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Its access is explicit.","$F_a$, $q$ & A state transition and the task observation it must preserve.","$x\\sim_{\\C}y$ & Equality of every task observation after every permitted finite word.","$V\\subseteq\\Omega$ & A set of candidate worlds, including the possible empty set.","$E\\in\\E$ & An admitted evidence event used to filter $V$.","$\\Sigma(V)$ & The complete signature of future task-valued corrections.","$U$, $S_a$ & A calibrated input subspace and a known operator restriction.","$U_\\infty$ & The maximal domain on which calibrated continuation is determined.","$I_Q$ & Initial vectors with common output continuations across fixed models.","$E_\\infty$ & Propagated span of all model-discrepancy directions.","$p_{i,y}$ & Joint probability of a likelihood profile and a target value.","\\section{A calibrated bit with an irrelevant hidden register}\\label{sec:sensor}\nConsider eight possible worlds\n\\[\n\\omega=(s,\\theta,k)\\in\\{0,1\\}^3.\n\\]\nThe physical bit is $s$. The sensor convention is $\\theta$, and $k$ is a nuisance bit that none of the admitted operations or questions uses. The display is\n\\[\ny=s\\oplus\\theta.\n\\]\nThe ordinary action $F$ flips $s$ and leaves the other bits fixed. Its effect on the display is exactly $y\\mapsto1-y$, whatever the unknown convention. Reading the current display therefore initializes a two-state representation that predicts every finite sequence of flips.","rrelevant hidden register}\\label{sec:sensor}\nConsider eight possible worlds\n\\[\n\\omega=(s,\\theta,k)\\in\\{0,1\\}^3.\n\\]\nThe physical bit is $s$. The sensor convention is $\\theta$, and $k$ is a nuisance bit that none of the admitted operations or questions uses. The display is\n\\[\ny=s\\oplus\\theta.\n\\]\nThe ordinary action $F$ flips $s$ and leaves the other bits fixed. Its effect on the display is exactly $y\\mapsto1-y$, whatever the unknown convention. Reading the current display therefore initializes a two-state representation that predicts every finite sequence of flips.","Neither $s$ nor $\\theta$ is identified separately. The two worlds $(0,0,k)$ and $(1,1,k)$ both display zero, and every flip sequence produces the same display in both. That ambiguity is harmless for the original display-prediction contract.","\\subsection{A new question}\nNow ask for the physical bit $s$ as well as the display. The old representation is insufficient. The two previously equivalent worlds require different answers. A permitted nondestructive reference measurement returning $\\theta$ makes the physical bit recoverable by $s=y\\oplus\\theta$. The enriched representation can retain $(y,\\theta)$, or equivalently $(s,\\theta)$. It has four states, not eight. The nuisance bit remains unidentified because no required continuation can detect it.","\\subsection{A state-changing measurement}\nA different operation $Z$ sets $s$ to zero and then reports the new display. The report is $\\theta$. If the old display $y_{\\mathrm{old}}$ was retained, then\n\\[\ns_{\\mathrm{old}}=y_{\\mathrm{old}}\\oplus y_{\\mathrm{after}\\ Z}.\n\\]\nThe current physical state is zero, not $s_{\\mathrm{old}}$. A correct account distinguishes the recovered historical value from the new state it will act on.","If the old display was discarded, worlds with opposite old physical bits but the same $\\theta$ reach the same current state and give the same future reports. The old physical bit is then irrecoverable through this unchanged interface. A known present state has replaced, rather than identified, the unknown past.","\\section{Laws, records, and reusable predictive state}\\label{sec:grounding}\n\\begin{definition}[Finite evidence contract]\\label{def:evidence}\nA finite evidence contract has a nonempty candidate set $\\Omega$, a family $\\Pi$ of permitted protocols, a finite record space $H_\\pi$ for each protocol, and kernels $K^\\pi(h\\mid\\omega)$. A target $t:\\Omega\\to Y$ states what is to be identified. Any hidden model index that is fixed during a protocol is included in $\\omega$. The protocol's action choices are functions of its permitted records and independent randomization, not of the unobserved world.\n\\end{definition}","el{sec:grounding}\n\\begin{definition}[Finite evidence contract]\\label{def:evidence}\nA finite evidence contract has a nonempty candidate set $\\Omega$, a family $\\Pi$ of permitted protocols, a finite record space $H_\\pi$ for each protocol, and kernels $K^\\pi(h\\mid\\omega)$. A target $t:\\Omega\\to Y$ states what is to be identified. Any hidden model index that is fixed during a protocol is included in $\\omega$. The protocol's action choices are functions of its permitted records and independent randomization, not of the unobserved world.\n\\end{definition}","Define $\\omega\\sim_\\Pi\\omega'$ when their record laws agree for every $\\pi\\in\\Pi$. This relation concerns the complete records, including action choices when those are observable. Equality of individual observation marginals is insufficient.","\\begin{proposition}[Law identification and finite-record identification]\\label{prop:law}\nThe target is a function of the complete family of protocol laws if and only if it is constant on the $\\sim_\\Pi$ classes. For a fixed protocol $\\pi$, an error-free rule $d:H_\\pi\\to Y$ exists if and only if $t$ is constant on every nonempty support set\n\\[\nB_\\pi(h)=\\{\\omega:K^\\pi(h\\mid\\omega)>0\\}.\n\\]\nAllowing a randomized output rule does not weaken this condition for probability-one correctness in every world.\n\\end{proposition}\n\\begin{proof}\nFor the first claim, factor $t$ through the map sending a world to its tuple of protocol laws. A factor exists precisely when worlds with the same tuple have the same target. For the second claim, a record with positive probability in two worlds must be given the same output. Distinct targets on its support therefore prevent certainty. Conversely, assign each atta","\\begin{proposition}[Law identification and finite-record identification]\\label{prop:law}\nThe target is a function of the complete family of protocol laws if and only if it is constant on the $\\sim_\\Pi$ classes. For a fixed protocol $\\pi$, an error-free rule $d:H_\\pi\\to Y$ exists if and only if $t$ is constant on every nonempty support set\n\\[\nB_\\pi(h)=\\{\\omega:K^\\pi(h\\mid\\omega)>0\\}.\n\\]\nAllowing a randomized output rule does not weaken this condition for probability-one correctness in every world.\n\\end{proposition}\n\\begin{proof}\nFor the first claim, factor $t$ through the map sending a world to its tuple of protocol laws. A factor exists precisely when worlds with the same tuple have the same target. For the second claim, a record with positive probability in two worlds must be given the same output. Distinct targets on its support therefore prevent certainty. Conversely, assign each attainable record its support's common t","\\begin{proposition}[Law identification and finite-record identification]\\label{prop:law}\nThe target is a function of the complete family of protocol laws if and only if it is constant on the $\\sim_\\Pi$ classes. For a fixed protocol $\\pi$, an error-free rule $d:H_\\pi\\to Y$ exists if and only if $t$ is constant on every nonempty support set\n\\[\nB_\\pi(h)=\\{\\omega:K^\\pi(h\\mid\\omega)>0\\}.\n\\]\nAllowing a randomized output rule does not weaken this condition for probability-one correctness in every world.\n\\end{proposition}\n\\begin{proof}\nFor the first claim, factor $t$ through the map sending a world to its tuple of protocol laws. A factor exists precisely when worlds with the same tuple have the same target. For the second claim, a record with positive probability in two worlds must be given the same output. Distinct targets on its support therefore prevent certainty. Conversely, assign each attainable record its support's common target. Every record generated in a","\\begin{proposition}[Law identification and finite-record identification]\\label{prop:law}\nThe target is a function of the complete family of protocol laws if and only if it is constant on the $\\sim_\\Pi$ classes. For a fixed protocol $\\pi$, an error-free rule $d:H_\\pi\\to Y$ exists if and only if $t$ is constant on every nonempty support set\n\\[\nB_\\pi(h)=\\{\\omega:K^\\pi(h\\mid\\omega)>0\\}.\n\\]\nAllowing a randomized output rule does not weaken this condition for probability-one correctness in every world.\n\\end{proposition}\n\\begin{proof}\nFor the first claim, factor $t$ through the map sending a world to its tuple of protocol laws. A factor exists precisely when worlds with the same tuple have the same target. For the second claim, a record with positive probability in two worlds must be given the same output. Distinct targets on its support therefore prevent certainty. Conversely, assign each attainable record its support's common target. Every record generated in a world receives that world'","\\begin{proposition}[Law identification and finite-record identification]\\label{prop:law}\nThe target is a function of the complete family of protocol laws if and only if it is constant on the $\\sim_\\Pi$ classes. For a fixed protocol $\\pi$, an error-free rule $d:H_\\pi\\to Y$ exists if and only if $t$ is constant on every nonempty support set\n\\[\nB_\\pi(h)=\\{\\omega:K^\\pi(h\\mid\\omega)>0\\}.\n\\]\nAllowing a randomized output rule does not weaken this condition for probability-one correctness in every world.\n\\end{proposition}\n\\begin{proof}\nFor the first claim, factor $t$ through the map sending a world to its tuple of protocol laws. A factor exists precisely when worlds with the same tuple have the same target. For the second claim, a record with positive probability in two worlds must be given the same output. Distinct targets on its support therefore prevent certainty. Conversely, assign each attainable record its support's common target. Every record generated in a world receives that world's target. A randomized error-free rule must place probability one on the required target","ly if it is constant on the $\\sim_\\Pi$ classes. For a fixed protocol $\\pi$, an error-free rule $d:H_\\pi\\to Y$ exists if and only if $t$ is constant on every nonempty support set\n\\[\nB_\\pi(h)=\\{\\omega:K^\\pi(h\\mid\\omega)>0\\}.\n\\]\nAllowing a randomized output rule does not weaken this condition for probability-one correctness in every world.\n\\end{proposition}\n\\begin{proof}\nFor the first claim, factor $t$ through the map sending a world to its tuple of protocol laws. A factor exists precisely when worlds with the same tuple have the same target. For the second claim, a record with positive probability in two worlds must be given the same output. Distinct targets on its support therefore prevent certainty. Conversely, assign each attainable record its support's common target. Every record generated in a world receives that world's target. A randomized error-free rule must place probability one on the required target at every positive-probability record. It cannot do so for two distinct targets at the same record.\n\\end{proof}","\\subsection{A canonical meaning for an operational state}\nLet $X$ be a finite nonempty set, let $\\A$ be a finite operation alphabet, and let each $F_a:X\\to X$ be total and deterministic. Let $q:X\\to Y$ be a finite vector of required observations. Different candidate models can be included by taking a disjoint union of their state sets and preserving each model tag under every action.","For a word $w=a_1\\cdots a_m$, $F_w$ applies $a_1$ first and $a_m$ last. The empty word is the identity. Define\n\\[\nx\\sim_\\C x'\\quad\\Longleftrightarrow\\quad\nq(F_wx)=q(F_wx')\\quad\\text{for every finite }w.\n\\]\nThis is a comparison in a common observation language. It does not require that different models share arbitrary latent coordinates. The behavioral class is a canonical response profile even when its implementation is not identifiable.","\\begin{definition}[Exact initialization from available evidence]\\label{def:initialization}\nGiven an available initial code $r:X\\to H$, an exact reusable predictor consists of a finite state set $Z$, an initialization $d:H\\to Z$, updates $G_a:Z\\to Z$, and an output map $\\bar q:Z\\to Y$ such that\n\\[\n\\bar q\\,G_w\\,d\\,r(x)=q(F_wx)\n\\]\nfor all $x$ and all finite words. After initialization the predictor receives the operation names, but no additional world observations or hidden state.\n\\end{definition}","\\begin{theorem}[Evidence sufficient for one exact predictive realization]\\label{thm:grounding}\nAn exact reusable predictor exists if and only if\n\\[\nr(x)=r(x')\\quad\\Longrightarrow\\quad x\\sim_\\C x'.\\tag{1}\n\\]\nWhen it exists, the behavioral quotient $X/{\\sim_\\C}$ supplies a predictor with the fewest initialized states. Every predictor initialized for all $x\\in X$ requires at least $|X/{\\sim_\\C}|$ distinct values in $d(r(X))$.\n\\end{theorem}\n\\begin{proof}\nEqual initial codes yield equal initialized states. The same operation sequence must then produce equal predicted outputs. Exactness forces equality of all true continuation outputs, proving necessity.","vidence sufficient for one exact predictive realization]\\label{thm:grounding}\nAn exact reusable predictor exists if and only if\n\\[\nr(x)=r(x')\\quad\\Longrightarrow\\quad x\\sim_\\C x'.\\tag{1}\n\\]\nWhen it exists, the behavioral quotient $X/{\\sim_\\C}$ supplies a predictor with the fewest initialized states. Every predictor initialized for all $x\\in X$ requires at least $|X/{\\sim_\\C}|$ distinct values in $d(r(X))$.\n\\end{theorem}\n\\begin{proof}\nEqual initial codes yield equal initialized states. The same operation sequence must then produce equal predicted outputs. Exactness forces equality of all true continuation outputs, proving necessity.","For sufficiency, note that $x\\sim_\\C x'$ implies $F_ax\\sim_\\C F_ax'$ because any subsequent word can be prefixed by $a$. The quotient consequently supports $G_a([x])=[F_ax]$ and $\\bar q([x])=q(x)$. Condition (1) makes $d(r(x))=[x]$ well defined on the attained initial codes. Arbitrary unused codes can be assigned any quotient state. Induction on word length gives the required equality. Finally, two behaviorally different states cannot have the same initialized predictor state, or a distinguishing continuation would receive the same prediction. Thus the number of initialized values is at least the quotient cardinality.\n\\end{proof}","\\subsection{Finite checking does not require an infinite list of words}\nBegin with the partition by $q$. Split any block whose members have successors in different current blocks under some action. A strict split increases the number of nonempty blocks, so at most $|X|-|q(X)|$ strict rounds occur. At a fixed point the quotient updates are well defined and preserve every later observation by induction. Any operation-stable refinement of the original output partition refines each round, so the terminal partition is the behavioral equivalence above. This supplies a finite computation when the complete candidate tables are given. It does not make their empirical acquisition free.","\\subsection{State identity is relative to the chosen questions}\nIn the sensor example, $r=y$ satisfies (1) for display prediction. The quotient has two states even though eight worlds remain in the candidate universe. Adding physical-bit output refines the quotient to four classes and invalidates (1) for that same initial code. Adding the nuisance bit to the target would refine again to eight, but no admitted sensor action reveals it.","immutable worlds, one with a permanently zero output and one with a permanently one output. The available initial record is the same symbol in both. Within each world separately, a one-state predictor exists: it simply prints that world's output. There is nevertheless no predictor that can be initialized from the common record and be correct in both. The obstruction is the order of quantifiers,\n\\[\n\\forall\\omega\\ \\exists P_\\omega\n\\qquad\\text{versus}\\qquad\n\\exists P\\ \\forall\\omega.\n\\]\nThe first does not provide a selector for the second. The behavioral quotient of the combined family has two states, but the evidence cannot choose between them. Adding an output observation repairs the initialization problem without changing the mathematical existence of either predictor.","We first separate these cases in a setting where observations do not interfere with one another. Let $\\Omega$ be finite, let $r:\\Omega\\to Z$ be retained evidence, and let $t:\\Omega\\to Y$ be the revised target. Let $\\E=\\{e_1,\\ldots,e_m\\}$ be a finite family of total, deterministic, nondestructive tests. Each test reads the same fixed world and leaves it unchanged. All tests remain available throughout the investigation. Equivalently, a reset may be allowed if it genuinely restores the same unknown world, including the relevant history. Resetting into a new known state does not satisfy this premise.","For $J\\subseteq\\{1,\\ldots,m\\}$, write\n\\[\nR_J(\\omega)=\\bigl(r(\\omega),(e_j(\\omega))_{j\\in J}\\bigr).\n\\]\nLet the unresolved target pairs be\n\\[\nD=\\{\\{\\omega,\\omega'\\}:r(\\omega)=r(\\omega'),\\ t(\\omega)\\ne t(\\omega')\\}.\n\\]\nA test covers such a pair when it gives different reports in the two worlds.","\\begin{theorem}[Exact revision trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim tha","\\begin{theorem}[Exact revision trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\en","\\begin{theorem}[Exact revision trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is e","\\begin{theorem}[Exact revision trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on eac","\\begin{theorem}[Exact revision trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exa","\\begin{theorem}[Exact revision trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old","trichotomy for a reusable test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains e…","e test interface]\\label{thm:trichotomy}\nExactly one of the following applies.\n\\begin{enumerate}\n\\item $D$ is empty. A new decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test rep…","decoder of the old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a","old code recovers $t$ without new observations.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a separating test","ons.