is $\\sigma^2\\big((1-\\rho)/n+\\rho\\big)$ by summing all diagonal and off-diagonal covariance entries. Mean squared error is variance plus squared bias. Taking the limit completes the proof.\n\\end{proof}","A claim about identity effects needs a design that can distinguish them from other differences. Hold the mathematical content fixed and let $Y_i(1)$ and $Y_i(0)$ be the fixed potential evaluations of reviewer $i$ under two labels. In a finite set of $N$ reviewers, assign exactly $n_1$ to the first label and $n_0=N-n_1$ to the second, uniformly over assignments with those sizes, where $n_0,n_1>0$. Assume each observed response equals the potential response under its assigned label and does not depend on other assignments. Then the difference of the two sample means is unbiased for $N^{-1}\\sum_i(Y_i(1)-Y_i(0))$. Indeed, each reviewer has inclusion probability $n_1/N$ in the first mean and $n_0/N$ in the second, so linearity of expectation gives that identity.","luations of reviewer $i$ under two labels. In a finite set of $N$ reviewers, assign exactly $n_1$ to the first label and $n_0=N-n_1$ to the second, uniformly over assignments with those sizes, where $n_0,n_1>0$. Assume each observed response equals the potential response under its assigned label and does not depend on other assignments. Then the difference of the two sample means is unbiased for $N^{-1}\\sum_i(Y_i(1)-Y_i(0))$. Indeed, each reviewer has inclusion probability $n_1/N$ in the first mean and $n_0/N$ in the second, so linearity of expectation gives that identity.","sign exactly $n_1$ to the first label and $n_0=N-n_1$ to the second, uniformly over assignments with those sizes, where $n_0,n_1>0$. Assume each observed response equals the potential response under its assigned label and does not depend on other assignments. Then the difference of the two sample means is unbiased for $N^{-1}\\sum_i(Y_i(1)-Y_i(0))$. Indeed, each reviewer has inclusion probability $n_1/N$ in the first mean and $n_0/N$ in the second, so linearity of expectation gives that identity.","t label and $n_0=N-n_1$ to the second, uniformly over assignments with those sizes, where $n_0,n_1>0$. Assume each observed response equals the potential response under its assigned label and does not depend on other assignments. Then the difference of the two sample means is unbiased for $N^{-1}\\sum_i(Y_i(1)-Y_i(0))$. Indeed, each reviewer has inclusion probability $n_1/N$ in the first mean and $n_0/N$ in the second, so linearity of expectation gives that identity.","Let $a_t\\geq0$ be candidate arrivals during audit cycle $t$, let $s_t\\geq0$ be service capacity, and let $B_t\\geq0$ be the unresolved backlog. Suppose arrivals in a cycle are available to that cycle's service. The basic deterministic queue is\n\\[\nB_{t+1}=\\max\\{0,B_t+a_t-s_t\\}.\n\\]","\\begin{theorem}[Constant-rate audit regimes]\n\\label{res:18:1}\n\\label{thm:audit-rates}\nSuppose $a_t\\equiv\\lambda$ and $s_t\\equiv\\mu$.\n\\begin{enumerate}[label=\\textup{(\\roman*)}]\n\\item If $\\lambda<\\mu$, every finite initial backlog reaches zero after finitely many cycles.\n\\item If $\\lambda=\\mu$, the backlog remains constant.\n\\item If $\\lambda>\\mu$, the backlog grows linearly with slope $\\lambda-\\mu$.\n\\end{enumerate}\n\\end{theorem}","\\begin{proof}\nBefore the truncation at zero, each cycle adds $\\lambda-\\mu$. When this quantity is negative, the backlog decreases by a fixed amount until it reaches zero. When it is zero, the backlog is unchanged. When it is positive, the truncation is never active after a nonnegative start, and the backlog is $B_0+t(\\lambda-\\mu)$.\n\\end{proof}","A simple audit priority score is\n\\[\nP(x)\n=\n\\frac{p_{\\mathrm{err}}(x)\\,c_{\\mathrm{consequence}}(x)\n\\bigl(1+r_{\\mathrm{dependency}}(x)\\bigr)}\n{c_{\\mathrm{audit}}(x)}.