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u_h(z)>0}","kind":"use","occurrences":[40]}],"contexts":["midrule\n$\\Gamma(F)$ & Compatible families in a specified system of views.","$r,\\pi,C$ & A record, a state quotient, and probability aggregation.","$F_a,K_{a,o}$ & Deterministic operations and joint stochastic instruments.","$\\mathcal E,\\Sigma$ & An evidence family and its correction signature.","$T_*,L_*,\\rho$ & Ending region, safe-running kernel, and first-entry rank.","$M,N,J$ & Old record, required new record, and migration map.","\\section{The joint data of a decision process}\nLet $X$ and $Z$ be finite nonempty state sets and let $\\pi:X\\to Z$ be surjective. Let $A$ and $O$ be finite nonempty action and report alphabets. At a fine state $x$, the set $D(x)\\subseteq A$ contains the admitted actions, and $\\ell(x)\\in Y$ is a protected label. For $a\\in D(x)$, the nonnegative numbers\n\\[\n K_{a,o}(x'\\mid x),\\qquad\n \\sum_{o\\in O}\\sum_{x'\\in X}K_{a,o}(x'\\mid x)=1\n\\]\nspecify the joint distribution of the next state and report. No transition is specified for a disabled action. An implementation may fill unused entries, but those values have no role in the contract.","For a nonnegative vector $u$ on $X$, write\n\\[\n (\\pi_*u)(z)=\\sum_{\\pi x=z}u(x).\n\\]\nThis notation applies to probability distributions and to unnormalized history masses. A coarse model has labels $\\bar\\ell$, enabled sets $\\bar D$, and joint instruments $\\bar K$ on its enabled actions.","\\begin{theorem}[Joint legal and stochastic quotient criterion]\\label{r:joint-quotient}\nThere are coarse data satisfying\n\\[\n D(x)=\\bar D(\\pi x),\\quad \\ell(x)=\\bar\\ell(\\pi x),\\quad\n \\sum_{\\pi x'=z'}K_{a,o}(x'\\mid x)=\\bar K_{a,o}(z'\\mid\\pi x)\n\\]\nfor every $x$, every $a\\in D(x)$, and every $o,z'$, if and only if states in the same fibre have identical labels and enabled sets and identical projected joint successor/report rows. Explicitly, $\\pi x=\\pi y$ must imply\n\\begin{align}\n D(x)&=D(y),\\qquad \\ell(x)=\\ell(y),\\label{eq:fibre-legal}","ic quotient criterion]\\label{r:joint-quotient}\nThere are coarse data satisfying\n\\[\n D(x)=\\bar D(\\pi x),\\quad \\ell(x)=\\bar\\ell(\\pi x),\\quad\n \\sum_{\\pi x'=z'}K_{a,o}(x'\\mid x)=\\bar K_{a,o}(z'\\mid\\pi x)\n\\]\nfor every $x$, every $a\\in D(x)$, and every $o,z'$, if and only if states in the same fibre have identical labels and enabled sets and identical projected joint successor/report rows. Explicitly, $\\pi x=\\pi y$ must imply\n\\begin{align}\n D(x)&=D(y),\\qquad \\ell(x)=\\ell(y),\\label{eq:fibre-legal}","quotient}\nThere are coarse data satisfying\n\\[\n D(x)=\\bar D(\\pi x),\\quad \\ell(x)=\\bar\\ell(\\pi x),\\quad\n \\sum_{\\pi x'=z'}K_{a,o}(x'\\mid x)=\\bar K_{a,o}(z'\\mid\\pi x)\n\\]\nfor every $x$, every $a\\in D(x)$, and every $o,z'$, if and only if states in the same fibre have identical labels and enabled sets and identical projected joint successor/report rows. Explicitly, $\\pi x=\\pi y$ must imply\n\\begin{align}\n D(x)&=D(y),\\qquad \\ell(x)=\\ell(y),\\label{eq:fibre-legal}\\\\\n \\sum_{\\pi x'=z'}K_{a,o}(x'\\mid x)\n &=\\sum_{\\pi x'=z'}K_{a,o}(x'\\mid y)\n \\quad(a\\in D(x),\\ o\\in O,\\ z'\\in Z).\\label{eq:fibre-joint}\n\\end{align}\nWhen these conditions hold, the coarse labels, enabled sets, and instruments on enabled actions are unique.\n\\end{theorem}\n\\begin{proof}\nThe factorization identities give necessity immediately. For sufficiency, choose a representative $x$ of each fibre. Define its coarse label and enabled set by their values at $x$, and define the coarse instrument by the displayed fibre sum. The hypotheses make all three choices independent of the representative. The new entries are nonnegative. Summing the…","\\sum_{\\pi x'=z'}K_{a,o}(x'\\mid x)\n &=\\sum_{\\pi x'=z'}K_{a,o}(x'\\mid y)\n \\quad(a\\in D(x),\\ o\\in O,\\ z'\\in Z).\\label{eq:fibre-joint}\n\\end{align}\nWhen these conditions hold, the coarse labels, enabled sets, and instruments on enabled actions are unique.\n\\end{theorem}\n\\begin{proof}\nThe factorization identities give necessity immediately. For sufficiency, choose a representative $x$ of each fibre. Define its coarse label and enabled set by their values at $x$, and define the coarse instrument by the displayed fibre sum. The hypotheses make all three choices independent of the representative. The new entries are nonnegative. Summing them over $o,z'$ sums every fine successor exactly once and therefore gives one. Surjectivity gives uniqueness at every coarse state. Uniqueness says nothing about padded values for disabled actions.\n\\end{proof}","a,o}(x'\\mid y)\n \\quad(a\\in D(x),\\ o\\in O,\\ z'\\in Z).\\label{eq:fibre-joint}\n\\end{align}\nWhen these conditions hold, the coarse labels, enabled sets, and instruments on enabled actions are unique.\n\\end{theorem}\n\\begin{proof}\nThe factorization identities give necessity immediately. For sufficiency, choose a representative $x$ of each fibre. Define its coarse label and enabled set by their values at $x$, and define the coarse instrument by the displayed fibre sum. The hypotheses make all three choices independent of the representative. The new entries are nonnegative. Summing them over $o,z'$ sums every fine successor exactly once and therefore gives one. Surjectivity gives uniqueness at every coarse state. Uniqueness says nothing about padded values for disabled actions.\n\\end{proof}","que.\n\\end{theorem}\n\\begin{proof}\nThe factorization identities give necessity immediately. For sufficiency, choose a representative $x$ of each fibre. Define its coarse label and enabled set by their values at $x$, and define the coarse instrument by the displayed fibre sum. The hypotheses make all three choices independent of the representative. The new entries are nonnegative. Summing them over $o,z'$ sums every fine successor exactly once and therefore gives one. Surjectivity gives uniqueness at every coarse state. Uniqueness says nothing about padded values for disabled actions.\n\\end{proof}","\\section{Observable histories and common admissibility}\nA history available to the policy is\n\\[\n h=(a_1,o_1,\\ldots,a_t,o_t).\n\\]\nThe fine hidden state and the coarse state are not additional policy inputs. Protected labels are recorded for comparison, but are not assumed observed merely because the theorem preserves them. An application that exposes a label must include it in the observation interface of both models.","Start from a fine prior $p_0$ and the coarse prior $\\bar p_0=\\pi_*p_0$. Let $u_h(x)$ be the unnormalized probability of history $h$ and current state $x$. A common policy assigns the same probability $\\alpha(a\\mid h)$ in both models. At every positive-mass history it may select only actions in\n\\[\n D_h=\\bigcap_{x:u_h(x)>0}D(x).\n\\]\nThe intersection is over a nonempty support. If it is empty, no policy can make another admitted action at that history. A stopping convention can be included as a shared terminal action, but is not silently supplied by a stochastic row.","\\begin{lemma}[Support and admissible actions agree]\\label{r:joint-support}\nIf $\\bar u_h=\\pi_*u_h$, then\n\\[\n \\operatorname{supp}\\bar u_h=\\pi(\\operatorname{supp}u_h),\\qquad\n \\bigcap_{x:u_h(x)>0}D(x)=\\bigcap_{z:\\bar u_h(z)>0}\\bar D(z)\n\\]\nwhenever the history has positive mass.\n\\end{lemma}\n\\begin{proof}\nA finite sum of nonnegative terms is positive exactly when at least one summand is positive. This gives the support identity. The enabled sets are constant on each fibre, so repeating a fibre's action set in an intersection changes nothing. The two intersections are therefore equal.\n\\end{proof}","cy preserves the law of every finite coarse-state/action/report path. Consequently it preserves every action/report/label history. At every positive-mass history it also preserves the aggregated posterior and the set of actions admitted throughout its support. These claims hold at every horizon for which the policy is defined on all positive-mass prefixes.\n\\end{theorem}\n\\begin{proof}\nFirst prove $\\bar u_h=\\pi_*u_h$ by induction. It holds initially by the prior assumption. At a positive history, Lemma~\\ref{r:joint-support} gives the same available actions on both sides. For any action with positive policy mass, every current fine state of positive mass enables that action. Thus\n\\[\n \\pi_*u_{hao}\n =\\pi_*\\bigl(\\alpha(a\\mid h)K_{a,o}u_h\\bigr)\n =\\alpha(a\\mid h)\\bar K_{a,o}\\pi_*u_h\n =\\bar u_{hao}.\n\\]\nHere multiplication uses only enabled columns. Values in all other columns are irrelevant because their input mass is zero. Zero-mass prefixes stay zero. Summation proves equality of history probabilities, and division by the common positive mass proves the posterior assertion.","horizon for which the policy is defined on all positive-mass prefixes.\n\\end{theorem}\n\\begin{proof}\nFirst prove $\\bar u_h=\\pi_*u_h$ by induction. It holds initially by the prior assumption. At a positive history, Lemma~\\ref{r:joint-support} gives the same available actions on both sides. For any action with positive policy mass, every current fine state of positive mass enables that action. Thus\n\\[\n \\pi_*u_{hao}\n =\\pi_*\\bigl(\\alpha(a\\mid h)K_{a,o}u_h\\bigr)\n =\\alpha(a\\mid h)\\bar K_{a,o}\\pi_*u_h\n =\\bar u_{hao}.\n\\]\nHere multiplication uses only enabled columns. Values in all other columns are irrelevant because their input mass is zero. Zero-mass prefixes stay zero. Summation proves equality of history probabilities, and division by the common positive mass proves the posterior assertion.","This induction alone records only the current state and the observed history. To retain the entire state path, fix a coarse prefix\n\\[\n (z_0,a_1,o_1,z_1,\\ldots,a_t,o_t,z_t).\n\\]\nSum fine path weights over all fine prefixes mapping to this prefix. For one more step, each fine current state in the last fibre contributes the same projected successor/report row by Equation~\\eqref{eq:fibre-joint}. The policy factor depends only on the already fixed action/report prefix. It can therefore be taken outside the fine-state sum. Induction gives the exact path mass\n\\[\n \\bar p_0(z_0)\\prod_{i=1}^{t}\n \\alpha(a_i\\mid a_1,o_1,\\ldots,a_{i-1},o_{i-1})\n \\bar K_{a_i,o_i}(z_i\\mid z_{i-1}).\n\\]\nAn inadmissible or zero-mass prefix contributes zero. This is precisely the coarse path law. Applying $\\bar\\ell$ to each $z_i$ gives the full label history, including its correlations across time","_t,o_t,z_t).\n\\]\nSum fine path weights over all fine prefixes mapping to this prefix. For one more step, each fine current state in the last fibre contributes the same projected successor/report row by Equation~\\eqref{eq:fibre-joint}. The policy factor depends only on the already fixed action/report prefix. It can therefore be taken outside the fine-state sum. Induction gives the exact path mass\n\\[\n \\bar p_0(z_0)\\prod_{i=1}^{t}\n \\alpha(a_i\\mid a_1,o_1,\\ldots,a_{i-1},o_{i-1})\n \\bar K_{a_i,o_i}(z_i\\mid z_{i-1}).\n\\]\nAn inadmissible or zero-mass prefix contributes zero. This is precisely the coarse path law. Applying $\\bar\\ell$ to each $z_i$ gives the full label history, including its correlations across time. Labels need not be disclosed to the policy for this pushforward comparison to be valid.\n\\end{proof}","ibre-joint}. The policy factor depends only on the already fixed action/report prefix. It can therefore be taken outside the fine-state sum. Induction gives the exact path mass\n\\[\n \\bar p_0(z_0)\\prod_{i=1}^{t}\n \\alpha(a_i\\mid a_1,o_1,\\ldots,a_{i-1},o_{i-1})\n \\bar K_{a_i,o_i}(z_i\\mid z_{i-1}).\n\\]\nAn inadmissible or zero-mass prefix contributes zero. This is precisely the coarse path law. Applying $\\bar\\ell$ to each $z_i$ gives the full label history, including its correlations across time. Labels need not be disclosed to the policy for this pushforward comparison to be valid.\n\\end{proof}","licy factor depends only on the already fixed action/report prefix. It can therefore be taken outside the fine-state sum. Induction gives the exact path mass\n\\[\n \\bar p_0(z_0)\\prod_{i=1}^{t}\n \\alpha(a_i\\mid a_1,o_1,\\ldots,a_{i-1},o_{i-1})\n \\bar K_{a_i,o_i}(z_i\\mid z_{i-1}).\n\\]\nAn inadmissible or zero-mass prefix contributes zero. This is precisely the coarse path law. Applying $\\bar\\ell$ to each $z_i$ gives the full label history, including its correlations across time. Labels need not be disclosed to the policy for this pushforward comparison to be valid.\n\\end{proof}","\\section{Why preserving two marginals is insufficient}\nConsider two fine states in one proposed coarse input fibre. From the first, the pairs $(z',o)$ are $(0,0)$ and $(1,1)$ with probability $1/2$ each. From the second, they are $(0,1)$ and $(1,0)$ with probability $1/2$ each. The successor marginal and the report marginal are both uniform in either case. Their association is different, however, so the same report leads to opposite posterior states. Equation~\\eqref{eq:fibre-joint} fails.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.87\\linewidth]{figures/joint_instrument_obstruction.pdf}\\caption{Equal marginals do not determine the joint instrument. Each table is an exact probability law. The report needed for correction is correlated with opposite successor states.}\\label{fig:joint-obstruction}\\end{figure}","\\section{An exact sensor realization}\nReturn to $x=(b,s,n)\\in\\{0,1\\}^3$ and retain $z=(b,s)$. The actions \\emph{hold} and \\emph{flip} are available everywhere. A \\emph{repair} action is available only where $b=0$ and leaves that target value fixed. The nuisance bit is retained by every fine transition but ignored by the coarse state.","After an action, report the new display $b\\oplus s$ correctly with probability $3/4$ and incorrectly with probability $1/4$. Both the label and the joint instrument ignore $n$. The ten coarse state/action rows satisfy the quotient conditions exactly.","From a uniform prior, a report leaves both target values possible, so repair is not uniformly admitted. A policy that reacts to those reports must continue with hold or flip. From a prior supported on $b=0$, repair is available until an action changes that support. The distinction exercises the support condition rather than assuming all actions are always available.","\\section{The controlled predecessor}\nLet $X,Z$ be finite nonempty sets and $\\pi:X\\to Z$ a surjection. Use a common finite action alphabet with separate running and ending catalogues $A^X_m(x)$ and $A^Z_m(z)$, where $m\\in\\{\\mathrm{run},\\mathrm{end}\\}$. Every admitted action has a nonempty successor set. The mode can be included in the action name, so the same printed action in two modes need not denote the same transition.","Assume the fine and coarse safe and goal sets agree by inverse image:\n\\[\n S_X=\\pi^{-1}(S_Z),\\qquad G_X=\\pi^{-1}(G_Z),\\qquad G_Z\\subseteq S_Z.\n\\]\nThis is a substantive restriction. A label that distinguishes a safe state from an unsafe state cannot be omitted from a safety-preserving quotient.","For $B\\subseteq X$, define the mode-specific predecessor\n\\[\n \\operatorname{CPre}^X_m(B)\n =\\{x:\\exists a\\in A^X_m(x),\\ \\operatorname{Post}_X(x,a)\\subseteq B\\}.\n\\]\nDefine $\\operatorname{CPre}^Z_m$ in the same way. These operators do not themselves include a safety test. Safety enters the ending recurrence through its intersection with $S$, and the running recurrence through its current set.","\\begin{lemma}[Predecessors commute with an exact controlled quotient]\\label{r:control-predecessor}\nSuppose, for each mode and fine state,\n\\begin{align*}\n A^X_m(x)&=A^Z_m(\\pi x),\\\\\n \\pi(\\operatorname{Post}_X(x,a))&=\\operatorname{Post}_Z(\\pi x,a)\n \\qquad(a\\in A^X_m(x)).\n\\end{align*}\nThen for every $B\\subseteq Z$,\n\\[\n \\operatorname{CPre}^X_m(\\pi^{-1}B)\n =\\pi^{-1}(\\operatorname{CPre}^Z_m(B)).\n\\]\n\\end{lemma}\n\\begin{proof}\nFor an admitted action,\n\\[\n \\operatorname{Post}_X(x,a)\\subseteq\\pi^{-1}B\n \\quad\\Longleftrightarrow\\quad\n \\pi(\\operatorname{Post}_X(x,a))\\subseteq B.\n\\]\nReplace the projected successor set by its assumed coarse value. The action ranges in the two existential quantifiers are identical, so existence of a witnessing action is equivalent on both sides.\n\\end{proof}","\\pi(\\operatorname{Post}_X(x,a))&=\\operatorname{Post}_Z(\\pi x,a)\n \\qquad(a\\in A^X_m(x)).\n\\end{align*}\nThen for every $B\\subseteq Z$,\n\\[\n \\operatorname{CPre}^X_m(\\pi^{-1}B)\n =\\pi^{-1}(\\operatorname{CPre}^Z_m(B)).\n\\]\n\\end{lemma}\n\\begin{proof}\nFor an admitted action,\n\\[\n \\operatorname{Post}_X(x,a)\\subseteq\\pi^{-1}B\n \\quad\\Longleftrightarrow\\quad\n \\pi(\\operatorname{Post}_X(x,a))\\subseteq B.\n\\]\nReplace the projected successor set by its assumed coarse value. The action ranges in the two existential quantifiers are identical, so existence of a witnessing action is equivalent on both sides.\n\\end{proof}","\\section{Transport of both fixed-point sequences}\nConstruct $T^X_k,L^X_k$ on the fine game and $T^Z_k,L^Z_k$ on the coarse game as in Chapter~\\ref{ch:endinganalysis}. In particular, the running sequence starts from the stable ending region in its own model.","\\begin{theorem}[Ending regions, running kernels, and ranks are preserved]\\label{r:control-transport}\nUnder the exact controlled-quotient assumptions,\n\\[\n T^X_k=\\pi^{-1}(T^Z_k),\\qquad\n T^X_*=\\pi^{-1}(T^Z_*),\\qquad\n L^X_k=\\pi^{-1}(L^Z_k),\\qquad\n L^X_*=\\pi^{-1}(L^Z_*)\n\\]\nfor every finite $k$. For every $x\\in T^X_*$, the first-entry ranks agree:\n\\[\n \\rho_X(x)=\\rho_Z(\\pi x).\n\\]\nA positional coarse policy for safe continuation and ending lifts to a fine policy with the same rank bound. To execute that lift, the controller needs the current coarse state and command mode, but not the remaining fine-state coordinates.\n\\end{theorem}\n\\begin{proof}\nThe goal-set assumption gives equality at $T_0$. Inverse images commute with union and intersection. Applying Lemma~\\ref{r:control-predecessor} to the ending predecessor gives the induction step\n\\begin{align*}\n T^X_{k+1}\n &=T^X_k\\cup\\bigl(S_X\\cap\\operatorname{CPre}^X_{\\rm end}(T^X_k)\\bigr)","$. For every $x\\in T^X_*$, the first-entry ranks agree:\n\\[\n \\rho_X(x)=\\rho_Z(\\pi x).\n\\]\nA positional coarse policy for safe continuation and ending lifts to a fine policy with the same rank bound. To execute that lift, the controller needs the current coarse state and command mode, but not the remaining fine-state coordinates.\n\\end{theorem}\n\\begin{proof}\nThe goal-set assumption gives equality at $T_0$. Inverse images commute with union and intersection. Applying Lemma~\\ref{r:control-predecessor} to the ending predecessor gives the induction step\n\\begin{align*}\n T^X_{k+1}\n &=T^X_k\\cup\\bigl(S_X\\cap\\operatorname{CPre}^X_{\\rm end}(T^X_k)\\bigr)","licy with the same rank bound. To execute that lift, the controller needs the current coarse state and command mode, but not the remaining fine-state coordinates.\n\\end{theorem}\n\\begin{proof}\nThe goal-set assumption gives equality at $T_0$. Inverse images commute with union and intersection. Applying Lemma~\\ref{r:control-predecessor} to the ending predecessor gives the induction step\n\\begin{align*}\n T^X_{k+1}\n &=T^X_k\\cup\\bigl(S_X\\cap\\operatorname{CPre}^X_{\\rm end}(T^X_k)\\bigr)\\\\\n &=\\pi^{-1}\\!\\left(T^Z_k\\cup\n       (S_Z\\cap\\operatorname{CPre}^Z_{\\rm end}(T^Z_k))\\right).\n\\end{align*}\nThe finite increasing sequences stabilize, so their stable sets also agree by inverse image. The first stage at which $x$ appears is exactly the first stage at which $\\pi x$ appears, proving rank equality.","&=\\pi^{-1}\\!\\left(T^Z_k\\cup\n       (S_Z\\cap\\operatorname{CPre}^Z_{\\rm end}(T^Z_k))\\right).\n\\end{align*}\nThe finite increasing sequences stabilize, so their stable sets also agree by inverse image. The first stage at which $x$ appears is exactly the first stage at which $\\pi x$ appears, proving rank equality.","At a coarse state in $L^Z_*$, choose a running action whose successors remain in that set. It is admitted at every fine representative, and all its fine successors lie in $\\pi^{-1}(L^Z_*)=L^X_*$. After the command, choose a coarse ending action that decreases rank. Its projected successors have smaller coarse rank, so every fine successor has smaller fine rank. These choices define the lifted policy.\n\\end{proof}","This does not give the controller an observation it did not have. A report-only policy from Chapter~\\ref{ch:jointcontract} need not know its current coarse state. Applying the policy-lifting conclusion requires that $\\pi x$ be observed or reconstructed from an available record. Equality of the abstract winning regions alone does not solve a partially observed control problem.","\\section{From stochastic rows to successor supports}\nSuppose a mode's game successors are exactly the support of the stochastic instrument:\n\\[\n \\operatorname{Post}_X(x,a)\n =\\{x':\\exists o,\\ K_{a,o}(x'\\mid x)>0\\}.\n\\]\nDefine the coarse successor set from $\\bar K$ in the same way.","\\begin{corollary}[An exact stochastic quotient preserves the support game]\\label{r:stochastic-support-game}\nThe joint quotient of Theorem~\\ref{r:joint-quotient} gives\n\\[\n \\pi(\\operatorname{Post}_X(x,a))=\\operatorname{Post}_Z(\\pi x,a)\n\\]\nfor every enabled action. If the mode catalogues and the safe and goal sets satisfy the preceding compatibility conditions, Theorem~\\ref{r:control-transport} applies to the resulting support games.\n\\end{corollary}\n\\begin{proof}\nFor each $o,z'$, the coarse entry is a finite sum of nonnegative fine entries. It is positive exactly when at least one fine successor in that fibre has a positive entry for that report. Taking the union over reports gives the support identity. Normalization makes the support of every enabled action nonempty. The remaining hypotheses are precisely those of Theorem~\\ref{r:control-transport}.\n\\end{proof}","label{r:stochastic-support-game}\nThe joint quotient of Theorem~\\ref{r:joint-quotient} gives\n\\[\n \\pi(\\operatorname{Post}_X(x,a))=\\operatorname{Post}_Z(\\pi x,a)\n\\]\nfor every enabled action. If the mode catalogues and the safe and goal sets satisfy the preceding compatibility conditions, Theorem~\\ref{r:control-transport} applies to the resulting support games.\n\\end{corollary}\n\\begin{proof}\nFor each $o,z'$, the coarse entry is a finite sum of nonnegative fine entries. It is positive exactly when at least one fine successor in that fibre has a positive entry for that report. Taking the union over reports gives the support identity. Normalization makes the support of every enabled action nonempty. The remaining hypotheses are precisely those of Theorem~\\ref{r:control-transport}.\n\\end{proof}","\\section{Conservative enclosures give one-sided certificates}\nExact equality is not needed to certify a policy conservatively. Suppose instead that a coarse envelope admits only actions available at every fine representative and includes all projected actual successors:\n\\[\n A^Z_m(\\pi x)\\subseteq A^X_m(x),\\qquad\n \\pi(\\operatorname{Post}_X(x,a))\n \\subseteq\\operatorname{Post}_Z(\\pi x,a)\n \\quad(a\\in A^Z_m(\\pi x)).\n\\]\nThe envelope may contain additional adverse successors and may omit some fine actions. Retain the exact inverse-image assumptions for safety and goals.","\\begin{proposition}[A winning envelope policy remains winning]\\label{r:control-envelope}\nUnder these one-sided assumptions,\n\\[\n \\pi^{-1}(T^Z_*)\\subseteq T^X_*,\\qquad\n \\pi^{-1}(L^Z_*)\\subseteq L^X_*.\n\\]\nA coarse ending rank bounds the ending time of the lifted policy. The same coarse policy works for every fine model satisfying the enclosure assumptions. The reverse inclusions need not hold.\n\\end{proposition}\n\\begin{proof}\nInduction on the coarse ending stages proves\n$\\pi^{-1}(T^Z_k)\\subseteq T^X_k$. A witnessing coarse action is available at each fine representative. Its actual successors project into the envelope's successors and hence into the previous coarse stage. The fine action therefore witnesses the next fine stage. The same argument proves rank decrease along every lifted ending path.","ontrol-envelope}\nUnder these one-sided assumptions,\n\\[\n \\pi^{-1}(T^Z_*)\\subseteq T^X_*,\\qquad\n \\pi^{-1}(L^Z_*)\\subseteq L^X_*.\n\\]\nA coarse ending rank bounds the ending time of the lifted policy. The same coarse policy works for every fine model satisfying the enclosure assumptions. The reverse inclusions need not hold.\n\\end{proposition}\n\\begin{proof}\nInduction on the coarse ending stages proves\n$\\pi^{-1}(T^Z_k)\\subseteq T^X_k$. A witnessing coarse action is available at each fine representative. Its actual successors project into the envelope's successors and hence into the previous coarse stage. The fine action therefore witnesses the next fine stage. The same argument proves rank decrease along every lifted ending path.","The inverse image of $L^Z_*$ lies inside $T^X_*$ by the first inclusion. Choose the coarse invariant running action at each coarse state. Every fine successor lies in the inverse image of $L^Z_*$. Thus that inverse image is a controlled invariant subset of the fine ending region. The greatest-invariant characterization places it inside $L^X_*$. The argument uses only the enclosure relations and therefore applies to every model satisfying them.\n\\end{proof}","\\section{Why a small probability error is not a sure guarantee}\nTake three states $s,g,b$, with safe set $\\{s,g\\}$ and goal $\\{g\\}$. Running holds $s$ or $g$ in place. The nominal ending action at $s$ moves to $g$ with probability one. The perturbed action instead moves to $g$ with probability $1-\\epsilon$ and to unsafe state $b$ with probability $\\epsilon$, where $0<\\epsilon<1$. A hold action is available at $g$.","small probability error is not a sure guarantee}\nTake three states $s,g,b$, with safe set $\\{s,g\\}$ and goal $\\{g\\}$. Running holds $s$ or $g$ in place. The nominal ending action at $s$ moves to $g$ with probability one. The perturbed action instead moves to $g$ with probability $1-\\epsilon$ and to unsafe state $b$ with probability $\\epsilon$, where $0<\\epsilon<1$. A hold action is available at $g$.","The row total variation distance is exactly $\\epsilon$. In the nominal game, $s$ belongs to both the ending region and the running kernel. In the perturbed support game, the only ending action from $s$ has an unsafe successor, so $s$ belongs to neither. This happens for every positive $\\epsilon$, however small.","\\chapter{From evidence to a reusable state}\\label{ch:evidence}\n\\question{The first question is not how much state to store. It is which distinctions are accessible.}\nLet $\\Omega$ be a finite nonempty set of possible worlds. A world may include a physical state, a fixed model index, and an origin tag. An observation record is a function $r:\\Omega\\to H$. A target is a function $t:\\Omega\\to Y$. All codomains used below are nonempty. We work on attained records $r(\\Omega)$, so no arbitrary interpretation of an impossible record is needed.","\\question{The first question is not how much state to store. It is which distinctions are accessible.}\nLet $\\Omega$ be a finite nonempty set of possible worlds. A world may include a physical state, a fixed model index, and an origin tag. An observation record is a function $r:\\Omega\\to H$. A target is a function $t:\\Omega\\to Y$. All codomains used below are nonempty. We work on attained records $r(\\Omega)$, so no arbitrary interpretation of an impossible record is needed.","\\begin{proposition}[A record determines a target exactly on its fibres]\\label{r:factor}\nThere exists $d:r(\\Omega)\\to Y$ with $d(r(\\omega))=t(\\omega)$ for every world if and only if\n\\[\nr(\\omega)=r(\\omega')\\ \\Longrightarrow\\ t(\\omega)=t(\\omega').\n\\]\n\\end{proposition}\n\\begin{proof}\nNecessity follows because a function assigns one value to one input. For sufficiency, define $d(h)$ to be $t(\\omega)$ for any world with $r(\\omega)=h$. Such a world exists for attained $h$, and the hypothesis makes the choice immaterial. The resulting decoder satisfies the required equation.\n\\end{proof}\n\\source{Grounded Abstraction [GA], Proposition 3.2 and Theorem 3.4. Machine Understanding [MU], the representation and sender-access criteria. This is the shared elementary factorization argument, stated once.}","}[A record determines a target exactly on its fibres]\\label{r:factor}\nThere exists $d:r(\\Omega)\\to Y$ with $d(r(\\omega))=t(\\omega)$ for every world if and only if\n\\[\nr(\\omega)=r(\\omega')\\ \\Longrightarrow\\ t(\\omega)=t(\\omega').