\n\\item $D$ is nonempty, but every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a separating test.","every pair in $D$ is separated by at least one admitted test. Executing all tests recovers $t$, and a selected nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a separating test.","nonadaptive set $J$ suffices exactly when it covers $D$.\n\\item Some pair in $D$ has identical reports under every admitted test. No finite adaptive protocol using this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a separating test.","this unchanged test interface and $r$ guarantees $t$ in every world.\n\\end{enumerate}\nIf test $j$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a separating test.","$ has nonnegative fixed cost $c_j$, the minimum cost of one fixed nonadaptive test set, used for all initial codes in case 2, is precisely\n\\[\n\\min_{J:\\,J\\text{ covers }D}\\ \\sum_{j\\in J}c_j.\\tag{2}\n\\]\nThis is not a claim that an optimal adaptive protocol has the same cost.\n\\end{theorem}\n\\begin{proof}\nIf $D$ is empty, the target is constant on each old-code fibre, giving a decoder. Otherwise a set $J$ permits a decoder exactly when no pair with the same old code and different target remains equal on all its test reports. That is exactly pair coverage. In particular, the full finite test family succeeds when every unresolved pair has a separating test.","\\subsection{Why experimentwise evidence need not compose}\n\\begin{counterexample}[Pairwise separation without a successful investigation]\\label{ce:destructive}\nThere are three possible initial cases $0,1,2$. Test $a$ reports whether the case is $0$, and test $b$ reports whether it is $1$. Either first test then destroys the accessible case state. Every later test returns a fixed zero report. The target remains the initial case, stored only as an inaccessible historical tag.","Every pair is distinguishable by some permitted first test. Yet beginning with $a$ leaves cases $1$ and $2$ indistinguishable whenever its answer is no. Beginning with $b$ leaves cases $0$ and $2$ indistinguishable. No finite adaptive protocol identifies every initial case. Randomizing the first test cannot guarantee success in every case either.\n\\end{counterexample}","Let $\\Omega$ be a finite model set and $t:\\Omega\\to Y$ a fixed target. For a candidate set $V\\subseteq\\Omega$, let\n\\[\nQ(V)=t(V)=\\{t(\\omega):\\omega\\in V\\}.\n\\]\nThe empty set is permitted. It represents a contradiction between the admitted model family and the evidence, not a successful identification. Let $\\E\\subseteq\\Pow(\\Omega)$ be a finite family of evidence events containing $\\Omega$ and closed under intersections. Filtering by $E$ sends $V$ to $V\\cap E$. The events are known, truthfully interpreted, and do not change the underlying model in this section.","and $t:\\Omega\\to Y$ a fixed target. For a candidate set $V\\subseteq\\Omega$, let\n\\[\nQ(V)=t(V)=\\{t(\\omega):\\omega\\in V\\}.\n\\]\nThe empty set is permitted. It represents a contradiction between the admitted model family and the evidence, not a successful identification. Let $\\E\\subseteq\\Pow(\\Omega)$ be a finite family of evidence events containing $\\Omega$ and closed under intersections. Filtering by $E$ sends $V$ to $V\\cap E$. The events are known, truthfully interpreted, and do not change the underlying model in this section.","ega\\to Y$ a fixed target. For a candidate set $V\\subseteq\\Omega$, let\n\\[\nQ(V)=t(V)=\\{t(\\omega):\\omega\\in V\\}.\n\\]\nThe empty set is permitted. It represents a contradiction between the admitted model family and the evidence, not a successful identification. Let $\\E\\subseteq\\Pow(\\Omega)$ be a finite family of evidence events containing $\\Omega$ and closed under intersections. Filtering by $E$ sends $V$ to $V\\cap E$. The events are known, truthfully interpreted, and do not change the underlying model in this section.","Y$ a fixed target. For a candidate set $V\\subseteq\\Omega$, let\n\\[\nQ(V)=t(V)=\\{t(\\omega):\\omega\\in V\\}.\n\\]\nThe empty set is permitted. It represents a contradiction between the admitted model family and the evidence, not a successful identification. Let $\\E\\subseteq\\Pow(\\Omega)$ be a finite family of evidence events containing $\\Omega$ and closed under intersections. Filtering by $E$ sends $V$ to $V\\cap E$. The events are known, truthfully interpreted, and do not change the underlying model in this section.","\\begin{definition}[Correction signature]\\label{def:signature}\nThe correction signature of $V$ is\n\\[\n\\Sigma_{\\E,t}(V)=\\bigl(Q(V\\cap E)\\bigr)_{E\\in\\E}.\\tag{3}\n\\]\nTwo candidate sets have the same signature when every admitted evidence filter leaves the same possible target values, including whether no candidate remains.\n\\end{definition}","\\begin{theorem}[The coarsest exact memory for admitted corrections]\\label{thm:correction}\nEquality of correction signatures is a congruence for all updates $V\\mapsto V\\cap E_0$. The signature therefore supports well-defined updates and a readout of $Q(V)$. It preserves that readout after every finite sequence of admitted filters.","Conversely, any retained representation with deterministic filter updates and a readout reproducing $Q$ after every such sequence must distinguish every two candidate sets with different signatures. Thus the image of $\\Sigma_{\\E,t}$ is the smallest exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","y, any retained representation with deterministic filter updates and a readout reproducing $Q$ after every such sequence must distinguish every two candidate sets with different signatures. Thus the image of $\\Sigma_{\\E,t}$ is the smallest exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","ined representation with deterministic filter updates and a readout reproducing $Q$ after every such sequence must distinguish every two candidate sets with different signatures. Thus the image of $\\Sigma_{\\E,t}$ is the smallest exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","y such sequence must distinguish every two candidate sets with different signatures. Thus the image of $\\Sigma_{\\E,t}$ is the smallest exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","must distinguish every two candidate sets with different signatures. Thus the image of $\\Sigma_{\\E,t}$ is the smallest exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","the image of $\\Sigma_{\\E,t}$ is the smallest exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","t exact state space, in cardinality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","inality, for this correction contract on all candidate subsets.\n\\end{theorem}\n\\begin{proof}\nIf $\\Sigma(V)=\\Sigma(W)$, then for every $E\\in\\E$,\n\\[\nQ((V\\cap E_0)\\cap E)=Q(V\\cap(E_0\\cap E))\n=Q(W\\cap(E_0\\cap E)).\n\\]\nIntersection closure puts $E_0\\cap E$ in $\\E$, so the equality is available from the original signatures. It proves $\\Sigma(V\\cap E_0)=\\Sigma(W\\cap E_0)$. The $E=\\Omega$ coordinate gives $Q(V)$, and induction gives all finite filter sequences.","If two signatures differ, one event $E$ produces different target sets. A retained representation that merged the two candidate sets would update identically on that event and return the same readout, contradicting exactness. Every distinct signature therefore requires a distinct retained value. The signature itself attains that count with the updates just constructed.\n\\end{proof}","\\subsection{An exact count when evidence identifies a partition cell}\nSuppose $A_1,\\ldots,A_m$ are a partition of $\\Omega$ into nonempty cells, and $\\E$ consists exactly of all unions of those cells. Define the attainable evidence--target pairs\n\\[\nD_t=\\{(i,y): y\\in t(A_i)\\},\\qquad r=|D_t|=\\sum_i|t(A_i)|.\n\\]\nA candidate set induces an occupancy record\n\\[\nO(V)=\\{(i,t(\\omega)):\\omega\\in V\\cap A_i\\}.\n\\]","\\begin{corollary}[Exact correction-memory size for partition evidence]\\label{cor:occupancy}\nTwo candidate sets have equal correction signatures if and only if their occupancy records agree. On the domain of all subsets $V\\subseteq\\Omega$, there are exactly $2^r$ distinct signatures. A fixed-length exact binary memory therefore requires at least $r$ bits, and an $r$-bit occupancy vector attains this bound.\n\\end{corollary}\n\\begin{proof}\nFiltering by cell $A_i$ returns exactly the target values occurring in the $i$th part of $O(V)$. Thus equal signatures imply equal occupancy. Conversely, the target set after filtering by any union of cells is the union of their occupied target values, so occupancy determines every signature coordinate.","on evidence]\\label{cor:occupancy}\nTwo candidate sets have equal correction signatures if and only if their occupancy records agree. On the domain of all subsets $V\\subseteq\\Omega$, there are exactly $2^r$ distinct signatures. A fixed-length exact binary memory therefore requires at least $r$ bits, and an $r$-bit occupancy vector attains this bound.\n\\end{corollary}\n\\begin{proof}\nFiltering by cell $A_i$ returns exactly the target values occurring in the $i$th part of $O(V)$. Thus equal signatures imply equal occupancy. Conversely, the target set after filtering by any union of cells is the union of their occupied target values, so occupancy determines every signature coordinate.","equal correction signatures if and only if their occupancy records agree. On the domain of all subsets $V\\subseteq\\Omega$, there are exactly $2^r$ distinct signatures. A fixed-length exact binary memory therefore requires at least $r$ bits, and an $r$-bit occupancy vector attains this bound.\n\\end{corollary}\n\\begin{proof}\nFiltering by cell $A_i$ returns exactly the target values occurring in the $i$th part of $O(V)$. Thus equal signatures imply equal occupancy. Conversely, the target set after filtering by any union of cells is the union of their occupied target values, so occupancy determines every signature coordinate.","ion signatures if and only if their occupancy records agree. On the domain of all subsets $V\\subseteq\\Omega$, there are exactly $2^r$ distinct signatures. A fixed-length exact binary memory therefore requires at least $r$ bits, and an $r$-bit occupancy vector attains this bound.\n\\end{corollary}\n\\begin{proof}\nFiltering by cell $A_i$ returns exactly the target values occurring in the $i$th part of $O(V)$. Thus equal signatures imply equal occupancy. Conversely, the target set after filtering by any union of cells is the union of their occupied target values, so occupancy determines every signature coordinate.","Every subset of $D_t$ is realized: for each included pair choose one world in its nonempty intersection $A_i\\cap t^{-1}(y)$, and choose no worlds for excluded pairs. These intersections are disjoint for different pairs. Hence the occupancy map has $2^r$ values. Distinct values need distinct memory states by the preceding theorem. Binary counting gives the lower bound, and direct occupancy bits attain it.\n\\end{proof}","The exact evidence family matters. Two worlds $a,b$ in one cell with the same target have identical singleton occupancies. Adding the event $\\{a\\}$ distinguishes them: one filtered singleton retains the target and the other is empty. A new event that splits a cell can therefore require finer memory.","\\subsection{Association can be the missing state}\n\\begin{counterexample}[Equal marginals with different corrected answers]\\label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","be the missing state}\n\\begin{counterexample}[Equal marginals with different corrected answers]\\label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","}[Equal marginals with different corrected answers]\\label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","inals with different corrected answers]\\label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","erent corrected answers]\\label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","ected answers]\\label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","label{ce:marginals}\nIn the sensor model, let\n\\[\nV_0=\\{(s,\\theta,k):s\\oplus\\theta=0\\},\\qquad\nV_1=\\{(s,\\theta,k):s\\oplus\\theta=1\\}.\n\\]\nTake the target to be $s$ and the evidence cells to be the two values of $\\theta$. Both candidate sets currently allow physical bits $\\{0,1\\}$, and both allow conventions $\\{0,1\\}$. Yet after the reference reports $\\theta=0$, $V_0$ requires $s=0$ and $V_1$ requires $s=1$.","Here the nuisance bit duplicates every attainable $(\\theta,s)$ pair. There are $2^8=256$ raw candidate subsets, but only $2^4=16$ correction signatures. Four occupancy bits preserve every admitted future correction. The present physical-target set alone has only four possibilities, requiring two bits, but that smaller code is inadequate for the correction contract.","If a future question asks for $k$, the occupancy is insufficient. When every available measurement ignores $k$, expanding the code cannot recover it. A finer signature can exist mathematically without being obtainable from the current interface.","For a fixed protocol, let two worlds with different targets generate finite transcript laws $P_0$ and $P_1$. An estimator sees only that transcript and independent randomness. Its average success under equal prior probabilities is bounded by the discrimination in the entire record.","\\begin{proposition}[A finite-record discrimination bound]\\label{prop:testing}\nFor two distinct target values and equal world probabilities, every transcript-based estimator has average success at most\n\\[\n\\frac{1+\\TV(P_0,P_1)}{2},\\qquad\n\\TV(P_0,P_1)=\\frac12\\sum_h|P_0(h)-P_1(h)|.\\tag{4}\n\\]\nA deterministic function or a common stochastic processing channel applied to the transcript cannot increase this upper bound.\n\\end{proposition}\n\\begin{proof}\nConvert the estimator into a binary test by reporting world zero when it outputs the first target and world one otherwise. A correct target output always gives a correct binary decision. If $f(h)\\in[0,1]$ is the test's chance of declaring world zero, its success is\n\\[\n\\frac12\\left(1+\\sum_h f(h)(P_0(h)-P_1(h))\\right).\n\\]\nThe sum is at most the sum of positive differences, which equals total variation. This proves (4). If $B(h,z)$ is a common processing channel and $g(z)\\in[0,1]$ a processed-record test, then $f(h)=\\sum_zB(h,z)g(z)$ i","(P_0,P_1)=\\frac12\\sum_h|P_0(h)-P_1(h)|.\\tag{4}\n\\]\nA deterministic function or a common stochastic processing channel applied to the transcript cannot increase this upper bound.\n\\end{proposition}\n\\begin{proof}\nConvert the estimator into a binary test by reporting world zero when it outputs the first target and world one otherwise. A correct target output always gives a correct binary decision. If $f(h)\\in[0,1]$ is the test's chance of declaring world zero, its success is\n\\[\n\\frac12\\left(1+\\sum_h f(h)(P_0(h)-P_1(h))\\right).\n\\]\nThe sum is at most the sum of positive differences, which equals total variation. This proves (4). If $B(h,z)$ is a common processing channel and $g(z)\\in[0,1]$ a processed-record test, then $f(h)=\\sum_zB(h,z)g(z)$ is a raw-record test with the same two expectations. Thus every processed decision was already available from the original record.\n\\end{proof}","a common stochastic processing channel applied to the transcript cannot increase this upper bound.\n\\end{proposition}\n\\begin{proof}\nConvert the estimator into a binary test by reporting world zero when it outputs the first target and world one otherwise. A correct target output always gives a correct binary decision. If $f(h)\\in[0,1]$ is the test's chance of declaring world zero, its success is\n\\[\n\\frac12\\left(1+\\sum_h f(h)(P_0(h)-P_1(h))\\right).\n\\]\nThe sum is at most the sum of positive differences, which equals total variation. This proves (4). If $B(h,z)$ is a common processing channel and $g(z)\\in[0,1]$ a processed-record test, then $f(h)=\\sum_zB(h,z)g(z)$ is a raw-record test with the same two expectations. Thus every processed decision was already available from the original record.\n\\end{proof}","o a binary test by reporting world zero when it outputs the first target and world one otherwise. A correct target output always gives a correct binary decision. If $f(h)\\in[0,1]$ is the test's chance of declaring world zero, its success is\n\\[\n\\frac12\\left(1+\\sum_h f(h)(P_0(h)-P_1(h))\\right).\n\\]\nThe sum is at most the sum of positive differences, which equals total variation. This proves (4). If $B(h,z)$ is a common processing channel and $g(z)\\in[0,1]$ a processed-record test, then $f(h)=\\sum_zB(h,z)g(z)$ is a raw-record test with the same two expectations. Thus every processed decision was already available from the original record.\n\\end{proof}","it outputs the first target and world one otherwise. A correct target output always gives a correct binary decision. If $f(h)\\in[0,1]$ is the test's chance of declaring world zero, its success is\n\\[\n\\frac12\\left(1+\\sum_h f(h)(P_0(h)-P_1(h))\\right).\n\\]\nThe sum is at most the sum of positive differences, which equals total variation. This proves (4). If $B(h,z)$ is a common processing channel and $g(z)\\in[0,1]$ a processed-record test, then $f(h)=\\sum_zB(h,z)g(z)$ is a raw-record test with the same two expectations. Thus every processed decision was already available from the original record.\n\\end{proof}","herwise. A correct target output always gives a correct binary decision. If $f(h)\\in[0,1]$ is the test's chance of declaring world zero, its success is\n\\[\n\\frac12\\left(1+\\sum_h f(h)(P_0(h)-P_1(h))\\right).\n\\]\nThe sum is at most the sum of positive differences, which equals total variation. This proves (4). If $B(h,z)$ is a common processing channel and $g(z)\\in[0,1]$ a processed-record test, then $f(h)=\\sum_zB(h,z)g(z)$ is a raw-record test with the same two expectations. Thus every processed decision was already available from the original record.\n\\end{proof}","\\begin{counterexample}[Distinct laws with no finite certain identification]\\label{ce:bernoulli}\nLet the unknown world be $\\omega\\in\\{0,1\\}$. A repeated observation is conditionally independent Bernoulli with success probability $1/4$ in world zero and $3/4$ in world one. The two one-observation laws differ, so the world is identified by its law. Every finite binary string nevertheless has positive probability under both worlds. No fixed finite sample supplies an error-free world classifier.","For odd $n$, majority voting is the likelihood-ratio rule. Its error in either world is\n\\[\ne_n=\\sum_{j=0}^{(n-1)/2}\\binom{n}{j}(3/4)^j(1/4)^{n-j}.