\n\\]\nHere $p_{\\mathrm{err}}(x)\\in[0,1]$, the consequence and dependency factors are nonnegative, and $c_{\\mathrm{audit}}(x)>0$. Zero-cost checks can be handled separately. This is a declared priority heuristic, not a probability of truth or a proved optimal scheduling rule.","A refinement preserves the certified target on its declared scope. Preservation of other tasks requires a separate recovery condition. Let $R_d:\\mathscr A\\to\\mathcal R_d$ be the representation scheme at depth $d$. A sufficient condition is a map\n\\[\n\\pi_d:R_{d+1}(\\mathscr A)\\longrightarrow R_d(\\mathscr A)\n\\qquad\\text{with}\\qquad R_d=\\pi_d\\circ R_{d+1}.\n\\]\nIf a task factors as $q=d_q\\circ R_d$, then\n\\[\nq=(d_q\\circ\\pi_d)\\circ R_{d+1}.\n\\]\nHence $R_d\\preceq_{\\Qcal}R_{d+1}$. Repeated refinements satisfying this condition give the recoverability chain. For a restricted task family, preserving its supported tasks directly can suffice without recovering the entire preceding representation.","efinement preserves the certified target on its declared scope. Preservation of other tasks requires a separate recovery condition. Let $R_d:\\mathscr A\\to\\mathcal R_d$ be the representation scheme at depth $d$. A sufficient condition is a map\n\\[\n\\pi_d:R_{d+1}(\\mathscr A)\\longrightarrow R_d(\\mathscr A)\n\\qquad\\text{with}\\qquad R_d=\\pi_d\\circ R_{d+1}.\n\\]\nIf a task factors as $q=d_q\\circ R_d$, then\n\\[\nq=(d_q\\circ\\pi_d)\\circ R_{d+1}.\n\\]\nHence $R_d\\preceq_{\\Qcal}R_{d+1}$. Repeated refinements satisfying this condition give the recoverability chain. For a restricted task family, preserving its supported tasks directly can suffice without recovering the entire preceding representation.","its declared scope. Preservation of other tasks requires a separate recovery condition. Let $R_d:\\mathscr A\\to\\mathcal R_d$ be the representation scheme at depth $d$. A sufficient condition is a map\n\\[\n\\pi_d:R_{d+1}(\\mathscr A)\\longrightarrow R_d(\\mathscr A)\n\\qquad\\text{with}\\qquad R_d=\\pi_d\\circ R_{d+1}.\n\\]\nIf a task factors as $q=d_q\\circ R_d$, then\n\\[\nq=(d_q\\circ\\pi_d)\\circ R_{d+1}.\n\\]\nHence $R_d\\preceq_{\\Qcal}R_{d+1}$. Repeated refinements satisfying this condition give the recoverability chain. For a restricted task family, preserving its supported tasks directly can suffice without recovering the entire preceding representation.","Let $r_d\\geq0$ be a declared unresolved-work score and $c_d\\geq0$ the cumulative cost. These are model inputs. A small $r_d$ is not, without another theorem, a certificate of a nearly valid proof. The word cube names the layered presentation, not a cubical complex supplied with additional geometric axioms.","\\begin{definition}[Goal-relative depth objective]\n\\label{res:19:2}\nFor goal $g$ with weights $\\alpha_g,\\beta_g\\ge0$, define\n\\[\nJ_g(d)=\\alpha_g r_d+\\beta_g c_d.\n\\]\nAn optimal expansion depth is any minimizer of $J_g$.\n\\end{definition}","\\begin{proof}\nTake a proof object with exactly two available depths $d_1<d_2$, with $r_{d_2}<r_{d_1}$ and $c_{d_2}>c_{d_1}$. Direct subtraction shows that $d_2$ is preferred exactly when\n\\[\n\\alpha_g(r_{d_1}-r_{d_2})>\\beta_g(c_{d_2}-c_{d_1}).\n\\]\nThe goal $(\\alpha_g,\\beta_g)=(1,0)$ strictly prefers $d_2$, whereas $(0,1)$ strictly prefers $d_1$. Since these are the only available depths, each preference is a global optimum. If both weights vanish, every depth is optimal. No division by a possibly zero weight is needed.\n\\end{proof}","A practical stopping rule selects the first depth satisfying\n\\[\nr_d\\le\\varepsilon\\qquad\\text{or}\\qquad\n\\frac{r_{d-1}-r_d}{c_d-c_{d-1}}<\\tau.