\n\\]\n\\end{proposition}\n\\begin{proof}\nNecessity follows because a function assigns one value to one input. For sufficiency, define $d(h)$ to be $t(\\omega)$ for any world with $r(\\omega)=h$. Such a world exists for attained $h$, and the hypothesis makes the choice immaterial. The resulting decoder satisfies the required equation.\n\\end{proof}\n\\source{Grounded Abstraction [GA], Proposition 3.2 and Theorem 3.4. Machine Understanding [MU], the representation and sender-access criteria. This is the shared elementary factorization argument, stated once.}","bres]\\label{r:factor}\nThere exists $d:r(\\Omega)\\to Y$ with $d(r(\\omega))=t(\\omega)$ for every world if and only if\n\\[\nr(\\omega)=r(\\omega')\\ \\Longrightarrow\\ t(\\omega)=t(\\omega').\n\\]\n\\end{proposition}\n\\begin{proof}\nNecessity follows because a function assigns one value to one input. For sufficiency, define $d(h)$ to be $t(\\omega)$ for any world with $r(\\omega)=h$. Such a world exists for attained $h$, and the hypothesis makes the choice immaterial. The resulting decoder satisfies the required equation.\n\\end{proof}\n\\source{Grounded Abstraction [GA], Proposition 3.2 and Theorem 3.4. Machine Understanding [MU], the representation and sender-access criteria. This is the shared elementary factorization argument, stated once.}","The proposition is an information statement, not a runtime bound. The table defining $d$ could be large. Nor does the mere existence of $d$ teach a learner how to find it. A proof that a suitable decoder exists and a justified account of its acquisition remain separate.","\\section{The sensor used throughout the book}\nWrite a world as\n\\[\n\\omega=(b,s,n)\\in\\{0,1\\}^3.\n\\]\nThe physical target bit is $b$. The sensor convention is $s$. The nuisance bit $n$ never affects the operations or outputs of the initial contract. The display is\n\\[\ny=b\\oplus s,\n\\]\nwhere $\\oplus$ is addition modulo two. A permitted control $f$ flips the physical bit and leaves $s,n$ unchanged. Consequently the next display is $1-y$.","roughout the book}\nWrite a world as\n\\[\n\\omega=(b,s,n)\\in\\{0,1\\}^3.\n\\]\nThe physical target bit is $b$. The sensor convention is $s$. The nuisance bit $n$ never affects the operations or outputs of the initial contract. The display is\n\\[\ny=b\\oplus s,\n\\]\nwhere $\\oplus$ is addition modulo two. A permitted control $f$ flips the physical bit and leaves $s,n$ unchanged. Consequently the next display is $1-y$.","An agent given only $y$ can predict every future display under repeated flips: after $k$ flips it is $y\\oplus(k\\bmod2)$. It cannot determine the current physical bit. Worlds $(0,0,0)$ and $(1,1,0)$ share display zero and have different targets. The obstruction is the fibre of the available record, not a shortage of computational effort.","Calibration is a separate, nondestructive observation $c(\\omega)=s$. When calibration is permitted and trustworthy, the joint record $(y,s)$ determines $b=y\\oplus s$. Adding a blank field called ``sensor convention'' does not provide this information. The observation operation, its interpretation, and its cost must be part of the contract.","midrule\n$y$ & 2 & Yes & No","$(y,s)$ & 4 & Yes & Yes","$(b,s,n)$ & 8 & Yes & Yes, with unused detail","\\section{One answer versus every permitted continuation}\nLet $X$ be a finite nonempty state space, and fix an available record $r:X\\to H$. Let $\\mathcal A$ be a finite alphabet of total deterministic operations $F_a:X\\to X$, and let $q:X\\to Y$ be the protected output. A word $w=a_1\\cdots a_k$ executes $a_1$ first. Thus $F_w=F_{a_k}\\cdots F_{a_1}$, and the empty word denotes the identity.","Define behavioral equivalence by\n\\[\nx\\sim x'\\quad\\Longleftrightarrow\\quad q(F_wx)=q(F_wx')\\text{ for every finite word }w.\n\\]\nThe empty word protects the present output. The other words protect its permitted future uses.","\\begin{theorem}[Evidence-initialized behavioral state]\\label{r:behavior}\nA finite predictor initialized only from $r(x)$, updated only by the operation names, and exact for all protected continuation outputs exists if and only if\n\\[\nr(x)=r(x')\\ \\Longrightarrow\\ x\\sim x'.\n\\]\nWhen it exists, $X/{\\sim}$ supplies a predictor with the smallest possible number of initialized states. Its updates are $\\bar F_a([x])=[F_ax]$ and its readout is $\\bar q([x])=q(x)$.\n\\end{theorem}\n\\begin{proof}\nIf two initial records are equal, a deterministic predictor has the same initialized state. Applying the same subsequent operations produces the same predicted outputs. Exactness therefore requires behavioral equivalence.","havioral state]\\label{r:behavior}\nA finite predictor initialized only from $r(x)$, updated only by the operation names, and exact for all protected continuation outputs exists if and only if\n\\[\nr(x)=r(x')\\ \\Longrightarrow\\ x\\sim x'.\n\\]\nWhen it exists, $X/{\\sim}$ supplies a predictor with the smallest possible number of initialized states. Its updates are $\\bar F_a([x])=[F_ax]$ and its readout is $\\bar q([x])=q(x)$.\n\\end{theorem}\n\\begin{proof}\nIf two initial records are equal, a deterministic predictor has the same initialized state. Applying the same subsequent operations produces the same predicted outputs. Exactness therefore requires behavioral equivalence.","For the converse, equivalent states have equivalent successors: an output after $F_a$ followed by $w$ is one of the outputs in the defining family. Hence $\\bar F_a$ is well defined. The empty word makes $\\bar q$ well defined. The fibre premise gives a well-defined initializer $d(r(x))=[x]$. Induction on word length proves exactness. Finally, behaviorally distinct states cannot initialize the same predictor state, since some continuation separates their outputs. Every exact predictor therefore needs at least one initialized value for each class, and the quotient attains that number.\n\\end{proof}\n\\source{[GA], Theorem 3.4, and [MU], Theorem 4.1. The two source versions are consolidated, not counted as independent discoveries.}","In the sensor, the display task has two behavioral states. Changing the protected output to $(y,b)$ yields four behavioral states. The old record still has only two values and cannot initialize the enlarged quotient. The new task has exposed a real information requirement.","\\section{Why a statistical law is not an observed record}\nA parameter can determine a distinct probability law while every finite sample remains possible under several parameters. For instance, Bernoulli parameters $1/3$ and $2/3$ assign positive probability to every finite binary string. No estimator based on one such string can be certainly correct for both parameters. The laws differ, but their finite supports overlap.","\\section{Exact packets and the hidden-return defect}\nLet $V,Z$ be vector spaces over a field. Let $A:V\\to Z$ be analysis and $S:Z\\to V$ synthesis, with $SA=I_V$. Then $AS$ is an idempotent on $Z$. The map $\\mathcal E(T)=ATS$ represents a full-state operator $T:V\\to V$ on resolved coordinates. Direct multiplication gives\n\\[\n\\mathcal E(T)\\mathcal E(U)=ATSAUS=ATUS=\\mathcal E(TU).\n\\]\nThe resolved identity is $AS$ on the compatible image $A(V)$, not necessarily $I_Z$ on arbitrary coordinates.","Exact packets and the hidden-return defect}\nLet $V,Z$ be vector spaces over a field. Let $A:V\\to Z$ be analysis and $S:Z\\to V$ synthesis, with $SA=I_V$. Then $AS$ is an idempotent on $Z$. The map $\\mathcal E(T)=ATS$ represents a full-state operator $T:V\\to V$ on resolved coordinates. Direct multiplication gives\n\\[\n\\mathcal E(T)\\mathcal E(U)=ATSAUS=ATUS=\\mathcal E(TU).\n\\]\nThe resolved identity is $AS$ on the compatible image $A(V)$, not necessarily $I_Z$ on arbitrary coordinates.","return defect}\nLet $V,Z$ be vector spaces over a field. Let $A:V\\to Z$ be analysis and $S:Z\\to V$ synthesis, with $SA=I_V$. Then $AS$ is an idempotent on $Z$. The map $\\mathcal E(T)=ATS$ represents a full-state operator $T:V\\to V$ on resolved coordinates. Direct multiplication gives\n\\[\n\\mathcal E(T)\\mathcal E(U)=ATSAUS=ATUS=\\mathcal E(TU).\n\\]\nThe resolved identity is $AS$ on the compatible image $A(V)$, not necessarily $I_Z$ on arbitrary coordinates.","be vector spaces over a field. Let $A:V\\to Z$ be analysis and $S:Z\\to V$ synthesis, with $SA=I_V$. Then $AS$ is an idempotent on $Z$. The map $\\mathcal E(T)=ATS$ represents a full-state operator $T:V\\to V$ on resolved coordinates. Direct multiplication gives\n\\[\n\\mathcal E(T)\\mathcal E(U)=ATSAUS=ATUS=\\mathcal E(TU).\n\\]\nThe resolved identity is $AS$ on the compatible image $A(V)$, not necessarily $I_Z$ on arbitrary coordinates.","For a compression, reverse the splitting. Let $C:V\\to Z$ and $R:Z\\to V$ satisfy $CR=I_Z$, and put $P=RC$. The compressed operator is $\\Phi(T)=CTR$.","\\begin{proposition}[Exact accounting of the discarded excursion]\\label{r:packet}\nFor any endomorphisms $T,U$ of $V$,\n\\[\n\\Phi(TU)-\\Phi(T)\\Phi(U)=CT(I-P)UR.\n\\]\nThus compression preserves this composition precisely when the discarded excursion on the right vanishes. Full split encoding $T\\mapsto ATS$ preserves composition without that extra condition.\n\\end{proposition}\n\\begin{proof}\nExpand the product $\\Phi(T)\\Phi(U)=CTRCUR=CTPUR$ and subtract it from $CTUR$. The identity for full split encoding follows from $SA=I_V$ as computed above. These computations make no approximation and require no orthogonality.\n\\end{proof}\n\\source{Sufficient Resolution [SR], Theorems 2.1 and 2.3. The present proof reviews only these algebraic identities, not the complete forty-seven-page source or its full formal implemen","\\begin{proposition}[Exact accounting of the discarded excursion]\\label{r:packet}\nFor any endomorphisms $T,U$ of $V$,\n\\[\n\\Phi(TU)-\\Phi(T)\\Phi(U)=CT(I-P)UR.\n\\]\nThus compression preserves this composition precisely when the discarded excursion on the right vanishes. Full split encoding $T\\mapsto ATS$ preserves composition without that extra condition.\n\\end{proposition}\n\\begin{proof}\nExpand the product $\\Phi(T)\\Phi(U)=CTRCUR=CTPUR$ and subtract it from $CTUR$. The identity for full split encoding follows from $SA=I_V$ as computed above. These computations make no approximation and require no orthogonality.\n\\end{proof}\n\\source{Sufficient Resolution [SR], Theorems 2.1 and 2.3. The present proof reviews only these algebraic identities, not the complete forty-seven-page source or its full formal implementation.","\\begin{proposition}[Exact accounting of the discarded excursion]\\label{r:packet}\nFor any endomorphisms $T,U$ of $V$,\n\\[\n\\Phi(TU)-\\Phi(T)\\Phi(U)=CT(I-P)UR.\n\\]\nThus compression preserves this composition precisely when the discarded excursion on the right vanishes. Full split encoding $T\\mapsto ATS$ preserves composition without that extra condition.\n\\end{proposition}\n\\begin{proof}\nExpand the product $\\Phi(T)\\Phi(U)=CTRCUR=CTPUR$ and subtract it from $CTUR$. The identity for full split encoding follows from $SA=I_V$ as computed above. These computations make no approximation and require no orthogonality.\n\\end{proof}\n\\source{Sufficient Resolution [SR], Theorems 2.1 and 2.3. The present proof reviews only these algebraic identities, not the complete forty-seven-page source or its full formal implementation.}","gin{proposition}[Exact accounting of the discarded excursion]\\label{r:packet}\nFor any endomorphisms $T,U$ of $V$,\n\\[\n\\Phi(TU)-\\Phi(T)\\Phi(U)=CT(I-P)UR.\n\\]\nThus compression preserves this composition precisely when the discarded excursion on the right vanishes. Full split encoding $T\\mapsto ATS$ preserves composition without that extra condition.\n\\end{proposition}\n\\begin{proof}\nExpand the product $\\Phi(T)\\Phi(U)=CTRCUR=CTPUR$ and subtract it from $CTUR$. The identity for full split encoding follows from $SA=I_V$ as computed above. These computations make no approximation and require no orthogonality.\n\\end{proof}\n\\source{Sufficient Resolution [SR], Theorems 2.1 and 2.3. The present proof reviews only these algebraic identities, not the complete forty-seven-page source or its full formal implementation.}","excursion]\\label{r:packet}\nFor any endomorphisms $T,U$ of $V$,\n\\[\n\\Phi(TU)-\\Phi(T)\\Phi(U)=CT(I-P)UR.\n\\]\nThus compression preserves this composition precisely when the discarded excursion on the right vanishes. Full split encoding $T\\mapsto ATS$ preserves composition without that extra condition.\n\\end{proposition}\n\\begin{proof}\nExpand the product $\\Phi(T)\\Phi(U)=CTRCUR=CTPUR$ and subtract it from $CTUR$. The identity for full split encoding follows from $SA=I_V$ as computed above. These computations make no approximation and require no orthogonality.\n\\end{proof}\n\\source{Sufficient Resolution [SR], Theorems 2.1 and 2.3. The present proof reviews only these algebraic identities, not the complete forty-seven-page source or its full formal implementation.}","$V$,\n\\[\n\\Phi(TU)-\\Phi(T)\\Phi(U)=CT(I-P)UR.\n\\]\nThus compression preserves this composition precisely when the discarded excursion on the right vanishes. Full split encoding $T\\mapsto ATS$ preserves composition without that extra condition.\n\\end{proposition}\n\\begin{proof}\nExpand the product $\\Phi(T)\\Phi(U)=CTRCUR=CTPUR$ and subtract it from $CTUR$. The identity for full split encoding follows from $SA=I_V$ as computed above. These computations make no approximation and require no orthogonality.\n\\end{proof}\n\\source{Sufficient Resolution [SR], Theorems 2.1 and 2.3. The present proof reviews only these algebraic identities, not the complete forty-seven-page source or its full formal implementation.}","\\begin{counterexample}[A discarded coordinate returns]\\label{ce:return}\nTake $V=\\mathbb R^2$, let $C(x_1,x_2)=x_1$, let $R(z)=(z,0)$, and let $T(x_1,x_2)=(x_2,x_1)$. The compressed one-step map is zero. Yet $T^2=I_V$, so $\\Phi(T^2)=I_Z$, while $\\Phi(T)^2=0$. The discarded second coordinate was not irrelevant to the two-step question.\n\\end{counterexample}","\\section{A method can be transported without reconstructing everything}\nSuppose source states $z\\in Z$ have operations $F_a$, target states $u\\in U$ have operations $G_{\\tau(a)}$, and $T:Z\\to U$ is a translation. The operation-name map $\\tau$ is fixed. Let $q$ and $\\bar q$ be matched readouts.","\\begin{theorem}[Forward transport of a reusable method]\\label{r:transport}\nIf the source domain is invariant, and\n\\[\nTF_a=G_{\\tau(a)}T,\\qquad q=\\bar qT,\n\\]\nthen for every finite word $w$,\n\\[\nTF_w=G_{\\tau(w)}T,\\qquad qF_w=\\bar qG_{\\tau(w)}T.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe empty word gives $TI_Z=I_UT$. Assume the first equality for a prefix. Invariance allows the next local equality to be applied on its successor, yielding\n$TF_aF_w=G_{\\tau(a)}TF_w=G_{\\tau(a)}G_{\\tau(w)}T$.\nInduction proves the first claim. Composition with the matched readout proves the second.\n\\end{proof}\n\\source{[MU], Theorem 6.1. The same local-to-continuation argument appears in OSL and Neuralese. No claim about a learned latent translator is imported.}\n\\boundary{The theorem is forward. It does not imply an inverse, matching of new target operations, cheap computation","\\begin{theorem}[Forward transport of a reusable method]\\label{r:transport}\nIf the source domain is invariant, and\n\\[\nTF_a=G_{\\tau(a)}T,\\qquad q=\\bar qT,\n\\]\nthen for every finite word $w$,\n\\[\nTF_w=G_{\\tau(w)}T,\\qquad qF_w=\\bar qG_{\\tau(w)}T.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe empty word gives $TI_Z=I_UT$. Assume the first equality for a prefix. Invariance allows the next local equality to be applied on its successor, yielding\n$TF_aF_w=G_{\\tau(a)}TF_w=G_{\\tau(a)}G_{\\tau(w)}T$.\nInduction proves the first claim. Composition with the matched readout proves the second.\n\\end{proof}\n\\source{[MU], Theorem 6.1. The same local-to-continuation argument appears in OSL and Neuralese. No claim about a learned latent translator is imported.}\n\\boundary{The theorem is forward. It does not imply an inverse, matching of new target operations, cheap computation of $T$, or that the source solve","\\begin{theorem}[Forward transport of a reusable method]\\label{r:transport}\nIf the source domain is invariant, and\n\\[\nTF_a=G_{\\tau(a)}T,\\qquad q=\\bar qT,\n\\]\nthen for every finite word $w$,\n\\[\nTF_w=G_{\\tau(w)}T,\\qquad qF_w=\\bar qG_{\\tau(w)}T.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe empty word gives $TI_Z=I_UT$. Assume the first equality for a prefix. Invariance allows the next local equality to be applied on its successor, yielding\n$TF_aF_w=G_{\\tau(a)}TF_w=G_{\\tau(a)}G_{\\tau(w)}T$.\nInduction proves the first claim. Composition with the matched readout proves the second.\n\\end{proof}\n\\source{[MU], Theorem 6.1. The same local-to-continuation argument appears in OSL and Neuralese. No claim about a learned latent translator is imported.}\n\\boundary{The theorem is forward. It does not imply an inverse, matching of new target operations, cheap computation of $T$, or that the source solver causally uses the represented state.}","ort of a reusable method]\\label{r:transport}\nIf the source domain is invariant, and\n\\[\nTF_a=G_{\\tau(a)}T,\\qquad q=\\bar qT,\n\\]\nthen for every finite word $w$,\n\\[\nTF_w=G_{\\tau(w)}T,\\qquad qF_w=\\bar qG_{\\tau(w)}T.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe empty word gives $TI_Z=I_UT$. Assume the first equality for a prefix. Invariance allows the next local equality to be applied on its successor, yielding\n$TF_aF_w=G_{\\tau(a)}TF_w=G_{\\tau(a)}G_{\\tau(w)}T$.\nInduction proves the first claim. Composition with the matched readout proves the second.\n\\end{proof}\n\\source{[MU], Theorem 6.1. The same local-to-continuation argument appears in OSL and Neuralese. No claim about a learned latent translator is imported.}\n\\boundary{The theorem is forward. It does not imply an inverse, matching of new target operations, cheap computation of $T$, or that the source solver causally uses the represented state.}","\\tau(a)}G_{\\tau(w)}T$.\nInduction proves the first claim. Composition with the matched readout proves the second.\n\\end{proof}\n\\source{[MU], Theorem 6.1. The same local-to-continuation argument appears in OSL and Neuralese. No claim about a learned latent translator is imported.}\n\\boundary{The theorem is forward. It does not imply an inverse, matching of new target operations, cheap computation of $T$, or that the source solver causally uses the represented state.}","The sensor representation $(y,s)$ translates to $(b,s)$ through $T(y,s)=(y\\oplus s,s)$. Flipping the physical bit complements the first coordinate in either representation. The square commutes, and so does every repeated flip. The nuisance bit is never reconstructed because it has no role in this contract.","\\section{One quotient for several possible dynamics}\nA stronger situation arises when the operator itself is uncertain. Let $V$ be a finite-dimensional vector space over a fixed field, and let $Y$ be a vector space over the same field. Let $T_a^\\theta:V\\to V$ be a finite family of candidate linear operators indexed by an action $a$ and a model $\\theta$. Choose one model $\\theta_0$ as the reference and put $A_a=T_a^{\\theta_0}$. A model index $\\theta$ is fixed during the initial comparison. Let $Q:V\\to Y$ be the same linear readout for every candidate.","ne quotient for several possible dynamics}\nA stronger situation arises when the operator itself is uncertain. Let $V$ be a finite-dimensional vector space over a fixed field, and let $Y$ be a vector space over the same field. Let $T_a^\\theta:V\\to V$ be a finite family of candidate linear operators indexed by an action $a$ and a model $\\theta$. Choose one model $\\theta_0$ as the reference and put $A_a=T_a^{\\theta_0}$. A model index $\\theta$ is fixed during the initial comparison. Let $Q:V\\to Y$ be the same linear readout for every candidate.","amics}\nA stronger situation arises when the operator itself is uncertain. Let $V$ be a finite-dimensional vector space over a fixed field, and let $Y$ be a vector space over the same field. Let $T_a^\\theta:V\\to V$ be a finite family of candidate linear operators indexed by an action $a$ and a model $\\theta$. Choose one model $\\theta_0$ as the reference and put $A_a=T_a^{\\theta_0}$. A model index $\\theta$ is fixed during the initial comparison. Let $Q:V\\to Y$ be the same linear readout for every candidate.","itself is uncertain. Let $V$ be a finite-dimensional vector space over a fixed field, and let $Y$ be a vector space over the same field. Let $T_a^\\theta:V\\to V$ be a finite family of candidate linear operators indexed by an action $a$ and a model $\\theta$. Choose one model $\\theta_0$ as the reference and put $A_a=T_a^{\\theta_0}$. A model index $\\theta$ is fixed during the initial comparison. Let $Q:V\\to Y$ be the same linear readout for every candidate.","Define $D_a^\\theta=T_a^\\theta-A_a$ and the propagated discrepancy space\n\\[\nE=\\Span\\{A_wD_a^\\theta v: v\\in V,\\ a,\\theta,\\ w\\text{ finite}\\}.\n\\]\nIt can be constructed by starting with the sum of the discrepancy ranges and repeatedly adding their images under the reference actions. In finite dimension this increasing sequence stops. Immediate invisibility $QD_a^\\theta=0$ alone does not guarantee $QE=0$.","\\begin{theorem}[A common observable quotient]\\label{r:common}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed candidate $\\theta$ and every word $w$.\n\\item $E\\subseteq\\ker Q$.\n\\item There is a vector space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and th","\\begin{theorem}[A common observable quotient]\\label{r:common}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed candidate $\\theta$ and every word $w$.\n\\item $E\\subseteq\\ker Q$.\n\\item There is a vector space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanni","\\begin{theorem}[A common observable quotient]\\label{r:common}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed candidate $\\theta$ and every word $w$.\n\\item $E\\subseteq\\ker Q$.\n\\item There is a vector space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","[A common observable quotient]\\label{r:common}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed candidate $\\theta$ and every word $w$.\n\\item $E\\subseteq\\ker Q$.\n\\item There is a vector space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","uotient]\\label{r:common}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed candidate $\\theta$ and every word $w$.\n\\item $E\\subseteq\\ker Q$.\n\\item There is a vector space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","e equivalent:\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed candidate $\\theta$ and every word $w$.\n\\item $E\\subseteq\\ker Q$.\n\\item There is a vector space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","r space $Z$, a surjective linear map $r:V\\to Z$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","$, common linear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","inear updates $G_a:Z\\to Z$, and a linear readout $\\bar Q:Z\\to Y$ with $rT_a^\\theta=G_ar$ and $Q=\\bar Qr$ for every $a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","$a,\\theta$.\n\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","\\end{enumerate}\nUnder these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","these conditions, $V/E$ is a common quotient and also predicts the output when the candidate index changes between operations.\n\\end{theorem}\n\\begin{proof}\nAssume the first condition. For every $a,\\theta,w$, equality as operators on all of $V$ gives\n\\[\nQA_wD_a^\\theta=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality applies the premise to the word executing $a$ and then $w$. Spanning proves $E\\subseteq\\ker Q$.","Conversely, $E$ is invariant under every $A_a$ by construction, and contains the entire range of each difference $D_a^\\theta$. Hence $T_a^\\theta E\\subseteq E$, and every candidate induces the same map $G_a([v])=[A_av]$ on $V/E$. Since $QE=0$, the readout descends. This proves the third condition. Composing its equalities yields the first condition. The same induction permits a different $\\theta$ at each step because the quotient map and update remain shared.\n\\end{proof}\n\\source{[GA], Theorem 13.1, ``Exact common quotient for structured model uncertainty.'' The all-initial-vectors hypothesis is retained.}","For example, let $Q(x_1,x_2,x_3)=x_1$ and let every candidate act as\n\\[\nT^\\theta(x_1,x_2,x_3)=(x_1+x_2,x_2,\\theta x_3).\n\\]\nThen $E$ is contained in the third coordinate and remains invisible. The two-coordinate method $(x_1,x_2)\\mapsto(x_1+x_2,x_2)$ is exact, although the hidden dynamics are not identified.","The conclusion would be different if a later action moved the third coordinate into the first. That change enlarges $E$ beyond $\\ker Q$. It invalidates the use of this quotient, not the theorem that justified the former one.","Let $\\mathcal K$ be a convex cone in a real vector space, and let $\\lambda$ be linear and strictly positive on every nonzero member of $\\mathcal K$. Its normalized states satisfy $v\\in\\mathcal K$ and $\\lambda v=1$. A coordinate change can send $\\mathcal K$ to a cone containing negative coordinates without changing those states' probability meaning. Conversely, an arbitrary signed component of a valid state need not itself be valid.","\\begin{proposition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete deco","\\begin{proposition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda","\\begin{proposition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing","\\begin{proposition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component","\\begin{proposition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture id","\\begin{proposition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","ition}[When complete packets are genuine probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","probability components]\\label{r:positive-packets}\nLet $P_1,\\ldots,P_m$ be linear maps with $\\sum_iP_i=I$. If every $P_i$ preserves $\\mathcal K$, each normalized $v\\in\\mathcal K$ has the mixture decomposition\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","on\n\\[\n v=\\sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","sum_{i:w_i>0}w_i v_i,\\qquad\n w_i=\\lambda P_iv\\ge0,\\quad \\sum_iw_i=1,\\quad v_i=P_iv/w_i\\quad(w_i>0).\n\\]\nEvery zero-weight component is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","onent is zero. If $\\mathcal K=\\mathbb R^d_{\\ge0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","0}$ and the $P_i$ are also idempotent, they must be disjoint diagonal coordinate selections. Orthogonal completeness alone does not imply a probability mixture.\n\\end{proposition}\n\\begin{proof}\nCone preservation places $P_iv$ in $\\mathcal K$, so its mass is nonnegative. Strict positivity says that a member of this cone with zero mass is zero. Applying $\\lambda$ to the complete decomposition gives $\\sum_iw_i=\\lambda v=1$. Dividing every nonzero component by its mass proves the mixture identity.","On the nonnegative coordinate cone, preservation is equivalent to nonnegative matrix entries, as is seen by applying a matrix to every standard basis vector. Because the matrices sum to $I$, their nonnegative off-diagonal entries must all vanish. An idempotent diagonal entry satisfies $x^2=x$, hence is zero or one. The diagonal sums are one, so each coordinate is selected by exactly one map. Finally, completeness and orthogonality without positivity do not give the first premise, as the following example shows.\n\\end{proof}\n\\source{Large Signal Model Foundations [LSM], Proposition 30, physical page 40. Only this finite packet argument and the joint-closure result below are admitted from that source at this stage.}","\\begin{counterexample}[An exact detail component has negative probability mass]\\label{ce:signed-packet}\nFor $p=(3/4,1/4)^\\top$, the orthogonal average and detail projectors are\n\\[\n P_{\\rm av}=\\tfrac12\\begin{pmatrix}1&1","\\begin{counterexample}[An exact detail component has negative probability mass]\\label{ce:signed-packet}\nFor $p=(3/4,1/4)^\\top$, the orthogonal average and detail projectors are\n\\[\n P_{\\rm av}=\\tfrac12\\begin{pmatrix}1&1\\\\1&1\\end{pmatrix},\\qquad\n P_{\\rm det}=\\tfrac12\\begin{pmatrix}1&-1\\\\-1&1\\end{pmatrix}.\n\\]\nThey sum to the identity and produce $(1/2,1/2)^\\top$ and $(1/4,-1/4)^\\top$. The components reconstruct $p$ exactly, but the detail has zero total mass without being zero. Normalizing it as a belief is undefined. Clipping its negative entry and normalizing the resulting sum would instead produce $(3/5,2/5)^\\top$, a different belief.\n\\end{counterexample}","-1&1\\end{pmatrix}.