\n\\]\nThe first values are $e_1=1/4$, $e_3=5/32$, and $e_5=53/512$. These are exact analytical probabilities, not observed error rates from an agent experiment.\n\\end{counterexample}","Let $v:\\Omega\\to\\mathbb R^d$ record a fixed finite collection of target quantities, where $d\\ge1$. These can be future predictions on specified operation words. For a nonempty $B\\subseteq\\Omega$, define\n\\[\n\\ell_j(B)=\\min_{\\omega\\in B}v_j(\\omega),\\qquad\nu_j(B)=\\max_{\\omega\\in B}v_j(\\omega),\n\\]\n\\[\nm_j(B)=\\frac{\\ell_j(B)+u_j(B)}2,\\qquad\n\\rho(B)=\\frac12\\max_j\\bigl(u_j(B)-\\ell_j(B)\\bigr).\\tag{5}\n\\]\nThe radius concerns coordinatewise prediction in a common frame. It is not an uncertainty measure invariant under arbitrary nonlinear reparameterization.","\\begin{theorem}[Ambiguity radius and an explicit prediction certificate]\\label{thm:radius}\nFor every nonempty finite $B$,\n\\[\n\\inf_{z\\in\\mathbb R^d}\\ \\max_{\\omega\\in B}\\|z-v(\\omega)\\|_\\infty=\\rho(B),\n\\]\nand $m(B)$ attains the infimum. If $B'\\subseteq B$ is nonempty then $\\rho(B')\\le\\rho(B)$.","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candi","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence betwee","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{pr","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\","Fix nonnegative tolerances $\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lo","$\\eta,\\varepsilon$ and $\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half o…","$\\delta_{\\mathrm{id}},\\delta_{\\mathrm{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's wi…","{model}}\\in[0,1]$. Suppose a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains…","se a data record $H$ produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion c…","produces a candidate set $B(H)$ with\n\\[\n\\Pr\\{\\omega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the co…","ega_*\\in B(H)\\}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","}\\ge1-\\delta_{\\mathrm{id}}.\n\\]\nSuppose a second event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","event, of probability at least $1-\\delta_{\\mathrm{model}}$, guarantees\n\\[\n\\sup_{\\omega\\in B(H)}\\|v(\\omega)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","a)-\\widehat v_H(\\omega)\\|_\\infty\\le\\eta.\n\\]\nRead the last supremum as zero when $B(H)$ is empty. When $B(H)$ is nonempty, form the midpoint $\\widehat m$ and radius $\\widehat\\rho$ from $\\widehat v_H$. If the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","the returned numerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","umerical vector $z$ satisfies $\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","$\\|z-\\widehat m\\|_\\infty\\le\\varepsilon$, then\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\\le\\widehat\\rho+\\eta+\\varepsilon\\tag{6}\n\\]\non an event of probability at least $1-\\delta_{\\mathrm{id}}-\\delta_{\\mathrm{model}}$. The rule abstains on an empty candidate set. No independence between the two stated events is required.\n\\end{theorem}\n\\begin{proof}\nFor each coordinate $j$, any point $z_j$ is at distance at least $(u_j-\\ell_j)/2$ from one of the two extremes. Taking the largest coordinate width gives the lower bound. The midpoint is within half of every coordinate's width of every candidate, so it attains that bound. Inclusion can only narrow the coordinate ranges.","On the intersection of the coverage and model-accuracy events, the true world lies in the nonempty set and\n\\[\n\\|z-v(\\omega_*)\\|_\\infty\n\\le\\|z-\\widehat m\\|_\\infty\n+\\|\\widehat m-\\widehat v_H(\\omega_*)\\|_\\infty\n+\\|\\widehat v_H(\\omega_*)-v(\\omega_*)\\|_\\infty.\n\\]\nThe three terms are bounded by $\\varepsilon$, $\\widehat\\rho$, and $\\eta$. The union bound gives the probability statement.\n\\end{proof}","The center is allowed to be any vector in $\\mathbb R^d$. Requiring it to be the prediction of one admissible model can increase the optimum. Nor does a coordinatewise midpoint automatically define a coherent stochastic process or preserve logical invariants. Those are additional constraints on the permitted approximation, not consequences of (5).","\\subsection{What compression does to the candidate set}\nFor a fixed raw-record channel, let $B(h)$ be its exact support set. If only $c(h)$ is retained, the correct support is\n\\[\nB_c(z)=\\bigcup_{h:c(h)=z}B(h).\n\\]\nIndeed, a world can produce the coarse record exactly when it can produce at least one raw record in that fibre. Thus $B(h)\\subseteq B_c(c(h))$, and the ambiguity radius cannot decrease for the same target vector. This is an information statement, not a command to retain every byte. The correction signature in Section~\\ref{sec:memory} identifies one principled way to compress while retaining the specified future uses.","Let $\\Omega$ be finite and nonempty, let $t:\\Omega\\to Y$ have finite image, and let $\\mathcal B$ be a finite collection of experiments. Experiment $e$ has a finite outcome set and specified likelihoods\n\\[\n\\ell_{e,o}(\\omega)\\ge0,\\qquad \\sum_o\\ell_{e,o}(\\omega)=1.\n\\]\nThe world is fixed. Every experiment remains available, and each new report has this conditional law given the world, the chosen experiment, and all earlier reports. This is a fresh conditional-observation premise, not a claim that copied readings become new samples. No experiment changes the physical state in this section.","Partition $\\Omega$ into the nonempty cells $A_1,\\ldots,A_m$ on which the complete likelihood vector $(\\ell_{e,o}(\\omega))_{e,o}$ is constant. Write its common value as $\\lambda_{e,o,i}$. The profile vectors of different cells are distinct by construction. For a prior $\\mu$, retain\n\\[\np_{i,y}(\\mu)=\\sum_{\\omega\\in A_i,\\ t(\\omega)=y}\\mu(\\omega),\n\\qquad (i,y)\\in D_t=\\{(i,y):A_i\\cap t^{-1}(y)\\ne\\varnothing\\}.\n\\]\nThese are joint masses, not the two separate marginals. There are $r=|D_t|$ coordinates with nonnegative entries summing to one. Exact real probabilities are not a fixed $r$-bit code. A bit-complexity claim would also have to account for their representation and growing denominators.","ete likelihood vector $(\\ell_{e,o}(\\omega))_{e,o}$ is constant. Write its common value as $\\lambda_{e,o,i}$. The profile vectors of different cells are distinct by construction. For a prior $\\mu$, retain\n\\[\np_{i,y}(\\mu)=\\sum_{\\omega\\in A_i,\\ t(\\omega)=y}\\mu(\\omega),\n\\qquad (i,y)\\in D_t=\\{(i,y):A_i\\cap t^{-1}(y)\\ne\\varnothing\\}.\n\\]\nThese are joint masses, not the two separate marginals. There are $r=|D_t|$ coordinates with nonnegative entries summing to one. Exact real probabilities are not a fixed $r$-bit code. A bit-complexity claim would also have to account for their representation and growing denominators.","i}$. The profile vectors of different cells are distinct by construction. For a prior $\\mu$, retain\n\\[\np_{i,y}(\\mu)=\\sum_{\\omega\\in A_i,\\ t(\\omega)=y}\\mu(\\omega),\n\\qquad (i,y)\\in D_t=\\{(i,y):A_i\\cap t^{-1}(y)\\ne\\varnothing\\}.\n\\]\nThese are joint masses, not the two separate marginals. There are $r=|D_t|$ coordinates with nonnegative entries summing to one. Exact real probabilities are not a fixed $r$-bit code. A bit-complexity claim would also have to account for their representation and growing denominators.","\\begin{theorem}[Exact target correction from likelihood-profile memory]\\label{thm:likelihoodmemory}\nUnder the fixed-world likelihood model, the record $p=(p_{i,y})$ supports every finite sequence of report predictions and target-posterior corrections. Its next-report probability and update are\n\\[\nz_{e,o}=\\sum_{i,y}\\lambda_{e,o,i}p_{i,y},\\qquad\np'_{i,y}=\\frac{\\lambda_{e,o,i}p_{i,y}}{z_{e,o}}\\quad(z_{e,o}>0),\n\\]\nand its target readout is $\\Pr(t=y)=\\sum_i p_{i,y}$. A zero-probability report is marked impossible rather than assigned a normalized posterior.","rom likelihood-profile memory]\\label{thm:likelihoodmemory}\nUnder the fixed-world likelihood model, the record $p=(p_{i,y})$ supports every finite sequence of report predictions and target-posterior corrections. Its next-report probability and update are\n\\[\nz_{e,o}=\\sum_{i,y}\\lambda_{e,o,i}p_{i,y},\\qquad\np'_{i,y}=\\frac{\\lambda_{e,o,i}p_{i,y}}{z_{e,o}}\\quad(z_{e,o}>0),\n\\]\nand its target readout is $\\Pr(t=y)=\\sum_i p_{i,y}$. A zero-probability report is marked impossible rather than assigned a normalized posterior.","Conversely, any deterministic memory, initialized for every prior on $\\Omega$, that preserves all next-report probabilities and all target posteriors after every finite history must distinguish priors with different joint-mass records. In fact, the joint probabilities of the target and all experiment-report words of length at most $m-1$ already determine the joint-mass record. The memory is therefore minimal in the sense of its induced equivalence on priors, not in machine words or physical storage.\n\\end{theorem}\n\\begin{proof}\nLikelihoods are constant on each profile cell, so summing the ordinary Bayesian update over $A_i\\cap t^{-1}(y)$ gives the displayed update. Its normalization and target readout are exact. Repeating this argument proves sufficiency for any fixe","Conversely, any deterministic memory, initialized for every prior on $\\Omega$, that preserves all next-report probabilities and all target posteriors after every finite history must distinguish priors with different joint-mass records. In fact, the joint probabilities of the target and all experiment-report words of length at most $m-1$ already determine the joint-mass record. The memory is therefore minimal in the sense of its induced equivalence on priors, not in machine words or physical storage.\n\\end{theorem}\n\\begin{proof}\nLikelihoods are constant on each profile cell, so summing the ordinary Bayesian update over $A_i\\cap t^{-1}(y)$ gives the displayed update. Its normalization and target readout are exact. Repeating this argument proves sufficiency for any fixed experiment sequence, and also for choices made from the retained report history without extra world-dependent information.","records. In fact, the joint probabilities of the target and all experiment-report words of length at most $m-1$ already determine the joint-mass record. The memory is therefore minimal in the sense of its induced equivalence on priors, not in machine words or physical storage.\n\\end{theorem}\n\\begin{proof}\nLikelihoods are constant on each profile cell, so summing the ordinary Bayesian update over $A_i\\cap t^{-1}(y)$ gives the displayed update. Its normalization and target readout are exact. Repeating this argument proves sufficiency for any fixed experiment sequence, and also for choices made from the retained report history without extra world-dependent information.","For necessity, a valid predictor and updater determine, for every word $w=((e_1,o_1),\\ldots,(e_k,o_k))$, the masses\n\\[\nM_{\\mu,y}(w)=\\sum_{t(\\omega)=y}\\mu(\\omega)\n                  \\prod_{j=1}^k\\ell_{e_j,o_j}(\\omega).\n\\]\nIndeed, multiply the successive predicted report probabilities by the final target posterior. If an earlier factor is zero, the joint mass is zero and no impossible conditional posterior is needed. The empty word gives the current target distribution.","For each cell $i$ and each $j\\ne i$, choose one likelihood coordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one","For each cell $i$ and each $j\\ne i$, choose one likelihood coordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argu","For each cell $i$ and each $j\\ne i$, choose one likelihood coordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two pri","For each cell $i$ and each $j\\ne i$, choose one likelihood coordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted j","For each cell $i$ and each $j\\ne i$, choose one likelihood coordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted joint masses, hence equal joint-mass records. This proves the claimed separat","For each cell $i$ and each $j\\ne i$, choose one likelihood coordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted joint masses, hence equal joint-mass records. This proves the claimed separation and finite word-length bound.\n\\end{proof}","ordinate $b(i,j)=(e,o)$ for which $\\lambda_{b(i,j),i}\\ne\\lambda_{b(i,j),j}$. Such a coordinate exists because profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted joint masses, hence equal joint-mass records. This proves the claimed separation and finite word-length bound.\n\\end{proof}","ause profile vectors are distinct. On $\\Omega$ the polynomial\n\\[\nP_i(\\omega)=\\prod_{j\\ne i}\n\\frac{\\ell_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted joint masses, hence equal joint-mass records. This proves the claimed separation and finite word-length bound.\n\\end{proof}","l_{b(i,j)}(\\omega)-\\lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted joint masses, hence equal joint-mass records. This proves the claimed separation and finite word-length bound.\n\\end{proof}","lambda_{b(i,j),j}}\n     {\\lambda_{b(i,j),i}-\\lambda_{b(i,j),j}}\n\\]\nis exactly $\\mathbf1_{A_i}(\\omega)$. It is a linear combination of likelihood monomials of degree at most $m-1$. Each monomial is the likelihood product of an admitted fixed experiment-report word. Consequently $p_{i,y}(\\mu)=\\sum_{t(\\omega)=y}\\mu(\\omega)P_i(\\omega)$ is a linear combination of the corresponding $M_{\\mu,y}(w)$. For $m=1$ the empty product is one and the same argument uses only the current target masses. Two priors merged by an exact memory have equal predicted joint masses, hence equal joint-mass records. This proves the claimed separation and finite word-length bound.\n\\end{proof}","The proof uses a finite interpolation argument. Its polynomial coefficients are an identification device, not negative-probability experiments that a learner must execute. All likelihood values are supplied exactly. Very similar profile vectors can make the interpolation numerically ill-conditioned, so the theorem does not provide a stable estimator of $p$ from noisy frequency estimates. Nor does it infer the true likelihood model from a narrow posterior. The distinction between sufficient experiments and compressed statistical evidence is classical~\\cite{blackwell}; the present contract additionally keeps the selected target and all future update operations explicit.","\\subsection{Two identical marginals, two opposite corrections}\nReturn to the sensor worlds $\\omega=(s,\\theta,n)\\in\\{0,1\\}^3$, where $s$ is the target bit, $\\theta$ is a convention, and $n$ is irrelevant. Suppose the only future experiment is a nondestructive noisy reference with\n\\[\n\\Pr(o=1\\mid\\theta=0)=\\tfrac14,\\qquad\n\\Pr(o=1\\mid\\theta=1)=\\tfrac34.\n\\]\nThere are two likelihood-profile cells, one for each convention, and four nonempty profile--target pairs. The eight-world prior can be reduced to four joint masses without losing any specified prediction or correction. The nuisance coordinate disappears.","Consider a prior $\\mu_+$ assigning mass $1/2$ to $(s,\\theta)=(0,0),(1,1)$ and a prior $\\mu_-$ assigning mass $1/2$ to $(1,0),(0,1)$, with any fixed nuisance convention. Both have uniform $s$ and uniform $\\theta$. Both predict the next report $1$ with probability $1/2$. After that report, however,\n\\[\n\\Pr_{\\mu_+}(s=1\\mid o=1)=\\tfrac34,\\qquad\n\\Pr_{\\mu_-}(s=1\\mid o=1)=\\tfrac14.\n\\]\nAfter $k$ fresh reports all equal to one, these become $3^k/(1+3^k)$ and $1/(1+3^k)$. The difference is not a numerical error. It is the association that the separate marginals failed to retain. At every finite $k$, both target values remain possible under either prior.","ass $1/2$ to $(s,\\theta)=(0,0),(1,1)$ and a prior $\\mu_-$ assigning mass $1/2$ to $(1,0),(0,1)$, with any fixed nuisance convention. Both have uniform $s$ and uniform $\\theta$. Both predict the next report $1$ with probability $1/2$. After that report, however,\n\\[\n\\Pr_{\\mu_+}(s=1\\mid o=1)=\\tfrac34,\\qquad\n\\Pr_{\\mu_-}(s=1\\mid o=1)=\\tfrac14.\n\\]\nAfter $k$ fresh reports all equal to one, these become $3^k/(1+3^k)$ and $1/(1+3^k)$. The difference is not a numerical error. It is the association that the separate marginals failed to retain. At every finite $k$, both target values remain possible under either prior.","theta)=(0,0),(1,1)$ and a prior $\\mu_-$ assigning mass $1/2$ to $(1,0),(0,1)$, with any fixed nuisance convention. Both have uniform $s$ and uniform $\\theta$. Both predict the next report $1$ with probability $1/2$. After that report, however,\n\\[\n\\Pr_{\\mu_+}(s=1\\mid o=1)=\\tfrac34,\\qquad\n\\Pr_{\\mu_-}(s=1\\mid o=1)=\\tfrac14.\n\\]\nAfter $k$ fresh reports all equal to one, these become $3^k/(1+3^k)$ and $1/(1+3^k)$. The difference is not a numerical error. It is the association that the separate marginals failed to retain. At every finite $k$, both target values remain possible under either prior.","nd uniform $\\theta$. Both predict the next report $1$ with probability $1/2$. After that report, however,\n\\[\n\\Pr_{\\mu_+}(s=1\\mid o=1)=\\tfrac34,\\qquad\n\\Pr_{\\mu_-}(s=1\\mid o=1)=\\tfrac14.\n\\]\nAfter $k$ fresh reports all equal to one, these become $3^k/(1+3^k)$ and $1/(1+3^k)$. The difference is not a numerical error. It is the association that the separate marginals failed to retain. At every finite $k$, both target values remain possible under either prior.","\\subsection{What a revised evidence language requires}\nLet $\\mathcal P_0$ be the partition of worlds into the nonempty old profile--target cells. Let $\\mathcal P_1$ be the corresponding partition after likelihoods or the target have changed, still on the same finite $\\Omega$. The old record stores the probability of each $\\mathcal P_0$ cell. The new record would store the probability of each $\\mathcal P_1$ cell.","\\begin{corollary}[Exact migration of a probability memory]\\label{cor:memorymigration}\nA deterministic map reconstructing the new record from the old record for every prior exists if and only if every $\\mathcal P_0$ cell is contained in a single $\\mathcal P_1$ cell. When this condition holds, summing the relevant old masses gives the new record. If the new evidence or target strictly splits an old cell, no such universal reconstruction exists.