\n\\]\nThe first condition uses a declared tolerance. The second applies only at $d\\geq1$ with $c_d>c_{d-1}$ and compares the most recent reduction with its extra cost. Zero-cost refinements require separate handling. These stopping rules do not generally minimize $J_g$ over all remaining depths.","ractical stopping rule selects the first depth satisfying\n\\[\nr_d\\le\\varepsilon\\qquad\\text{or}\\qquad\n\\frac{r_{d-1}-r_d}{c_d-c_{d-1}}<\\tau.\n\\]\nThe first condition uses a declared tolerance. The second applies only at $d\\geq1$ with $c_d>c_{d-1}$ and compares the most recent reduction with its extra cost. Zero-cost refinements require separate handling. These stopping rules do not generally minimize $J_g$ over all remaining depths.","The field \\texttt{examples/and\\_dag.json} has required premise $A$, a definition $D$, lemmas $L_1$ and $L_2$, and target $B$. It also contains a worked example, a failed route, and a boundary case. The unique verification core is\n\\[\n\\{A,D,L_1,L_2,B\\}.\n\\]\nThe optional nodes have zero declared verification value for $B$ and positive discovery, meaning, or bridge value. This is the executable instance behind Figure~\\ref{fig:loss}.","The field \\texttt{examples/alternative\\_proofs.json} contains a short route and a richer explanatory route. The short route has storage cost $1$, while the richer route has storage cost $3$ and larger meaning and discovery values. Exact certificate search chooses\n\\[\nA\\longrightarrow L_{\\mathrm{short}}\\longrightarrow B\n\\]\nwhen the objective is storage cost alone. A teaching compiler may rationally choose the other route because it uses a different objective.","The field \\texttt{examples/set\\_cover\\_certificate.json} contains four optional source lemmas and four obligations. Exact search returns the certificate using $S_1$ and $S_3$, with total cost $2$. The particular optimum is not important. The example demonstrates how local alternative derivations create a global combinatorial selection problem.","The bridge experiment uses ten candidate nodes and sixteen weighted future targets. For budgets zero through five, the script computes both the exact optimum and the greedy solution. The generated data are stored in \\texttt{data/retention\\_comparison.csv}. On this instance the two curves coincide. The theorem guarantees only the worst-case $(1-1/e)$ factor.","For one hundred equicorrelated reviewers with $\\rho=0.1$,\n\\[\nn_{\\mathrm{eff}}\n=\n\\frac{100}{1+99(0.1)}\n\\approx 9.17.\n\\]\nThe audit example begins with backlog $10$. Arrivals $4$ and capacity $5$ clear the queue. Equal rates hold the backlog at $10$. Arrivals $6$ and capacity $5$ increase it by one per cycle.","Alternative field & minimum cost $1$ & exact finite search","Set Cover field & minimum cost $2$ & Theorem~\\ref{thm:minimum-certificate} encoding","Review model & $n_{\\mathrm{eff}}$ formula & Theorem~\\ref{thm:review-count}","\\subsection{An expansion-depth calculation}\n\\label{sec:depth-example}\nFor depths $0$ through $6$, take unresolved-work scores\n$(1,0.55,0.30,0.18,0.12,0.09,0.075)$ and costs\n$(0,0.25,0.65,1.2,2.1,3.5,5.5)$. The declared weights\n$(\\alpha,\\beta)=(1,0.75)$, $(2.5,0.35)$, and $(5,0.15)$ give\nminimizing depths $1$, $3$, and $4$, respectively. Their minimum objective\nvalues are $0.7375$, $0.87$, and $0.915$. These follow by evaluating\n$\\alpha r_d+\\beta c_d$ at the seven available depths. The labels certificate,\nteaching, and archive name these three supplied objectives rather than\nmeasured preferences of actual readers.","lculation}\n\\label{sec:depth-example}\nFor depths $0$ through $6$, take unresolved-work scores\n$(1,0.55,0.30,0.18,0.12,0.09,0.075)$ and costs\n$(0,0.25,0.65,1.2,2.1,3.5,5.5)$. The declared weights\n$(\\alpha,\\beta)=(1,0.75)$, $(2.5,0.35)$, and $(5,0.15)$ give\nminimizing depths $1$, $3$, and $4$, respectively. Their minimum objective\nvalues are $0.7375$, $0.87$, and $0.915$. These follow by evaluating\n$\\alpha r_d+\\beta c_d$ at the seven available depths. The labels certificate,\nteaching, and archive name these three supplied objectives rather than\nmeasured preferences of actual readers.","otient partial order.