\n\\]\nThey sum to the identity and produce $(1/2,1/2)^\\top$ and $(1/4,-1/4)^\\top$. The components reconstruct $p$ exactly, but the detail has zero total mass without being zero. Normalizing it as a belief is undefined. Clipping its negative entry and normalizing the resulting sum would instead produce $(3/5,2/5)^\\top$, a different belief.\n\\end{counterexample}","\\chapter{Memory that remains open to correction}\\label{ch:memory}\n\\question{A useful memory preserves the distinctions by which its own conclusions can be revised.}\nLet $t:\\Omega\\to Y$ be the current target. A candidate set $V\\subseteq\\Omega$ records worlds still possible. The empty set is allowed and means that the represented candidates contradict the admitted evidence. It is not a successful identification.","Let $\\mathcal E\\subseteq\\Pow(\\Omega)$ contain $\\Omega$ and be closed under intersection. In this chapter the world does not change when an evidence event is received. The update is $V\\mapsto V\\cap E$. Physical actions and their effects will return in Chapter~\\ref{ch:inquiry}.","\\section{A state sufficient for every later filter}\nDefine the correction signature\n\\[\n\\Sigma(V)=\\bigl(t(V\\cap E)\\bigr)_{E\\in\\mathcal E}.\n\\]\nIt records what each admitted future filter would leave of the possible target values.","\\begin{theorem}[Minimal exact memory for a correction contract]\\label{r:memory}\nEquality of $\\Sigma$ is a congruence for every update $V\\mapsto V\\cap E_0$. The signature supports the current target set and its exact update after every finite evidence sequence. Any deterministic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finit","\\begin{theorem}[Minimal exact memory for a correction contract]\\label{r:memory}\nEquality of $\\Sigma$ is a congruence for every update $V\\mapsto V\\cap E_0$. The signature supports the current target set and its exact update after every finite evidence sequence. Any deterministic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","ct memory for a correction contract]\\label{r:memory}\nEquality of $\\Sigma$ is a congruence for every update $V\\mapsto V\\cap E_0$. The signature supports the current target set and its exact update after every finite evidence sequence. Any deterministic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","every update $V\\mapsto V\\cap E_0$. The signature supports the current target set and its exact update after every finite evidence sequence. Any deterministic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","e signature supports the current target set and its exact update after every finite evidence sequence. Any deterministic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","ature supports the current target set and its exact update after every finite evidence sequence. Any deterministic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","ic retained state supporting the same readouts and updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","updates on all candidate subsets must distinguish every two distinct signatures. Thus the image of $\\Sigma$ has the least possible cardinality for this contract.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for each $E$,\n\\[\nt((V\\cap E_0)\\cap E)=t(V\\cap(E_0\\cap E))=t(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure supplies the coordinate $E_0\\cap E$, so the updated signatures agree. The $\\Omega$ coordinate is the present target set. Induction gives every finite sequence.","\\section{An exact count for partition evidence}\nSuppose $C_1,\\ldots,C_m$ partition $\\Omega$ into nonempty cells and the admitted events are \\emph{exactly} their unions. Let\n\\[\nD_t=\\{(i,y): C_i\\cap t^{-1}(y)\\ne\\varnothing\\},\\qquad r=|D_t|.\n\\]\nThe occupancy memory retains whether a candidate exists in each attainable cell--target pair.","\\begin{corollary}[The size of partition correction memory]\\label{r:occupancy}\nOn all subsets of $\\Omega$, correction signatures are exactly distinguished by these occupancy records. There are $2^r$ signatures, and $r$ fixed binary bits are necessary and sufficient to encode them.\n\\end{corollary}\n\\begin{proof}\nAn individual-cell filter reveals the target values occupied in that cell, so the signature determines occupancy. Conversely, occupancy determines the target values left by any union of cells. Every subset of $D_t$ occurs: select one world from each included nonempty cell--target intersection and none from excluded intersections. Those intersections are disjoint. Thus there are precisely $2^r$ possible records. The preceding theorem and binary counting give necessity, while the occupancy","\\begin{corollary}[The size of partition correction memory]\\label{r:occupancy}\nOn all subsets of $\\Omega$, correction signatures are exactly distinguished by these occupancy records. There are $2^r$ signatures, and $r$ fixed binary bits are necessary and sufficient to encode them.\n\\end{corollary}\n\\begin{proof}\nAn individual-cell filter reveals the target values occupied in that cell, so the signature determines occupancy. Conversely, occupancy determines the target values left by any union of cells. Every subset of $D_t$ occurs: select one world from each included nonempty cell--target intersection and none from excluded intersections. Those intersections are disjoint. Thus there are precisely $2^r$ possible records. The preceding theorem and binary counting give necessity, while the occupancy vector supplies sufficiency.\n\\end{proof}\n\\source{[GA], Corollary 5.3. The word ``exactly'' i","\\begin{corollary}[The size of partition correction memory]\\label{r:occupancy}\nOn all subsets of $\\Omega$, correction signatures are exactly distinguished by these occupancy records. There are $2^r$ signatures, and $r$ fixed binary bits are necessary and sufficient to encode them.\n\\end{corollary}\n\\begin{proof}\nAn individual-cell filter reveals the target values occupied in that cell, so the signature determines occupancy. Conversely, occupancy determines the target values left by any union of cells. Every subset of $D_t$ occurs: select one world from each included nonempty cell--target intersection and none from excluded intersections. Those intersections are disjoint. Thus there are precisely $2^r$ possible records. The preceding theorem and binary counting give necessity, while the occupancy vector supplies sufficiency.\n\\end{proof}\n\\source{[GA], Corollary 5.3. The word ``exactly'' in the evidence premi","atures are exactly distinguished by these occupancy records. There are $2^r$ signatures, and $r$ fixed binary bits are necessary and sufficient to encode them.\n\\end{corollary}\n\\begin{proof}\nAn individual-cell filter reveals the target values occupied in that cell, so the signature determines occupancy. Conversely, occupancy determines the target values left by any union of cells. Every subset of $D_t$ occurs: select one world from each included nonempty cell--target intersection and none from excluded intersections. Those intersections are disjoint. Thus there are precisely $2^r$ possible records. The preceding theorem and binary counting give necessity, while the occupancy vector supplies sufficiency.\n\\end{proof}\n\\source{[GA], Corollary 5.3. The word ``exactly'' in the evidence premise is essential.}","{proof}\nAn individual-cell filter reveals the target values occupied in that cell, so the signature determines occupancy. Conversely, occupancy determines the target values left by any union of cells. Every subset of $D_t$ occurs: select one world from each included nonempty cell--target intersection and none from excluded intersections. Those intersections are disjoint. Thus there are precisely $2^r$ possible records. The preceding theorem and binary counting give necessity, while the occupancy vector supplies sufficiency.\n\\end{proof}\n\\source{[GA], Corollary 5.3. The word ``exactly'' in the evidence premise is essential.}","For this sensor contract, the four cells are fixed pairs $(b,s)$ with either nuisance value. We admit the entire Boolean algebra generated by calibration and display events, including every union of its cells. This is a choice of evidence language, not a claim that a single calibration physically implements every union event. The target is $b$, so there are four attainable cell--target pairs. Candidate membership uses eight bits for eight worlds. Correction memory uses four bits for sixteen possible occupancies. The current target set alone has four possibilities and needs only two bits, but is not sufficient for later calibration.","For example,\n\\[\nV=\\{(0,0,0),(1,1,0)\\},\\qquad W=\\{(0,1,0),(1,0,0)\\}\n\\]\nhave the same possible target set $\\{0,1\\}$. After the event $s=0$, one leaves target zero and the other target one. Their different associations are operationally relevant.","\\begin{counterexample}[A new evidence event changes the required memory]\\label{ce:split}\nTake one cell $\\Omega=\\{u,v\\}$ and a constant target. The singleton candidate sets have the same old occupancy. If the event $\\{u\\}$ is added, filtering one singleton gives a target value while filtering the other gives contradiction. A record sufficient for the old event algebra is insufficient for the larger one.\n\\end{counterexample}","Let $\\mathcal M$ be a finite nonempty model set with prior $p(m)$. The sampled model remains fixed during the experiment. For each $m$, let $L_m$ be a probability distribution on a finite nonempty observation alphabet $\\mathcal O$. Let $\\mathcal Z$ be a finite nonempty report alphabet. A stochastic reporting channel $B(o,z)$, conditionally independent of the model given $o$, creates a retained report $Z\\in\\mathcal Z$. The joint law is\n\\[\n\\Pr(m,o,z)=p(m)L_m(o)B(o,z),\\qquad Q_m(z)=\\sum_oL_m(o)B(o,z).\n\\]\nPosterior distributions are defined only on positive-probability events.","\\mathcal M$ be a finite nonempty model set with prior $p(m)$. The sampled model remains fixed during the experiment. For each $m$, let $L_m$ be a probability distribution on a finite nonempty observation alphabet $\\mathcal O$. Let $\\mathcal Z$ be a finite nonempty report alphabet. A stochastic reporting channel $B(o,z)$, conditionally independent of the model given $o$, creates a retained report $Z\\in\\mathcal Z$. The joint law is\n\\[\n\\Pr(m,o,z)=p(m)L_m(o)B(o,z),\\qquad Q_m(z)=\\sum_oL_m(o)B(o,z).\n\\]\nPosterior distributions are defined only on positive-probability events.","el set with prior $p(m)$. The sampled model remains fixed during the experiment. For each $m$, let $L_m$ be a probability distribution on a finite nonempty observation alphabet $\\mathcal O$. Let $\\mathcal Z$ be a finite nonempty report alphabet. A stochastic reporting channel $B(o,z)$, conditionally independent of the model given $o$, creates a retained report $Z\\in\\mathcal Z$. The joint law is\n\\[\n\\Pr(m,o,z)=p(m)L_m(o)B(o,z),\\qquad Q_m(z)=\\sum_oL_m(o)B(o,z).\n\\]\nPosterior distributions are defined only on positive-probability events.","\\begin{theorem}[When a report preserves every Bayesian correction]\\label{r:bayes}\nThe following are equivalent:\n\\begin{enumerate}\n\\item For every prior, the posterior after $o$ equals that after $z$ on every positive-probability pair $(o,z)$.\n\\item The equality holds for one prior positive on every model.\n\\item There is a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubst","\\begin{theorem}[When a report preserves every Bayesian correction]\\label{r:bayes}\nThe following are equivalent:\n\\begin{enumerate}\n\\item For every prior, the posterior after $o$ equals that after $z$ on every positive-probability pair $(o,z)$.\n\\item The equality holds for one prior positive on every model.\n\\item There is a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula","\\begin{theorem}[When a report preserves every Bayesian correction]\\label{r:bayes}\nThe following are equivalent:\n\\begin{enumerate}\n\\item For every prior, the posterior after $o$ equals that after $z$ on every positive-probability pair $(o,z)$.\n\\item The equality holds for one prior positive on every model.\n\\item There is a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the f","\\begin{theorem}[When a report preserves every Bayesian correction]\\label{r:bayes}\nThe following are equivalent:\n\\begin{enumerate}\n\\item For every prior, the posterior after $o$ equals that after $z$ on every positive-probability pair $(o,z)$.\n\\item The equality holds for one prior positive on every model.\n\\item There is a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weigh","\\begin{theorem}[When a report preserves every Bayesian correction]\\label{r:bayes}\nThe following are equivalent:\n\\begin{enumerate}\n\\item For every prior, the posterior after $o$ equals that after $z$ on every positive-probability pair $(o,z)$.\n\\item The equality holds for one prior positive on every model.\n\\item There is a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every mo…","positive on every model.\n\\item There is a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","s a stochastic kernel $\\rho(o\\mid z)$ such that\n\\[\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","id z)\\quad\\text{for all }m,o,z.\n\\]\n\\end{enumerate}\nThe reconstruction in the last condition refers to the actual joint experiment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","ment, not an unrelated simulation with matching marginals.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","of}\nThe first condition implies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","mplies the second. Under a positive prior, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","r, define $\\rho(o\\mid z)=\\Pr(o\\mid z)$ whenever $z$ has positive probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","probability. For a positive pair, posterior equality yields\n\\[\np(m)L_m(o)B(o,z)=\\Pr(o,z)p(m\\mid o)\n=\\rho(o\\mid z)\\Pr(z)p(m\\mid z).\n\\]\nSubstituting Bayes' formula and cancelling positive $p(m)$ gives the factorization. A zero-probability pair under a positive prior has $L_m(o)B(o,z)=0$ for all $m$. Its conditional weight is zero. If $z$ is impossible for every model, assign any probability row to $\\rho(\\cdot\\mid z)$ because both sides vanish.","Conversely, the factorization makes the likelihood of $(o,z)$ a model-independent multiple of the likelihood of $z$. Normalization under any prior therefore gives $p(m\\mid o,z)=p(m\\mid z)$. Since the common channel only reads $o$, the same positive-pair posterior equals $p(m\\mid o)$. This proves the first condition.\n\\end{proof}\n\\source{[MU], Theorem 12.1. The full-support and zero-probability cases are included in the proof.}","The theorem permits removal of an independent nuisance flag. It does not permit removal of every association. Consider two priors on $(b,s)$, with rows indexed by $b$ and columns by $s$:\n\\[\nP=\\begin{pmatrix}1/2&0","The theorem permits removal of an independent nuisance flag. It does not permit removal of every association. Consider two priors on $(b,s)$, with rows indexed by $b$ and columns by $s$:\n\\[\nP=\\begin{pmatrix}1/2&0\\\\0&1/2\\end{pmatrix},\\qquad\nP'=\\begin{pmatrix}0&1/2\\\\1/2&0\\end{pmatrix}.\n\\]\nBoth individual marginals are uniform. A noisy calibration reports $s$ correctly with probability $3/4$, independently conditional on the fixed world. The probability of report one is $1/2$ under both priors. After that report, however, the target posteriors are $3/4$ and $1/4$. Separate target and sensor marginals cannot support this correction. Their association matters even though neither marginal changed.","1/2&0\\end{pmatrix}.\n\\]\nBoth individual marginals are uniform. A noisy calibration reports $s$ correctly with probability $3/4$, independently conditional on the fixed world. The probability of report one is $1/2$ under both priors. After that report, however, the target posteriors are $3/4$ and $1/4$. Separate target and sensor marginals cannot support this correction. Their association matters even though neither marginal changed.","First consider a known finite state set $X$ with row-stochastic transition matrices $K_a(x,x')$. We retain column-vector probability distributions. A partition $\\pi$ has aggregation $C_\\pi(z,x)=1_{\\{\\pi(x)=z\\}}$, so prediction is $p\\mapsto K_a^\\top p$. For each permitted observation symbol $o$, let $\\ell_o(x)>0$ be its specified likelihood. When these are a complete report alphabet, assume also $\\sum_o\\ell_o(x)=1$.","A mandatory partition $\\pi_0$ with at least two blocks protects the target distinctions. We seek a refinement that supports every prediction and each separate Bayesian update for every prior. The observation is an operation available in its own right. This is stronger than inspecting only a chosen combined transition-and-report schedule.","\\begin{theorem}[Least refinement closed under prediction and correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition","\\begin{theorem}[Least refinement closed under prediction and correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top","\\begin{theorem}[Least refinement closed under prediction and correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top C_\\pi$, first on point masse","\\begin{theorem}[Least refinement closed under prediction and correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top C_\\pi$, first on point masses and then by linearity. Likelihoods are constant within final blocks. Grouping the numerat","\\begin{theorem}[Least refinement closed under prediction and correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top C_\\pi$, first on point masses and then by linearity. Likelihoods are constant within final blocks. Grouping the numerator and norm","\\begin{theorem}[Least refinement closed under prediction and correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top C_\\pi$, first on point masses and then by linearity. Likelihoods are constant within final blocks. Grouping the numerator and normalizer by those blocks therefore commutes with every Bayesian update.","nd correction]\\label{r:joint-refinement}\nFirst split $\\pi_0$ by the complete vectors $(\\ell_o(x))_o$, obtaining $\\pi_{\\rm init}$. Then repeatedly split current blocks by\n\\[\n \\left(\\sum_{x'\\in B}K_a(x,x')\\right)_{a,B},\n\\]\nwhere $B$ runs over the current blocks. The process stops after at most $|X|-|\\pi_{\\rm init}|$ strict refinement rounds. Its final partition is the unique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top C_\\pi$, first on point masses and then by linearity. Likelihoods are constant within final blocks. Grouping the numerator and normalizer by those blocks therefore commutes with every Bayesian update.","nique coarsest refinement of $\\pi_0$ for which all specified predictions admit stochastic coarse transitions and all separate Bayesian updates admit fixed strictly positive coarse likelihoods for every prior.\n\\end{theorem}\n\\begin{proof}\nAt a stable partition the probability of reaching each coarse block depends only on the current block. These sums define a stochastic coarse transition and yield $C_\\pi K_a^\\top=\\bar K_a^\\top C_\\pi$, first on point masses and then by linearity. Likelihoods are constant within final blocks. Grouping the numerator and normalizer by those blocks therefore commutes with every Bayesian update.","For minimality, take any admissible refinement $\\eta$ of $\\pi_0$. It has at least two blocks. Fix a strictly positive fine likelihood $\\ell$ and a strictly positive coarse likelihood $\\bar\\ell$ inducing its assumed update. For states $x,y$ in one block and $z$ in a different block, compare priors assigning half their mass to $x,z$ and to $y,z$. Their coarse priors agree. Equality of the posterior odds gives\n\\[\n \\ell(x)/\\ell(z)=\\bar\\ell(\\eta(x))/\\bar\\ell(\\eta(z))=\\ell(y)/\\ell(z).\n\\]\nThus $\\ell(x)=\\ell(y)$, and $\\eta$ refines the first likelihood split. Coarse likelihoods may differ by a positive scale for each observation without changing this conclusion.","y, take any admissible refinement $\\eta$ of $\\pi_0$. It has at least two blocks. Fix a strictly positive fine likelihood $\\ell$ and a strictly positive coarse likelihood $\\bar\\ell$ inducing its assumed update. For states $x,y$ in one block and $z$ in a different block, compare priors assigning half their mass to $x,z$ and to $y,z$. Their coarse priors agree. Equality of the posterior odds gives\n\\[\n \\ell(x)/\\ell(z)=\\bar\\ell(\\eta(x))/\\bar\\ell(\\eta(z))=\\ell(y)/\\ell(z).\n\\]\nThus $\\ell(x)=\\ell(y)$, and $\\eta$ refines the first likelihood split. Coarse likelihoods may differ by a positive scale for each observation without changing this conclusion.","Fix a strictly positive fine likelihood $\\ell$ and a strictly positive coarse likelihood $\\bar\\ell$ inducing its assumed update. For states $x,y$ in one block and $z$ in a different block, compare priors assigning half their mass to $x,z$ and to $y,z$. Their coarse priors agree. Equality of the posterior odds gives\n\\[\n \\ell(x)/\\ell(z)=\\bar\\ell(\\eta(x))/\\bar\\ell(\\eta(z))=\\ell(y)/\\ell(z).\n\\]\nThus $\\ell(x)=\\ell(y)$, and $\\eta$ refines the first likelihood split. Coarse likelihoods may differ by a positive scale for each observation without changing this conclusion.","e fine likelihood $\\ell$ and a strictly positive coarse likelihood $\\bar\\ell$ inducing its assumed update. For states $x,y$ in one block and $z$ in a different block, compare priors assigning half their mass to $x,z$ and to $y,z$. Their coarse priors agree. Equality of the posterior odds gives\n\\[\n \\ell(x)/\\ell(z)=\\bar\\ell(\\eta(x))/\\bar\\ell(\\eta(z))=\\ell(y)/\\ell(z).\n\\]\nThus $\\ell(x)=\\ell(y)$, and $\\eta$ refines the first likelihood split. Coarse likelihoods may differ by a positive scale for each observation without changing this conclusion.","If $\\eta$ refines a current partition, each current block is a union of $\\eta$ blocks. Exact prediction under $\\eta$ makes the probability of reaching each such union constant inside every $\\eta$ block. Hence $\\eta$ refines the next split too. Induction proves that every admissible refinement refines the terminal one. Each strict round increases the block count, proving the bound. Coarseness determines the partition uniquely, up to naming its blocks.\n\\end{proof}\n\\source{[LSM], Proposition 9 and Proposition 12, physical pages 11--14. Bayesian necessity also appears in PRIQ [PRIQ], Theorem 5.6. The proof includes the two-block and strict-positivity premises rather than importing them by a cross-refere","If $\\eta$ refines a current partition, each current block is a union of $\\eta$ blocks. Exact prediction under $\\eta$ makes the probability of reaching each such union constant inside every $\\eta$ block. Hence $\\eta$ refines the next split too. Induction proves that every admissible refinement refines the terminal one. Each strict round increases the block count, proving the bound. Coarseness determines the partition uniquely, up to naming its blocks.\n\\end{proof}\n\\source{[LSM], Proposition 9 and Proposition 12, physical pages 11--14. Bayesian necessity also appears in PRIQ [PRIQ], Theorem 5.6. The proof includes the two-block and strict-positivity premises rather than importing them by a cross-reference.}","For the sensor, the mandatory display partition has two blocks. A noisy calibration that reports $s$ correctly with probability $3/4$ splits it into the four $(b,s)$ blocks. The nuisance bit remains merged. The physical flip maps these four blocks among themselves, so the transition stage introduces no further split. The representation is now closed for both prediction and correction, not merely for today's display.","Refinement can also propagate backward through dynamics. In a separate four-state example, the mandatory partition is $\\{0,1,2\\}\\mid\\{3\\}$ and the deterministic successors are $(0,2,3,3)$. A constant likelihood adds no information. Transition signatures first isolate state 2, then separate 0 from 1. The final partition is discrete. The reason for the second split is a future observable difference, not a newly sampled measurement.","\\section{Retain the joint law of the report and the next state}\nA general experiment can change the state and report correlated information about the change. Describing its transition and report marginals separately is then insufficient. Let\n\\[\n K_{a,o}(x',x)=\\Pr(X_{t+1}=x',O_{t+1}=o\\mid X_t=x,A_t=a)\n\\]\nbe a nonnegative finite instrument with $\\sum_{o,x'}K_{a,o}(x',x)=1$ for every $a,x$. This notation is column-oriented: $K_{a,o}p$ is an unnormalized post-experiment distribution. It specifies an experiment's physical change and its observation in one object.","law of the report and the next state}\nA general experiment can change the state and report correlated information about the change. Describing its transition and report marginals separately is then insufficient. Let\n\\[\n K_{a,o}(x',x)=\\Pr(X_{t+1}=x',O_{t+1}=o\\mid X_t=x,A_t=a)\n\\]\nbe a nonnegative finite instrument with $\\sum_{o,x'}K_{a,o}(x',x)=1$ for every $a,x$. This notation is column-oriented: $K_{a,o}p$ is an unnormalized post-experiment distribution. It specifies an experiment's physical change and its observation in one object.","For a surjection $\\pi:X\\to Z$, let $C$ sum probabilities over its fibres. Write\n\\[\n z_{a,o}(p)=\\mathbf1^\\top K_{a,o}p,\\qquad\n \\Psi_{a,o}(p)=K_{a,o}p/z_{a,o}(p)\n\\]\nwhen the denominator is positive. Zero mass means an impossible report under that prior, not an arbitrarily assigned posterior.","\\begin{theorem}[One quotient for reports, posteriors, and feedback]\\label{r:instrument}\nA nonnegative normalized coarse instrument $\\bar K_{a,o}$ satisfies\n\\[\n CK_{a,o}=\\bar K_{a,o}C\\quad\\text{for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same histor","\\begin{theorem}[One quotient for reports, posteriors, and feedback]\\label{r:instrument}\nA nonnegative normalized coarse instrument $\\bar K_{a,o}$ satisfies\n\\[\n CK_{a,o}=\\bar K_{a,o}C\\quad\\text{for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}","\\begin{theorem}[One quotient for reports, posteriors, and feedback]\\label{r:instrument}\nA nonnegative normalized coarse instrument $\\bar K_{a,o}$ satisfies\n\\[\n CK_{a,o}=\\bar K_{a,o}C\\quad\\text{for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a","\\begin{theorem}[One quotient for reports, posteriors, and feedback]\\label{r:instrument}\nA nonnegative normalized coarse instrument $\\bar K_{a,o}$ satisfies\n\\[\n CK_{a,o}=\\bar K_{a,o}C\\quad\\text{for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a point mass proves necessity of the fibre","\\begin{theorem}[One quotient for reports, posteriors, and feedback]\\label{r:instrument}\nA nonnegative normalized coarse instrument $\\bar K_{a,o}$ satisfies\n\\[\n CK_{a,o}=\\bar K_{a,o}C\\quad\\text{for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a point mass proves necessity of the fibre condition. Conversel","ormalized coarse instrument $\\bar K_{a,o}$ satisfies\n\\[\n CK_{a,o}=\\bar K_{a,o}C\\quad\\text{for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a point mass proves necessity of the fibre condition. Conversely, use the common sums to define each column of $\\bar K_{a,o}$. They are nonnegative. Summing across $z',o$ partitions the fine successor set and gives column mass one. The definition proves the matri…","for every }a,o\n\\]\nif and only if, for all $a,o,z'$,\n\\[\n \\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\n\\]\nis constant as $x$ varies within any one input fibre. Such a coarse instrument is unique. Equivalently, it preserves for every prior both the report probabilities and the aggregated positive-probability posteriors:\n\\[\n z_{a,o}(p)=\\bar z_{a,o}(Cp),\\qquad C\\Psi_{a,o}(p)=\\bar\\Psi_{a,o}(Cp).\n\\]\nStarting from $Cp_0$, these identities preserve every finite joint action/report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a point mass proves necessity of the fibre condition. Conversely, use the common sums to define each column of $\\bar K_{a,o}$. They are nonnegative. Summing across $z',o$ partitions the fine successor set and gives column mass one. The definition proves the matrix equality on point masses, hence on every vector. Surjectivity gives uniqueness.","report law and every corresponding coarse posterior under one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a point mass proves necessity of the fibre condition. Conversely, use the common sums to define each column of $\\bar K_{a,o}$. They are nonnegative. Summing across $z',o$ partitions the fine successor set and gives column mass one. The definition proves the matrix equality on point masses, hence on every vector. Surjectivity gives uniqueness.","nder one common policy that uses only the previously observed action/report history. This includes randomized policies with the same history-conditional action probabilities.