\n\\end{corollary}\n\\begin{proof}\nWhen every old cell lies inside a new one, each new mass is the sum of the old-cell masses it contains. Conversely, suppose two worlds lie in the same old cell but in different new cells. Point-mass priors on those worlds yield the same old record and distinct new records. No deterministic map can send that one old record to both required new records.\n\\end{proof}","\\begin{counterexample}[An old exact memory does not upgrade itself]\\label{ce:memorymigration}\nTwo worlds have the same target and the same old report law, so point masses on them give an identical old record. A newly admitted experiment reports one with probabilities $1/4$ and $3/4$ in the respective worlds. The old record cannot predict its probability, although it was exact for every old question and update. Adding a field for the new likelihood profile does not determine its contents.\n\\end{counterexample}","Let $V$ be a finite-dimensional vector space over a field, and let $\\A$ be a finite nonempty set of operation names. The unknown operations are linear maps $T_a:V\\to V$. Calibration supplies input vectors $x_1,\\ldots,x_m$ and exact outputs $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such exte","Let $V$ be a finite-dimensional vector space over a field, and let $\\A$ be a finite nonempty set of operation names. The unknown operations are linear maps $T_a:V\\to V$. Calibration supplies input vectors $x_1,\\ldots,x_m$ and exact outputs $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such extensions are admissible. There is no hidden norm bound, symmetry,","Let $V$ be a finite-dimensional vector space over a field, and let $\\A$ be a finite nonempty set of operation names. The unknown operations are linear maps $T_a:V\\to V$. Calibration supplies input vectors $x_1,\\ldots,x_m$ and exact outputs $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such extensions are admissible. There is no hidden norm bound, symmetry, invertibility condition, or preferred matrix completion.","to V$. Calibration supplies input vectors $x_1,\\ldots,x_m$ and exact outputs $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such extensions are admissible. There is no hidden norm bound, symmetry, invertibility condition, or preferred matrix completion.","supplies input vectors $x_1,\\ldots,x_m$ and exact outputs $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such extensions are admissible. There is no hidden norm bound, symmetry, invertibility condition, or preferred matrix completion.","$x_1,\\ldots,x_m$ and exact outputs $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such extensions are admissible. There is no hidden norm bound, symmetry, invertibility condition, or preferred matrix completion.","uts $y_{a,1},\\ldots,y_{a,m}$ for every $a$. The operator family is fixed, rather than being reselected after each action. Let\n\\[\nU=\\Span\\{x_1,\\ldots,x_m\\}.\n\\]\nThe calibration is consistent precisely when every linear relation among the inputs is respected by the outputs, for each action. Under that condition it defines maps $S_a:U\\to V$ with $S_ax_i=y_{a,i}$. Every $S_a$ has linear extensions to $V$ by extending a basis. Unless stated otherwise, \\emph{all} such extensions are admissible. There is no hidden norm bound, symmetry, invertibility condition, or preferred matrix completion.","\\begin{proposition}[One-step identification from exact calibration]\\label{prop:onestep}\nFor a fixed action $a$, the value $T_av$ is the same for every linear extension of $S_a$ if and only if $v\\in U$. The common value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","egin{proposition}[One-step identification from exact calibration]\\label{prop:onestep}\nFor a fixed action $a$, the value $T_av$ is the same for every linear extension of $S_a$ if and only if $v\\in U$. The common value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","from exact calibration]\\label{prop:onestep}\nFor a fixed action $a$, the value $T_av$ is the same for every linear extension of $S_a$ if and only if $v\\in U$. The common value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","{prop:onestep}\nFor a fixed action $a$, the value $T_av$ is the same for every linear extension of $S_a$ if and only if $v\\in U$. The common value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","For a fixed action $a$, the value $T_av$ is the same for every linear extension of $S_a$ if and only if $v\\in U$. The common value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","for every linear extension of $S_a$ if and only if $v\\in U$. The common value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","value is then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","then $S_av$.\n\\end{proposition}\n\\begin{proof}\nIf $v\\in U$, all extensions agree by definition. If $v\\notin U$, choose a linear functional $\\ell$ that vanishes on $U$ and satisfies $\\ell(v)=1$, and choose a nonzero vector $b\\in V$. Given any extension $T_a$, the map $T_a'(x)=T_a(x)+\\ell(x)b$ is another extension, but $T_a'v=T_av+b\\ne T_av$. Such a functional exists by extending a basis of $U$ with $v$.\n\\end{proof}","The proposition uses the complete unrestricted extension class. A justified structural prior can determine more. For example, knowing an operator belongs to a smaller parametric family can constrain its action outside $U$. Those constraints must be included as prior information, not inferred from the calibration span alone.","\\subsection{A known first step can leave the known domain}\nEven if $v\\in U$, the vector $S_av$ can lie outside $U$. A second action then acts at an input whose response was not calibrated. One-step accuracy is therefore insufficient for reusable operator meaning.","Define a descending sequence of subspaces\n\\[\nU_0=U,\\qquad\nU_{j+1}=\\{v\\in U:S_av\\in U_j\\text{ for every }a\\in\\A\\}.\\tag{7}\n\\]\nThe sequence is descending by induction. Every strict inclusion decreases dimension, so a fixed point is reached after at most $\\dim U$ strict drops. Denote it by $U_\\infty$.","\\begin{theorem}[The maximal calibrated continuation domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are conse","\\begin{theorem}[The maximal calibrated continuation domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated.","\\begin{theorem}[The maximal calibrated continuation domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are","\\begin{theorem}[The maximal calibrated continuation domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of ever","\\begin{theorem}[The maximal calibrated continuation domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","mal calibrated continuation domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","domain]\\label{thm:core}\nFor every integer $n\\ge1$ and every $v\\in V$, the values $T_wv$ for all operation words of length at most $n$ are independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","independent of the chosen compatible extension family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","sion family if and only if\n\\[\nv\\in U_{n-1}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","}.\n\\]\nConsequently, every finite full-state continuation is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","is identifiable exactly on $U_\\infty$. This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","This is the largest subspace $L\\subseteq U$ such that $S_aL\\subseteq L$ for every $a$. On this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","n this subspace the known restrictions themselves realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","realize the complete operation family.\n\\end{theorem}\n\\begin{proof}\nMembership in $U_{n-1}$ means that every path of at most $n-1$ calibrated operations stays inside $U$. This follows by induction from (7), beginning with $U_0=U$. All inputs on which the first $n$ operations must act are consequently calibrated. Their outputs are determined by the $S_a$, independently of every completion outside $U$.","For necessity, suppose $v\\notin U_{n-1}$. If $v\\notin U$, the one-step proposition supplies two extensions that disagree within one action. Otherwise the recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the e","For necessity, suppose $v\\notin U_{n-1}$. If $v\\notin U$, the one-step proposition supplies two extensions that disagree within one action. Otherwise the recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followe","For necessity, suppose $v\\notin U_{n-1}$. If $v\\notin U$, the one-step proposition supplies two extensions that disagree within one action. Otherwise the recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","ppose $v\\notin U_{n-1}$. If $v\\notin U$, the one-step proposition supplies two extensions that disagree within one action. Otherwise the recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","tin U$, the one-step proposition supplies two extensions that disagree within one action. Otherwise the recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","o extensions that disagree within one action. Otherwise the recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","he recursive definition gives a word of length $j\\le n-1$ with a first exit from $U$. Its intermediate vectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","ectors before exit lie in $U$, so its exit vector $z\\notin U$ is the same under every compatible extension family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","family. Choose any next action $b$. Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.",". Two extensions of $S_b$ can be made to disagree on $z$ by adding a linear map that vanishes on $U$. Keep the other action maps the same. Even if action $b$ occurred earlier, that modification has no effect before the exit because the earlier inputs lie in $U$. The two fixed extension families therefore produce different vectors after the exit word followed by $b$, whose total length is at most $n$. This contradicts uniform continuation identification.","At a fixed point, (7) gives $S_aU_\\infty\\subseteq U_\\infty$. If $L\\subseteq U$ is invariant under every restriction, induction gives $L\\subseteq U_j$ for every $j$, so $L\\subseteq U_\\infty$. Applying the finite-horizon equivalence at all $n$ proves the all-word assertion.\n\\end{proof}","\\subsection{A two-dimensional calibration with a one-dimensional reusable core}\n\\begin{counterexample}[Calibration does not close under continuation]\\label{ce:linear}\nLet $V$ have basis $e_1,e_2,e_3$, let $U=\\Span(e_1,e_2)$, and let the single calibrated operation satisfy\n\\[\nSe_1=e_1,\\qquad Se_2=e_3.\n\\]\nThe first continuation is known on all of $U$. The next calibrated domain is $U_1=\\Span(e_1)$, which is already invariant. One completion has $T_0e_3=0$ and another has $T_1e_3=e_1$. Both agree on every supplied calibration. Starting from $e_2$, their first outputs agree at $e_3$, but their second outputs are respectively zero and $e_1$.\n\\end{counterexample}","one-dimensional reusable core}\n\\begin{counterexample}[Calibration does not close under continuation]\\label{ce:linear}\nLet $V$ have basis $e_1,e_2,e_3$, let $U=\\Span(e_1,e_2)$, and let the single calibrated operation satisfy\n\\[\nSe_1=e_1,\\qquad Se_2=e_3.\n\\]\nThe first continuation is known on all of $U$. The next calibrated domain is $U_1=\\Span(e_1)$, which is already invariant. One completion has $T_0e_3=0$ and another has $T_1e_3=e_1$. Both agree on every supplied calibration. Starting from $e_2$, their first outputs agree at $e_3$, but their second outputs are respectively zero and $e_1$.\n\\end{counterexample}","ore}\n\\begin{counterexample}[Calibration does not close under continuation]\\label{ce:linear}\nLet $V$ have basis $e_1,e_2,e_3$, let $U=\\Span(e_1,e_2)$, and let the single calibrated operation satisfy\n\\[\nSe_1=e_1,\\qquad Se_2=e_3.\n\\]\nThe first continuation is known on all of $U$. The next calibrated domain is $U_1=\\Span(e_1)$, which is already invariant. One completion has $T_0e_3=0$ and another has $T_1e_3=e_1$. Both agree on every supplied calibration. Starting from $e_2$, their first outputs agree at $e_3$, but their second outputs are respectively zero and $e_1$.\n\\end{counterexample}","ion]\\label{ce:linear}\nLet $V$ have basis $e_1,e_2,e_3$, let $U=\\Span(e_1,e_2)$, and let the single calibrated operation satisfy\n\\[\nSe_1=e_1,\\qquad Se_2=e_3.\n\\]\nThe first continuation is known on all of $U$. The next calibrated domain is $U_1=\\Span(e_1)$, which is already invariant. One completion has $T_0e_3=0$ and another has $T_1e_3=e_1$. Both agree on every supplied calibration. Starting from $e_2$, their first outputs agree at $e_3$, but their second outputs are respectively zero and $e_1$.\n\\end{counterexample}","asis $e_1,e_2,e_3$, let $U=\\Span(e_1,e_2)$, and let the single calibrated operation satisfy\n\\[\nSe_1=e_1,\\qquad Se_2=e_3.\n\\]\nThe first continuation is known on all of $U$. The next calibrated domain is $U_1=\\Span(e_1)$, which is already invariant. One completion has $T_0e_3=0$ and another has $T_1e_3=e_1$. Both agree on every supplied calibration. Starting from $e_2$, their first outputs agree at $e_3$, but their second outputs are respectively zero and $e_1$.\n\\end{counterexample}","ngle calibrated operation satisfy\n\\[\nSe_1=e_1,\\qquad Se_2=e_3.\n\\]\nThe first continuation is known on all of $U$. The next calibrated domain is $U_1=\\Span(e_1)$, which is already invariant. One completion has $T_0e_3=0$ and another has $T_1e_3=e_1$. Both agree on every supplied calibration. Starting from $e_2$, their first outputs agree at $e_3$, but their second outputs are respectively zero and $e_1$.\n\\end{counterexample}","Every continuation from $e_1$ is nevertheless determined. The system can ground a genuinely reusable operation on that line while remaining unable to predict repeated operations from $e_2$. A new exact calibration of $e_3$ expands the known input span to $V$ and identifies the whole operator. Obtaining that calibration is a new experiment, not an algebraic consequence of the old data.","\\begin{figure}[tbp]\\centering\n\\includegraphics[width=.95\\linewidth]{figures/calibrated_core.pdf}\n\\caption{A partial calibration can support unlimited exact continuation on a proper invariant core. The known image of $e_2$ leaves the calibrated input subspace, so its next image is unrestricted. Agreement with every supplied one-step calibration is not agreement on two-step behavior.}\\label{fig:core}\n\\end{figure}","The initial vector $v$ is known to the predictor in this section. If the actual input is hidden behind an evidence code, its initialization must satisfy an additional factorization condition like (1). A calibrated operation and an unobserved initial state are different missing inputs. They cannot be exchanged merely because both appear as uncertainty about the next output.","\\section{Why small local error needs an ambient growth bound}\\label{sec:approx}\nExact calibration can justify a sharp domain. Approximate calibration needs a quantitative statement about errors that leave it. Let $V$ be a normed vector space and $L\\subseteq V$ a linear subspace invariant under known bounded linear maps $S_a:L\\to L$. Suppose the actual bounded linear maps $T_a:V\\to V$ satisfy\n\\[\n\\|T_a\\|\\le A,\\qquad \\|S_a\\|\\le B,\\qquad\n\\|(T_a-S_a)v\\|\\le\\eta\\|v\\|\\quad(v\\in L),\\tag{8}\n\\]\nfor common nonnegative bounds $A,B,\\eta$. The norm on $T_a$ is an ambient bound. Its purpose is to control propagation of errors outside $L$, not only behavior on calibrated inputs.","main. Approximate calibration needs a quantitative statement about errors that leave it. Let $V$ be a normed vector space and $L\\subseteq V$ a linear subspace invariant under known bounded linear maps $S_a:L\\to L$. Suppose the actual bounded linear maps $T_a:V\\to V$ satisfy\n\\[\n\\|T_a\\|\\le A,\\qquad \\|S_a\\|\\le B,\\qquad\n\\|(T_a-S_a)v\\|\\le\\eta\\|v\\|\\quad(v\\in L),\\tag{8}\n\\]\nfor common nonnegative bounds $A,B,\\eta$. The norm on $T_a$ is an ambient bound. Its purpose is to control propagation of errors outside $L$, not only behavior on calibrated inputs.","ation needs a quantitative statement about errors that leave it. Let $V$ be a normed vector space and $L\\subseteq V$ a linear subspace invariant under known bounded linear maps $S_a:L\\to L$. Suppose the actual bounded linear maps $T_a:V\\to V$ satisfy\n\\[\n\\|T_a\\|\\le A,\\qquad \\|S_a\\|\\le B,\\qquad\n\\|(T_a-S_a)v\\|\\le\\eta\\|v\\|\\quad(v\\in L),\\tag{8}\n\\]\nfor common nonnegative bounds $A,B,\\eta$. The norm on $T_a$ is an ambient bound. Its purpose is to control propagation of errors outside $L$, not only behavior on calibrated inputs.","d vector space and $L\\subseteq V$ a linear subspace invariant under known bounded linear maps $S_a:L\\to L$. Suppose the actual bounded linear maps $T_a:V\\to V$ satisfy\n\\[\n\\|T_a\\|\\le A,\\qquad \\|S_a\\|\\le B,\\qquad\n\\|(T_a-S_a)v\\|\\le\\eta\\|v\\|\\quad(v\\in L),\\tag{8}\n\\]\nfor common nonnegative bounds $A,B,\\eta$. The norm on $T_a$ is an ambient bound. Its purpose is to control propagation of errors outside $L$, not only behavior on calibrated inputs.","\\begin{proposition}[Approximate continuation with explicit leakage control]\\label{prop:approx}\nFor every $v\\in L$ and every word $w$ of length $n\\ge1$,\n\\[\n\\|T_wv-S_wv\\|\n\\le\\eta\\|v\\|\\sum_{j=0}^{n-1}A^{n-1-j}B^j.\\tag{9}\n\\]\nFor the empty word the discrepancy is zero. Integer zero powers in the displayed finite sum have their usual value one. The conclusion does not require the actual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}","oposition}[Approximate continuation with explicit leakage control]\\label{prop:approx}\nFor every $v\\in L$ and every word $w$ of length $n\\ge1$,\n\\[\n\\|T_wv-S_wv\\|\n\\le\\eta\\|v\\|\\sum_{j=0}^{n-1}A^{n-1-j}B^j.\\tag{9}\n\\]\nFor the empty word the discrepancy is zero. Integer zero powers in the displayed finite sum have their usual value one. The conclusion does not require the actual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}","plicit leakage control]\\label{prop:approx}\nFor every $v\\in L$ and every word $w$ of length $n\\ge1$,\n\\[\n\\|T_wv-S_wv\\|\n\\le\\eta\\|v\\|\\sum_{j=0}^{n-1}A^{n-1-j}B^j.\\tag{9}\n\\]\nFor the empty word the discrepancy is zero. Integer zero powers in the displayed finite sum have their usual value one. The conclusion does not require the actual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}","rox}\nFor every $v\\in L$ and every word $w$ of length $n\\ge1$,\n\\[\n\\|T_wv-S_wv\\|\n\\le\\eta\\|v\\|\\sum_{j=0}^{n-1}A^{n-1-j}B^j.