\n\\item Verification loss, discovery loss, meaning loss, bridge loss, and audit loss can diverge.\n\\item Expected reconstruction debt decreases monotonically as more required material is retained.\n\\item Bridge criticality is domination in the declared future-reach graph.\n\\item Weighted future reach is monotone and submodular.\n\\item Greedy cardinality retention has the standard $(1-1/e)$ guarantee.\n\\item Append-only certified updates preserve existing derivability.\n\\item The forward dependency closure is the least set containing the changed nodes and closed under outgoing dependency edges. It marks potential review, not automatic falsehood.\n\\item Positive review dependence lowers the variance-equivalent count. Under a positive definite covariance model, generalized least squares gives the minimum-variance unbiased linear aggregate.\n\\item Constant-rate audit queues have an exact stable, critical, or unstable classification.\n\\item Optimal proof expansion depth depends on the presentation goal.\n\\end{enumerate}","The final thesis is therefore:\n\\[\n\\boxed{\\begin{minipage}{0.88\\linewidth}\nA proof should be published as a certificate and maintained as a versioned proof field. The certificate preserves deductive validity. The larger field preserves the dependencies, alternatives, examples, failures, provenance, and bridges from which later mathematics may grow.\n\\end{minipage}}\n\\]","Letters are local to a section. In particular, $A$ denotes required premises in a proof field, but a matrix in the binary examples. The type declarations, not the letter alone, determine the object.","\\midrule\n\\endhead\n$\\Pcal$, $V$, $\\Jcal$ & A declared proof field, its nodes, and its justifications.","$A$, $L$, $T_0$ & Required premises, admissible leaves, and requested targets, with $A\\subseteq L$.","$(K,H)$ & Selected nodes and certified justification records. Only usable rules contribute to derivability.","$D_{\\mathrm{cert}}(K,H)$ & The least certified closure in the selected record.","$\\Anc_L(T_0)$ & Backward ancestry expanded only outside the admissible leaves.","$\\mathcal M_F(U)$ & Local models on patch $U$ satisfying all certified rules supported entirely in $U$.","$\\Gamma(\\mathcal M_F)$ & Coherent families of those local models under restriction.","$\\operatorname{cl}_F(S)$ & The least declared rule model containing the positive observations $S$ and all admitted leaves.","$\\operatorname{Ext}_F(S,N)$ & Rule models containing $S$ and avoiding the negative observations $N$.","$F_{\\mathcal U}$ & The same field with only rules whose entire supports appear in a covering patch.","$Q(a)=(q_i(a))_i$ & The joint task representation, least in retained distinctions for its task family.","$\\mathscr A$, $C$, $q$ & A class of possible archives, an encoding scheme, and a task on that class.","$P_K(S)=S\\cap K$ & Pure retention of records selected by a fixed mask.","$D:E\\to R$, $p:E\\to S$ & A linear defect and a protected observable. The repair condition is $Dy=0$, $p(y)=p(x)$.","$H=\\ker D\\cap\\ker p$ & The ambiguity space of a nonempty protected fiber. This $H$ is not a set of justifications.","$U_M(S)$ & Weighted target reach from the mandatory starts and a selected supplement, in a fixed directed bridge graph.","$n_{\\mathrm{eff}}$ & A real, variance-equivalent review count in the specified covariance model.","[0pt]\nPPT-015 & Greedy cardinality retention achieves at least $1-1/e$ for positive feasible cardinality budgets.\\newline{\\footnotesize\\color{KomorebiMuted}greedy and exact solvers} & {\\sffamily\\small\\color{KomorebiTeal}Written}"]}