\n\\end{theorem}\n\\begin{proof}\nApplying the matrix identity to a point mass proves necessity of the fibre condition. Conversely, use the common sums to define each column of $\\bar K_{a,o}$. They are nonnegative. Summing across $z',o$ partitions the fine successor set and gives column mass one. The definition proves the matrix equality on point masses, hence on every vector. Surjectivity gives uniqueness.","Because $\\mathbf1_Z^\\top C=\\mathbf1_X^\\top$, the matrix equality gives equal report masses. Dividing it by a common positive mass gives equal aggregated posteriors. Conversely, equality of masses and positive-mass posteriors gives the unnormalized matrix equality on every prior. If the common mass is zero, both unnormalized vectors are zero by nonnegativity. Point masses finish that direction.","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the project","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the projected-law construction in AVOC [AVOC], Theor","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the projected-law construction in AVOC [AVOC], Theorem 4.5. The joint instrument and feedback formu","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the projected-law construction in AVOC [AVOC], Theorem 4.5. The joint instrument and feedback formulation is proved here. No e","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the projected-law construction in AVOC [AVOC], Theorem 4.5. The joint instrument and feedback formulation is proved here. No entire nonlinear source theorem or Lean imp","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the projected-law construction in AVOC [AVOC], Theorem 4.5. The joint instrument and feedback formulation is proved here. No entire nonlinear source theorem or Lean implementation is imp","For a history $h$, keep unnormalized states $u_h,\\bar u_h$. Initially $\\bar u_\\varnothing=Cu_\\varnothing$. After choosing action $a$ with common probability $\\gamma(a\\mid h)$ and receiving $o$, the next states are $\\gamma(a\\mid h)K_{a,o}u_h$ and $\\gamma(a\\mid h)\\bar K_{a,o}\\bar u_h$. The intertwining identity preserves their aggregation. Induction gives equal mass for every history, including zero-mass histories. Normalization on positive histories gives the posterior statement. Stopping decisions based on the same history can be included as matching terminal choices.\n\\end{proof}\n\\source{Selected synthesis of the positive-transfer and filtering arguments in [PRIQ], Theorems 5.2, 6.1 and 6.3, and the projected-law construction in AVOC [AVOC], Theorem 4.5. The joint instrument and feedback formulation is proved here. No entire nonlinear source theorem or Lean implementation is imported.}","\\begin{counterexample}[Correct marginals with an incorrect correction]\\label{ce:joint-marginals}\nTwo fine current states have the same retained label. From the first, the next retained state and report are $(0,0)$ or $(1,1)$, each with probability $1/2$. From the second they are $(0,1)$ or $(1,0)$, again equally likely. In both cases the next-state marginal is uniform and the report marginal is uniform. After report one, however, the next retained state is certainly one in the first case and certainly zero in the second. No common filter based only on the old label can realize both. The joint fibre condition detects exactly the lost association.\n\\end{counterexample}","Preserving only normalized posteriors is weaker still. A report coin independent of the hidden state leaves every prior unchanged whether its success probability is $1/4$ or $3/4$. Those two instruments have identical posterior maps but different report laws. A planner that prices later branches needs those probabilities. In the exact instrument theorem the evidence scale is fixed, rather than silently discarded during normalization.","\\section{Small law error and a rare conditional history}\nApproximate models must be compared before declaring their posteriors close. Let $\\widetilde K$ be a normalized instrument on the same coarse state and report alphabets. Suppose, at step $t$, every relevant action and fine state satisfy\n\\[\n \\TV\\left(\\left(\\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\\right)_{o,z'},\n             (\\widetilde K_{a,o}(z',\\pi(x)))_{o,z'}\\right)\\le\\epsilon_t.\n\\]\nThis is a bound on a complete report/next-state distribution, not on an isolated matrix entry. Assume $\\TV(Cp_0,\\widetilde p_0)\\le\\delta_0$ with nonnegative $\\delta_0,\\epsilon_t$. Both models run the same history-based policy. Coarse states recorded in the comparison below are analytical coordinates, not extra observations supplied to that policy.","idetilde K$ be a normalized instrument on the same coarse state and report alphabets. Suppose, at step $t$, every relevant action and fine state satisfy\n\\[\n \\TV\\left(\\left(\\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\\right)_{o,z'},\n             (\\widetilde K_{a,o}(z',\\pi(x)))_{o,z'}\\right)\\le\\epsilon_t.\n\\]\nThis is a bound on a complete report/next-state distribution, not on an isolated matrix entry. Assume $\\TV(Cp_0,\\widetilde p_0)\\le\\delta_0$ with nonnegative $\\delta_0,\\epsilon_t$. Both models run the same history-based policy. Coarse states recorded in the comparison below are analytical coordinates, not extra observations supplied to that policy.","se state and report alphabets. Suppose, at step $t$, every relevant action and fine state satisfy\n\\[\n \\TV\\left(\\left(\\sum_{x':\\pi(x')=z'}K_{a,o}(x',x)\\right)_{o,z'},\n             (\\widetilde K_{a,o}(z',\\pi(x)))_{o,z'}\\right)\\le\\epsilon_t.\n\\]\nThis is a bound on a complete report/next-state distribution, not on an isolated matrix entry. Assume $\\TV(Cp_0,\\widetilde p_0)\\le\\delta_0$ with nonnegative $\\delta_0,\\epsilon_t$. Both models run the same history-based policy. Coarse states recorded in the comparison below are analytical coordinates, not extra observations supplied to that policy.","\\begin{proposition}[A whole-history bound with an explicit conditioning cost]\\label{r:history-bound}\nAt horizon $H$, the distributions $P_H,\\widetilde P_H$ of the full coarse-state and action/report histories satisfy\n\\[\n \\TV(P_H,\\widetilde P_H)\\le d_H:=\\min\\left\\{1,\\delta_0+\\sum_{t=0}^{H-1}\\epsilon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with p","\\begin{proposition}[A whole-history bound with an explicit conditioning cost]\\label{r:history-bound}\nAt horizon $H$, the distributions $P_H,\\widetilde P_H$ of the full coarse-state and action/report histories satisfy\n\\[\n \\TV(P_H,\\widetilde P_H)\\le d_H:=\\min\\left\\{1,\\delta_0+\\sum_{t=0}^{H-1}\\epsilon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden cur","\\begin{proposition}[A whole-history bound with an explicit conditioning cost]\\label{r:history-bound}\nAt horizon $H$, the distributions $P_H,\\widetilde P_H$ of the full coarse-state and action/report histories satisfy\n\\[\n \\TV(P_H,\\widetilde P_H)\\le d_H:=\\min\\left\\{1,\\delta_0+\\sum_{t=0}^{H-1}\\epsilon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden current-state distribution is a mixture of states in the last coarse fibre. Convexity of total variation makes the true next report/coarse-state row differ from","n}[A whole-history bound with an explicit conditioning cost]\\label{r:history-bound}\nAt horizon $H$, the distributions $P_H,\\widetilde P_H$ of the full coarse-state and action/report histories satisfy\n\\[\n \\TV(P_H,\\widetilde P_H)\\le d_H:=\\min\\left\\{1,\\delta_0+\\sum_{t=0}^{H-1}\\epsilon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden current-state distribution is a mixture of states in the last coarse fibre. Convexity of total variation makes the true next report/coarse-state row differ from the declared row by at most $\\epsilon_t$. Mixing over a common action kernel preserves this bound. Extend","d with an explicit conditioning cost]\\label{r:history-bound}\nAt horizon $H$, the distributions $P_H,\\widetilde P_H$ of the full coarse-state and action/report histories satisfy\n\\[\n \\TV(P_H,\\widetilde P_H)\\le d_H:=\\min\\left\\{1,\\delta_0+\\sum_{t=0}^{H-1}\\epsilon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden current-state distribution is a mixture of states in the last coarse fibre. Convexity of total variation makes the true next report/coarse-state row differ from the declared row by at most $\\epsilon_t$. Mixing over a common action kernel preserves this bound. Extend the two old history la…","e distributions $P_H,\\widetilde P_H$ of the full coarse-state and action/report histories satisfy\n\\[\n \\TV(P_H,\\widetilde P_H)\\le d_H:=\\min\\left\\{1,\\delta_0+\\sum_{t=0}^{H-1}\\epsilon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden current-state distribution is a mixture of states in the last coarse fibre. Convexity of total variation makes the true next report/coarse-state row differ from the declared row by at most $\\epsilon_t$. Mixing over a common action kernel preserves this bound. Extend the two old history laws first through the same declared row. Total variation cannot increase under a…","lon_t\\right\\}.\n\\]\nThe same bound holds after forgetting any of those history coordinates. For an observed history event $E$ with probabilities $p_E,\\widetilde p_E>0$, the conditional final coarse-state distributions obey\n\\[\n \\TV\\bigl(P_H(Z_H\\in\\cdot\\mid E),\\widetilde P_H(Z_H\\in\\cdot\\mid E)\\bigr)\n \\le\\min\\left\\{1,\\frac{d_H}{\\max(p_E,\\widetilde p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden current-state distribution is a mixture of states in the last coarse fibre. Convexity of total variation makes the true next report/coarse-state row differ from the declared row by at most $\\epsilon_t$. Mixing over a common action kernel preserves this bound. Extend the two old history laws first through the same declared row. Total variation cannot increase under a common stochastic kernel, by the triangle inequality and row sums. Replacing those rows with the true conditional rows adds at most $\\epsilon_t$, weighted by a total history ma…","p_E)}\\right\\}.\n\\]\nIf either event probability exceeds $d_H$, the other is necessarily positive.\n\\end{proposition}\n\\begin{proof}\nCondition on a complete coarse history with positive fine probability. Its hidden current-state distribution is a mixture of states in the last coarse fibre. Convexity of total variation makes the true next report/coarse-state row differ from the declared row by at most $\\epsilon_t$. Mixing over a common action kernel preserves this bound. Extend the two old history laws first through the same declared row. Total variation cannot increase under a common stochastic kernel, by the triangle inequality and row sums. Replacing those rows with the true conditional rows adds at most $\\epsilon_t$, weighted by a total history mass of one. Hence the history distance at the next step is at most the old distance plus $\\epsilon_t$. Induction and the cap one prove the first claim. Summing over forgotten coordinates is also a stochastic map, proving the second.","on makes the true next report/coarse-state row differ from the declared row by at most $\\epsilon_t$. Mixing over a common action kernel preserves this bound. Extend the two old history laws first through the same declared row. Total variation cannot increase under a common stochastic kernel, by the triangle inequality and row sums. Replacing those rows with the true conditional rows adds at most $\\epsilon_t$, weighted by a total history mass of one. Hence the history distance at the next step is at most the old distance plus $\\epsilon_t$. Induction and the cap one prove the first claim. Summing over forgotten coordinates is also a stochastic map, proving the second.","nel preserves this bound. Extend the two old history laws first through the same declared row. Total variation cannot increase under a common stochastic kernel, by the triangle inequality and row sums. Replacing those rows with the true conditional rows adds at most $\\epsilon_t$, weighted by a total history mass of one. Hence the history distance at the next step is at most the old distance plus $\\epsilon_t$. Induction and the cap one prove the first claim. Summing over forgotten coordinates is also a stochastic map, proving the second.","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made expli","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made explicit. The symmetric conditioning fo","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made explicit. The symmetric conditioning form follows by applying the same elementary boun","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made explicit. The symmetric conditioning form follows by applying the same elementary bound in both directions. [PRIQ], Section","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made explicit. The symmetric conditioning form follows by applying the same elementary bound in both directions. [PRIQ], Section 9 supplies the complementary stepwise posterior-er","For conditioning, let $u,v$ be the restrictions of the two history laws to $E$, with masses $p_E,\\widetilde p_E$. The difference on the complement has $\\ell^1$ norm at least $|p_E-\\widetilde p_E|$. Thus\n\\[\n \\|u-v\\|_1+|p_E-\\widetilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made explicit. The symmetric conditioning form follows by applying the same elementary bound in both directions. [PRIQ], Section 9 supplies the complementary stepwise posterior-error perspective.}","ilde p_E|\\le2d_H.\n\\]\nAdding and subtracting $v/p_E$ bounds the conditional distance by $d_H/p_E$. Interchanging the laws gives $d_H/\\widetilde p_E$, so the smaller of these bounds has the displayed maximum denominator. Projection to the final state cannot increase the distance. Finally, the probability of any event changes by at most total variation, which proves positivity when one mass exceeds $d_H$.\n\\end{proof}\n\\source{Finite probability-instrument specialization of [LSM], Propositions 43--44, with the joint retained-state record made explicit. The symmetric conditioning form follows by applying the same elementary bound in both directions. [PRIQ], Section 9 supplies the complementary stepwise posterior-error perspective.}","\\begin{counterexample}[A rare report carries the entire discrepancy]\\label{ce:rare-history}\nTwo joint laws place mass $1-\\delta$ on the same ordinary outcome. On a report event $E$ of mass $\\delta>0$, one puts all mass on retained state zero and the other on retained state one. Their whole-law distance is $\\delta$, while their conditional distance after $E$ is one. The factor involving event probability is therefore essential. This is a finite-law statement, not an empirical estimate of a sensor error.\n\\end{counterexample}","\\section{One quantitative decision contract}\nLet $p$ and $\\widehat p$ be distributions on the same finite world set. For each world let $v_\\omega$ be a response and $\\widehat v_\\omega$ its approximation in one metric space. Assume\n\\[\n\\TV(p,\\widehat p)\\le e,\\qquad\n\\sup_\\omega d(v_\\omega,\\widehat v_\\omega)\\le d_0.\n\\]\nFor each action $a$ in a finite nonempty set, let the loss $\\ell(a,v)$ lie in $[0,1]$ and be $K$-Lipschitz in $v$. Define the model risks\n\\[\nR(a)=\\sum_\\omega p(\\omega)\\ell(a,v_\\omega),\\qquad\n\\widehat R(a)=\\sum_\\omega\\widehat p(\\omega)\\ell(a,\\widehat v_\\omega).\n\\]","e quantitative decision contract}\nLet $p$ and $\\widehat p$ be distributions on the same finite world set. For each world let $v_\\omega$ be a response and $\\widehat v_\\omega$ its approximation in one metric space. Assume\n\\[\n\\TV(p,\\widehat p)\\le e,\\qquad\n\\sup_\\omega d(v_\\omega,\\widehat v_\\omega)\\le d_0.\n\\]\nFor each action $a$ in a finite nonempty set, let the loss $\\ell(a,v)$ lie in $[0,1]$ and be $K$-Lipschitz in $v$. Define the model risks\n\\[\nR(a)=\\sum_\\omega p(\\omega)\\ell(a,v_\\omega),\\qquad\n\\widehat R(a)=\\sum_\\omega\\widehat p(\\omega)\\ell(a,\\widehat v_\\omega).\n\\]","cision contract}\nLet $p$ and $\\widehat p$ be distributions on the same finite world set. For each world let $v_\\omega$ be a response and $\\widehat v_\\omega$ its approximation in one metric space. Assume\n\\[\n\\TV(p,\\widehat p)\\le e,\\qquad\n\\sup_\\omega d(v_\\omega,\\widehat v_\\omega)\\le d_0.\n\\]\nFor each action $a$ in a finite nonempty set, let the loss $\\ell(a,v)$ lie in $[0,1]$ and be $K$-Lipschitz in $v$. Define the model risks\n\\[\nR(a)=\\sum_\\omega p(\\omega)\\ell(a,v_\\omega),\\qquad\n\\widehat R(a)=\\sum_\\omega\\widehat p(\\omega)\\ell(a,\\widehat v_\\omega).\n\\]","$\\widehat p$ be distributions on the same finite world set. For each world let $v_\\omega$ be a response and $\\widehat v_\\omega$ its approximation in one metric space. Assume\n\\[\n\\TV(p,\\widehat p)\\le e,\\qquad\n\\sup_\\omega d(v_\\omega,\\widehat v_\\omega)\\le d_0.\n\\]\nFor each action $a$ in a finite nonempty set, let the loss $\\ell(a,v)$ lie in $[0,1]$ and be $K$-Lipschitz in $v$. Define the model risks\n\\[\nR(a)=\\sum_\\omega p(\\omega)\\ell(a,v_\\omega),\\qquad\n\\widehat R(a)=\\sum_\\omega\\widehat p(\\omega)\\ell(a,\\widehat v_\\omega).\n\\]","\\begin{theorem}[A decision bound for one actual approximation chain]\\label{r:risk}\nPut $\\xi=\\min\\{1,e+Kd_0\\}$. For every action, $|R(a)-\\widehat R(a)|\\le\\xi$. If $\\widehat a$ minimizes the approximate risk, then\n\\[\nR(\\widehat a)-\\min_a R(a)\\le\\min\\{1,2\\xi\\}.\n\\]\nIf the approximate gap between its unique minimizer and every alternative exceeds $2\\xi$, that action is also the unique true model-risk minimizer.\n\\end{theorem}\n\\begin{proof}\nAdd and subtract $\\sum_\\omega p(\\omega)\\ell(a,\\widehat v_\\omega)$. The response term has absolute value at most $Kd_0$. The expectation of a function in $[0,1]$ changes by at most total variation, giving the other term at most $e$. Risks lie in $[0,1]$, so their difference is also at most one.","ation chain]\\label{r:risk}\nPut $\\xi=\\min\\{1,e+Kd_0\\}$. For every action, $|R(a)-\\widehat R(a)|\\le\\xi$. If $\\widehat a$ minimizes the approximate risk, then\n\\[\nR(\\widehat a)-\\min_a R(a)\\le\\min\\{1,2\\xi\\}.\n\\]\nIf the approximate gap between its unique minimizer and every alternative exceeds $2\\xi$, that action is also the unique true model-risk minimizer.\n\\end{theorem}\n\\begin{proof}\nAdd and subtract $\\sum_\\omega p(\\omega)\\ell(a,\\widehat v_\\omega)$. The response term has absolute value at most $Kd_0$. The expectation of a function in $[0,1]$ changes by at most total variation, giving the other term at most $e$. Risks lie in $[0,1]$, so their difference is also at most one.","le\\xi$. If $\\widehat a$ minimizes the approximate risk, then\n\\[\nR(\\widehat a)-\\min_a R(a)\\le\\min\\{1,2\\xi\\}.\n\\]\nIf the approximate gap between its unique minimizer and every alternative exceeds $2\\xi$, that action is also the unique true model-risk minimizer.\n\\end{theorem}\n\\begin{proof}\nAdd and subtract $\\sum_\\omega p(\\omega)\\ell(a,\\widehat v_\\omega)$. The response term has absolute value at most $Kd_0$. The expectation of a function in $[0,1]$ changes by at most total variation, giving the other term at most $e$. Risks lie in $[0,1]$, so their difference is also at most one.","roximate risk, then\n\\[\nR(\\widehat a)-\\min_a R(a)\\le\\min\\{1,2\\xi\\}.\n\\]\nIf the approximate gap between its unique minimizer and every alternative exceeds $2\\xi$, that action is also the unique true model-risk minimizer.\n\\end{theorem}\n\\begin{proof}\nAdd and subtract $\\sum_\\omega p(\\omega)\\ell(a,\\widehat v_\\omega)$. The response term has absolute value at most $Kd_0$. The expectation of a function in $[0,1]$ changes by at most total variation, giving the other term at most $e$. Risks lie in $[0,1]$, so their difference is also at most one.","he approximate gap between its unique minimizer and every alternative exceeds $2\\xi$, that action is also the unique true model-risk minimizer.\n\\end{theorem}\n\\begin{proof}\nAdd and subtract $\\sum_\\omega p(\\omega)\\ell(a,\\widehat v_\\omega)$. The response term has absolute value at most $Kd_0$. The expectation of a function in $[0,1]$ changes by at most total variation, giving the other term at most $e$. Risks lie in $[0,1]$, so their difference is also at most one.","between its unique minimizer and every alternative exceeds $2\\xi$, that action is also the unique true model-risk minimizer.\n\\end{theorem}\n\\begin{proof}\nAdd and subtract $\\sum_\\omega p(\\omega)\\ell(a,\\widehat v_\\omega)$. The response term has absolute value at most $Kd_0$. The expectation of a function in $[0,1]$ changes by at most total variation, giving the other term at most $e$. Risks lie in $[0,1]$, so their difference is also at most one.","For a true minimizer $a_*$,\n\\[\nR(\\widehat a)\\le\\widehat R(\\widehat a)+\\xi\n\\le\\widehat R(a_*)+\\xi\\le R(a_*)+2\\xi.\n\\]\nThe loss range supplies the cap one. Finally, for any other $a$, subtract the two error bounds to obtain\n$R(a)-R(\\widehat a)\\ge\\widehat R(a)-\\widehat R(\\widehat a)-2\\xi>0$.\n\\end{proof}\n\\source{[GA], ``A common quantitative grounding certificate'' and [MU], ``A joint posterior and recipient decision certificate.'' This is a specialization to a common finite candidate set and bounded loss.}","For the noisy calibration example, use zero-one loss for selecting the target bit. Under $P$ and report one, choosing one has risk $1/4$, while choosing zero has risk $3/4$. An approximation with total variation at most $1/10$ and no response error preserves the unique decision because the gap $1/2$ exceeds $2/10$. Under $P'$ the correct decision reverses. A valid error bound around the wrong prior is not a substitute for retaining the correct association.","\\section{An explanation belongs to a recipient}\nAn artifact is not an explanation merely because its producer can interpret it. Fix a finite case set $X$, a sender-access map $\\alpha:X\\to H$, and a finite artifact catalogue $\\mathcal C$. Recipient $i$ has a background record $\\beta_i(x)$, a demanded output $t_i(x)$, and a fixed bounded procedure $L_i$. A timeout, malformed artifact, wrong output, or budget overrun counts as failure.","Let $\\mathcal C(x)$ be the artifacts that succeed for every recipient in case $x$. For an attained sender value $h$, put\n\\[\n\\mathcal C_h=\\bigcap_{x:\\alpha(x)=h}\\mathcal C(x).\n\\]","\\begin{theorem}[A common artifact for fixed recipients]\\label{r:recipient}\nA sender restricted to $\\alpha(x)$ can provide one exact artifact for all recipients if and only if every $\\mathcal C_h$ is nonempty. When feasible, the smallest possible number of distinct transmitted artifacts is\n\\[\n\\min\\{|C|:C\\subseteq\\mathcal C,\\ C\\cap\\mathcal C_h\\ne\\varnothing\n\\text{ for every attained }h\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nA sender seeing $h$ must choose the same artifact for all cases in that fibre. It must therefore select a member of $\\mathcal C_h$, proving necessity. Choosing one member of each nonempty finite intersection gives a valid encoder. The image of any encoder intersects every $\\mathcal C_h$. Conversely, any catalogue meeting all those sets supplies such an encoder by choosing from the int","\\begin{theorem}[A common artifact for fixed recipients]\\label{r:recipient}\nA sender restricted to $\\alpha(x)$ can provide one exact artifact for all recipients if and only if every $\\mathcal C_h$ is nonempty. When feasible, the smallest possible number of distinct transmitted artifacts is\n\\[\n\\min\\{|C|:C\\subseteq\\mathcal C,\\ C\\cap\\mathcal C_h\\ne\\varnothing\n\\text{ for every attained }h\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nA sender seeing $h$ must choose the same artifact for all cases in that fibre. It must therefore select a member of $\\mathcal C_h$, proving necessity. Choosing one member of each nonempty finite intersection gives a valid encoder. The image of any encoder intersects every $\\mathcal C_h$. Conversely, any catalogue meeting all those sets supplies such an encoder by choosing from the intersections. Finite minimization gives the formula.\n\\end{proof}\n\\source{[MU], ``The fix","\\begin{theorem}[A common artifact for fixed recipients]\\label{r:recipient}\nA sender restricted to $\\alpha(x)$ can provide one exact artifact for all recipients if and only if every $\\mathcal C_h$ is nonempty. When feasible, the smallest possible number of distinct transmitted artifacts is\n\\[\n\\min\\{|C|:C\\subseteq\\mathcal C,\\ C\\cap\\mathcal C_h\\ne\\varnothing\n\\text{ for every attained }h\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nA sender seeing $h$ must choose the same artifact for all cases in that fibre. It must therefore select a member of $\\mathcal C_h$, proving necessity. Choosing one member of each nonempty finite intersection gives a valid encoder. The image of any encoder intersects every $\\mathcal C_h$. Conversely, any catalogue meeting all those sets supplies such an encoder by choosing from the intersections. Finite minimization gives the formula.\n\\end{proof}\n\\source{[MU], ``The fixed-recipient covering criterion.'' The earlier designed-decoder coloring criterion is not substituted for the installed-recipient problem.}","fixed recipients]\\label{r:recipient}\nA sender restricted to $\\alpha(x)$ can provide one exact artifact for all recipients if and only if every $\\mathcal C_h$ is nonempty. When feasible, the smallest possible number of distinct transmitted artifacts is\n\\[\n\\min\\{|C|:C\\subseteq\\mathcal C,\\ C\\cap\\mathcal C_h\\ne\\varnothing\n\\text{ for every attained }h\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nA sender seeing $h$ must choose the same artifact for all cases in that fibre. It must therefore select a member of $\\mathcal C_h$, proving necessity. Choosing one member of each nonempty finite intersection gives a valid encoder. The image of any encoder intersects every $\\mathcal C_h$. Conversely, any catalogue meeting all those sets supplies such an encoder by choosing from the intersections. Finite minimization gives the formula.\n\\end{proof}\n\\source{[MU], ``The fixed-recipient covering criterion.'' The earlier designed-decoder coloring criterion is not substituted for the installed-recipient problem.}","ct for all recipients if and only if every $\\mathcal C_h$ is nonempty. When feasible, the smallest possible number of distinct transmitted artifacts is\n\\[\n\\min\\{|C|:C\\subseteq\\mathcal C,\\ C\\cap\\mathcal C_h\\ne\\varnothing\n\\text{ for every attained }h\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nA sender seeing $h$ must choose the same artifact for all cases in that fibre. It must therefore select a member of $\\mathcal C_h$, proving necessity. Choosing one member of each nonempty finite intersection gives a valid encoder. The image of any encoder intersects every $\\mathcal C_h$. Conversely, any catalogue meeting all those sets supplies such an encoder by choosing from the intersections. Finite minimization gives the formula.\n\\end{proof}\n\\source{[MU], ``The fixed-recipient covering criterion.'' The earlier designed-decoder coloring criterion is not substituted for the installed-recipient problem.}","in\\{|C|:C\\subseteq\\mathcal C,\\ C\\cap\\mathcal C_h\\ne\\varnothing\n\\text{ for every attained }h\\}.\n\\]\n\\end{theorem}\n\\begin{proof}\nA sender seeing $h$ must choose the same artifact for all cases in that fibre. It must therefore select a member of $\\mathcal C_h$, proving necessity. Choosing one member of each nonempty finite intersection gives a valid encoder. The image of any encoder intersects every $\\mathcal C_h$. Conversely, any catalogue meeting all those sets supplies such an encoder by choosing from the intersections. Finite minimization gives the formula.\n\\end{proof}\n\\source{[MU], ``The fixed-recipient covering criterion.'' The earlier designed-decoder coloring criterion is not substituted for the installed-recipient problem.}","\\begin{counterexample}[Individual acceptability is not a common message]\\label{ce:message}\nTwo cases have the same sender record. In one, the sole acceptable artifact is $c_0$. In the other it is $c_1\\ne c_0$. Each case has a valid explanation if its hidden identity is known. Their intersection is empty, so the sender cannot select a valid common explanation from its actual information.\n\\end{counterexample}","In the sensor, a sender observing $y$ can transmit one bit to a recipient holding $s$. The recipient's fixed XOR procedure returns $b=y\\oplus s$. The sender need not know $b$: its message combines with the recipient's background.","\\chapter{One investigation, one resource budget}\\label{ch:inquiry}\n\\question{A possible answer, an affordable observation, and a small program must belong to the same workflow.