\\tag{9}\n\\]\nFor the empty word the discrepancy is zero. Integer zero powers in the displayed finite sum have their usual value one. The conclusion does not require the actual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}","\\ge1$,\n\\[\n\\|T_wv-S_wv\\|\n\\le\\eta\\|v\\|\\sum_{j=0}^{n-1}A^{n-1-j}B^j.\\tag{9}\n\\]\nFor the empty word the discrepancy is zero. Integer zero powers in the displayed finite sum have their usual value one. The conclusion does not require the actual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}",". The conclusion does not require the actual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}","tual trajectory to remain in $L$.\n\\end{proposition}\n\\begin{proof}\nWrite $z_j=S_{a_j}\\cdots S_{a_1}v\\in L$ and $x_j=T_{a_j}\\cdots T_{a_1}v$. For the next operation,\n\\[\n\\|x_{j+1}-z_{j+1}\\|\n\\le A\\|x_j-z_j\\|+\\eta\\|z_j\\|\n\\le A\\|x_j-z_j\\|+\\eta B^j\\|v\\|.\n\\]\nThe initial discrepancy is zero. Iterating this scalar recurrence gives (9). Ambient boundedness applies to $x_j-z_j$ even when that error has left $L$.\n\\end{proof}","\\begin{counterexample}[An uncalibrated direction amplifies a small residual]\\label{ce:growth}\nLet $L=\\Span(e_1)$, let $Se_1=e_1$, and in the $\\ell^1$ norm define\n\\[\nT=\\begin{pmatrix}1&0","\\begin{counterexample}[An uncalibrated direction amplifies a small residual]\\label{ce:growth}\nLet $L=\\Span(e_1)$, let $Se_1=e_1$, and in the $\\ell^1$ norm define\n\\[\nT=\\begin{pmatrix}1&0\\\\ \\eta&M\\end{pmatrix},\\qquad \\eta>0,\\ M\\ge0.\n\\]\nThe one-step error on the unit calibrated vector is $\\eta$, independent of $M$. After two steps the error is $\\eta(1+M)$. It can be arbitrarily large while the local calibration error is fixed. For $n$ steps the second coordinate is $\\eta\\sum_{j=0}^{n-1}M^j$.\n\\end{counterexample}","\\eta&M\\end{pmatrix},\\qquad \\eta>0,\\ M\\ge0.\n\\]\nThe one-step error on the unit calibrated vector is $\\eta$, independent of $M$. After two steps the error is $\\eta(1+M)$. It can be arbitrarily large while the local calibration error is fixed. For $n$ steps the second coordinate is $\\eta\\sum_{j=0}^{n-1}M^j$.\n\\end{counterexample}","Let $Q:V\\to Y$ be a fixed linear observation. The initial vector $v$ remains known. All linear extensions of every calibrated restriction $S_a$ remain permitted, independently across action names but fixed throughout an execution.","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action b","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ pre","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibratio","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed outp","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disa","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen ane","\\begin{proposition}[No enlargement under unrestricted linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during e","linear completion]\\label{prop:noenlargement}\nIf $Q\\ne0$, then for every $h\\ge1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","1$, the vectors $QT_wv$ are independent of the compatible extension family for every word of length at most $h$ if and only if $v\\in U_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","_{h-1}$. In particular, the all-word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","word output-identifiable domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","able domain is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","in is exactly $U_\\infty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","ty$. If $Q=0$, every initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","very initial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","ial vector is trivially output-identifiable.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","able.\n\\end{proposition}\n\\begin{proof}\nOn $U_{h-1}$ the full vectors are determined by Theorem~\\ref{thm:core}, so their $Q$ outputs are determined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","termined. Outside that subspace, follow the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","the same first-exit construction as in that proof. At an uncalibrated input $z$, choose a linear functional $\\ell$ vanishing on $U$ with $\\ell(z)=1$. Because $Q\\ne0$, choose $b$ with $Qb\\ne0$. Replacing the next action by $T_a+\\ell(\\cdot)b$ preserves all calibration data and every earlier step of the first-exit path, but changes the next observed output by $Qb$. This produces disagreement within $h$ actions. Each modified action map is fixed, not chosen anew during execution. The zero-output case follows immediately.\n\\end{proof}","\\section{A finite test for agreement under a fixed unknown model}\\label{sec:pairtest}\nLet $V$ be $n$-dimensional over a field, let $\\Theta$ be a finite nonempty model set, and let $T_a^\\theta:V\\to V$ be specified linear maps for each model $\\theta$ and action $a$ in a finite nonempty alphabet. All words are allowed. Choose a reference model $0\\in\\Theta$ and write $A_a=T_a^0$. The shared linear observation is $Q:V\\to Y$. Define\n\\[\nI_Q=\\{v\\in V:QT_w^\\theta v=QA_wv\\text{ for every }\\theta\\in\\Theta\\text{ and every word }w\\}.\n\\]\nThis is an initial-condition subspace. It is not assumed invariant under the physical maps. The common coordinates are part of this linear comparison model. With unrelated latent carriers, a justified identification of the carriers would be an additional premise","\\section{A finite test for agreement under a fixed unknown model}\\label{sec:pairtest}\nLet $V$ be $n$-dimensional over a field, let $\\Theta$ be a finite nonempty model set, and let $T_a^\\theta:V\\to V$ be specified linear maps for each model $\\theta$ and action $a$ in a finite nonempty alphabet. All words are allowed. Choose a reference model $0\\in\\Theta$ and write $A_a=T_a^0$. The shared linear observation is $Q:V\\to Y$. Define\n\\[\nI_Q=\\{v\\in V:QT_w^\\theta v=QA_wv\\text{ for every }\\theta\\in\\Theta\\text{ and every word }w\\}.\n\\]\nThis is an initial-condition subspace. It is not assumed invariant under the physical maps. The common coordinates are part of this linear comparison model. With unrelated latent carriers, a justified identification of the carriers would be an additional premise.","nite test for agreement under a fixed unknown model}\\label{sec:pairtest}\nLet $V$ be $n$-dimensional over a field, let $\\Theta$ be a finite nonempty model set, and let $T_a^\\theta:V\\to V$ be specified linear maps for each model $\\theta$ and action $a$ in a finite nonempty alphabet. All words are allowed. Choose a reference model $0\\in\\Theta$ and write $A_a=T_a^0$. The shared linear observation is $Q:V\\to Y$. Define\n\\[\nI_Q=\\{v\\in V:QT_w^\\theta v=QA_wv\\text{ for every }\\theta\\in\\Theta\\text{ and every word }w\\}.\n\\]\nThis is an initial-condition subspace. It is not assumed invariant under the physical maps. The common coordinates are part of this linear comparison model. With unrelated latent carriers, a justified identification of the carriers would be an additional premise.","ent under a fixed unknown model}\\label{sec:pairtest}\nLet $V$ be $n$-dimensional over a field, let $\\Theta$ be a finite nonempty model set, and let $T_a^\\theta:V\\to V$ be specified linear maps for each model $\\theta$ and action $a$ in a finite nonempty alphabet. All words are allowed. Choose a reference model $0\\in\\Theta$ and write $A_a=T_a^0$. The shared linear observation is $Q:V\\to Y$. Define\n\\[\nI_Q=\\{v\\in V:QT_w^\\theta v=QA_wv\\text{ for every }\\theta\\in\\Theta\\text{ and every word }w\\}.\n\\]\nThis is an initial-condition subspace. It is not assumed invariant under the physical maps. The common coordinates are part of this linear comparison model. With unrelated latent carriers, a justified identification of the carriers would be an additional premise.","For each $\\theta$, form the paired maps and difference observation\n\\[\nB_a^\\theta=\\begin{pmatrix}T_a^\\theta&0","For each $\\theta$, form the paired maps and difference observation\n\\[\nB_a^\\theta=\\begin{pmatrix}T_a^\\theta&0\\\\0&A_a\\end{pmatrix},\\qquad\nC=\\begin{pmatrix}Q&-Q\\end{pmatrix},\\qquad\n\\Delta v=(v,v).\n\\]\nThe pair is evolved by the \\emph{same} action word in both components. Define\n\\[\nK_0^\\theta=\\ker C,\\qquad\nK_{j+1}^\\theta=K_j^\\theta\\cap\\bigcap_a(B_a^\\theta)^{-1}K_j^\\theta.\n\\]","0&A_a\\end{pmatrix},\\qquad\nC=\\begin{pmatrix}Q&-Q\\end{pmatrix},\\qquad\n\\Delta v=(v,v).\n\\]\nThe pair is evolved by the \\emph{same} action word in both components. Define\n\\[\nK_0^\\theta=\\ker C,\\qquad\nK_{j+1}^\\theta=K_j^\\theta\\cap\\bigcap_a(B_a^\\theta)^{-1}K_j^\\theta.\n\\]","\\begin{theorem}[Finite output-agreement test with persistent model tags]\\label{thm:pairtest}\nThe descending paired-kernel sequence stabilizes in at most $2n$ strict dimension drops. Its terminal subspace is\n\\[\nK_\\infty^\\theta=\\bigcap_w\\ker(CB_w^\\theta),\\qquad\nI_Q=\\bigcap_{\\theta\\in\\Theta}\\Delta^{-1}K_\\infty^\\theta.\n\\]\nThus exact all-word output agreement for the finite candidate family is decidable by finite linear algebra. If an initial vector is excluded, a disagreeing candidate and a distinguishing word can be extracted. Words of length at most $2n$ suffice for this test. The conclusion compares fixed candidate models; it does not admit arbitrary switching between candidates during a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$","\\begin{theorem}[Finite output-agreement test with persistent model tags]\\label{thm:pairtest}\nThe descending paired-kernel sequence stabilizes in at most $2n$ strict dimension drops. Its terminal subspace is\n\\[\nK_\\infty^\\theta=\\bigcap_w\\ker(CB_w^\\theta),\\qquad\nI_Q=\\bigcap_{\\theta\\in\\Theta}\\Delta^{-1}K_\\infty^\\theta.\n\\]\nThus exact all-word output agreement for the finite candidate family is decidable by finite linear algebra. If an initial vector is excluded, a disagreeing candidate and a distinguishing word can be extracted. Words of length at most $2n$ suffice for this test. The conclusion compares fixed candidate models; it does not admit arbitrary switching between candidates during a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$. The initial case is the empty word. Membership in the next intersection says that the present difference is zero and that applying any first action leaves a pa","n$ strict dimension drops. Its terminal subspace is\n\\[\nK_\\infty^\\theta=\\bigcap_w\\ker(CB_w^\\theta),\\qquad\nI_Q=\\bigcap_{\\theta\\in\\Theta}\\Delta^{-1}K_\\infty^\\theta.\n\\]\nThus exact all-word output agreement for the finite candidate family is decidable by finite linear algebra. If an initial vector is excluded, a disagreeing candidate and a distinguishing word can be extracted. Words of length at most $2n$ suffice for this test. The conclusion compares fixed candidate models; it does not admit arbitrary switching between candidates during a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$. The initial case is the empty word. Membership in the next intersection says that the present difference is zero and that applying any first action leaves a pair whose next $j$ actions have zero difference. This is exactly the length-$j+1$ condition.","t agreement for the finite candidate family is decidable by finite linear algebra. If an initial vector is excluded, a disagreeing candidate and a distinguishing word can be extracted. Words of length at most $2n$ suffice for this test. The conclusion compares fixed candidate models; it does not admit arbitrary switching between candidates during a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$. The initial case is the empty word. Membership in the next intersection says that the present difference is zero and that applying any first action leaves a pair whose next $j$ actions have zero difference. This is exactly the length-$j+1$ condition.","cluded, a disagreeing candidate and a distinguishing word can be extracted. Words of length at most $2n$ suffice for this test. The conclusion compares fixed candidate models; it does not admit arbitrary switching between candidates during a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$. The initial case is the empty word. Membership in the next intersection says that the present difference is zero and that applying any first action leaves a pair whose next $j$ actions have zero difference. This is exactly the length-$j+1$ condition.","does not admit arbitrary switching between candidates during a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$. The initial case is the empty word. Membership in the next intersection says that the present difference is zero and that applying any first action leaves a pair whose next $j$ actions have zero difference. This is exactly the length-$j+1$ condition.","a word.\n\\end{theorem}\n\\begin{proof}\nBy induction, $K_j^\\theta$ consists precisely of paired states with zero difference output for all words of length at most $j$. The initial case is the empty word. Membership in the next intersection says that the present difference is zero and that applying any first action leaves a pair whose next $j$ actions have zero difference. This is exactly the length-$j+1$ condition.","Every strict inclusion lowers dimension in $V\\oplus V$, which has dimension $2n$. Once two successive subspaces agree, the subspace is invariant under every $B_a^\\theta$ and lies in $\\ker C$, so all later intersections agree. This proves the all-word formula. At the diagonal initial pair, $CB_w^\\theta\\Delta v=QT_w^\\theta v-QA_wv$, giving the paired-kernel formula. Row-space construction retains the operation word producing each tested row. If a vector fails the terminal kernel test, one such row is nonzero on it and supplies a witness. The dimension argument gives the stated conservative length bound.\n\\end{proof}","\\subsection{Why the unknown model must not change between steps}\n\\begin{counterexample}[A switching adversary creates a nonexistent failure]\\label{ce:switching}\nOver $\\mathbb Q$, let there be one action and two possible fixed matrices\n\\[\nA=\\begin{pmatrix}0&1","\\subsection{Why the unknown model must not change between steps}\n\\begin{counterexample}[A switching adversary creates a nonexistent failure]\\label{ce:switching}\nOver $\\mathbb Q$, let there be one action and two possible fixed matrices\n\\[\nA=\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix},\\qquad\nB=\\begin{pmatrix}0&0\\\\1&0\\end{pmatrix},\\qquad\nQ=\\begin{pmatrix}1&1\\end{pmatrix},\\qquad v=\\binom11.\n\\]\nBoth fixed systems have output sequence $2,1,0,0,\\ldots$ from $v$, since $A^2=B^2=0$. But applying $A$ first and $B$ second gives $QBAv=1$, not zero. That hybrid trajectory belongs to neither candidate model. Treating fixed uncertainty as a freely changing matrix would incorrectly reject the valid output prediction.\n\\end{counterexample}","1&0\\end{pmatrix},\\qquad\nQ=\\begin{pmatrix}1&1\\end{pmatrix},\\qquad v=\\binom11.\n\\]\nBoth fixed systems have output sequence $2,1,0,0,\\ldots$ from $v$, since $A^2=B^2=0$. But applying $A$ first and $B$ second gives $QBAv=1$, not zero. That hybrid trajectory belongs to neither candidate model. Treating fixed uncertainty as a freely changing matrix would incorrectly reject the valid output prediction.\n\\end{counterexample}","Here $I_Q=\\Span\\{(1,1)\\}$, obtained already from the one-step equality. It is not invariant under $A$ or $B$: $Av=(1,0)$ and $Bv=(0,1)$ leave that line. Those new physical vectors cannot be treated as fresh initial conditions with a forgotten model history. For example, starting both models at $(1,0)$ gives different next outputs. The original fixed-model executions nevertheless remain in output agreement.","This is why the identifiable initial set is not automatically a common physical invariant core. A reference predictor may still work: it follows $A$ from the initial vector and reports its outputs. It represents the common response sequence, not a claim that the unknown physical state always equals the reference state.","\\begin{figure}[tbp]\\centering\n\\includegraphics[width=.95\\linewidth]{figures/fixed_model_outputs.pdf}\n\\caption{Two fixed candidate systems produce the same complete output history from $(1,1)$ even though their hidden states differ. Switching from one candidate to the other after the first step creates an inadmissible history with a different second output. The fixed-model tag is information about the allowed trajectories.}\\label{fig:fixedmodels}\n\\end{figure}","\\subsection{More data restricts the model family, not the logical quantifier}\nIf new exact evidence removes candidates while preserving the reference model, the intersection defining $I_Q$ can only enlarge. Its practical value depends on whether the learner can justify that removal and initialize the resulting predictor. Adding a new candidate has the opposite effect. Neither operation warrants replacing the model label at each step. A candidate family that omits the true system can still yield an impressive but inapplicable agreement subspace.","\\section{When uncertain dynamics admit one common observable quotient}\\label{sec:commonquotient}\nAgreement on selected initial conditions is weaker than agreement on the whole space. The stronger problem has a useful exact solution. Retain the finite family and reference matrices of Section~\\ref{sec:pairtest}, and define the difference maps\n\\[\nD_a^\\theta=T_a^\\theta-A_a.\n\\]\nTheir ranges contain all possible one-step discrepancies. Let\n\\[\nE_0=\\sum_{a,\\theta}\\im D_a^\\theta,\\qquad\nE_{j+1}=E_j+\\sum_a A_aE_j,\\qquad E_\\infty=\\bigcup_jE_j.\n\\]\nThis increasing sequence stabilizes after at most $n$ strict dimension increases. Its limit is the smallest subspace invariant under all reference actions and containing every discrepancy range. It includes discrepancies that are invisible now but return to the output later.","ne common observable quotient}\\label{sec:commonquotient}\nAgreement on selected initial conditions is weaker than agreement on the whole space. The stronger problem has a useful exact solution. Retain the finite family and reference matrices of Section~\\ref{sec:pairtest}, and define the difference maps\n\\[\nD_a^\\theta=T_a^\\theta-A_a.\n\\]\nTheir ranges contain all possible one-step discrepancies. Let\n\\[\nE_0=\\sum_{a,\\theta}\\im D_a^\\theta,\\qquad\nE_{j+1}=E_j+\\sum_a A_aE_j,\\qquad E_\\infty=\\bigcup_jE_j.\n\\]\nThis increasing sequence stabilizes after at most $n$ strict dimension increases. Its limit is the smallest subspace invariant under all reference actions and containing every discrepancy range. It includes discrepancies that are invisible now but return to the output later.","ger problem has a useful exact solution. Retain the finite family and reference matrices of Section~\\ref{sec:pairtest}, and define the difference maps\n\\[\nD_a^\\theta=T_a^\\theta-A_a.\n\\]\nTheir ranges contain all possible one-step discrepancies. Let\n\\[\nE_0=\\sum_{a,\\theta}\\im D_a^\\theta,\\qquad\nE_{j+1}=E_j+\\sum_a A_aE_j,\\qquad E_\\infty=\\bigcup_jE_j.\n\\]\nThis increasing sequence stabilizes after at most $n$ strict dimension increases. Its limit is the smallest subspace invariant under all reference actions and containing every discrepancy range. It includes discrepancies that are invisible now but return to the output later.","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outpu","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the cand","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between a","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agree","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\b","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition","\\begin{theorem}[Exact common quotient for structured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement","ctured model uncertainty]\\label{thm:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=…","m:commonquotient}\nThe following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n…","following conditions are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality i…","s are equivalent.\n\\begin{enumerate}\n\\item $QT_w^\\theta=QA_w$ on all of $V$, for every fixed model $\\theta$ and every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for t…","every word $w$.\n\\item $E_\\infty\\subseteq\\ker Q$.\n\\item There exists a subspace $N\\subseteq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is …","eq\\ker Q$, invariant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","riant under every $A_a$, containing every $\\im D_a^\\theta$.\n\\item There is a surjective linear representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","ar representation $r:V\\to Z$, shared linear updates $G_a$, and a readout $\\bar Q$ such that\n\\[\nrT_a^\\theta=G_ar\\quad\\text{for all }a,\\theta,\\qquad Q=\\bar Qr.\n\\]\n\\end{enumerate}\nWhen the conditions hold, $V/E_\\infty$ gives one such quotient. The smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","he smallest-dimensional common linear quotient is $V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","$V/N_0$, where $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","here $N_0=\\bigcap_w\\ker(QA_w)$ is the reference system's unobservable subspace. Its dimension is $n-\\dim N_0$. The shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","shared quotient also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","nt also preserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","eserves outputs if the candidate model is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","del is changed between actions, a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","a stronger conclusion than fixed-model agreement on a restricted initial set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","al set.\n\\end{theorem}\n\\begin{proof}\nAssume condition 1. For any $a,\\theta,w$, agreement as linear maps on all of $V$ gives\n\\[\nQA_wD_a^\\theta\n=QT_w^\\theta T_a^\\theta-QA_wA_a=0.\n\\]\nThe last equality is condition 1 for the word that executes $a$ and then $w$. Thus $Q$ annihilates the range of every $A_wD_a^\\theta$, whose span is $E_\\infty$, proving condition 2. Condition 2 supplies condition 3 with $N=E_\\infty$.","Under condition 3, $T_a^\\theta N\\subseteq N$ because $T_a^\\theta=A_a+D_a^\\theta$, $A_aN\\subseteq N$, and the entire range of $D_a^\\theta$ lies in $N$. Every candidate therefore induces the same map $G_a([v])=[A_av]$ on $V/N$. Since $QN=0$, the output factors through that quotient. This proves condition 4. Composing its commuting equations proves condition 1. The same composition works with a different model index at each action because $r$ and $G_a$ are shared.","N$ because $T_a^\\theta=A_a+D_a^\\theta$, $A_aN\\subseteq N$, and the entire range of $D_a^\\theta$ lies in $N$. Every candidate therefore induces the same map $G_a([v])=[A_av]$ on $V/N$. Since $QN=0$, the output factors through that quotient. This proves condition 4. Composing its commuting equations proves condition 1. The same composition works with a different model index at each action because $r$ and $G_a$ are shared.","use $T_a^\\theta=A_a+D_a^\\theta$, $A_aN\\subseteq N$, and the entire range of $D_a^\\theta$ lies in $N$. Every candidate therefore induces the same map $G_a([v])=[A_av]$ on $V/N$. Since $QN=0$, the output factors through that quotient. This proves condition 4. Composing its commuting equations proves condition 1. The same composition works with a different model index at each action because $r$ and $G_a$ are shared.","The subspace $N_0$ is invariant under every $A_a$ and lies in $\\ker Q$. Every invariant subspace in $\\ker Q$, including $E_\\infty$ when condition 2 holds, lies in $N_0$. Hence $N_0$ gives a valid common quotient. Conversely, the kernel of any representation in condition 4 is invariant under the reference actions and contained in $\\ker Q$, so it lies in $N_0$. Rank-nullity gives the lower bound $\\dim Z\\ge n-\\dim N_0$, attained by $V/N_0$.\n\\end{proof}","der every $A_a$ and lies in $\\ker Q$. Every invariant subspace in $\\ker Q$, including $E_\\infty$ when condition 2 holds, lies in $N_0$. Hence $N_0$ gives a valid common quotient. Conversely, the kernel of any representation in condition 4 is invariant under the reference actions and contained in $\\ker Q$, so it lies in $N_0$. Rank-nullity gives the lower bound $\\dim Z\\ge n-\\dim N_0$, attained by $V/N_0$.\n\\end{proof}","\\subsection{Large hidden variation with exact visible behavior}\nFor a structured example over $\\mathbb Q$, take either of the fixed maps\n\\[\nT_0=\\begin{pmatrix}1&1&0","\\subsection{Large hidden variation with exact visible behavior}\nFor a structured example over $\\mathbb Q$, take either of the fixed maps\n\\[\nT_0=\\begin{pmatrix}1&1&0\\\\0&1&0\\\\0&0&0\\end{pmatrix},\\qquad\nT_1=\\begin{pmatrix}1&1&0\\\\0&1&0\\\\2&-1&3\\end{pmatrix},\\qquad\nQ=\\begin{pmatrix}1&0&0\\end{pmatrix}.\n\\]\nThe hidden third-coordinate dynamics are different, and $T_1$ can make that coordinate grow. Nevertheless the discrepancy range is $\\Span\\{e_3\\}$, which the reference leaves inside itself and $Q$ annihilates. The common quotient retains $(x_1,x_2)$, with update\n\\[\nG(x_1,x_2)=(x_1+x_2,x_2).\n\\]\nThe output at step $k$ is $x_1+kx_2$ under both models, and even under switching between them. The output alone is not an initialized reusable state: equal $x_1$ with different $x_2$ gives different next outputs. Ignoring $x_3$ is justified by the no-return structure, not by its small size or its being temporarily unobserved.","2&-1&3\\end{pmatrix},\\qquad\nQ=\\begin{pmatrix}1&0&0\\end{pmatrix}.\n\\]\nThe hidden third-coordinate dynamics are different, and $T_1$ can make that coordinate grow. Nevertheless the discrepancy range is $\\Span\\{e_3\\}$, which the reference leaves inside itself and $Q$ annihilates. The common quotient retains $(x_1,x_2)$, with update\n\\[\nG(x_1,x_2)=(x_1+x_2,x_2).\n\\]\nThe output at step $k$ is $x_1+kx_2$ under both models, and even under switching between them. The output alone is not an initialized reusable state: equal $x_1$ with different $x_2$ gives different next outputs. Ignoring $x_3$ is justified by the no-return structure, not by its small size or its being temporarily unobserved.","and $T_1$ can make that coordinate grow. Nevertheless the discrepancy range is $\\Span\\{e_3\\}$, which the reference leaves inside itself and $Q$ annihilates. The common quotient retains $(x_1,x_2)$, with update\n\\[\nG(x_1,x_2)=(x_1+x_2,x_2).\n\\]\nThe output at step $k$ is $x_1+kx_2$ under both models, and even under switching between them. The output alone is not an initialized reusable state: equal $x_1$ with different $x_2$ gives different next outputs. Ignoring $x_3$ is justified by the no-return structure, not by its small size or its being temporarily unobserved.","hat coordinate grow. Nevertheless the discrepancy range is $\\Span\\{e_3\\}$, which the reference leaves inside itself and $Q$ annihilates. The common quotient retains $(x_1,x_2)$, with update\n\\[\nG(x_1,x_2)=(x_1+x_2,x_2).\n\\]\nThe output at step $k$ is $x_1+kx_2$ under both models, and even under switching between them. The output alone is not an initialized reusable state: equal $x_1$ with different $x_2$ gives different next outputs. Ignoring $x_3$ is justified by the no-return structure, not by its small size or its being temporarily unobserved.","ancy range is $\\Span\\{e_3\\}$, which the reference leaves inside itself and $Q$ annihilates. The common quotient retains $(x_1,x_2)$, with update\n\\[\nG(x_1,x_2)=(x_1+x_2,x_2).\n\\]\nThe output at step $k$ is $x_1+kx_2$ under both models, and even under switching between them. The output alone is not an initialized reusable state: equal $x_1$ with different $x_2$ gives different next outputs. Ignoring $x_3$ is justified by the no-return structure, not by its small size or its being temporarily unobserved.","\\begin{counterexample}[An invisible difference returns one step later]\\label{ce:hiddenreturn}\nLet\n\\[\nA=\\begin{pmatrix}0&1\\\\0&0\\end{pmatrix},\\qquad\nD=\\begin{pmatrix}0&0\\\\1&0\\end{pmatrix},\\qquad Q=\\begin{pmatrix}1&0\\end{pmatrix}.\n\\]\nThe candidates are $A$ and $A+D$. Since $QD=0$, every one-step output agrees for every input. But $QAD=(1,0)\\ne0$. Starting at $e_1$, the second outputs are zero and one. The initial discrepancy range is invisible, while its propagated span $E_\\infty=V$ is not. Checking only $QD=0$ misses the failure.\n\\end{counterexample}","1&0\\end{pmatrix},\\qquad Q=\\begin{pmatrix}1&0\\end{pmatrix}.\n\\]\nThe candidates are $A$ and $A+D$. Since $QD=0$, every one-step output agrees for every input. But $QAD=(1,0)\\ne0$. Starting at $e_1$, the second outputs are zero and one. The initial discrepancy range is invisible, while its propagated span $E_\\infty=V$ is not. Checking only $QD=0$ misses the failure.\n\\end{counterexample}","\\subsection{What the criterion does not learn}\nThe family $\\{T_a^\\theta\\}$, its shared carrier, and $Q$ are given. The closure in the discrepancy-span recursion uses exact linear algebra and decides whether those models share the stated quotient. It neither discovers the model class from finite observations nor proves that the true mechanism belongs to it. A structural prior such as an invariant hidden block can justify a smaller uncertainty family, but that prior must be carried as evidence or an explicit assumption.","Let $L\\subseteq I_Q$ be a specified linear initial-condition subspace and let the available initial record be a linear map $R:L\\to H$. Define the reference reachable span and its unobservable part by\n\\[\nW=\\Span\\{A_wv:v\\in L,\\ w\\text{ any word}\\},\\qquad\nN=\\{z\\in W:QA_wz=0\\text{ for every word }w\\}.\n\\]\nBoth $W$ and $N$ are invariant under the reference actions, and $N\\subseteq\\ker Q$. These spaces can be computed by finite reachable-span and row-kernel iterations. They concern a reference predictor, not a claim that every physical candidate remains inside $W$.","able span and its unobservable part by\n\\[\nW=\\Span\\{A_wv:v\\in L,\\ w\\text{ any word}\\},\\qquad\nN=\\{z\\in W:QA_wz=0\\text{ for every word }w\\}.\n\\]\nBoth $W$ and $N$ are invariant under the reference actions, and $N\\subseteq\\ker Q$. These spaces can be computed by finite reachable-span and row-kernel iterations. They concern a reference predictor, not a claim that every physical candidate remains inside $W$.","\\begin{theorem}[Evidence-initialized minimal common linear prediction]\\label{thm:linearpredictor}\nThere is a finite-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=","\\begin{theorem}[Evidence-initialized minimal common linear prediction]\\label{thm:linearpredictor}\nThere is a finite-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration g","\\begin{theorem}[Evidence-initialized minimal common linear prediction]\\label{thm:linearpredictor}\nThere is a finite-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees wi","\\begin{theorem}[Evidence-initialized minimal common linear prediction]\\label{thm:linearpredictor}\nThere is a finite-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","rediction]\\label{thm:linearpredictor}\nThere is a finite-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","tor}\nThere is a finite-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","-dimensional linear predictor with initialization $d:R(L)\\to Z$, updates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","pdates $G_a:Z\\to Z$, and readout $\\bar Q:Z\\to Y$ satisfying\n\\[\n\\bar QG_wdR(v)=QT_w^\\theta v\n\\quad\\text{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","t{for every }v\\in L,\\theta\\in\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","\\Theta,w\n\\]\nif and only if $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","f $\\ker R\\subseteq N$. When it exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","exists, the smallest possible state-space dimension among such \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","uch \\emph{linear} predictors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","ctors is\n\\[\n\\dim W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","W-\\dim N.\n\\]\nThe quotient $W/N$, initialized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","lized by $d(Rv)=[v]$, attains that minimum.\n\\end{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","nd{theorem}\n\\begin{proof}\nIf $v\\in\\ker R$, linear initialization gives the same state as input zero. Exactness therefore requires $QA_wv=0$ for every word, so $v\\in N$. Conversely, if $\\ker R\\subseteq N$, the rule $d(Rv)=[v]$ is well defined and linear. Invariance supplies $G_a([z])=[A_az]$, and $QN=0$ supplies $\\bar Q([z])=Qz$. Their iteration gives $QA_wv$, which agrees with every candidate by $L\\subseteq I_Q$.","For minimality, let any competing linear predictor be given. Its reachable span is generated by vectors $G_wdR(v)$. Associate such a vector with $[A_wv]\\in W/N$ and extend by linearity. To see this association is well defined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}",", let any competing linear predictor be given. Its reachable span is generated by vectors $G_wdR(v)$. Associate such a vector with $[A_wv]\\in W/N$ and extend by linearity. To see this association is well defined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}","ar predictor be given. Its reachable span is generated by vectors $G_wdR(v)$. Associate such a vector with $[A_wv]\\in W/N$ and extend by linearity. To see this association is well defined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}","s generated by vectors $G_wdR(v)$. Associate such a vector with $[A_wv]\\in W/N$ and extend by linearity. To see this association is well defined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}","$[A_wv]\\in W/N$ and extend by linearity. To see this association is well defined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}","n is well defined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}","fined, suppose a finite linear combination of the predictor vectors is zero. Applying any later word and the predictor readout shows that the matching combination of reference vectors has zero $QA_u$ output for every $u$. That combination therefore belongs to $N$. The induced map from the competitor's reachable span onto $W/N$ is a well-defined linear surjection, since the classes of $A_wv$ span $W/N$. Its dimension cannot be smaller than the displayed quotient dimension. The constructed quotient attains equality.\n\\end{proof}","\\subsection{Initial dimension, dynamic memory, and hidden state}\nIn Counterexample~\\ref{ce:switching}, set $L=\\Span\\{(1,1)\\}$ and $R(x)=x_1$. The reference reaches $(1,1)$ and $(1,0)$, so $W=\\mathbb Q^2$. The rows $Q=(1,1)$ and $QA=(0,1)$ have trivial common kernel, giving $N=0$. A two-dimensional linear predictor is therefore necessary even for this one-dimensional initial family: it must reproduce $2t,t,0,0,\\ldots$ for every multiplier $t$.","section{Initial dimension, dynamic memory, and hidden state}\nIn Counterexample~\\ref{ce:switching}, set $L=\\Span\\{(1,1)\\}$ and $R(x)=x_1$. The reference reaches $(1,1)$ and $(1,0)$, so $W=\\mathbb Q^2$. The rows $Q=(1,1)$ and $QA=(0,1)$ have trivial common kernel, giving $N=0$. A two-dimensional linear predictor is therefore necessary even for this one-dimensional initial family: it must reproduce $2t,t,0,0,\\ldots$ for every multiplier $t$.","ry, and hidden state}\nIn Counterexample~\\ref{ce:switching}, set $L=\\Span\\{(1,1)\\}$ and $R(x)=x_1$. The reference reaches $(1,1)$ and $(1,0)$, so $W=\\mathbb Q^2$. The rows $Q=(1,1)$ and $QA=(0,1)$ have trivial common kernel, giving $N=0$. A two-dimensional linear predictor is therefore necessary even for this one-dimensional initial family: it must reproduce $2t,t,0,0,\\ldots$ for every multiplier $t$.","For the three-coordinate structured example, $L=V$, $W=V$, and $N=\\Span\\{e_3\\}$. Two predictor coordinates suffice. The record $(x_1,x_2)$ initializes them; $x_1$ alone does not, since its kernel contains $e_2$, which becomes visible after one action. Physical dimension, initial-family dimension, and minimum predictive dimension are different quantities.","\\subsection{A concrete probability record}\nFor an integer $k\\ge1$, write $\\Delta_k$ for the probability simplex on $k$ cells. Fix a denominator $N\\ge1$. A retained record has the form\n\\[\n \\widehat p=(n_1/N,\\ldots,n_k/N),\\qquad n_i\\in\\{0,\\ldots,N\\},\\quad \\sum_i n_i=N.