}\nLet $X$ be a finite nonempty state space. It includes every origin tag, permission, background record, and history variable that affects the stated behavior. An experiment $e$ is allowed on a domain $D_e\\subseteq X$, has a deterministic successor $f_e:D_e\\to X$, a report $o_e:D_e\\to O_e$, and a fixed nonnegative integer resource vector $c_e$.",", one resource budget}\\label{ch:inquiry}\n\\question{A possible answer, an affordable observation, and a small program must belong to the same workflow.}\nLet $X$ be a finite nonempty state space. It includes every origin tag, permission, background record, and history variable that affects the stated behavior. An experiment $e$ is allowed on a domain $D_e\\subseteq X$, has a deterministic successor $f_e:D_e\\to X$, a report $o_e:D_e\\to O_e$, and a fixed nonnegative integer resource vector $c_e$.","bel{ch:inquiry}\n\\question{A possible answer, an affordable observation, and a small program must belong to the same workflow.}\nLet $X$ be a finite nonempty state space. It includes every origin tag, permission, background record, and history variable that affects the stated behavior. An experiment $e$ is allowed on a domain $D_e\\subseteq X$, has a deterministic successor $f_e:D_e\\to X$, a report $o_e:D_e\\to O_e$, and a fixed nonnegative integer resource vector $c_e$.","tion, and a small program must belong to the same workflow.}\nLet $X$ be a finite nonempty state space. It includes every origin tag, permission, background record, and history variable that affects the stated behavior. An experiment $e$ is allowed on a domain $D_e\\subseteq X$, has a deterministic successor $f_e:D_e\\to X$, a report $o_e:D_e\\to O_e$, and a fixed nonnegative integer resource vector $c_e$.","A current support $V\\ne\\varnothing$ records possible states. The experiment is permitted only if $V\\subseteq D_e$. After report $o$, the possible \\emph{post-experiment} states are\n\\[\nV_{e,o}=\\{f_e(x):x\\in V,\\ o_e(x)=o\\}.\n\\]\nOnly nonempty report branches are retained. If the target concerns a past value, that value must survive as an immutable tag. Resetting the current state is not recovering its history.","The experiment catalogue and the terminal-artifact catalogue are finite, and the resource vector has a fixed finite dimension. A terminal artifact $a$ is acceptable only if every fixed recipient succeeds in every state of $V$. Its worst-case terminal cost is $g_a(V)$, computed componentwise. A controller is a literal finite tree with one record per experiment or terminal node. Stored memory counts all branches, including those not visited in one execution.","\\section{The complete finite frontier}\nLet $\\mathcal P_h(V)$ be the nondominated pairs $(v,m)$ of worst-case online resource vectors and stored-record counts attainable by successful trees with at most $h$ experiment steps. At depth zero the available pairs are\n\\[\n\\operatorname{Pareto}\\{(g_a(V),1):a\\text{ succeeds throughout }V\\}.\n\\]\nAt greater depth retain those stops and, for each uniformly permitted experiment and one feasible child pair $(v_o,m_o)$ for every nonempty branch, form\n\\[\n\\left(c_e+\\max_o v_o,\\quad1+\\sum_o m_o\\right).\n\\]\nA profile dominates another when it is no larger in every resource coordinate and strictly smaller in at least one. The maximum is componentwise. An impossible child contributes no tree. It is not a zero-cost branch.","athcal P_h(V)$ be the nondominated pairs $(v,m)$ of worst-case online resource vectors and stored-record counts attainable by successful trees with at most $h$ experiment steps. At depth zero the available pairs are\n\\[\n\\operatorname{Pareto}\\{(g_a(V),1):a\\text{ succeeds throughout }V\\}.\n\\]\nAt greater depth retain those stops and, for each uniformly permitted experiment and one feasible child pair $(v_o,m_o)$ for every nonempty branch, form\n\\[\n\\left(c_e+\\max_o v_o,\\quad1+\\sum_o m_o\\right).\n\\]\nA profile dominates another when it is no larger in every resource coordinate and strictly smaller in at least one. The maximum is componentwise. An impossible child contributes no tree. It is not a zero-cost branch.","m)$ of worst-case online resource vectors and stored-record counts attainable by successful trees with at most $h$ experiment steps. At depth zero the available pairs are\n\\[\n\\operatorname{Pareto}\\{(g_a(V),1):a\\text{ succeeds throughout }V\\}.\n\\]\nAt greater depth retain those stops and, for each uniformly permitted experiment and one feasible child pair $(v_o,m_o)$ for every nonempty branch, form\n\\[\n\\left(c_e+\\max_o v_o,\\quad1+\\sum_o m_o\\right).\n\\]\nA profile dominates another when it is no larger in every resource coordinate and strictly smaller in at least one. The maximum is componentwise. An impossible child contributes no tree. It is not a zero-cost branch.","ined pair. Removing dominated child profiles does not change feasibility under any componentwise budget.\n\\end{theorem}\n\\begin{proof}\nAt zero depth the only possibilities are acceptable terminals, whose profiles are exactly the stated stops. At positive depth a successful tree either stops or begins with one uniformly allowed experiment. Every possible report needs a successful child on precisely $V_{e,o}$. By induction those children have profiles in the lower-depth attainable set. Worst-case online cost is the root cost plus their componentwise maximum. Complete tree storage is one plus the sum of all child records.","ima do not form one controller]\\label{ce:pareto}\nThere are four immutable cases, each requiring its own terminal answer. A direct observation of the case costs four units and gives a five-record tree including four terminals. Unit-cost equality tests can identify the case in at most three tests with seven records. Terminal answers have zero online cost in this particular example. The frontier is $(4,5)$ and $(3,7)$. The mixed budget $(3,5)$ is infeasible: an all-equality successful tree needs four leaves and at least seven records. Any tree using the direct observation on a reachable path costs at least four.\n\\end{counterexample}","orm one controller]\\label{ce:pareto}\nThere are four immutable cases, each requiring its own terminal answer. A direct observation of the case costs four units and gives a five-record tree including four terminals. Unit-cost equality tests can identify the case in at most three tests with seven records. Terminal answers have zero online cost in this particular example. The frontier is $(4,5)$ and $(3,7)$. The mixed budget $(3,5)$ is infeasible: an all-equality successful tree needs four leaves and at least seven records. Any tree using the direct observation on a reachable path costs at least four.\n\\end{counterexample}","ce:pareto}\nThere are four immutable cases, each requiring its own terminal answer. A direct observation of the case costs four units and gives a five-record tree including four terminals. Unit-cost equality tests can identify the case in at most three tests with seven records. Terminal answers have zero online cost in this particular example. The frontier is $(4,5)$ and $(3,7)$. The mixed budget $(3,5)$ is infeasible: an all-equality successful tree needs four leaves and at least seven records. Any tree using the direct observation on a reachable path costs at least four.\n\\end{counterexample}","\\section{When a checked candidate can end the search}\nConsider $n\\ge1$ immutable cases with total nondestructive binary unit-cost queries. Every unit-cost repair succeeds in exactly one case, and every case has a repair. A controller must succeed certainly in every case.","\\begin{theorem}[A candidate attaining both lower bounds is optimal]\\label{r:lowerbound}\nEvery successful controller has worst-case work at least $1+\\lceil\\log_2n\\rceil$ and literal storage at least $2n-1$. A checked controller attaining both bounds and the requested horizon realizes the entire nondominated work/storage frontier.\n\\end{theorem}\n\\begin{proof}\nDistinct cases cannot end at the same successful repair leaf, so at least $n$ leaves are reachable. A binary tree of query depth $d$ has at most $2^d$ leaves, giving $d\\ge\\lceil\\log_2n\\rceil$. The terminal repair adds one unit.","ing both lower bounds is optimal]\\label{r:lowerbound}\nEvery successful controller has worst-case work at least $1+\\lceil\\log_2n\\rceil$ and literal storage at least $2n-1$. A checked controller attaining both bounds and the requested horizon realizes the entire nondominated work/storage frontier.\n\\end{theorem}\n\\begin{proof}\nDistinct cases cannot end at the same successful repair leaf, so at least $n$ leaves are reachable. A binary tree of query depth $d$ has at most $2^d$ leaves, giving $d\\ge\\lceil\\log_2n\\rceil$. The terminal repair adds one unit.","very successful controller has worst-case work at least $1+\\lceil\\log_2n\\rceil$ and literal storage at least $2n-1$. A checked controller attaining both bounds and the requested horizon realizes the entire nondominated work/storage frontier.\n\\end{theorem}\n\\begin{proof}\nDistinct cases cannot end at the same successful repair leaf, so at least $n$ leaves are reachable. A binary tree of query depth $d$ has at most $2^d$ leaves, giving $d\\ge\\lceil\\log_2n\\rceil$. The terminal repair adds one unit.","controller has worst-case work at least $1+\\lceil\\log_2n\\rceil$ and literal storage at least $2n-1$. A checked controller attaining both bounds and the requested horizon realizes the entire nondominated work/storage frontier.\n\\end{theorem}\n\\begin{proof}\nDistinct cases cannot end at the same successful repair leaf, so at least $n$ leaves are reachable. A binary tree of query depth $d$ has at most $2^d$ leaves, giving $d\\ge\\lceil\\log_2n\\rceil$. The terminal repair adds one unit.","case work at least $1+\\lceil\\log_2n\\rceil$ and literal storage at least $2n-1$. A checked controller attaining both bounds and the requested horizon realizes the entire nondominated work/storage frontier.\n\\end{theorem}\n\\begin{proof}\nDistinct cases cannot end at the same successful repair leaf, so at least $n$ leaves are reachable. A binary tree of query depth $d$ has at most $2^d$ leaves, giving $d\\ge\\lceil\\log_2n\\rceil$. The terminal repair adds one unit.","Let $l$ be the number of leaves and $i_1,i_2$ the counts of unary and binary internal nodes. Counting tree edges gives $i_1+2i_2=i_1+i_2+l-1$, hence $i_2=l-1$. Total records are $2l-1+i_1\\ge2n-1$. A successful candidate at both coordinatewise lower bounds dominates every other attainable cost profile. Its validity and horizon check make it an actual member of the frontier, leaving that pair as the whole nondominated frontier.\n\\end{proof}\n\\source{[SII], Theorem 11.1. The assumptions exclude destructive resets and repairs that succeed on several cases.}","This eight-cell example is separate from the eight-world sensor. The sensor has only four action-relevant $(b,s)$ pairs and an irrelevant nuisance coordinate. Confusing hidden-world cardinality with the number of required repair distinctions would import a false lower bound.","Use the finite instrument of Theorem~\\ref{r:instrument}, with fine belief $p$ and coarse belief $q=Cp$. At $q$, a finite set of experiments $A(q)$ is permitted, and a nonempty finite set of terminal decisions $D(q)$ is available. These catalogues must be identical for every fine belief with the same $q$. Stopping with decision $d$ has loss $\\ell_d(\\pi(x))$ in fine state $x$, so its conditional risk is $\\langle q,\\ell_d\\rangle$. An experiment has a finite nonnegative charge $c_a(q)$, identical on the fine and coarse side. All history variables affecting permissions, charges, or targets must be included in the declared state.","e finite instrument of Theorem~\\ref{r:instrument}, with fine belief $p$ and coarse belief $q=Cp$. At $q$, a finite set of experiments $A(q)$ is permitted, and a nonempty finite set of terminal decisions $D(q)$ is available. These catalogues must be identical for every fine belief with the same $q$. Stopping with decision $d$ has loss $\\ell_d(\\pi(x))$ in fine state $x$, so its conditional risk is $\\langle q,\\ell_d\\rangle$. An experiment has a finite nonnegative charge $c_a(q)$, identical on the fine and coarse side. All history variables affecting permissions, charges, or targets must be included in the declared state.","nd coarse belief $q=Cp$. At $q$, a finite set of experiments $A(q)$ is permitted, and a nonempty finite set of terminal decisions $D(q)$ is available. These catalogues must be identical for every fine belief with the same $q$. Stopping with decision $d$ has loss $\\ell_d(\\pi(x))$ in fine state $x$, so its conditional risk is $\\langle q,\\ell_d\\rangle$. An experiment has a finite nonnegative charge $c_a(q)$, identical on the fine and coarse side. All history variables affecting permissions, charges, or targets must be included in the declared state.","Let $J_h(p)$ be the least expected charge plus terminal loss attainable using at most $h$ further experiments, with terminal decisions permitted at any step. It is defined by finite observation trees, not by a hidden-state controller. Its coarse recurrence is\n\\begin{align*}\n \\bar J_0(q)&=\\min_{d\\in D(q)}\\langle q,\\ell_d\\rangle,","Let $J_h(p)$ be the least expected charge plus terminal loss attainable using at most $h$ further experiments, with terminal decisions permitted at any step. It is defined by finite observation trees, not by a hidden-state controller. Its coarse recurrence is\n\\begin{align*}\n \\bar J_0(q)&=\\min_{d\\in D(q)}\\langle q,\\ell_d\\rangle,\\\\\n \\mathcal Q_h(q,a)&=c_a(q)+\\sum_{o:\\bar z_{a,o}(q)>0}\n       \\bar z_{a,o}(q)\\bar J_{h-1}\\bigl(\\bar\\Psi_{a,o}(q)\\bigr),\\\\\n \\bar J_h(q)&=\\min\\left\\{\\bar J_0(q),\\ \\min_{a\\in A(q)}\\mathcal Q_h(q,a)\\right\\}.\n\\end{align*}\nAn empty experiment catalogue leaves only stopping. Impossible reports do not receive invented posteriors.","\\begin{theorem}[The same compressed belief supports the same inquiry]\\label{r:belief-planning}\nUnder the exact instrument quotient and the common loss, charge, and catalogue assumptions above,\n\\[\n J_h(p)=\\bar J_h(Cp)\\quad\\text{for every prior and every finite }h.\n\\]\nEvery first-action value and optimal first-action set agrees. A minimizing coarse-belief policy produces a fine-model optimal history-based policy, including its stopping decision. Finite minimizing trees exist. The result supplies neither an efficient synthesis algorithm nor a worst-case storage or resource guarantee.\n\\end{theorem}\n\\begin{proof}\nAt zero horizon the terminal risks agree by fibrewise summation. Suppose the equality holds at $h-1$. For any permitted first experiment, Theorem~\\ref{r:instrument} gives the same probability of each report and the same coarse posterior on every positive branch. The induction hypothesis gives equal optimal continuation costs on each such branch.","tion set agrees. A minimizing coarse-belief policy produces a fine-model optimal history-based policy, including its stopping decision. Finite minimizing trees exist. The result supplies neither an efficient synthesis algorithm nor a worst-case storage or resource guarantee.\n\\end{theorem}\n\\begin{proof}\nAt zero horizon the terminal risks agree by fibrewise summation. Suppose the equality holds at $h-1$. For any permitted first experiment, Theorem~\\ref{r:instrument} gives the same probability of each report and the same coarse posterior on every positive branch. The induction hypothesis gives equal optimal continuation costs on each such branch. The first charge agrees, so the entire action value agrees. Stopping has the same value as well. Taking the minimum over the identical catalogues proves both value equality and equality of minimizing first choices.","\\section{The sensor chooses when another calibration is worth its cost}\nReturn to the sensor after display $y=0$. Conditional on that display, assign equal probability to $(b,s)=(0,0)$ and $(1,1)$, with any nuisance distribution inside each pair. Only noisy calibration and terminal bit selection are offered in this example. A calibration reports $s$ correctly with probability $3/4$, with fresh conditionally independent noise and no change to the hidden world. It costs $1/20$ in the same stated loss units as a wrong terminal choice, whose loss is one. A correct terminal choice has loss zero.","turn to the sensor after display $y=0$. Conditional on that display, assign equal probability to $(b,s)=(0,0)$ and $(1,1)$, with any nuisance distribution inside each pair. Only noisy calibration and terminal bit selection are offered in this example. A calibration reports $s$ correctly with probability $3/4$, with fresh conditionally independent noise and no change to the hidden world. It costs $1/20$ in the same stated loss units as a wrong terminal choice, whose loss is one. A correct terminal choice has loss zero.","The retained belief on the four $(b,s)$ pairs is an exact quotient of the eight-world belief. Its two diagonal masses initially equal $1/2$. After one calibration the favored target has probability $3/4$. A second matching report raises that probability to $9/10$, while conflicting reports restore $1/2$.","midrule\n0 & $1/2$ & Stop and select either bit.","1 & $3/10$ & Calibrate once, then select.","2 & $3/10$ & Calibrate once, then stop.","3 & $11/40$ & Calibrate, then use an adaptive continuation.","With three reports available, the optimal policy takes a second report after the first. If the first two agree, it stops. If they disagree, it takes the third. The disagreement probability is $3/8$. Thus the expected report count is $2+3/8=19/8$, the expected charge is $19/160$, and the error probability is $5/32$. The total is $19/160+5/32=11/40$.","\\section{An approximate model supports a different, quantitative claim}\nExact policy transport and approximate planning must not share one unqualified guarantee. Fix a finite horizon $H$ and one nonempty finite class $\\mathfrak P$ of admitted action/report trees, shared by the true and approximate models. The complete loss of a tree is the same function of its coarse-state and action/report path in both models, taking values in $[0,B]$. For example, a terminal loss in $[0,1]$ and at most $H$ charges bounded by $c_{\\max}$ give $B=1+Hc_{\\max}$.","ports a different, quantitative claim}\nExact policy transport and approximate planning must not share one unqualified guarantee. Fix a finite horizon $H$ and one nonempty finite class $\\mathfrak P$ of admitted action/report trees, shared by the true and approximate models. The complete loss of a tree is the same function of its coarse-state and action/report path in both models, taking values in $[0,B]$. For example, a terminal loss in $[0,1]$ and at most $H$ charges bounded by $c_{\\max}$ give $B=1+Hc_{\\max}$.","act policy transport and approximate planning must not share one unqualified guarantee. Fix a finite horizon $H$ and one nonempty finite class $\\mathfrak P$ of admitted action/report trees, shared by the true and approximate models. The complete loss of a tree is the same function of its coarse-state and action/report path in both models, taking values in $[0,B]$. For example, a terminal loss in $[0,1]$ and at most $H$ charges bounded by $c_{\\max}$ give $B=1+Hc_{\\max}$.","and approximate planning must not share one unqualified guarantee. Fix a finite horizon $H$ and one nonempty finite class $\\mathfrak P$ of admitted action/report trees, shared by the true and approximate models. The complete loss of a tree is the same function of its coarse-state and action/report path in both models, taking values in $[0,B]$. For example, a terminal loss in $[0,1]$ and at most $H$ charges bounded by $c_{\\max}$ give $B=1+Hc_{\\max}$.","ng must not share one unqualified guarantee. Fix a finite horizon $H$ and one nonempty finite class $\\mathfrak P$ of admitted action/report trees, shared by the true and approximate models. The complete loss of a tree is the same function of its coarse-state and action/report path in both models, taking values in $[0,B]$. For example, a terminal loss in $[0,1]$ and at most $H$ charges bounded by $c_{\\max}$ give $B=1+Hc_{\\max}$.","e one unqualified guarantee. Fix a finite horizon $H$ and one nonempty finite class $\\mathfrak P$ of admitted action/report trees, shared by the true and approximate models. The complete loss of a tree is the same function of its coarse-state and action/report path in both models, taking values in $[0,B]$. For example, a terminal loss in $[0,1]$ and at most $H$ charges bounded by $c_{\\max}$ give $B=1+Hc_{\\max}$.","Assume Theorem~\\ref{r:history-bound}'s bound $d_H$ holds uniformly over every tree in that class. An early stop is padded by a common absorbing terminal step for the history comparison. State-dependent charges are permitted when included in the same coarse-path loss. A model-dependent posterior score used as an uncharged surrogate loss would be a different problem.","\\begin{proposition}[Uniform model error controls a changed policy]\\label{r:policy-error}\nLet $J(\\pi)$ and $\\widetilde J(\\pi)$ be a tree's expected losses in the true and approximate models. Then\n\\[\n |J(\\pi)-\\widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\wid","\\begin{proposition}[Uniform model error controls a changed policy]\\label{r:policy-error}\nLet $J(\\pi)$ and $\\widetilde J(\\pi)$ be a tree's expected losses in the true and approximate models. Then\n\\[\n |J(\\pi)-\\widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+","\\begin{proposition}[Uniform model error controls a changed policy]\\label{r:policy-error}\nLet $J(\\pi)$ and $\\widetilde J(\\pi)$ be a tree's expected losses in the true and approximate models. Then\n\\[\n |J(\\pi)-\\widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself acros","\\begin{proposition}[Uniform model error controls a changed policy]\\label{r:policy-error}\nLet $J(\\pi)$ and $\\widetilde J(\\pi)$ be a tree's expected losses in the true and approximate models. Then\n\\[\n |J(\\pi)-\\widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before app","\\begin{proposition}[Uniform model error controls a changed policy]\\label{r:policy-error}\nLet $J(\\pi)$ and $\\widetilde J(\\pi)$ be a tree's expected losses in the true and approximate models. Then\n\\[\n |J(\\pi)-\\widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before applying optimality to different controllers.","\\begin{proposition}[Uniform model error controls a changed policy]\\label{r:policy-error}\nLet $J(\\pi)$ and $\\widetilde J(\\pi)$ be a tree's expected losses in the true and approximate models. Then\n\\[\n |J(\\pi)-\\widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before applying optimality to different controllers. A zero-probability bad event in one model can have positive probability in the…","widetilde J(\\pi)|\\le Bd_H\\quad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before applying optimality to different controllers. A zero-probability bad event in one model can have positive probability in the other, so the average bound does not certify sure feasibility.\n\\end{proof}\n\\source{Direct combination of the whole-history bound with the selection argument already used in Theorem~\\ref{r:risk}. [AVOC], Sect…","uad(\\pi\\in\\mathfrak P).\n\\]\nIf $\\widetilde\\pi$ minimizes the approximate loss and $\\pi_*$ minimizes the true loss over this same class, then\n\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before applying optimality to different controllers. A zero-probability bad event in one model can have positive probability in the other, so the average bound does not certify sure feasibility.\n\\end{proof}\n\\source{Direct combination of the whole-history bound with the selection argument already used in Theorem~\\ref{r:risk}. [AVOC], Sections 8--10, provide the sou…","\\[\n J(\\widetilde\\pi)-J(\\pi_*)\\le\\min\\{B,2Bd_H\\}.\n\\]\nThese conclusions do not certify a zero-risk constraint or preservation of admissibility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before applying optimality to different controllers. A zero-probability bad event in one model can have positive probability in the other, so the average bound does not certify sure feasibility.\n\\end{proof}\n\\source{Direct combination of the whole-history bound with the selection argument already used in Theorem~\\ref{r:risk}. [AVOC], Sections 8--10, provide the source context for guarded planning and model-error value bounds. This finite fixed-policy-class formulation is stated and proved in full here.…","ility outside the common class.\n\\end{proposition}\n\\begin{proof}\nThe expectation of a function in $[0,B]$ changes by at most $B$ times total variation. Apply the common-policy history bound separately to each fixed tree, proving the first assertion uniformly. Now\n\\[\n J(\\widetilde\\pi)\\le\\widetilde J(\\widetilde\\pi)+Bd_H\n \\le\\widetilde J(\\pi_*)+Bd_H\\le J(\\pi_*)+2Bd_H.\n\\]\nThe loss range gives the cap $B$. Each tree is first compared with itself across models, before applying optimality to different controllers. A zero-probability bad event in one model can have positive probability in the other, so the average bound does not certify sure feasibility.\n\\end{proof}\n\\source{Direct combination of the whole-history bound with the selection argument already used in Theorem~\\ref{r:risk}. [AVOC], Sections 8--10, provide the source context for guarded planning and model-error value bounds. This finite fixed-policy-class formulation is stated and proved in full here.}","With true calibration accuracy $37/50$ and planning accuracy $3/4$, the one-step defect is $1/100$. Taking $H=3$ and $B=23/20$ gives the uniform regret bound $69/1000$. A computed performance value does not replace that guarantee. Expected-risk and hard-budget claims remain distinct.","\\chapter{Improving discovery without changing its meaning}\\label{ch:discovery}\n\\question{Removing redundant work is an improvement only when the removed work could not change the justified result.}\nLet $\\mathcal G=(p_0,\\ldots,p_{N-1})$ be a finite ordered grammar of pure, total candidate programs. Let $D$ be a finite context set and $t:D\\to Y$ a fixed labeling. A teacher either accepts a candidate that agrees everywhere or returns a context where it disagrees. The teacher's work is part of the development account.","\\begin{proposition}[What finite counterexamples establish]\\label{r:cegis}\nThe learner never submits the same rejected candidate twice. Within at most $N$ teacher calls, it returns the earliest globally correct candidate in the declared grammar or exhausts that grammar. A timeout before this process finishes is neither acceptance nor exhaustion.\n\\end{proposition}\n\\begin{proof}\nEach counterexample refutes the submitted program and remains a constraint forever. That program cannot be submitted again. A globally correct candidate never violates a truthful counterexample. With only $N$ candidates, either an accepted one is found or all are ruled out. If the accepted candidate were preceded by a globally correct candidate, that earlier candidate would still be consistent and would have been selected first. This proves the ordering claim. Interrup","arliest globally correct candidate in the declared grammar or exhausts that grammar. A timeout before this process finishes is neither acceptance nor exhaustion.\n\\end{proposition}\n\\begin{proof}\nEach counterexample refutes the submitted program and remains a constraint forever. That program cannot be submitted again. A globally correct candidate never violates a truthful counterexample. With only $N$ candidates, either an accepted one is found or all are ruled out. If the accepted candidate were preceded by a globally correct candidate, that earlier candidate would still be consistent and would have been selected first. This proves the ordering claim. Interruption changes none of these logical implications and cannot supply a completed search result.\n\\end{proof}\n\\source{[SII], Theorem 15.1. This is finite exact synthesis in a supplied language, not a statistical generalization guarantee.}","Agreement on $D$ is not validity outside $D$. That distinction is especially important when a candidate is used to prune future searches. A wrong proposal can be rejected by a final checker. A wrong pruning rule may remove the only candidate the checker ought to see.","\\section{A learned search representation can preserve the exact investigation}\nSuppose a deterministic normalizing map $n$ is sound in the following direction:\n\\[\nn(p)=n(q)\\ \\Longrightarrow\\ p(z)=q(z)\\quad\\text{for every admitted context }z.\n\\]\nThe map need not detect every semantic equivalence or produce a mathematically canonical normal form. Each successful contextual rewrite must be justified under the actual pure, total semantics. Keep the earliest original AST for every attained normal form, not a replacement AST with an uncharged new interpreter.","\\begin{counterexample}[A probe-equivalent equation is not a general identity]\\label{ce:probe}\nThe expressions $\\max(x,1)$ and $x$ agree for probes $x\\in\\{1,2,3\\}$. At zero they differ. Grouping them by those probe values alone can remove a needed candidate. The identity $\\max(x,0)=x$ is valid over nonnegative integers, but not over all integers. The semantic domain is part of the equation's certificate.\n\\end{counterexample}","\\section{A reusable rule for stopping an unpromising read}\nA diagnostic compiler inspects a binary query on $m$ immutable cases. In the prefix already read, $a$ cases give zero and $b$ give one. Each child is allowed at most $C$ cases because the remaining binary depth has that leaf capacity. Assume integers\n\\[\nC\\ge1,\\quad 2\\le m\\le2C,\\quad a,b\\ge0,\\quad a+b\\le m.\n\\]","\\begin{proposition}[A sound prefix exclusion]\\label{r:prefix}\nThe unread reports can be completed with both final branch counts at most $C$ if and only if $\\max(a,b)\\le C$. A scan may therefore stop when one count first exceeds $C$. In a fixed ordered depth-first compiler with identical scope, terminal, and fallback rules, replacing full root scans by this early rejection preserves the first returned tree or the completed depth-bounded failure result.\n\\end{proposition}\n\\begin{proof}\nCounts only increase, so an overfull child cannot be rescued. Otherwise the two remaining capacities are $C-a$ and $C-b$, both nonnegative. Their sum $2C-a-b$ is at least the number $m-a-b$ of unread rows. Distributing the unread cases between those capacities gives a feasible completion.","stop when one count first exceeds $C$. In a fixed ordered depth-first compiler with identical scope, terminal, and fallback rules, replacing full root scans by this early rejection preserves the first returned tree or the completed depth-bounded failure result.\n\\end{proposition}\n\\begin{proof}\nCounts only increase, so an overfull child cannot be rescued. Otherwise the two remaining capacities are $C-a$ and $C-b$, both nonnegative. Their sum $2C-a-b$ is at least the number $m-a-b$ of unread rows. Distributing the unread cases between those capacities gives a feasible completion.","one count first exceeds $C$. In a fixed ordered depth-first compiler with identical scope, terminal, and fallback rules, replacing full root scans by this early rejection preserves the first returned tree or the completed depth-bounded failure result.\n\\end{proposition}\n\\begin{proof}\nCounts only increase, so an overfull child cannot be rescued. Otherwise the two remaining capacities are $C-a$ and $C-b$, both nonnegative. Their sum $2C-a-b$ is at least the number $m-a-b$ of unread rows. Distributing the unread cases between those capacities gives a feasible completion.","ixed ordered depth-first compiler with identical scope, terminal, and fallback rules, replacing full root scans by this early rejection preserves the first returned tree or the completed depth-bounded failure result.\n\\end{proposition}\n\\begin{proof}\nCounts only increase, so an overfull child cannot be rescued. Otherwise the two remaining capacities are $C-a$ and $C-b$, both nonnegative. Their sum $2C-a-b$ is at least the number $m-a-b$ of unread rows. Distributing the unread cases between those capacities gives a feasible completion.","r with identical scope, terminal, and fallback rules, replacing full root scans by this early rejection preserves the first returned tree or the completed depth-bounded failure result.\n\\end{proposition}\n\\begin{proof}\nCounts only increase, so an overfull child cannot be rescued. Otherwise the two remaining capacities are $C-a$ and $C-b$, both nonnegative. Their sum $2C-a-b$ is at least the number $m-a-b$ of unread rows. Distributing the unread cases between those capacities gives a feasible completion.","For $C=4$, a fifth zero already closes a root. Reading more rows cannot reduce that branch. By contrast, a four-versus-four split merely survives this necessary test. If no remaining query distinguishes either child, the root still fails.","\\section{Reuse follows semantic dependencies}\nLet an artifact state be $a=(a_1,\\ldots,a_d)$. A finite directed acyclic graph represents claimed guarantees. For vertex $v$, let $D_v$ be its direct artifact dependencies and let $\\operatorname{pa}(v)$ be its parent guarantees. The declaration is semantically complete if the meaning of its claim and local verification rule is determined by those components and parent claims. External assumptions are included explicitly as premises.","After changing $a$ to $a'$, mark a vertex stale when a direct input changes and propagate staleness to all descendants. Call the resulting set $\\Delta$.","\\begin{theorem}[Selective reuse with complete dependencies]\\label{r:dependencies}\nAssume the graph is finite and acyclic, direct dependencies are semantically complete, local rules are sound, and the applicable external premises hold. Assume also that every old witness established its old claim before the revision. Old witnesses still establish every claim outside $\\Delta$. Re-establishing affected claims in topological order, using revised values and established parents, restores the corresponding revised claims. Without an affected parent's re-establishment, this reuse rule supplies no justification for its dependent claim.\n\\end{theorem}\n\\begin{proof}\nA vertex outside $\\Delta$ has unchanged direct values and no stale parent. In topological order, induction shows that every parent retains its meaning and proof. Completeness of the dependency declaration leaves the local argument applicable. For a stale vertex, once its revised parents are established, soundness of the newly checked local rule gives the revised conclusion. A second topological induction prov","d its old claim before the revision. Old witnesses still establish every claim outside $\\Delta$. Re-establishing affected claims in topological order, using revised values and established parents, restores the corresponding revised claims. Without an affected parent's re-establishment, this reuse rule supplies no justification for its dependent claim.\n\\end{theorem}\n\\begin{proof}\nA vertex outside $\\Delta$ has unchanged direct values and no stale parent. In topological order, induction shows that every parent retains its meaning and proof. Completeness of the dependency declaration leaves the local argument applicable. For a stale vertex, once its revised parents are established, soundness of the newly checked local rule gives the revised conclusion. A second topological induction proves restoration. An unchecked parent leaves a missing premise, so the rule alone cannot justify a dependent conclusion. This is not a proof that the conclusion is false.\n\\end{proof}\n\\source{[SII], Theorem 24.1. It is a theorem about a declared complete dependency graph, not a method that infers complete de…","\\begin{counterexample}[A missing dependency defeats a perfect replay]\\label{ce:dependency}\nA stored claim is $u=v$, initially with both values zero. Its registry records only $u$. Change $v$ to one. A flawless comparison with the recorded $u$ still passes, while the claim is false. A hash or replay cannot recover the premise the registry omitted.\n\\end{counterexample}","\\section{Semantic improvement and resource improvement are separate}\nFix named additive event types. Let $K\\ge0$ be an extra preparation vector, let $s_i$ be the old-minus-new per-use vector for workload type $i$, and let $n_i$ count its uses. The net event difference is\n\\[\n\\Delta=\\sum_i n_i s_i-K.\n\\]\nA nonnegative price vector $w$ converts these events into one declared additive measure. Storage capacity, peak memory, and elapsed time need their own models unless a justified conversion is specified.","\\begin{proposition}[Which gains survive every admitted price?]\\label{r:prices}\nThe saving $w\\cdot\\Delta$ is nonnegative for every nonnegative price vector if and only if every coordinate of $\\Delta$ is nonnegative. If prices lie in the convex hull of a finite nonempty set $W$, their minimum saving equals $\\min_{w\\in W}w\\cdot\\Delta$.\n\\end{proposition}\n\\begin{proof}\nA componentwise nonnegative vector has nonnegative inner product with every nonnegative vector. If coordinate $j$ is negative, assigning price one only to that coordinate gives a negative result. This proves the first assertion. A point in the stated convex hull is a convex combination of its listed prices, so its saving is the same convex combination of their savings. It cannot be smaller than the minimum listed value, and a listed","\\begin{proposition}[Which gains survive every admitted price?]\\label{r:prices}\nThe saving $w\\cdot\\Delta$ is nonnegative for every nonnegative price vector if and only if every coordinate of $\\Delta$ is nonnegative. If prices lie in the convex hull of a finite nonempty set $W$, their minimum saving equals $\\min_{w\\in W}w\\cdot\\Delta$.\n\\end{proposition}\n\\begin{proof}\nA componentwise nonnegative vector has nonnegative inner product with every nonnegative vector. If coordinate $j$ is negative, assigning price one only to that coordinate gives a negative result. This proves the first assertion. A point in the stated convex hull is a convex combination of its listed prices, so its saving is the same convex combination of their savings. It cannot be smaller than the minimum listed value, and a listed minimizer attains that value.\n\\end{proof}\n\\source{[SII], Proposition 25.1. No hardware perfor","\\begin{proposition}[Which gains survive every admitted price?]\\label{r:prices}\nThe saving $w\\cdot\\Delta$ is nonnegative for every nonnegative price vector if and only if every coordinate of $\\Delta$ is nonnegative. If prices lie in the convex hull of a finite nonempty set $W$, their minimum saving equals $\\min_{w\\in W}w\\cdot\\Delta$.\n\\end{proposition}\n\\begin{proof}\nA componentwise nonnegative vector has nonnegative inner product with every nonnegative vector. If coordinate $j$ is negative, assigning price one only to that coordinate gives a negative result. This proves the first assertion. A point in the stated convex hull is a convex combination of its listed prices, so its saving is the same convex combination of their savings. It cannot be smaller than the minimum listed value, and a listed minimizer attains that value.\n\\end{proof}\n\\source{[SII], Proposition 25.1. No hardware performance law follows from this additive accounting identity.}","The saving $w\\cdot\\Delta$ is nonnegative for every nonnegative price vector if and only if every coordinate of $\\Delta$ is nonnegative. If prices lie in the convex hull of a finite nonempty set $W$, their minimum saving equals $\\min_{w\\in W}w\\cdot\\Delta$.\n\\end{proposition}\n\\begin{proof}\nA componentwise nonnegative vector has nonnegative inner product with every nonnegative vector. If coordinate $j$ is negative, assigning price one only to that coordinate gives a negative result. This proves the first assertion. A point in the stated convex hull is a convex combination of its listed prices, so its saving is the same convex combination of their savings. It cannot be smaller than the minimum listed value, and a listed minimizer attains that value.\n\\end{proof}\n\\source{[SII], Proposition 25.1. No hardware performance law follows from this additive accounting identity.}","For the source's checked weighted-search comparison, the event difference is $(1552,1552,-1868)$ in reads, rational additions, and comparisons. Equal prices give $1236$ saved units. Prices $(1,0,1)$ give a loss of $316$. Both conclusions concern the same valid controller. The main text uses this as a transparently rechecked source calculation, not as a new measured timing result.","\\begin{proposition}[A conditional error budget for a lineage]\\label{r:lineage}\nLet $\\mathcal H_{j-1}$ contain the information available before attempt $j$. Suppose the event $A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes","\\begin{proposition}[A conditional error budget for a lineage]\\label{r:lineage}\nLet $\\mathcal H_{j-1}$ contain the information available before attempt $j$. Suppose the event $A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes attempted events and cannot enlarge their union.\n\\en","\\begin{proposition}[A conditional error budget for a lineage]\\label{r:lineage}\nLet $\\mathcal H_{j-1}$ contain the information available before attempt $j$. Suppose the event $A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes attempted events and cannot enlarge their union.\n\\end{proof}\n\\source{[SII], P","\\begin{proposition}[A conditional error budget for a lineage]\\label{r:lineage}\nLet $\\mathcal H_{j-1}$ contain the information available before attempt $j$. Suppose the event $A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes attempted events and cannot enlarge their union.\n\\end{proof}\n\\source{[SII], Proposition 26.1. The mathematical input is conditional validity of each actual attempt. A numerical error schedule alone does not establish that","\\begin{proposition}[A conditional error budget for a lineage]\\label{r:lineage}\nLet $\\mathcal H_{j-1}$ contain the information available before attempt $j$. Suppose the event $A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes attempted events and cannot enlarge their union.\n\\end{proof}\n\\source{[SII], Proposition 26.1. The mathematical input is conditional validity of each actual attempt. A numerical error schedule alone does not establish that premise.}","rror budget for a lineage]\\label{r:lineage}\nLet $\\mathcal H_{j-1}$ contain the information available before attempt $j$. Suppose the event $A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes attempted events and cannot enlarge their union.\n\\end{proof}\n\\source{[SII], Proposition 26.1. The mathematical input is conditional validity of each actual attempt. A numerical error schedule alone does not establish that premise.}","$A_j$ of accepting a false claim satisfies\n\\[\n\\Pr(A_j\\mid\\mathcal H_{j-1})\\le\\delta_j\\quad\\text{almost surely},\n\\qquad\\sum_{j\\ge1}\\delta_j\\le\\delta,\n\\]\nfor deterministic nonnegative $\\delta_j$. Then the probability of ever accepting any false claim is at most $\\delta$, even when candidates and the stopping time depend on earlier history.\n\\end{proposition}\n\\begin{proof}\nTaking expectations gives $\\Pr(A_j)\\le\\delta_j$. A union bound gives the result for every finite set of attempts. The finite unions increase to the event of any false acceptance, so continuity of probability gives the infinite statement. Stopping removes attempted events and cannot enlarge their union.\n\\end{proof}\n\\source{[SII], Proposition 26.1. The mathematical input is conditional validity of each actual attempt. A numerical error schedule alone does not establish that premise.}","\\section{What several accessible records can reconstruct}\nLet $V$ be a finite-dimensional vector space over a field. Participant $i$ has a linear record $A_i v$. Stack the records into $Av=(A_1v,\\ldots,A_mv)$, and let $Qv$ be the protected linear target.","\\begin{proposition}[Target-relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{p","\\begin{proposition}[Target-relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\sourc","\\begin{proposition}[Target-relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the","\\begin{proposition}[Target-relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficien","\\begin{proposition}[Target-relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]","\\begin{proposition}[Target-relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}","relative joint reconstruction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}","ction]\\label{r:reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}",":reconstruction}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}","ion}\nThere is a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}","a linear decoder $D$ with $Q=DA$ if and only if $\\ker A\\subseteq\\ker Q$. Full-state reconstruction is possible exactly when $A$ is injective. The intersection identity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}","ity $\\ker A=\\bigcap_i\\ker A_i$ determines which ambiguities sharing removes.\n\\end{proposition}\n\\begin{proof}\nIf $Q=DA$, a vector annihilated by $A$ is also annihilated by $Q$. Conversely, define $D(Av)=Qv$ on $\\im A$. If $Av=Aw$, then $v-w\\in\\ker A\\subseteq\\ker Q$, so this definition is independent of the representative. It is linear and extends to the full record space by extending a basis. For $Q=I_V$ the kernel condition is injectivity. A stacked record vanishes exactly when each component record vanishes, giving the intersection identity.\n\\end{proof}\n\\source{Target-relative specialization of the reconstruction and factorization machinery in Sufficient Resolution [SR] and Stratified Relational Workspace Theory [SRWT]. The proof here does not import their complete noise, dynamics, or identity theories.}","Over $\\mathbb F_2$, Agent A observes $A_A(b,s)=b+s$ and Agent B observes $A_B(b,s)=s$. The joint matrix is invertible. Either agent alone lacks a direction needed to recover $b$. The nuisance bit may remain unknown because the target map ignores it.","\\section{Permissions are part of an operational state}\nNow let a finite institutional state $x$ have a protected label $\\ell(x)$, an enabled-action set $D(x)$ within a fixed finite alphabet, and a deterministic successor $F(x,a)$ for enabled actions. Labels may include resource balances, outstanding obligations, and offices, but need not retain every historical byte.","\\begin{theorem}[A quotient that preserves legal operations]\\label{r:governance}\nFor a surjection $\\pi:X\\to Z$, quotient labels, enabled sets, and successors satisfying\n\\[\n\\ell=\\bar\\ell\\pi,\\qquad D=\\bar D\\pi,\\qquad\n\\pi F(x,a)=\\bar F(\\pi x,a)\n\\]\nexist exactly when equal fibres have equal labels, equal enabled sets, and equal projected successors for every enabled action. The quotient labels, enabled sets, and successors on enabled actions are unique. They preserve every legal finite action and label sequence. Values assigned by a totalized implementation to disabled actions are outside this uniqueness statement.\n\\end{theorem}\n\\begin{proof}\nFactoring each quantity proves necessity. For sufficiency choose a representative of each fibre to define its label, enabled set, and successor. The three equalit","\\begin{theorem}[A quotient that preserves legal operations]\\label{r:governance}\nFor a surjection $\\pi:X\\to Z$, quotient labels, enabled sets, and successors satisfying\n\\[\n\\ell=\\bar\\ell\\pi,\\qquad D=\\bar D\\pi,\\qquad\n\\pi F(x,a)=\\bar F(\\pi x,a)\n\\]\nexist exactly when equal fibres have equal labels, equal enabled sets, and equal projected successors for every enabled action. The quotient labels, enabled sets, and successors on enabled actions are unique. They preserve every legal finite action and label sequence. Values assigned by a totalized implementation to disabled actions are outside this uniqueness statement.\n\\end{theorem}\n\\begin{proof}\nFactoring each quantity proves necessity. For sufficiency choose a representative of each fibre to define its label, enabled set, and successor. The three equalities make those definitions independent of the representative. Surjectivity gives uniqueness. Equal enabled sets admit the same first","\\section{Continuing should not erase the ability to finish}\nUse a finite \\emph{observed} configuration space $Z$. Under partial observation, the configurations would instead have to be justified public possibility states. At $z$ there are finite catalogues of running actions and ending actions, each with a nonempty successor set. A declared safe set is $S$, and an ending target is $G\\subseteq S$. A safe hold action is available at $G$.","the ability to finish}\nUse a finite \\emph{observed} configuration space $Z$. Under partial observation, the configurations would instead have to be justified public possibility states. At $z$ there are finite catalogues of running actions and ending actions, each with a nonempty successor set. A declared safe set is $S$, and an ending target is $G\\subseteq S$. A safe hold action is available at $G$.","A stop command is assumed authentic and observable before the next action is selected. Its delivery and authentication are premises, not consequences of the following fixed-point calculation. Define\n\\begin{align*}\nT_0&=G,\\\\\nT_{k+1}&=T_k\\cup\\{z\\in S:\\exists a\\in A_{\\rm end}(z),\\ \\operatorname{Post}(z,a)\\subseteq T_k\\}.\n\\end{align*}\nLet $T_*$ be the stable set. Then define\n\\[\nL_0=T_*,\\qquad\nL_{k+1}=\\{z\\in L_k:\\exists a\\in A_{\\rm run}(z),\\ \\operatorname{Post}(z,a)\\subseteq L_k\\},\n\\]\nand let $L_*$ be its stable value. Both sequences stabilize because $Z$ is finite.","T_{k+1}&=T_k\\cup\\{z\\in S:\\exists a\\in A_{\\rm end}(z),\\ \\operatorname{Post}(z,a)\\subseteq T_k\\}.\n\\end{align*}\nLet $T_*$ be the stable set. Then define\n\\[\nL_0=T_*,\\qquad\nL_{k+1}=\\{z\\in L_k:\\exists a\\in A_{\\rm run}(z),\\ \\operatorname{Post}(z,a)\\subseteq L_k\\},\n\\]\nand let $L_*$ be its stable value. Both sequences stabilize because $Z$ is finite.","\\begin{theorem}[Safe continuation that retains an ending strategy]\\label{r:ending}\nA starting configuration belongs to $L_*$ exactly when one policy can run safely indefinitely without a stop command and, after a stop command at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a succes","\\begin{theorem}[Safe continuation that retains an ending strategy]\\label{r:ending}\nA starting configuration belongs to $L_*$ exactly when one policy can run safely indefinitely without a stop command and, after a stop command at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","e continuation that retains an ending strategy]\\label{r:ending}\nA starting configuration belongs to $L_*$ exactly when one policy can run safely indefinitely without a stop command and, after a stop command at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","ng}\nA starting configuration belongs to $L_*$ exactly when one policy can run safely indefinitely without a stop command and, after a stop command at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","afely indefinitely without a stop command and, after a stop command at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","stop command at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","nd at any decision epoch, reach $G$ on every possible successor path using only the admitted ending actions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","ions. Before the command a policy preserving $L_*$ suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","suffices. After the command it can reach $G$ within the first-entry rank of the current state in the $T_k$ sequence.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","nce.\n\\end{theorem}\n\\begin{proof}\nInduction shows that membership in $T_k$ is equivalent to an ending-only policy reaching $G$ within $k$ transitions while staying safe. The chosen action moves every successor to a smaller rank. Outside $T_*$, every ending action has a successor outside $T_*$ or outside safety. Choosing such a successor at every step defeats any purported surely safe ending. Thus $T_*$ is exactly the sure-ending region.","At $L_*$ a running action keeps every successor inside $L_*$. Repeatedly using such actions is safe and retains membership in $T_*$. A command can therefore switch to the decreasing-rank ending policy at any epoch.","Conversely, if a policy has the stated guarantee, every pre-command reachable configuration must admit an ending immediately and hence lie in $T_*$. For any finite number of further running steps, it must keep all paths in that set. Induction in the decreasing construction places the initial state in every $L_k$, hence in $L_*$. No action is chosen after secretly observing its successor.\n\\end{proof}\n\\source{Fully observed finite specialization of [MS], ``Continued operation with an available authorized ending.'' The more general public-possibility construction is not silently assumed in this version.}","\\begin{counterexample}[Safe inactivity without an ending]\\label{ce:ending}\nA state $z$ has a safe idle self-loop and no ending action reaching $G$. It is safe forever but outside $T_*$. In a second state $u$, an ending is available, while the only running action spends its reserved resource and moves to $z$. Then $u$ belongs to $T_*$ but not $L_*$. Being able to stop now does not establish that ordinary operation preserves that ability.\n\\end{counterexample}","Begin with eight candidate worlds $(b,s,n)$. The agent's initial display $y$ determines its two-state predictive class under flips, but not the physical target bit. The first chapter proves both claims. The second gives the operator on the retained class and the coordinate map that becomes usable after calibration. Neither requires the nuisance bit.","\\section{A worked end-to-end record}\nTake the actual illustrative world $(b,s,n)=(0,1,1)$. The complete protocol begins with all eight worlds possible:\n\\begin{center}\\begin{tabular}{p{0.25\\linewidth}p{0.27\\linewidth}p{0.35\\linewidth}}\\toprule\nEvent & Remaining worlds & What is justified","Display $y=1$ & Four worlds & Predict every later display flip.","Calibration $s=1$ & Two worlds differing only in $n$ & Physical target $b=0$.","Send $y$ to a recipient holding $s$ & Same two worlds & Recipient computes $b=0$ through its fixed XOR decoder.","\\section{Finite game data and the command boundary}\nFix a finite, fully observed configuration set $Z$. Each $z\\in Z$ has finite collections $A_{\\rm run}(z)$ of running actions and $A_{\\rm end}(z)$ of ending actions. For each admitted action $a$ there is a nonempty set $\\operatorname{Post}(z,a)\\subseteq Z$ of possible successors. A controller chooses an action \\emph{before} the successor is determined. The quantifier over all successors is adversarial and is stronger than an expected-value objective.","Let $S\\subseteq Z$ be a safe set and $G\\subseteq S$ an ending target. At every state of $G$, a hold operation with successors contained in $G$ is available after ending. Running forever means that the controller can choose an admitted action at every pre-command epoch. Deadlock is not counted as successful indefinite operation. A stop command is authenticated and observed before the next action is selected. Whether such an authentication protocol exists in an external institution is not proved by the game calculation. It is part of the model supplied to this theorem.","A strategy may inspect the finite history of observed states and chosen actions. Before the stop command it selects running actions. After the command it selects ending actions. The desired guarantee consists of two clauses: every running path without a command remains safe forever, and every path following a command at any decision epoch reaches $G$ in finitely many ending steps while remaining safe. The strategy must satisfy both clauses under the same command-conditioned policy.","\\section{The sure-ending attractor}\nDefine\n\\begin{align*}\nT_0&=G,\\\\\nT_{k+1}&=T_k\\cup\\left\\{z\\in S:\\text{some }a\\in A_{\\rm end}(z)\n\\text{ satisfies }\\operatorname{Post}(z,a)\\subseteq T_k\\right\\}.\n\\end{align*}\nThe sequence is increasing and every term is contained in $S$. Because the state set is finite, it becomes constant after finitely many strict additions. Let $T_*$ denote its stable value. For $z\\in T_*$, define the first-entry rank $\\rho(z)$ as the least $k$ with $z\\in T_k$.","T_{k+1}&=T_k\\cup\\left\\{z\\in S:\\text{some }a\\in A_{\\rm end}(z)\n\\text{ satisfies }\\operatorname{Post}(z,a)\\subseteq T_k\\right\\}.\n\\end{align*}\nThe sequence is increasing and every term is contained in $S$. Because the state set is finite, it becomes constant after finitely many strict additions. Let $T_*$ denote its stable value. For $z\\in T_*$, define the first-entry rank $\\rho(z)$ as the least $k$ with $z\\in T_k$.","}&=T_k\\cup\\left\\{z\\in S:\\text{some }a\\in A_{\\rm end}(z)\n\\text{ satisfies }\\operatorname{Post}(z,a)\\subseteq T_k\\right\\}.