\n\\]\nThe cell identities, denominator, and decoding rule are fixed background. The $k$ integer fields can be stored literally in $k\\lceil\\log_2(N+1)\\rceil$ bits. This is an upper bound for this specified encoding, not a minimum description length. It excludes the likelihood table, program, temporary arithmetic, and any retained raw observation history.","Define $\\mathcal Q_N(p)$ by first setting $n_i=\\lfloor Np_i\\rfloor$. Distribute the remaining $N-\\sum_i n_i$ units to the largest fractional remainders, with ties broken by a fixed cell order. This makes the normalization exact after every rounding. It is not independent coordinate rounding followed by an unspecified repair.","For a likelihood vector $\\ell=(\\ell_i)$ define the Bayes map\n\\[\n \\mathcal B_\\ell(p)_i=\\frac{p_i\\ell_i}{\\sum_jp_j\\ell_j}\n\\]\nwhenever its denominator is positive.","\\begin{proposition}[Rounding and positive-likelihood update bounds]\\label{prop:roundbayes}\nFor every $p\\in\\Delta_k$,\n\\[\n \\TV(p,\\mathcal Q_N(p))\\le\\frac{k}{2N}.\n\\]\nLet $p,q\\in\\Delta_k$, let $0<\\beta\\le1$, and let every entry of $\\ell$ and $\\widetilde\\ell$ lie in $[\\beta,1]$. If $\\max_i|\\ell_i-\\widetilde\\ell_i|\\le\\varepsilon$, then\n\\[\n \\TV\\bigl(\\mathcal B_\\ell(p),\\mathcal B_{\\widetilde\\ell}(q)\\bigr)\n \\le \\frac{\\TV(p,q)+\\varepsilon}{\\beta}.\n\\]\nEach right-hand side may also be replaced by its minimum with one. The probability-field storage bound above applies to the retained rounded record, not to all working memory.\n\\end{proposition}\n\\begin{proof}\nThe remainder rule produces nonnegative integers summing to $N$. Each rounded coordinate differs from the original by at most $1/N$, so half the sum of absolute","\\begin{proposition}[Rounding and positive-likelihood update bounds]\\label{prop:roundbayes}\nFor every $p\\in\\Delta_k$,\n\\[\n \\TV(p,\\mathcal Q_N(p))\\le\\frac{k}{2N}.\n\\]\nLet $p,q\\in\\Delta_k$, let $0<\\beta\\le1$, and let every entry of $\\ell$ and $\\widetilde\\ell$ lie in $[\\beta,1]$. If $\\max_i|\\ell_i-\\widetilde\\ell_i|\\le\\varepsilon$, then\n\\[\n \\TV\\bigl(\\mathcal B_\\ell(p),\\mathcal B_{\\widetilde\\ell}(q)\\bigr)\n \\le \\frac{\\TV(p,q)+\\varepsilon}{\\beta}.\n\\]\nEach right-hand side may also be replaced by its minimum with one. The probability-field storage bound above applies to the retained rounded record, not to all working memory.\n\\end{proposition}\n\\begin{proof}\nThe remainder rule produces nonnegative integers summing to $N$. Each rounded coordinate differs from the original by at most $1/N$, so half the sum of absolute differences is at most $k/(2N)$.","varepsilon$, then\n\\[\n \\TV\\bigl(\\mathcal B_\\ell(p),\\mathcal B_{\\widetilde\\ell}(q)\\bigr)\n \\le \\frac{\\TV(p,q)+\\varepsilon}{\\beta}.\n\\]\nEach right-hand side may also be replaced by its minimum with one. The probability-field storage bound above applies to the retained rounded record, not to all working memory.\n\\end{proposition}\n\\begin{proof}\nThe remainder rule produces nonnegative integers summing to $N$. Each rounded coordinate differs from the original by at most $1/N$, so half the sum of absolute differences is at most $k/(2N)$.","etilde\\ell}(q)\\bigr)\n \\le \\frac{\\TV(p,q)+\\varepsilon}{\\beta}.\n\\]\nEach right-hand side may also be replaced by its minimum with one. The probability-field storage bound above applies to the retained rounded record, not to all working memory.\n\\end{proposition}\n\\begin{proof}\nThe remainder rule produces nonnegative integers summing to $N$. Each rounded coordinate differs from the original by at most $1/N$, so half the sum of absolute differences is at most $k/(2N)$.","a}.\n\\]\nEach right-hand side may also be replaced by its minimum with one. The probability-field storage bound above applies to the retained rounded record, not to all working memory.\n\\end{proposition}\n\\begin{proof}\nThe remainder rule produces nonnegative integers summing to $N$. Each rounded coordinate differs from the original by at most $1/N$, so half the sum of absolute differences is at most $k/(2N)$.","For the update, first hold $\\ell$ fixed. Given $0\\le f_i\\le1$, put $a=\\sum_i\\mathcal B_\\ell(q)_if_i$ and $Z_p=\\sum_ip_i\\ell_i$. Then\n\\[\n \\sum_i\\bigl(\\mathcal B_\\ell(p)_i-\\mathcal B_\\ell(q)_i\\bigr)f_i\n =\\frac{\\sum_i(p_i-q_i)\\ell_i(f_i-a)}{Z_p}.\n\\]\nEvery value $\\ell_i(f_i-a)$ lies in $[-a,1-a]$, an interval of length one. Since $p-q$ has total mass zero, its pairing with a function of oscillation at most one is bounded in absolute value by $\\TV(p,q)$. Also $Z_p\\ge\\beta$. Taking the supremum over $f$ bounds the fixed-likelihood difference by $\\TV(p,q)/\\beta$.","ven $0\\le f_i\\le1$, put $a=\\sum_i\\mathcal B_\\ell(q)_if_i$ and $Z_p=\\sum_ip_i\\ell_i$. Then\n\\[\n \\sum_i\\bigl(\\mathcal B_\\ell(p)_i-\\mathcal B_\\ell(q)_i\\bigr)f_i\n =\\frac{\\sum_i(p_i-q_i)\\ell_i(f_i-a)}{Z_p}.\n\\]\nEvery value $\\ell_i(f_i-a)$ lies in $[-a,1-a]$, an interval of length one. Since $p-q$ has total mass zero, its pairing with a function of oscillation at most one is bounded in absolute value by $\\TV(p,q)$. Also $Z_p\\ge\\beta$. Taking the supremum over $f$ bounds the fixed-likelihood difference by $\\TV(p,q)/\\beta$.","$, put $a=\\sum_i\\mathcal B_\\ell(q)_if_i$ and $Z_p=\\sum_ip_i\\ell_i$. Then\n\\[\n \\sum_i\\bigl(\\mathcal B_\\ell(p)_i-\\mathcal B_\\ell(q)_i\\bigr)f_i\n =\\frac{\\sum_i(p_i-q_i)\\ell_i(f_i-a)}{Z_p}.\n\\]\nEvery value $\\ell_i(f_i-a)$ lies in $[-a,1-a]$, an interval of length one. Since $p-q$ has total mass zero, its pairing with a function of oscillation at most one is bounded in absolute value by $\\TV(p,q)$. Also $Z_p\\ge\\beta$. Taking the supremum over $f$ bounds the fixed-likelihood difference by $\\TV(p,q)/\\beta$.","and $Z_p=\\sum_ip_i\\ell_i$. Then\n\\[\n \\sum_i\\bigl(\\mathcal B_\\ell(p)_i-\\mathcal B_\\ell(q)_i\\bigr)f_i\n =\\frac{\\sum_i(p_i-q_i)\\ell_i(f_i-a)}{Z_p}.\n\\]\nEvery value $\\ell_i(f_i-a)$ lies in $[-a,1-a]$, an interval of length one. Since $p-q$ has total mass zero, its pairing with a function of oscillation at most one is bounded in absolute value by $\\TV(p,q)$. Also $Z_p\\ge\\beta$. Taking the supremum over $f$ bounds the fixed-likelihood difference by $\\TV(p,q)/\\beta$.","gl(\\mathcal B_\\ell(p)_i-\\mathcal B_\\ell(q)_i\\bigr)f_i\n =\\frac{\\sum_i(p_i-q_i)\\ell_i(f_i-a)}{Z_p}.\n\\]\nEvery value $\\ell_i(f_i-a)$ lies in $[-a,1-a]$, an interval of length one. Since $p-q$ has total mass zero, its pairing with a function of oscillation at most one is bounded in absolute value by $\\TV(p,q)$. Also $Z_p\\ge\\beta$. Taking the supremum over $f$ bounds the fixed-likelihood difference by $\\TV(p,q)/\\beta$.","Next hold $q$ fixed and set $a=\\sum_i\\mathcal B_{\\widetilde\\ell}(q)_if_i$. The corresponding numerator is $\\sum_iq_i(\\ell_i-\\widetilde\\ell_i)(f_i-a)$, whose absolute value is at most $\\varepsilon$. Its denominator is at least $\\beta$. A triangle inequality proves the displayed update bound. The cap by one is the universal range of total variation. The storage assertion counts the stated integer fields only.\n\\end{proof}","\\subsection{Approximate likelihood profiles}\nLet $\\Omega$ be a finite static world set with prior $\\mu_0$. A fixed partition $C_1,\\ldots,C_k$ refines the target: $t(\\omega)=t_i$ throughout $C_i$. Each admitted experiment $e$ is nondestructive and has a finite report alphabet. Its actual conditional probabilities are $\\ell_{e,o}(\\omega)$. Reports are freshly generated from those likelihoods conditional on the fixed world and the chosen experiments.","Choose one valid representative report law $\\bar\\ell_{e,o,i}$ per cell. Assume, for every admitted experiment, report, cell, and world in that cell,\n\\[\n \\beta\\le \\ell_{e,o}(\\omega),\\bar\\ell_{e,o,i}\\le1,\n \\qquad |\\ell_{e,o}(\\omega)-\\bar\\ell_{e,o,i}|\\le\\varepsilon. \\tag{10}\n\\]\nBoth the actual and representative report laws sum to one over reports. Exact profile cells satisfy this condition with $\\varepsilon=0$ when their likelihoods have the positive floor. A coarser partition can be used with a nonzero $\\varepsilon$, but target-constant cells are still required for the target readout.","t, report, cell, and world in that cell,\n\\[\n \\beta\\le \\ell_{e,o}(\\omega),\\bar\\ell_{e,o,i}\\le1,\n \\qquad |\\ell_{e,o}(\\omega)-\\bar\\ell_{e,o,i}|\\le\\varepsilon. \\tag{10}\n\\]\nBoth the actual and representative report laws sum to one over reports. Exact profile cells satisfy this condition with $\\varepsilon=0$ when their likelihoods have the positive floor. A coarser partition can be used with a nonzero $\\varepsilon$, but target-constant cells are still required for the target readout.","After a history, let $p_t$ be the cell masses of the exact world posterior. Initialize the retained record by $\\widehat p_0=\\mathcal Q_N(p_0)$. For the next selected experiment and observed report use\n\\[\n \\widehat p_{t+1}=\\mathcal Q_N\\!\\left(\\mathcal B_{\\bar\\ell_{e,o}}(\\widehat p_t)\\right).\n\\]\nThe reference posterior and rounded record are compared along the same actual action/report history. Experiment selection may depend on earlier retained records and independent randomness. It may not introduce a hidden world-dependent selector whose information is omitted from that history. The likelihood condition applies to every experiment the policy can select.","\\begin{theorem}[Finite retained memory for a declared correction horizon]\\label{thm:finiteprecision}\nUnder (10), set $\\eta_N=k/(2N)$, $\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis suffic","\\begin{theorem}[Finite retained memory for a declared correction horizon]\\label{thm:finiteprecision}\nUnder (10), set $\\eta_N=k/(2N)$, $\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and c","\\begin{theorem}[Finite retained memory for a declared correction horizon]\\label{thm:finiteprecision}\nUnder (10), set $\\eta_N=k/(2N)$, $\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at mo","\\begin{theorem}[Finite retained memory for a declared correction horizon]\\label{thm:finiteprecision}\nUnder (10), set $\\eta_N=k/(2N)$, $\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sum","\\begin{theorem}[Finite retained memory for a declared correction horizon]\\label{thm:finiteprecision}\nUnder (10), set $\\eta_N=k/(2N)$, $\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record gua","\\begin{theorem}[Finite retained memory for a declared correction horizon]\\label{thm:finiteprecision}\nUnder (10), set $\\eta_N=k/(2N)$, $\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable hist…","$\\gamma=\\beta^{-1}$, and let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ …","let $e_0$ be any valid bound on $\\TV(p_0,\\widehat p_0)$. For every possible action/report history of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It rema…","story of length $t$,\n\\[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any …","[\n \\TV(p_t,\\widehat p_t)\\le e_t\n :=\\min\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bo…","\\!\\left\\{1,\\ \\gamma^t e_0+(\\gamma\\varepsilon+\\eta_N)\n                         \\sum_{j=0}^{t-1}\\gamma^j\\right\\}. \\tag{11}\n\\]\nThe same bound holds after mapping cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. …","cells to target values. With the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. Thus the exact next cell mass is a Bayes update with such an effective likelihood vector, even when the worlds within a cell do not have identical laws.","the generic initialization bound $e_0\\le\\eta_N$, define\n\\[\n A_H=\\sum_{j=0}^{H}\\gamma^j,\\qquad\n B_H=\\gamma\\sum_{j=0}^{H-1}\\gamma^j.\n\\]\nFor any horizon $H\\ge0$ and tolerance $\\tau>\\varepsilon B_H$, every integer\n\\[\n N\\ge \\max\\!\\left\\{1,\\left\\lceil\n       \\frac{kA_H}{2(\\tau-\\varepsilon B_H)}\\right\\rceil\\right\\} \\tag{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. Thus the exact next cell mass is a Bayes update with such an effective likelihood vector, even when the worlds within a cell do not have identical laws.","{12}\n\\]\nis sufficient for target and cell error at most $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. Thus the exact next cell mass is a Bayes update with such an effective likelihood vector, even when the worlds within a cell do not have identical laws.","st $\\tau$ at every time through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. Thus the exact next cell mass is a Bayes update with such an effective likelihood vector, even when the worlds within a cell do not have identical laws.","through $H$. Empty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. Thus the exact next cell mass is a Bayes update with such an effective likelihood vector, even when the worlds within a cell do not have identical laws.","ty sums are zero. This is a finite-horizon retained-record guarantee, not an unbounded-time or total-computation guarantee.\n\\end{theorem}\n\\begin{proof}\nCondition on a particular attainable history and its next experiment. In each positive-mass cell, the exact next likelihood is the conditional average of $\\ell_{e,o}(\\omega)$ within that cell. It remains in $[\\beta,1]$ and within $\\varepsilon$ of $\\bar\\ell_{e,o,i}$. For a zero-mass cell any value obeying these bounds gives the same zero contribution. Thus the exact next cell mass is a Bayes update with such an effective likelihood vector, even when the worlds within a cell do not have identical laws.","Proposition~\\ref{prop:roundbayes} and the rounding bound give\n\\[\n \\TV(p_{t+1},\\widehat p_{t+1})\n \\le\\gamma\\TV(p_t,\\widehat p_t)+\\gamma\\varepsilon+\\eta_N.\n\\]\nIterating the uncapped scalar bound proves (11); total variation is also at most one. Aggregating cell coordinates into target coordinates cannot increase total variation, by the triangle inequality. Substituting $e_0\\le\\eta_N$ gives $\\eta_N A_t+\\varepsilon B_t$. These quantities are nondecreasing in $t$ because $\\gamma\\ge1$ and all terms are nonnegative. Inequality (12) makes the horizon-$H$ value at most $\\tau$. It therefore suffices at every earlier step as well. Every report has positive probability under both updates by the floor assumption, so no zero denominator is silently normalized.\n\\end{proof}","e\n\\[\n \\TV(p_{t+1},\\widehat p_{t+1})\n \\le\\gamma\\TV(p_t,\\widehat p_t)+\\gamma\\varepsilon+\\eta_N.\n\\]\nIterating the uncapped scalar bound proves (11); total variation is also at most one. Aggregating cell coordinates into target coordinates cannot increase total variation, by the triangle inequality. Substituting $e_0\\le\\eta_N$ gives $\\eta_N A_t+\\varepsilon B_t$. These quantities are nondecreasing in $t$ because $\\gamma\\ge1$ and all terms are nonnegative. Inequality (12) makes the horizon-$H$ value at most $\\tau$. It therefore suffices at every earlier step as well. Every report has positive probability under both updates by the floor assumption, so no zero denominator is silently normalized.\n\\end{proof}","{t+1},\\widehat p_{t+1})\n \\le\\gamma\\TV(p_t,\\widehat p_t)+\\gamma\\varepsilon+\\eta_N.\n\\]\nIterating the uncapped scalar bound proves (11); total variation is also at most one. Aggregating cell coordinates into target coordinates cannot increase total variation, by the triangle inequality. Substituting $e_0\\le\\eta_N$ gives $\\eta_N A_t+\\varepsilon B_t$. These quantities are nondecreasing in $t$ because $\\gamma\\ge1$ and all terms are nonnegative. Inequality (12) makes the horizon-$H$ value at most $\\tau$. It therefore suffices at every earlier step as well. Every report has positive probability under both updates by the floor assumption, so no zero denominator is silently normalized.\n\\end{proof}","_N.\n\\]\nIterating the uncapped scalar bound proves (11); total variation is also at most one. Aggregating cell coordinates into target coordinates cannot increase total variation, by the triangle inequality. Substituting $e_0\\le\\eta_N$ gives $\\eta_N A_t+\\varepsilon B_t$. These quantities are nondecreasing in $t$ because $\\gamma\\ge1$ and all terms are nonnegative. Inequality (12) makes the horizon-$H$ value at most $\\tau$. It therefore suffices at every earlier step as well. Every report has positive probability under both updates by the floor assumption, so no zero denominator is silently normalized.\n\\end{proof}","he uncapped scalar bound proves (11); total variation is also at most one. Aggregating cell coordinates into target coordinates cannot increase total variation, by the triangle inequality. Substituting $e_0\\le\\eta_N$ gives $\\eta_N A_t+\\varepsilon B_t$. These quantities are nondecreasing in $t$ because $\\gamma\\ge1$ and all terms are nonnegative. Inequality (12) makes the horizon-$H$ value at most $\\tau$. It therefore suffices at every earlier step as well. Every report has positive probability under both updates by the floor assumption, so no zero denominator is silently normalized.\n\\end{proof}","Equation (12) is a sufficient precision requirement, not an exact minimum. An infeasible or large bound does not prove that the particular rounded filter is inaccurate. It means that this uniform certificate is insufficient. Conversely, increasing precision cannot eliminate the model term $\\varepsilon B_H$. A false likelihood model does not become correct because its probabilities are stored with more bits.","\\begin{counterexample}[A tiny rounded-away mass can control a later posterior]\\label{ce:rareprecision}\nTake two worlds with prior $(1-d,d)$, where $0<d<1/(2N)$. The remainder rule gives the retained record $(1,0)$. A binary report has likelihoods $(d^2,1)$ for outcome one. After that outcome, the actual posterior mass of the second world is\n\\[\n \\frac{d}{d+(1-d)d^2}=\\frac{1}{1+d-d^2},\n\\]\nwhich tends to one as $d$ tends to zero. The rounded posterior remains zero on that world. The initial total-variation error is only $d$. There is no common positive likelihood floor across this family: the first likelihood tends to zero, and the other report is impossible in the second world. No nontrivial uniform continuation conclusion follows from the small initial rounding error alone.\n\\end{counterexample}","rexample}[A tiny rounded-away mass can control a later posterior]\\label{ce:rareprecision}\nTake two worlds with prior $(1-d,d)$, where $0<d<1/(2N)$. The remainder rule gives the retained record $(1,0)$. A binary report has likelihoods $(d^2,1)$ for outcome one. After that outcome, the actual posterior mass of the second world is\n\\[\n \\frac{d}{d+(1-d)d^2}=\\frac{1}{1+d-d^2},\n\\]\nwhich tends to one as $d$ tends to zero. The rounded posterior remains zero on that world. The initial total-variation error is only $d$. There is no common positive likelihood floor across this family: the first likelihood tends to zero, and the other report is impossible in the second world. No nontrivial uniform continuation conclusion follows from the small initial rounding error alone.