\n\\end{align*}\nThe sequence is increasing and every term is contained in $S$. Because the state set is finite, it becomes constant after finitely many strict additions. Let $T_*$ denote its stable value. For $z\\in T_*$, define the first-entry rank $\\rho(z)$ as the least $k$ with $z\\in T_k$.","\\begin{lemma}[Finite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ ret","\\begin{lemma}[Finite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strateg","\\begin{lemma}[Finite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or","\\begin{lemma}[Finite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first e","\\begin{lemma}[Finite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continua","\\begin{lemma}[Finite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible succes","inite-horizon ending criterion]\\label{r:end-horizon}\nA configuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At th…","figuration $z$ belongs to $T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends e…","$T_k$ exactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\r","xactly when there exists an ending-only strategy that, whatever the successors, reaches $G$ within at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1…","hin at most $k$ transitions while remaining in $S$. In particular, if $\\rho(z)>0$, there is an ending action whose every successor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","ccessor has strictly smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","y smaller first-entry rank.\n\\end{lemma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","mma}\n\\begin{proof}\nThe assertion at $k=0$ says precisely that the initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","he initial state is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","e is already in $G$. Assume it holds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","lds for $k$. If $z\\in T_{k+1}\\setminus T_k$, the construction supplies an ending action with all successors in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","rs in $T_k$. After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}",". After the action the controller uses the strategy given by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","n by induction for the observed successor. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","or. Every path reaches $G$ within another $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","other $k$ transitions. States already in $T_k$ retain their strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","r strategy. Conversely, a policy that reaches $G$ in at most $k+1$ steps must either start in $G$, or select some first ending action such that the continuation from every possible successor reaches $G$ within $k$ steps. By induction every such successor belongs to $T_k$, hence the initial state lies in $T_{k+1}$. At the first-entry stage $\\rho(z)$, the selected action sends every successor into $T_{\\rho(z)-1}$, establishing strict rank decrease.\n\\end{proof}","\\begin{lemma}[Adversarial completeness of the ending region]\\label{r:ending-complete}\nFrom a configuration outside $T_*$, no strategy, including one depending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at e","\\begin{lemma}[Adversarial completeness of the ending region]\\label{r:ending-complete}\nFrom a configuration outside $T_*$, no strategy, including one depending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the t","\\begin{lemma}[Adversarial completeness of the ending region]\\label{r:ending-complete}\nFrom a configuration outside $T_*$, no strategy, including one depending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches","\\begin{lemma}[Adversarial completeness of the ending region]\\label{r:ending-complete}\nFrom a configuration outside $T_*$, no strategy, including one depending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any s","of the ending region]\\label{r:ending-complete}\nFrom a configuration outside $T_*$, no strategy, including one depending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history…","a configuration outside $T_*$, no strategy, including one depending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history-dependent policies and requires nonempty successor …","ending on arbitrary finite observed history, can guarantee safe eventual arrival at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history-dependent policies and requires nonempty successor sets so that the proposed actions represent actual transition…","l at $G$ along every successor path.\n\\end{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history-dependent policies and requires nonempty successor sets so that the proposed actions represent actual transitions.\n\\end{proof}","d{lemma}\n\\begin{proof}\nIf the initial configuration is outside $S$, safety has already failed. Otherwise, because $T_*$ is a fixed point, no ending action at a state $z\\in S\\setminus T_*$ can have all its successors in $T_*$. An adversary can therefore select a successor outside $T_*$ after every proposed ending action. If the selected successor is outside $S$, safety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history-dependent policies and requires nonempty successor sets so that the proposed actions represent actual transitions.\n\\end{proof}","ety fails. If it remains within $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history-dependent policies and requires nonempty successor sets so that the proposed actions represent actual transitions.\n\\end{proof}","thin $S\\setminus T_*$, the same argument can be repeated. The adversary does not need to know the controller's future choices, only its chosen action at each current state. If the controller reaches a non-goal state with no ending action, it deadlocks before reaching the target and loses. Otherwise, either safety fails at a finite step or an infinite path never reaches $G$, which is contained in $T_*$. This defeats any sure-ending guarantee. The argument permits history-dependent policies and requires nonempty successor sets so that the proposed actions represent actual transitions.\n\\end{proof}","These lemmas show that $T_*$ is not merely the region reached by a convenient algorithm. It is \\emph{exactly} the set from which an ending strategy can guarantee success under all modeled successor choices. The rank bounds the required ending transitions by a finite number no greater than the number of strict expansion rounds.","\\section{A safety invariant that retains the ending region}\nThe ending region is not necessarily closed under ordinary running. Define the decreasing sequence\n\\[\nL_0=T_*,\\qquad\nL_{k+1}=\\left\\{z\\in L_k:\\exists a\\in A_{\\rm run}(z),\\ \\operatorname{Post}(z,a)\\subseteq L_k\\right\\}.\n\\]\nLet $L_*$ be its stable value. Each stage removes states for which ordinary operation cannot be guaranteed to remain in the preceding stage.","\\begin{lemma}[Greatest controlled invariant inside the ending region]\\label{r:running-invariant}\nThe set $L_*$ is the greatest subset of $T_*$ admitting a running action at every member with all successors again in that subset. A state belongs to $L_k$ exactly when a strategy can safely take $k$ running actions while leaving a sure-ending strategy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial","\\begin{lemma}[Greatest controlled invariant inside the ending region]\\label{r:running-invariant}\nThe set $L_*$ is the greatest subset of $T_*$ admitting a running action at every member with all successors again in that subset. A state belongs to $L_k$ exactly when a strategy can safely take $k$ running actions while leaving a sure-ending strategy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the i","\\begin{lemma}[Greatest controlled invariant inside the ending region]\\label{r:running-invariant}\nThe set $L_*$ is the greatest subset of $T_*$ admitting a running action at every member with all successors again in that subset. A state belongs to $L_k$ exactly when a strategy can safely take $k$ running actions while leaving a sure-ending strategy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely","\\begin{lemma}[Greatest controlled invariant inside the ending region]\\label{r:running-invariant}\nThe set $L_*$ is the greatest subset of $T_*$ admitting a running action at every member with all successors again in that subset. A state belongs to $L_k$ exactly when a strategy can safely take $k$ running actions while leaving a sure-ending strategy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and t","set of $T_*$ admitting a running action at every member with all successors again in that subset. A state belongs to $L_k$ exactly when a strategy can safely take $k$ running actions while leaving a sure-ending strategy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","exactly when a strategy can safely take $k$ running actions while leaving a sure-ending strategy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","tegy available at every reached decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","hed decision epoch, including the initial one.\n\\end{lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","lemma}\n\\begin{proof}\nThe bounded-horizon claim follows by induction. At horizon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","zon zero only membership in $T_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","_*$ is needed. For the next step, a first running action must send every successor to a state that permits the remaining $k$ running steps and an ending option at every intermediate epoch. Truncating a successful $(k+1)$-step strategy after $k$ steps also shows that its initial state lies in $L_k$. Thus the initial state and all successors of its first action lie in $L_k$, exactly as required by $L_{k+1}$. Conversely, membership supplies the first action and the inductive strategies from its successors.","At stabilization, the defining equation gives one running action for each state in $L_*$ whose successors stay in $L_*$. Selecting these actions proves controlled invariance. If $B\\subseteq T_*$ is any other subset with the same property, induction shows $B\\subseteq L_k$ for every $k$: inclusion at zero is given. The witnessing running action for any $z\\in B$ has all successors in $B\\subseteq L_k$, so $z\\in L_{k+1}$. Thus $B\\subseteq L_*$. No larger controlled invariant within $T_*$ exists.\n\\end{proof}","bilization, the defining equation gives one running action for each state in $L_*$ whose successors stay in $L_*$. Selecting these actions proves controlled invariance. If $B\\subseteq T_*$ is any other subset with the same property, induction shows $B\\subseteq L_k$ for every $k$: inclusion at zero is given. The witnessing running action for any $z\\in B$ has all successors in $B\\subseteq L_k$, so $z\\in L_{k+1}$. Thus $B\\subseteq L_*$. No larger controlled invariant within $T_*$ exists.\n\\end{proof}","ing equation gives one running action for each state in $L_*$ whose successors stay in $L_*$. Selecting these actions proves controlled invariance. If $B\\subseteq T_*$ is any other subset with the same property, induction shows $B\\subseteq L_k$ for every $k$: inclusion at zero is given. The witnessing running action for any $z\\in B$ has all successors in $B\\subseteq L_k$, so $z\\in L_{k+1}$. Thus $B\\subseteq L_*$. No larger controlled invariant within $T_*$ exists.\n\\end{proof}","$L_*$ whose successors stay in $L_*$. Selecting these actions proves controlled invariance. If $B\\subseteq T_*$ is any other subset with the same property, induction shows $B\\subseteq L_k$ for every $k$: inclusion at zero is given. The witnessing running action for any $z\\in B$ has all successors in $B\\subseteq L_k$, so $z\\in L_{k+1}$. Thus $B\\subseteq L_*$. No larger controlled invariant within $T_*$ exists.\n\\end{proof}","\\begin{theorem}[Safe running with an available ending at every epoch]\\label{r:joint-ending}\nA state $z$ belongs to $L_*$ if and only if a single command-conditioned policy can run safely forever in the absence of a command and, upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration.","\\begin{theorem}[Safe running with an available ending at every epoch]\\label{r:joint-ending}\nA state $z$ belongs to $L_*$ if and only if a single command-conditioned policy can run safely forever in the absence of a command and, upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop comman","\\begin{theorem}[Safe running with an available ending at every epoch]\\label{r:joint-ending}\nA state $z$ belongs to $L_*$ if and only if a single command-conditioned policy can run safely forever in the absence of a command and, upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases s","\\begin{theorem}[Safe running with an available ending at every epoch]\\label{r:joint-ending}\nA state $z$ belongs to $L_*$ if and only if a single command-conditioned policy can run safely forever in the absence of a command and, upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$","rem}[Safe running with an available ending at every epoch]\\label{r:joint-ending}\nA state $z$ belongs to $L_*$ if and only if a single command-conditioned policy can run safely forever in the absence of a command and, upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At …","unning with an available ending at every epoch]\\label{r:joint-ending}\nA state $z$ belongs to $L_*$ if and only if a single command-conditioned policy can run safely forever in the absence of a command and, upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero t…",", upon receiving an authentic stop command at any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","any decision epoch, can reach $G$ on every possible successor path using only ending actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","actions. The stopping phase from a state $w$ reaches $G$ within $\\rho(w)$ transitions. The combined policy can be chosen positional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","ional in the pair consisting of the current state and the command mode.\n\\end{theorem}\n\\begin{proof}\nFor sufficiency, at each running state in $L_*$ choose an action with all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","h all successors in $L_*$. Such actions exist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","xist by the preceding lemma and keep every running path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","ing path within $L_*\\subseteq T_*\\subseteq S$ for any duration. If a stop command arrives at a decision epoch, the current state still lies in $T_*$. Switch modes and choose a first-entry-rank-decreasing ending action. The rank is a nonnegative integer, decreases strictly after every ending transition outside $G$, and must reach zero in no more than $\\rho(w)$ steps. At rank zero the system is in $G$ and can hold safely. This is one policy with the command as its observable mode switch, not two incompatible policies chosen after seeing a future path.","For necessity, consider any policy with the stated guarantee, including one that depends on the entire available history. The stop command may arrive at the present epoch, so by the adversarial completeness lemma the starting state must belong to $T_*$. It may instead arrive after any number $k$ of running steps. Every state reachable immediately before such a command must therefore remain in $T_*$. The running prefix of length $k$ consequently satisfies the bounded-horizon condition of the preceding lemma, so the initial state is in $L_k$. This holds for every $k$. The decreasing sequence stabilizes on a finite set, hence the state belongs to its eventual value $L_*$. No operation is selected with advance knowledge of which successor an adversary will return.\n\\end{proof}","y with the stated guarantee, including one that depends on the entire available history. The stop command may arrive at the present epoch, so by the adversarial completeness lemma the starting state must belong to $T_*$. It may instead arrive after any number $k$ of running steps. Every state reachable immediately before such a command must therefore remain in $T_*$. The running prefix of length $k$ consequently satisfies the bounded-horizon condition of the preceding lemma, so the initial state is in $L_k$. This holds for every $k$. The decreasing sequence stabilizes on a finite set, hence the state belongs to its eventual value $L_*$. No operation is selected with advance knowledge of which successor an adversary will return.\n\\end{proof}","y arrive at the present epoch, so by the adversarial completeness lemma the starting state must belong to $T_*$. It may instead arrive after any number $k$ of running steps. Every state reachable immediately before such a command must therefore remain in $T_*$. The running prefix of length $k$ consequently satisfies the bounded-horizon condition of the preceding lemma, so the initial state is in $L_k$. This holds for every $k$. The decreasing sequence stabilizes on a finite set, hence the state belongs to its eventual value $L_*$. No operation is selected with advance knowledge of which successor an adversary will return.\n\\end{proof}","h, so by the adversarial completeness lemma the starting state must belong to $T_*$. It may instead arrive after any number $k$ of running steps. Every state reachable immediately before such a command must therefore remain in $T_*$. The running prefix of length $k$ consequently satisfies the bounded-horizon condition of the preceding lemma, so the initial state is in $L_k$. This holds for every $k$. The decreasing sequence stabilizes on a finite set, hence the state belongs to its eventual value $L_*$. No operation is selected with advance knowledge of which successor an adversary will return.\n\\end{proof}","ive after any number $k$ of running steps. Every state reachable immediately before such a command must therefore remain in $T_*$. The running prefix of length $k$ consequently satisfies the bounded-horizon condition of the preceding lemma, so the initial state is in $L_k$. This holds for every $k$. The decreasing sequence stabilizes on a finite set, hence the state belongs to its eventual value $L_*$. No operation is selected with advance knowledge of which successor an adversary will return.\n\\end{proof}","\\section{A three-state obstruction to naive safety}\nConsider the safe configurations $S=\\{u,z,g\\}$ with ending target $G=\\{g\\}$. The state $g$ has a hold action in both modes, $z$ has a safe running self-loop but cannot end, and $u$ can end immediately at $g$. Yet the only running action at $u$ moves to $z$. This gives\n\\[\nT_*=\\{u,g\\},\\qquad L_*=\\{g\\}.\n\\]\nIt is possible to end from $u$ now, but not to run from $u$ while retaining the option to end later. The obstruction is about \\emph{future control access}, not immediate safety.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.95\\linewidth]{figures/safe_ending_kernel.pdf}\\caption{Exact finite safe-running and sure-ending example. The initial running action from $u$ loses the ending option. The sets were checked against explicit str","\\section{A three-state obstruction to naive safety}\nConsider the safe configurations $S=\\{u,z,g\\}$ with ending target $G=\\{g\\}$. The state $g$ has a hold action in both modes, $z$ has a safe running self-loop but cannot end, and $u$ can end immediately at $g$. Yet the only running action at $u$ moves to $z$. This gives\n\\[\nT_*=\\{u,g\\},\\qquad L_*=\\{g\\}.\n\\]\nIt is possible to end from $u$ now, but not to run from $u$ while retaining the option to end later. The obstruction is about \\emph{future control access}, not immediate safety.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.95\\linewidth]{figures/safe_ending_kernel.pdf}\\caption{Exact finite safe-running and sure-ending example. The initial running action from $u$ loses the ending option. The sets were checked against explicit strategy enumeration.}\\label{fig","\\section{A three-state obstruction to naive safety}\nConsider the safe configurations $S=\\{u,z,g\\}$ with ending target $G=\\{g\\}$. The state $g$ has a hold action in both modes, $z$ has a safe running self-loop but cannot end, and $u$ can end immediately at $g$. Yet the only running action at $u$ moves to $z$. This gives\n\\[\nT_*=\\{u,g\\},\\qquad L_*=\\{g\\}.\n\\]\nIt is possible to end from $u$ now, but not to run from $u$ while retaining the option to end later. The obstruction is about \\emph{future control access}, not immediate safety.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.95\\linewidth]{figures/safe_ending_kernel.pdf}\\caption{Exact finite safe-running and sure-ending example. The initial running action from $u$ loses the ending option. The sets were checked against explicit strategy enumeration.}\\label{fig:ending-kernel}","\\section{A three-state obstruction to naive safety}\nConsider the safe configurations $S=\\{u,z,g\\}$ with ending target $G=\\{g\\}$. The state $g$ has a hold action in both modes, $z$ has a safe running self-loop but cannot end, and $u$ can end immediately at $g$. Yet the only running action at $u$ moves to $z$. This gives\n\\[\nT_*=\\{u,g\\},\\qquad L_*=\\{g\\}.\n\\]\nIt is possible to end from $u$ now, but not to run from $u$ while retaining the option to end later. The obstruction is about \\emph{future control access}, not immediate safety.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.95\\linewidth]{figures/safe_ending_kernel.pdf}\\caption{Exact finite safe-running and sure-ending example. The initial running action from $u$ loses the ending option. The sets were checked against explicit strategy enumeration.}\\label{fig:ending-kernel}\\end{figure}","ee-state obstruction to naive safety}\nConsider the safe configurations $S=\\{u,z,g\\}$ with ending target $G=\\{g\\}$. The state $g$ has a hold action in both modes, $z$ has a safe running self-loop but cannot end, and $u$ can end immediately at $g$. Yet the only running action at $u$ moves to $z$. This gives\n\\[\nT_*=\\{u,g\\},\\qquad L_*=\\{g\\}.\n\\]\nIt is possible to end from $u$ now, but not to run from $u$ while retaining the option to end later. The obstruction is about \\emph{future control access}, not immediate safety.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.95\\linewidth]{figures/safe_ending_kernel.pdf}\\caption{Exact finite safe-running and sure-ending example. The initial running action from $u$ loses the ending option. The sets were checked against explicit strategy enumeration.}\\label{fig:ending-kernel}\\end{figure}","=\\{u,g\\},\\qquad L_*=\\{g\\}.\n\\]\nIt is possible to end from $u$ now, but not to run from $u$ while retaining the option to end later. The obstruction is about \\emph{future control access}, not immediate safety.\n\\begin{figure}[htbp]\\centering\\includegraphics[width=.95\\linewidth]{figures/safe_ending_kernel.pdf}\\caption{Exact finite safe-running and sure-ending example. The initial running action from $u$ loses the ending option. The sets were checked against explicit strategy enumeration.}\\label{fig:ending-kernel}\\end{figure}","The companion verifier compares the fixed-point calculation with explicit positional-policy enumeration and checks the rank-decreasing witnesses. The exact input domains and results appear in the verification guide. The calculation treats arrival at $G$ as termination. The stipulated hold concerns subsequent behavior and does not change the reachability calculation.","\\section{What changes with partial observation}\nThe proof assumes the actual configuration is observed when an action is chosen. If only a public report is available, strategies generally act on a belief or possibility state whose update must be justified by the observation protocol. Replacing $Z$ with arbitrary hidden-state names does not produce a valid partially observed ending theorem. One would need a common public-state construction, exact reporting and successor semantics, and a proof that the stop command is observed before action selection. The finite theorem is a solid component of that larger problem without silently solving it.","which smaller state to construct from sparse experience. Learning such a state requires a declared hypothesis family, a policy for obtaining observations, and conditions under which incompatible hypotheses can be separated. Statistical identifiability is not certain finite identification. Every finite binary sequence can occur with positive probability under two distinct Bernoulli parameters in $(0,1)$.","\\section{A learned semantic map is not a written commuting equation}\nConditional method transport has the form $TF_a=G_{\\tau(a)}T$. This is meaningful when $T$, the operator translation $\\tau$, and the applicable domains are already specified. A learned system must acquire them, ensure they cover its actual state trajectories, and show the diagrams continue to commute under the relevant interventions. A correct table for a finite declared task is a useful positive control, but it cannot certify that a large recurrent network learned a stable operator language across changed domains.","\\section{A migration map exists exactly on constant fibres}\nLet $H$ be a finite nonempty set of possible source histories. A history may include a model tag or a past observation whose value must remain available for later correction. Let\n\\[\n M:H\\to U,\\qquad N:H\\to V\n\\]\nbe the old record and the required new record. Work on their attained images $U_0=M(H)$ and $V_0=N(H)$. The new record must be specified before sufficiency is claimed. It can be an answer, a correction signature, or the complete profile of future outputs in a declared finite contract.","\\begin{theorem}[Exact migration criterion]\\label{r:migration}\nThere is a map $J:U_0\\to V_0$ satisfying $JM=N$ if and only if\n\\[\n M(h)=M(h')\\quad\\Longrightarrow\\quad N(h)=N(h')\n \\qquad(h,h'\\in H).\n\\]\nWhen it exists, $J$ is unique on attained old records.\n\\end{theorem}\n\\begin{proof}\nNecessity follows by applying $J$ to the equal old records. Conversely, define $J(M(h))=N(h)$. The fibre condition makes the definition independent of the representative. Every $u\\in U_0$ has such a representative, so the same equation determines $J$ uniquely.\n\\end{proof}","}\nThere is a map $J:U_0\\to V_0$ satisfying $JM=N$ if and only if\n\\[\n M(h)=M(h')\\quad\\Longrightarrow\\quad N(h)=N(h')\n \\qquad(h,h'\\in H).\n\\]\nWhen it exists, $J$ is unique on attained old records.\n\\end{theorem}\n\\begin{proof}\nNecessity follows by applying $J$ to the equal old records. Conversely, define $J(M(h))=N(h)$. The fibre condition makes the definition independent of the representative. Every $u\\in U_0$ has such a representative, so the same equation determines $J$ uniquely.\n\\end{proof}","(h)=M(h')\\quad\\Longrightarrow\\quad N(h)=N(h')\n \\qquad(h,h'\\in H).\n\\]\nWhen it exists, $J$ is unique on attained old records.\n\\end{theorem}\n\\begin{proof}\nNecessity follows by applying $J$ to the equal old records. Conversely, define $J(M(h))=N(h)$. The fibre condition makes the definition independent of the representative. Every $u\\in U_0$ has such a representative, so the same equation determines $J$ uniquely.\n\\end{proof}","\\section{What additional evidence must provide}\nSuppose an allowed observation produces $E:H\\to W$. A translator can use the pair $(M(h),E(h))$ rather than $M(h)$ alone.","\\begin{corollary}[Evidence-assisted migration]\\label{r:migration-evidence}\nAn exact translator from the available pair $(M,E)$ to $N$ exists if and only if\n\\[\n M(h)=M(h'),\\ E(h)=E(h')\\quad\\Longrightarrow\\quad N(h)=N(h').\n\\]\nIf a collision remains after the observation, that particular evidence interface cannot support the requested migration.\n\\end{corollary}\n\\begin{proof}\nApply Theorem~\\ref{r:migration} to the record $h\\mapsto(M(h),E(h))$.\n\\end{proof}","ence-assisted migration]\\label{r:migration-evidence}\nAn exact translator from the available pair $(M,E)$ to $N$ exists if and only if\n\\[\n M(h)=M(h'),\\ E(h)=E(h')\\quad\\Longrightarrow\\quad N(h)=N(h').\n\\]\nIf a collision remains after the observation, that particular evidence interface cannot support the requested migration.\n\\end{corollary}\n\\begin{proof}\nApply Theorem~\\ref{r:migration} to the record $h\\mapsto(M(h),E(h))$.\n\\end{proof}","\\section{The least joint refinement}\nWhen both the old and new uses must remain available, retain\n\\[\n P(h)=(M(h),N(h))\\in P(H).\n\\]\nThe projections recover each record. No ordering between the old and new information is required.","\\begin{proposition}[Universal property of the joint record]\\label{r:joint-refinement-record}\nIf a record $R:H\\to W$ determines both $M$ and $N$, then it determines $P$. In particular, $|R(H)|\\ge |P(H)|$. The pair record attains this lower bound among records required to preserve both uses.\n\\end{proposition}\n\\begin{proof}\nLet $fR=M$ and $gR=N$ on $H$. Then $w\\mapsto(f(w),g(w))$ maps every attained value of $R$ to the corresponding attained pair. This map is onto $P(H)$, since every pair comes from a source history. Finiteness gives the cardinality inequality. The record $P$ itself has the required two projection maps.\n\\end{proof}","position}[Universal property of the joint record]\\label{r:joint-refinement-record}\nIf a record $R:H\\to W$ determines both $M$ and $N$, then it determines $P$. In particular, $|R(H)|\\ge |P(H)|$. The pair record attains this lower bound among records required to preserve both uses.\n\\end{proposition}\n\\begin{proof}\nLet $fR=M$ and $gR=N$ on $H$. Then $w\\mapsto(f(w),g(w))$ maps every attained value of $R$ to the corresponding attained pair. This map is onto $P(H)$, since every pair comes from a source history. Finiteness gives the cardinality inequality. The record $P$ itself has the required two projection maps.\n\\end{proof}",":joint-refinement-record}\nIf a record $R:H\\to W$ determines both $M$ and $N$, then it determines $P$. In particular, $|R(H)|\\ge |P(H)|$. The pair record attains this lower bound among records required to preserve both uses.\n\\end{proposition}\n\\begin{proof}\nLet $fR=M$ and $gR=N$ on $H$. Then $w\\mapsto(f(w),g(w))$ maps every attained value of $R$ to the corresponding attained pair. This map is onto $P(H)$, since every pair comes from a source history. Finiteness gives the cardinality inequality. The record $P$ itself has the required two projection maps.\n\\end{proof}","cular, $|R(H)|\\ge |P(H)|$. The pair record attains this lower bound among records required to preserve both uses.\n\\end{proposition}\n\\begin{proof}\nLet $fR=M$ and $gR=N$ on $H$. Then $w\\mapsto(f(w),g(w))$ maps every attained value of $R$ to the corresponding attained pair. This map is onto $P(H)$, since every pair comes from a source history. Finiteness gives the cardinality inequality. The record $P$ itself has the required two projection maps.\n\\end{proof}","\\begin{theorem}[Minimum fixed-length supplementary information]\\label{r:migration-bits}\nFor $u\\in U_0$, put\n\\[\n k(u)=|N(M^{-1}\\{u\\})|,\\qquad k_* =\\max_{u\\in U_0}k(u).\n\\]\nSuppose an auxiliary encoder may inspect the complete source history and sends one fixed-length binary word in addition to the old record. Exact recovery of $N$ is possible with $b$ supplementary bits if and only if $2^b\\ge k_*$. Thus the minimum is $\\lceil\\log_2 k_*\\rceil$.\n\\end{theorem}\n\\begin{proof}\nFor a fixed old record $u$, distinct required new values must receive distinct supplementary words. Otherwise the decoder would receive identical inputs with different required outputs. Hence $2^b\\ge k(u)$ for every $u$.","um fixed-length supplementary information]\\label{r:migration-bits}\nFor $u\\in U_0$, put\n\\[\n k(u)=|N(M^{-1}\\{u\\})|,\\qquad k_* =\\max_{u\\in U_0}k(u).\n\\]\nSuppose an auxiliary encoder may inspect the complete source history and sends one fixed-length binary word in addition to the old record. Exact recovery of $N$ is possible with $b$ supplementary bits if and only if $2^b\\ge k_*$. Thus the minimum is $\\lceil\\log_2 k_*\\rceil$.\n\\end{theorem}\n\\begin{proof}\nFor a fixed old record $u$, distinct required new values must receive distinct supplementary words. Otherwise the decoder would receive identical inputs with different required outputs. Hence $2^b\\ge k(u)$ for every $u$.","U_0$, put\n\\[\n k(u)=|N(M^{-1}\\{u\\})|,\\qquad k_* =\\max_{u\\in U_0}k(u).\n\\]\nSuppose an auxiliary encoder may inspect the complete source history and sends one fixed-length binary word in addition to the old record. Exact recovery of $N$ is possible with $b$ supplementary bits if and only if $2^b\\ge k_*$. Thus the minimum is $\\lceil\\log_2 k_*\\rceil$.\n\\end{theorem}\n\\begin{proof}\nFor a fixed old record $u$, distinct required new values must receive distinct supplementary words. Otherwise the decoder would receive identical inputs with different required outputs. Hence $2^b\\ge k(u)$ for every $u$.","inary word in addition to the old record. Exact recovery of $N$ is possible with $b$ supplementary bits if and only if $2^b\\ge k_*$. Thus the minimum is $\\lceil\\log_2 k_*\\rceil$.\n\\end{theorem}\n\\begin{proof}\nFor a fixed old record $u$, distinct required new values must receive distinct supplementary words. Otherwise the decoder would receive identical inputs with different required outputs. Hence $2^b\\ge k(u)$ for every $u$.","o the old record. Exact recovery of $N$ is possible with $b$ supplementary bits if and only if $2^b\\ge k_*$. Thus the minimum is $\\lceil\\log_2 k_*\\rceil$.\n\\end{theorem}\n\\begin{proof}\nFor a fixed old record $u$, distinct required new values must receive distinct supplementary words. Otherwise the decoder would receive identical inputs with different required outputs. Hence $2^b\\ge k(u)$ for every $u$.","Conversely, within each old fibre assign distinct binary words to the distinct attained new values, using at most $k_*\\le 2^b$ words. The same word can be reused in different old fibres because the decoder receives $u$ as well. Define the encoder by this fibre-specific table and decode by the inverse table. Every source history is recovered correctly.\n\\end{proof}","The encoder's access assumption is essential. The bound does not allow a controller holding only $M(h)$ to manufacture the missing bits. It describes the amount of supplementary information when an encoder actually has the source information. With restricted observations, Corollary~\\ref{r:migration-evidence} must be checked first. Code tables, acquisition, transmission, and validation also carry costs not counted by the fixed binary payload.","\\section{Migration must commute with subsequent operations}\nRecovering one new record is not yet preserving a method. Let a common finite action alphabet act on $H$ through total maps $F_a:H\\to H$. Suppose both representations close under those same source operations:\n\\[\n M F_a=P_a M,\\qquad N F_a=Q_a N.\n\\]\nThe maps $P_a$ and $Q_a$ preserve the attained record sets. These equations refer to the same source action, not merely to operations sharing a name.","\\begin{theorem}[A valid migration intertwines the common operations]\\label{r:migration-operations}\nIf $JM=N$, then on $U_0$,\n\\[\n JP_a=Q_aJ.\n\\]\nConsequently $JP_w=Q_wJ$ for every finite action word $w$, with the execution-order convention of Chapter~\\ref{ch:evidence}. Any protected readout that factors through $N$ agrees along the two migrated executions.\n\\end{theorem}\n\\begin{proof}\nTake $u=M(h)$. Then\n\\[\n JP_a(u)=JM(F_a h)=N(F_a h)=Q_aN(h)=Q_aJ(u).\n\\]\nEvery attained $u$ has a representative, so the equality holds throughout $U_0$. Closure keeps later values within that domain. Induction on the word length gives the continuation identity. Composing with the protected readout gives equality of outputs.\n\\end{proof}","theorem}[A valid migration intertwines the common operations]\\label{r:migration-operations}\nIf $JM=N$, then on $U_0$,\n\\[\n JP_a=Q_aJ.\n\\]\nConsequently $JP_w=Q_wJ$ for every finite action word $w$, with the execution-order convention of Chapter~\\ref{ch:evidence}. Any protected readout that factors through $N$ agrees along the two migrated executions.\n\\end{theorem}\n\\begin{proof}\nTake $u=M(h)$. Then\n\\[\n JP_a(u)=JM(F_a h)=N(F_a h)=Q_aN(h)=Q_aJ(u).\n\\]\nEvery attained $u$ has a representative, so the equality holds throughout $U_0$. Closure keeps later values within that domain. Induction on the word length gives the continuation identity. Composing with the protected readout gives equality of outputs.\n\\end{proof}","el{r:migration-operations}\nIf $JM=N$, then on $U_0$,\n\\[\n JP_a=Q_aJ.\n\\]\nConsequently $JP_w=Q_wJ$ for every finite action word $w$, with the execution-order convention of Chapter~\\ref{ch:evidence}. Any protected readout that factors through $N$ agrees along the two migrated executions.\n\\end{theorem}\n\\begin{proof}\nTake $u=M(h)$. Then\n\\[\n JP_a(u)=JM(F_a h)=N(F_a h)=Q_aN(h)=Q_aJ(u).\n\\]\nEvery attained $u$ has a representative, so the equality holds throughout $U_0$. Closure keeps later values within that domain. Induction on the word length gives the continuation identity. Composing with the protected readout gives equality of outputs.\n\\end{proof}","_a=Q_aJ.\n\\]\nConsequently $JP_w=Q_wJ$ for every finite action word $w$, with the execution-order convention of Chapter~\\ref{ch:evidence}. Any protected readout that factors through $N$ agrees along the two migrated executions.\n\\end{theorem}\n\\begin{proof}\nTake $u=M(h)$. Then\n\\[\n JP_a(u)=JM(F_a h)=N(F_a h)=Q_aN(h)=Q_aJ(u).\n\\]\nEvery attained $u$ has a representative, so the equality holds throughout $U_0$. Closure keeps later values within that domain. Induction on the word length gives the continuation identity. Composing with the protected readout gives equality of outputs.\n\\end{proof}","\\section{A worked change of task}\nUse the eight histories $(b,s,n)\\in\\{0,1\\}^3$ and the old display record $M=b\\oplus s$. For display prediction under physical-bit flips, this record is sufficient. Suppose the new task requires both the display and the physical target, so $N=(b\\oplus s,b)$.","There are two possible new records inside each old display fibre. Direct migration fails, but $k_*=2$, so one supplementary bit is information-theoretically sufficient. Calibration supplies $E=s$, and\n\\[\n b=M\\oplus E\n\\]\ngives the actual decoder. The joint old/new record has four attained states. The nuisance bit remains discarded because the new task still does not use it.","Now require the nuisance bit as well. The new record $N'=(b\\oplus s,b,n)$ has four possible values inside each old fibre, giving a two-bit supplementary lower bound. Calibration alone no longer suffices. Two histories with equal $b,s$ and different $n$ collide under $(M,E)$ but have different $N'$. The lower bound identifies a missing distinction rather than a computational bottleneck.","\\section{Certificates after a change of contract}\nA migration map transports data. A certificate authorizes a conclusion only relative to its premises. To reuse a certificate across a revision, specify an old predicate $\\varphi$ on $U_0$, a new predicate $\\psi$ on $V_0$, and a proved implication\n\\[\n \\varphi(u)\\ \\Longrightarrow\\ \\psi(Ju).\n\\]\nAn old witness for $\\varphi(u)$ and this implication then yield the new conclusion. The implication may express semantic preservation, a stronger guard, or a deliberately weaker postcondition. It cannot be replaced by equality of file names or by an unchanged serialization format.","\\section{A finite field with explicitly admitted rules}\nLet $V=\\{0,\\ldots,n-1\\}$ be a finite set of claim addresses. Let $A\\subseteq V$ be admitted leaves, $S\\subseteq V$ positive observations, and $N\\subseteq V$ forbidden consequences. These sets need not be disjoint. A finite rule list $R$ has rows\n\\[\n r_j=(P_j,c_j),\\qquad P_j\\subseteq V,\\quad c_j\\in V.\n\\]\nOnly admitted rules are included. Identical rows may occupy different addresses. A conjectural rule elsewhere in the discovery archive is not admitted merely because it has a convenient conclusion.","A model of the field is a set $B\\subseteq V$ containing $A$ and satisfying every implication $P_j\\subseteq B\\Rightarrow c_j\\in B$. A protected extension also contains $S$ and avoids $N$. Define\n\\[\n C_0=A\\cup S,\\qquad\n C_{k+1}=C_k\\cup\\{c_j:P_j\\subseteq C_k\\}.\n\\]\nThe terms increase and lie in $V$, so the sequence stabilizes. Write $C$ for its stable value.","\\begin{proposition}[The least seeded model]\\label{r:field-closure}\nThe set $C$ is the least model containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no prot","\\begin{proposition}[The least seeded model]\\label{r:field-closure}\nThe set $C$ is the least model containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof","\\begin{proposition}[The least seeded model]\\label{r:field-closure}\nThe set $C$ is the least model containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","osition}[The least seeded model]\\label{r:field-closure}\nThe set $C$ is the least model containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","eded model]\\label{r:field-closure}\nThe set $C$ is the least model containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","The set $C$ is the least model containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","containing $S$. There are at most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","most $n-|A\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","\\cup S|$ strict expansion rounds. A protected extension exists if and only if $C\\cap N=\\varnothing$.\n\\end{proposition}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","tion}\n\\begin{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","n{proof}\nAt a fixed point every enabled rule has its conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","conclusion in $C$, so $C$ is a model containing the seeds. If $B$ is any other model containing $S$, induction gives $C_k\\subseteq B$: the seeds lie in $B$, and every rule enabled on $C_k$ has its conclusion in $B$. Hence $C\\subseteq B$. Each strict expansion adds at least one of the finitely many nonseed nodes, giving the bound. If $C$ avoids $N$, it is itself a protected extension. If it meets $N$, every seeded model contains that forbidden node, so no protected extension exists.\n\\end{proof}","A cycle does not create a seed. With rules $u\\to v$ and $v\\to u$ and no admitted leaves or observations, the least model is empty. The set $\\{u,v\\}$ is closed, but its members have no derivation from the empty seed. This is why closedness alone is not an adequate proof certificate.","\\section{Replay checks a finite causal sequence}\nA trace $\\sigma=(j_1,\\ldots,j_m)$ lists rule addresses. Start from $K_0=A\\cup S$. At each step, reject an unknown rule address or a rule whose premises are not all in the current known set. Otherwise adjoin its conclusion. \\emph{Strict replay} also rejects a conclusion already known. \\emph{Causal replay} allows it.","\\begin{theorem}[Replay is sound and compositional]\\label{r:field-replay}\nIf replay accepts a trace from a set $K\\subseteq C$ and returns $K'$, then\n\\[\n K\\subseteq K'\\subseteq C.\n\\]\nFor either fixed replay mode, replay of a concatenated trace is replay of its first part followed by replay of its second part from the returned set. Failure in either part is failure of the concatenated replay.\n\\end{theorem}\n\\begin{proof}\nThe empty trace changes nothing. At a successful next step, all premises of an admitted rule already lie in $C$. Closedness of $C$ places its conclusion in $C$. Adjoining it preserves the old known set. Induction proves the inclusions. A strict novelty test can reject a step, but does not alter this reasoning for an accepted step.","eturns $K'$, then\n\\[\n K\\subseteq K'\\subseteq C.\n\\]\nFor either fixed replay mode, replay of a concatenated trace is replay of its first part followed by replay of its second part from the returned set. Failure in either part is failure of the concatenated replay.\n\\end{theorem}\n\\begin{proof}\nThe empty trace changes nothing. At a successful next step, all premises of an admitted rule already lie in $C$. Closedness of $C$ places its conclusion in $C$. Adjoining it preserves the old known set. Induction proves the inclusions. A strict novelty test can reject a step, but does not alter this reasoning for an accepted step.","[\n K\\subseteq K'\\subseteq C.\n\\]\nFor either fixed replay mode, replay of a concatenated trace is replay of its first part followed by replay of its second part from the returned set. Failure in either part is failure of the concatenated replay.\n\\end{theorem}\n\\begin{proof}\nThe empty trace changes nothing. At a successful next step, all premises of an admitted rule already lie in $C$. Closedness of $C$ places its conclusion in $C$. Adjoining it preserves the old known set. Induction proves the inclusions. A strict novelty test can reject a step, but does not alter this reasoning for an accepted step.","\\]\nFor either fixed replay mode, replay of a concatenated trace is replay of its first part followed by replay of its second part from the returned set. Failure in either part is failure of the concatenated replay.\n\\end{theorem}\n\\begin{proof}\nThe empty trace changes nothing. At a successful next step, all premises of an admitted rule already lie in $C$. Closedness of $C$ places its conclusion in $C$. Adjoining it preserves the old known set. Induction proves the inclusions. A strict novelty test can reject a step, but does not alter this reasoning for an accepted step.","\\begin{theorem}[Replay and saturation identify exact closure]\\label{r:field-exact}\nA complete-closure certificate consists of a strict trace and a reported set $B$. Accept it only when replay from $A\\cup S$ succeeds with output $B$ and $B$ satisfies every admitted rule. Acceptance implies $B=C$.\n\\end{theorem}\n\\begin{proof}\nReplay soundness gives $B\\subseteq C$. Its extensivity ensures that $B$ contains both seed sets. The saturation check makes $B$ a seeded model, so leastness gives $C\\subseteq B$. The two inclusions prove equality.\n\\end{proof}","ct closure]\\label{r:field-exact}\nA complete-closure certificate consists of a strict trace and a reported set $B$. Accept it only when replay from $A\\cup S$ succeeds with output $B$ and $B$ satisfies every admitted rule. Acceptance implies $B=C$.\n\\end{theorem}\n\\begin{proof}\nReplay soundness gives $B\\subseteq C$. Its extensivity ensures that $B$ contains both seed sets. The saturation check makes $B$ a seeded model, so leastness gives $C\\subseteq B$. The two inclusions prove equality.\n\\end{proof}","lete-closure certificate consists of a strict trace and a reported set $B$. Accept it only when replay from $A\\cup S$ succeeds with output $B$ and $B$ satisfies every admitted rule. Acceptance implies $B=C$.\n\\end{theorem}\n\\begin{proof}\nReplay soundness gives $B\\subseteq C$. Its extensivity ensures that $B$ contains both seed sets. The saturation check makes $B$ a seeded model, so leastness gives $C\\subseteq B$. The two inclusions prove equality.\n\\end{proof}","\\section{A complete negative answer has a witness}\nAn obstruction certificate contains a forbidden target $v\\in N$ and a causal trace deriving $v$ from $A\\cup S$.","\\begin{theorem}[Finite obstruction witnesses are sound and complete]\\label{r:field-obstruction}\nAn accepted obstruction certificate exists exactly when no protected extension exists. No acyclicity assumption on the rule graph is required.\n\\end{theorem}\n\\begin{proof}\nAn accepted certificate derives a node of $C\\cap N$, so Proposition~\\ref{r:field-closure} excludes a protected extension.","Conversely, if there is no extension, choose $v\\in C\\cap N$. Every node in $C$ is reached at a finite stage. Starting with the seeds, process the rules that add the nodes of each successive stage, appending a rule only when its conclusion is not yet known. All its premises belong to an earlier stage and are already known. This yields a finite strict trace reaching the whole closure, and in particular $v$. Its prefix ending when $v$ is first present is also accepted in causal mode. If $v$ is a seed, the empty trace suffices. Cycles cause no difficulty because only rules with already known premises are applied.\n\\end{proof}","rsely, if there is no extension, choose $v\\in C\\cap N$. Every node in $C$ is reached at a finite stage. Starting with the seeds, process the rules that add the nodes of each successive stage, appending a rule only when its conclusion is not yet known. All its premises belong to an earlier stage and are already known. This yields a finite strict trace reaching the whole closure, and in particular $v$. Its prefix ending when $v$ is first present is also accepted in causal mode. If $v$ is a seed, the empty trace suffices. Cycles cause no difficulty because only rules with already known premises are applied.\n\\end{proof}","ion, choose $v\\in C\\cap N$. Every node in $C$ is reached at a finite stage. Starting with the seeds, process the rules that add the nodes of each successive stage, appending a rule only when its conclusion is not yet known. All its premises belong to an earlier stage and are already known. This yields a finite strict trace reaching the whole closure, and in particular $v$. Its prefix ending when $v$ is first present is also accepted in causal mode. If $v$ is a seed, the empty trace suffices. Cycles cause no difficulty because only rules with already known premises are applied.\n\\end{proof}","at a finite stage. Starting with the seeds, process the rules that add the nodes of each successive stage, appending a rule only when its conclusion is not yet known. All its premises belong to an earlier stage and are already known. This yields a finite strict trace reaching the whole closure, and in particular $v$. Its prefix ending when $v$ is first present is also accepted in causal mode. If $v$ is a seed, the empty trace suffices. Cycles cause no difficulty because only rules with already known premises are applied.\n\\end{proof}","\\section{A worked four-node certificate}\nLet $V=\\{0,1,2,3\\}$, $A=\\{0\\}$, $S=\\varnothing$, and let the two rules be $0\\to1$ and $1\\to2$. The exact closure is $\\{0,1,2\\}$.","The trace containing only the first rule is a valid partial derivation of $\\{0,1\\}$, but it fails saturation. The set $\\{0,1,2,3\\}$ is saturated, but node $3$ cannot be produced by replay. The trace of both rules and the exact three-node set pass both tests.","With forbidden set $N=\\{3\\}$, the exact closure is a positive feasibility certificate. With $N=\\{2\\}$, the same closure and the trace to node $2$ certify infeasibility. The request changed, even though the derivation did not. A receiving checker must therefore be given the intended request separately from the producer's proposed answer.","\\section{Logical interpretation is an additional map}\nLet $\\mu(v)$ be the proposition represented by claim address $v$. Suppose every admitted leaf and positive observation is true under $\\mu$, and each admitted rule satisfies\n\\[\n \\left(\\bigwedge_{p\\in P_j}\\mu(p)\\right)\\Longrightarrow\\mu(c_j).\n\\]\nThen every node in $C$ is true under $\\mu$. Indeed, induction over the closure stages propagates truth from the seeds through the admitted implications.","The finite checker proves derivability relative to those inputs. It does not establish the truth of an empirical observation merely because that observation is placed in $S$. A node excluded by a protected request is likewise not necessarily a logically false proposition. Its appearance shows incompatibility with that request, whose meaning must be stated separately.","\\section*{Views, restrictions, and a coherent realization}\nLet $\\mathcal V$ be a small category of views. Associate a set $F_v$ to each view $v$ and a restriction $f^*:F_w\\to F_v$ to each arrow $f:v\\to w$. Identity and composition satisfy $(1_v)^*=\\mathrm{id}$ and $(g\\circ f)^*=f^*\\circ g^*$. Thus $F$ is a contravariant set-valued functor. A coherent realization is a family of elements $x_v\\in F_v$ satisfying all the restriction equations:\n\\[\n\\Gamma(F)=\\left\\{(x_v)_v\\in\\prod_v F_v:\\ f^*(x_w)=x_v\\text{ for every }f:v\\to w\\right\\}.\n\\]\nThe set $\\Gamma(F)$ is the strict realization set. It may be empty. A nonempty fibre at every individual view does not force a coherent family to exist.","realization}\nLet $\\mathcal V$ be a small category of views. Associate a set $F_v$ to each view $v$ and a restriction $f^*:F_w\\to F_v$ to each arrow $f:v\\to w$. Identity and composition satisfy $(1_v)^*=\\mathrm{id}$ and $(g\\circ f)^*=f^*\\circ g^*$. Thus $F$ is a contravariant set-valued functor. A coherent realization is a family of elements $x_v\\in F_v$ satisfying all the restriction equations:\n\\[\n\\Gamma(F)=\\left\\{(x_v)_v\\in\\prod_v F_v:\\ f^*(x_w)=x_v\\text{ for every }f:v\\to w\\right\\}.\n\\]\nThe set $\\Gamma(F)$ is the strict realization set. It may be empty. A nonempty fibre at every individual view does not force a coherent family to exist.","a restriction $f^*:F_w\\to F_v$ to each arrow $f:v\\to w$. Identity and composition satisfy $(1_v)^*=\\mathrm{id}$ and $(g\\circ f)^*=f^*\\circ g^*$. Thus $F$ is a contravariant set-valued functor. A coherent realization is a family of elements $x_v\\in F_v$ satisfying all the restriction equations:\n\\[\n\\Gamma(F)=\\left\\{(x_v)_v\\in\\prod_v F_v:\\ f^*(x_w)=x_v\\text{ for every }f:v\\to w\\right\\}.\n\\]\nThe set $\\Gamma(F)$ is the strict realization set. It may be empty. A nonempty fibre at every individual view does not force a coherent family to exist.","For example, take the one-object category whose automorphisms form the two-element group, and let its nontrivial arrow exchange the two elements of $F_v=\\{0,1\\}$. A coherent family would be a fixed point of that exchange. There is none. No amount of relabeling can turn either local element into a global fixed point without changing the specified action. This illustrates a basic obstruction: local admissibility is weaker than coherent realization.","The evaluation $\\operatorname{ev}_v:\\Gamma(F)\\to F_v$ records the information visible in one view. When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving","The evaluation $\\operatorname{ev}_v:\\Gamma(F)\\to F_v$ records the information visible in one view. When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coheren","The evaluation $\\operatorname{ev}_v:\\Gamma(F)\\to F_v$ records the information visible in one view. When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibr","The evaluation $\\operatorname{ev}_v:\\Gamma(F)\\to F_v$ records the information visible in one view. When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","he evaluation $\\operatorname{ev}_v:\\Gamma(F)\\to F_v$ records the information visible in one view. When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","v$ records the information visible in one view. When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}",". When $\\Gamma(F)$ is nonempty and a target $q:\\Gamma(F)\\to Y$ is specified, that view is sufficient exactly when the target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","target is constant on its fibres.\n\\begin{proposition}[Target factorization through a realization view]\\label{r:viewfactor}\nSuppose $\\Gamma(F)$ and $Y$ are nonempty. There is a function $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","n $d:\\operatorname{ev}_v(\\Gamma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","mma(F))\\to Y$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","$ such that $q=d\\circ\\operatorname{ev}_v$ if and only if\n\\[\n\\operatorname{ev}_v(x)=\\operatorname{ev}_v(x')\\quad\\Longrightarrow\\quad q(x)=q(x')\n\\]\nfor all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","r all coherent realizations $x,x'\\in\\Gamma(F)$.\n\\end{proposition}\n\\begin{proof}\nA function of one attained record must assign the same value to equal records, proving necessity. Conversely, each attained record $a$ has a coherent preimage $x$. Define $d(a)=q(x)$. The fibre hypothesis makes this independent of the chosen preimage and supplies the required factorization. There is no need to define $d$ on unattained, possibly impossible view values.\n\\end{proof}","\\section*{Protected operations and what a quotient must retain}\nFor an observed deterministic system, let $X$ be a set of realizations, $r:X\\to Z$ a retained record, $F_a:X\\to X$ an action, and $q:X\\to Y$ a protected output. The coarse description is closed under action $a$ only if there is $\\bar F_a:Z\\to Z$ satisfying $rF_a=\\bar F_a r$ on the attained record domain. Preservation of the current output $q=\\bar q r$ is a separate condition. The first protects subsequent computation. The second protects present interpretation.","on*{Protected operations and what a quotient must retain}\nFor an observed deterministic system, let $X$ be a set of realizations, $r:X\\to Z$ a retained record, $F_a:X\\to X$ an action, and $q:X\\to Y$ a protected output. The coarse description is closed under action $a$ only if there is $\\bar F_a:Z\\to Z$ satisfying $rF_a=\\bar F_a r$ on the attained record domain. Preservation of the current output $q=\\bar q r$ is a separate condition. The first protects subsequent computation. The second protects present interpretation.","When these conditions hold for each action, induction transports every finite operation word. When they fail, the mismatch can be recorded as a residual rather than silently suppressed. Linear resolutions make the defect explicit. Let $C$ compress a state, let $R$ reconstruct a retained coordinate with $CR=I$, and let $T,U$ be two linear operations. The excursion $CT(I-RC)UR$ measures how discarded information can return to an observed coordinate after composition. Chapter~\\ref{ch:operations} derives this identity."]}