\n\\end{counterexample}","prior $(1-d,d)$, where $0<d<1/(2N)$. The remainder rule gives the retained record $(1,0)$. A binary report has likelihoods $(d^2,1)$ for outcome one. After that outcome, the actual posterior mass of the second world is\n\\[\n \\frac{d}{d+(1-d)d^2}=\\frac{1}{1+d-d^2},\n\\]\nwhich tends to one as $d$ tends to zero. The rounded posterior remains zero on that world. The initial total-variation error is only $d$. There is no common positive likelihood floor across this family: the first likelihood tends to zero, and the other report is impossible in the second world. No nontrivial uniform continuation conclusion follows from the small initial rounding error alone.\n\\end{counterexample}","The counterexample concerns evidence amplification, not a failure of the remainder arithmetic. Replacing the likelihood one by $1-d^2$ makes both outcomes possible in both worlds while preserving the limiting phenomenon. The minimum likelihood still tends to zero. Exact support, approximate probability, and justified abstention should therefore remain distinct data.","Write the physical state as $(v_t,h_t)$ in a direct sum of two finite-dimensional normed spaces. The required output is the visible component $v_t$. Along an admitted execution suppose\n\\[\n \\begin{pmatrix}v_{t+1}","Write the physical state as $(v_t,h_t)$ in a direct sum of two finite-dimensional normed spaces. The required output is the visible component $v_t$. Along an admitted execution suppose\n\\[\n \\begin{pmatrix}v_{t+1}\\\\h_{t+1}\\end{pmatrix}\n =\\begin{pmatrix}G_t&B_t\\\\C_t&H_t\\end{pmatrix}\n  \\begin{pmatrix}v_t\\\\h_t\\end{pmatrix},\n \\qquad \\bar v_{t+1}=G_t\\bar v_t. \\tag{13}\n\\]\nThe reduced predictor knows the chosen $G_t$ and receives the same action sequence. It does not observe $h_t$. The blocks may come from a fixed unknown candidate or from an explicitly admitted varying family. Assume common operator-norm bounds\n\\[\n \\|G_t\\|\\le g,\\quad \\|B_t\\|\\le b,\\quad\n \\|C_t\\|\\le c,\\quad \\|H_t\\|\\le d,\n \\qquad g,b,c,d\\ge0.\n\\]\nLet $\\|v_0\\|\\le V_0$, $\\|h_0\\|\\le H_0$, and $\\|v_0-\\bar v_0\\|\\le e_0$. These initial bounds are supplied evidence or declared model assumptions. Put\n\\[\n M=\\begin{pmatrix}g&b","h_t\\end{pmatrix},\n \\qquad \\bar v_{t+1}=G_t\\bar v_t. \\tag{13}\n\\]\nThe reduced predictor knows the chosen $G_t$ and receives the same action sequence. It does not observe $h_t$. The blocks may come from a fixed unknown candidate or from an explicitly admitted varying family. Assume common operator-norm bounds\n\\[\n \\|G_t\\|\\le g,\\quad \\|B_t\\|\\le b,\\quad\n \\|C_t\\|\\le c,\\quad \\|H_t\\|\\le d,\n \\qquad g,b,c,d\\ge0.\n\\]\nLet $\\|v_0\\|\\le V_0$, $\\|h_0\\|\\le H_0$, and $\\|v_0-\\bar v_0\\|\\le e_0$. These initial bounds are supplied evidence or declared model assumptions. Put\n\\[\n M=\\begin{pmatrix}g&b","ix},\n \\qquad \\bar v_{t+1}=G_t\\bar v_t. \\tag{13}\n\\]\nThe reduced predictor knows the chosen $G_t$ and receives the same action sequence. It does not observe $h_t$. The blocks may come from a fixed unknown candidate or from an explicitly admitted varying family. Assume common operator-norm bounds\n\\[\n \\|G_t\\|\\le g,\\quad \\|B_t\\|\\le b,\\quad\n \\|C_t\\|\\le c,\\quad \\|H_t\\|\\le d,\n \\qquad g,b,c,d\\ge0.\n\\]\nLet $\\|v_0\\|\\le V_0$, $\\|h_0\\|\\le H_0$, and $\\|v_0-\\bar v_0\\|\\le e_0$. These initial bounds are supplied evidence or declared model assumptions. Put\n\\[\n M=\\begin{pmatrix}g&b","v_{t+1}=G_t\\bar v_t. \\tag{13}\n\\]\nThe reduced predictor knows the chosen $G_t$ and receives the same action sequence. It does not observe $h_t$. The blocks may come from a fixed unknown candidate or from an explicitly admitted varying family. Assume common operator-norm bounds\n\\[\n \\|G_t\\|\\le g,\\quad \\|B_t\\|\\le b,\\quad\n \\|C_t\\|\\le c,\\quad \\|H_t\\|\\le d,\n \\qquad g,b,c,d\\ge0.\n\\]\nLet $\\|v_0\\|\\le V_0$, $\\|h_0\\|\\le H_0$, and $\\|v_0-\\bar v_0\\|\\le e_0$. These initial bounds are supplied evidence or declared model assumptions. Put\n\\[\n M=\\begin{pmatrix}g&b","tag{13}\n\\]\nThe reduced predictor knows the chosen $G_t$ and receives the same action sequence. It does not observe $h_t$. The blocks may come from a fixed unknown candidate or from an explicitly admitted varying family. Assume common operator-norm bounds\n\\[\n \\|G_t\\|\\le g,\\quad \\|B_t\\|\\le b,\\quad\n \\|C_t\\|\\le c,\\quad \\|H_t\\|\\le d,\n \\qquad g,b,c,d\\ge0.\n\\]\nLet $\\|v_0\\|\\le V_0$, $\\|h_0\\|\\le H_0$, and $\\|v_0-\\bar v_0\\|\\le e_0$. These initial bounds are supplied evidence or declared model assumptions. Put\n\\[\n M=\\begin{pmatrix}g&b","observe $h_t$. The blocks may come from a fixed unknown candidate or from an explicitly admitted varying family. Assume common operator-norm bounds\n\\[\n \\|G_t\\|\\le g,\\quad \\|B_t\\|\\le b,\\quad\n \\|C_t\\|\\le c,\\quad \\|H_t\\|\\le d,\n \\qquad g,b,c,d\\ge0.\n\\]\nLet $\\|v_0\\|\\le V_0$, $\\|h_0\\|\\le H_0$, and $\\|v_0-\\bar v_0\\|\\le e_0$. These initial bounds are supplied evidence or declared model assumptions. Put\n\\[\n M=\\begin{pmatrix}g&b\\\\c&d\\end{pmatrix},\\qquad\n u_0=\\begin{pmatrix}V_0\\\\H_0\\end{pmatrix},\\qquad u_j=M^ju_0.\n\\]\nPowers of this nonnegative comparison matrix are bounds on norms. They are not a replacement physical dynamics inferred from measurements.","\\begin{theorem}[Visible-error bounds from a common block comparison]\\label{thm:blockleakage}\nEvery admitted execution satisfying (13) obeys\n\\[\n \\|v_t-\\bar v_t\\|\\le d_t\n :=g^te_0+b\\sum_{j=0}^{t-1}g^{t-1-j}(u_j)_2. \\tag{14}\n\\]\nIn particular, when $b=0$ and $e_0=0$, the visible prediction is exact for every finite time, without a stability requirement on $d$ or $c$.","If $g<1$, $d<1$, and\n\\[\n D=(1-g)(1-d)-bc>0,\n\\]\nthen the total visible discrepancy is bounded by\n\\[\n \\sum_{t=0}^{\\infty}\\|v_t-\\bar v_t\\|\n \\le \\frac{e_0}{1-g}\n       +\\frac{b\\bigl(cV_0+(1-g)H_0\\bigr)}{(1-g)D}. \\tag{15}\n\\]\nThese are sufficient common bounds. Failure of the inequality for $D$ is not a proof that a particular signed system is unstable. Conversely, the two diagonal conditions alone are insufficient for a uniform conclusion over the stated block class.\n\\end{theorem}\n\\begin{proof}\nThe norm triangle inequality and the common block bounds give\n\\[\n \\begin{pmatrix}\\|v_{t+1}\\|","+\\frac{b\\bigl(cV_0+(1-g)H_0\\bigr)}{(1-g)D}. \\tag{15}\n\\]\nThese are sufficient common bounds. Failure of the inequality for $D$ is not a proof that a particular signed system is unstable. Conversely, the two diagonal conditions alone are insufficient for a uniform conclusion over the stated block class.\n\\end{theorem}\n\\begin{proof}\nThe norm triangle inequality and the common block bounds give\n\\[\n \\begin{pmatrix}\\|v_{t+1}\\|\\\\\\|h_{t+1}\\|\\end{pmatrix}\n \\le M\\begin{pmatrix}\\|v_t\\|\\\\\\|h_t\\|\\end{pmatrix}\n \\le M^{t+1}u_0\n\\]\ncomponentwise. The error satisfies\n\\[\n v_{t+1}-\\bar v_{t+1}=G_t(v_t-\\bar v_t)+B_th_t.\n\\]\nInduction gives (14), including its empty sum at $t=0$. If $b=0$ and the initial error is zero, the recurrence has no forcing term and remains zero, regardless of hidden growth.","|h_t\\|\\end{pmatrix}\n \\le M^{t+1}u_0\n\\]\ncomponentwise. The error satisfies\n\\[\n v_{t+1}-\\bar v_{t+1}=G_t(v_t-\\bar v_t)+B_th_t.\n\\]\nInduction gives (14), including its empty sum at $t=0$. If $b=0$ and the initial error is zero, the recurrence has no forcing term and remains zero, regardless of hidden growth.","Under the three strict inequalities, $(I-M)^{-1}=D^{-1}\\begin{pmatrix}1-d&b\\\\c&1-g\\end{pmatrix}$ has nonnegative entries. Let $a=(I-M)^{-1}(1,1)^T$. Both coordinates are positive and $Ma=a-(1,1)^T<a$. Hence $M$ is a strict contraction in the weighted maximum norm with weights $a$, and its geometric series converges to $(I-M)^{-1}$. Summing the nonnegative right-hand side of (14) gives\n\\[\n \\frac{e_0}{1-g}+\\frac{b}{1-g}\\sum_{j\\ge0}(M^ju_0)_2.\n\\]\nThe second coordinate of $(I-M)^{-1}u_0$ is $(cV_0+(1-g)H_0)/D$, proving (15). The common comparison applies at each step and therefore also covers switching when that switching is explicitly allowed by the contract. It need not cover a family supplied only with unrelated per-model stability certificates.\n\\end{proof}","c&1-g\\end{pmatrix}$ has nonnegative entries. Let $a=(I-M)^{-1}(1,1)^T$. Both coordinates are positive and $Ma=a-(1,1)^T<a$. Hence $M$ is a strict contraction in the weighted maximum norm with weights $a$, and its geometric series converges to $(I-M)^{-1}$. Summing the nonnegative right-hand side of (14) gives\n\\[\n \\frac{e_0}{1-g}+\\frac{b}{1-g}\\sum_{j\\ge0}(M^ju_0)_2.\n\\]\nThe second coordinate of $(I-M)^{-1}u_0$ is $(cV_0+(1-g)H_0)/D$, proving (15). The common comparison applies at each step and therefore also covers switching when that switching is explicitly allowed by the contract. It need not cover a family supplied only with unrelated per-model stability certificates.\n\\end{proof}","x}$ has nonnegative entries. Let $a=(I-M)^{-1}(1,1)^T$. Both coordinates are positive and $Ma=a-(1,1)^T<a$. Hence $M$ is a strict contraction in the weighted maximum norm with weights $a$, and its geometric series converges to $(I-M)^{-1}$. Summing the nonnegative right-hand side of (14) gives\n\\[\n \\frac{e_0}{1-g}+\\frac{b}{1-g}\\sum_{j\\ge0}(M^ju_0)_2.\n\\]\nThe second coordinate of $(I-M)^{-1}u_0$ is $(cV_0+(1-g)H_0)/D$, proving (15). The common comparison applies at each step and therefore also covers switching when that switching is explicitly allowed by the contract. It need not cover a family supplied only with unrelated per-model stability certificates.\n\\end{proof}","The result connects exact and approximate abstraction without collapsing them. Exact no-return can protect an output even when the hidden coordinate grows rapidly. With nonzero return, hidden growth and two-way feedback enter the error account. A small $b$ is useful only in relation to the horizon, the initial hidden bound, and the other blocks.","\\begin{counterexample}[An arbitrarily small return channel exposes an unstable hidden mode]\\label{ce:observableleak}\nLet $v_0=\\bar v_0=0$, $h_0=1$, and use scalar blocks $G=1/2$, $B=\\delta>0$, $C=0$, $H=2$. The reduced output remains zero, while\n\\[\n v_t=\\delta\\sum_{j=0}^{t-1}(1/2)^{t-1-j}2^j\n     =\\frac{\\delta(2^t-2^{-t})}{3/2}.\n\\]\nThe first error is $\\delta$, yet the later error is unbounded for every fixed positive $\\delta$. At $\\delta=0$, every output is exactly zero. Thus a tiny positive coupling is not uniformly equivalent to exact no-return on an unbounded horizon.\n\\end{counterexample}","[An arbitrarily small return channel exposes an unstable hidden mode]\\label{ce:observableleak}\nLet $v_0=\\bar v_0=0$, $h_0=1$, and use scalar blocks $G=1/2$, $B=\\delta>0$, $C=0$, $H=2$. The reduced output remains zero, while\n\\[\n v_t=\\delta\\sum_{j=0}^{t-1}(1/2)^{t-1-j}2^j\n     =\\frac{\\delta(2^t-2^{-t})}{3/2}.\n\\]\nThe first error is $\\delta$, yet the later error is unbounded for every fixed positive $\\delta$. At $\\delta=0$, every output is exactly zero. Thus a tiny positive coupling is not uniformly equivalent to exact no-return on an unbounded horizon.\n\\end{counterexample}","ly small return channel exposes an unstable hidden mode]\\label{ce:observableleak}\nLet $v_0=\\bar v_0=0$, $h_0=1$, and use scalar blocks $G=1/2$, $B=\\delta>0$, $C=0$, $H=2$. The reduced output remains zero, while\n\\[\n v_t=\\delta\\sum_{j=0}^{t-1}(1/2)^{t-1-j}2^j\n     =\\frac{\\delta(2^t-2^{-t})}{3/2}.\n\\]\nThe first error is $\\delta$, yet the later error is unbounded for every fixed positive $\\delta$. At $\\delta=0$, every output is exactly zero. Thus a tiny positive coupling is not uniformly equivalent to exact no-return on an unbounded horizon.\n\\end{counterexample}","\\begin{counterexample}[Stable diagonal blocks can have unstable feedback]\\label{ce:feedbackgain}\nTake scalar blocks $G=H=1/2$ and $B=C=3/4$. Both diagonal subsystems are contractions. The full matrix sends $(1,1)$ to $(5/4)(1,1)$ and consequently grows on that direction. Here $D=1/4-9/16<0$. A certificate that checks only the two diagonal bounds misses the feedback loop.\n\\end{counterexample}","\\begin{figure}[tbp]\\centering\n\\includegraphics[width=.93\\linewidth]{figures/observable_leakage.pdf}\n\\caption{Exact scalar output discrepancies with the same small return coupling $1/64$ and visible multiplier $1/2$. The stable hidden multiplier $1/2$ gives $t2^{-(t-1)}/64$; the hidden multiplier $2$ gives increasing error. The vertical axis is logarithmic. At exactly zero return coupling, the discrepancy is zero at every time in both cases and is not plotted on the log scale. These are explicit linear models, not neural activation traces.}\\label{fig:visibleleakage}\n\\end{figure}","Fix an actual evidence history. Let $\\mu$ be the exact posterior on the stipulated finite world set and let $p_i=\\mu(C_i)$. Suppose the retained distribution is $\\widehat p$ with $\\TV(p,\\widehat p)\\le e$. For a declared later action word or response horizon, let $y(\\omega)$ be the actual model response and let $\\bar y_i$ be the response used for cell $i$. Assume\n\\[\n \\sup_{\\omega\\in C_i}\\|y(\\omega)-\\bar y_i\\|\\le d\n \\quad\\text{for every cell }i. \\tag{16}\n\\]\nThe cell need not identify the physical state or candidate operator. It must support this common response allowance. A block bound such as (14) can supply it when the initial records, common visible updates, and hidden-state bounds match that cell.","Let the permitted decision set $\\mathcal D$ be finite and nonempty. Its loss $L(y,a)$ takes values in $[0,1]$ and is $K$-Lipschitz in $y$, uniformly over $a$. Define\n\\[\n R(a)=\\sum_{\\omega}\\mu(\\omega)L(y(\\omega),a),\\qquad\n \\widehat R(a)=\\sum_i\\widehat p_iL(\\bar y_i,a).\n\\]\nThese are risks under the stated model posterior. They are not frequentist coverage probabilities or guarantees on the loss of a single realized world.","\\begin{theorem}[A common quantitative grounding certificate]\\label{thm:jointrisk}\nUnder (16), put $\\xi=\\min\\{1,e+Kd\\}$. Then\n\\[\n |R(a)-\\widehat R(a)|\\le\\xi\\quad\\text{for every }a\\in\\mathcal D.\n\\]\nIf $\\widehat a$ minimizes $\\widehat R$, its true posterior regret is at most $2\\xi$. If it is the unique approximate minimizer and\n\\[\n \\widehat R(a)-\\widehat R(\\widehat a)>2\\xi\\quad\\text{for every }a\\ne\\widehat a,\n\\]\nthen it is the unique true minimizer. A decision issued only when $\\widehat R(a)+\\xi\\le\\tau$ has true posterior risk at most the declared threshold $\\tau$.","}\nUnder (16), put $\\xi=\\min\\{1,e+Kd\\}$. Then\n\\[\n |R(a)-\\widehat R(a)|\\le\\xi\\quad\\text{for every }a\\in\\mathcal D.\n\\]\nIf $\\widehat a$ minimizes $\\widehat R$, its true posterior regret is at most $2\\xi$. If it is the unique approximate minimizer and\n\\[\n \\widehat R(a)-\\widehat R(\\widehat a)>2\\xi\\quad\\text{for every }a\\ne\\widehat a,\n\\]\nthen it is the unique true minimizer. A decision issued only when $\\widehat R(a)+\\xi\\le\\tau$ has true posterior risk at most the declared threshold $\\tau$.","ext{for every }a\\in\\mathcal D.\n\\]\nIf $\\widehat a$ minimizes $\\widehat R$, its true posterior regret is at most $2\\xi$. If it is the unique approximate minimizer and\n\\[\n \\widehat R(a)-\\widehat R(\\widehat a)>2\\xi\\quad\\text{for every }a\\ne\\widehat a,\n\\]\nthen it is the unique true minimizer. A decision issued only when $\\widehat R(a)+\\xi\\le\\tau$ has true posterior risk at most the declared threshold $\\tau$.","The conclusion can combine (11) and (14) only when they refer to the same history, cell meanings, future action word, initialization, and admitted candidate family. No guarantee that the true external process belongs to that family is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","story, cell meanings, future action word, initialization, and admitted candidate family. No guarantee that the true external process belongs to that family is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","e action word, initialization, and admitted candidate family. No guarantee that the true external process belongs to that family is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","tion, and admitted candidate family. No guarantee that the true external process belongs to that family is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","ly. No guarantee that the true external process belongs to that family is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","al process belongs to that family is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","ly is supplied by this composition.\n\\end{theorem}\n\\begin{proof}\nInsert the intermediate risk $\\sum_ip_iL(\\bar y_i,a)$. Its difference from $R(a)$ is at most $Kd$ by (16), the Lipschitz condition, and summing posterior masses. Its difference from $\\widehat R(a)$ is at most $e$ because the function $i\\mapsto L(\\bar y_i,a)$ lies in $[0,1]$. A triangle inequality gives $e+Kd$; both risks also lie in $[0,1]$.","Let $a_*$ minimize $R$. Then\n\\[\n R(\\widehat a)\\le\\widehat R(\\widehat a)+\\xi\n \\le\\widehat R(a_*)+\\xi\\le R(a_*)+2\\xi.\n\\]\nFor any other decision, the approximate gap loses at most $2\\xi$ when the two risks are replaced by their true values, proving the strict-gap assertion. The threshold assertion is the upper risk bound itself. Each step uses the same posterior and surrogate responses, which is why separately valid certificates about different histories or targets cannot be substituted.\n\\end{proof}","\\subsection{A complete sensor-to-predictor example}\nUse the eight worlds $(s,\\theta,n)\\in\\{0,1\\}^3$. The four joint $(s,\\theta)$ masses in the order $(0,0),(0,1),(1,0),(1,1)$ are initially\n\\[\n p_0=(3,1,1,3)/8,\n\\]\nand $n$ is conditionally uniform. A noisy nondestructive reference has probability $1/4$ of report one under $\\theta=0$ and $3/4$ under $\\theta=1$. It does not depend on $n$. After three fresh reports all equal to one, the exact target probability is $\\Pr(s=1\\mid111)=41/56$.","worlds $(s,\\theta,n)\\in\\{0,1\\}^3$. The four joint $(s,\\theta)$ masses in the order $(0,0),(0,1),(1,0),(1,1)$ are initially\n\\[\n p_0=(3,1,1,3)/8,\n\\]\nand $n$ is conditionally uniform. A noisy nondestructive reference has probability $1/4$ of report one under $\\theta=0$ and $3/4$ under $\\theta=1$. It does not depend on $n$. After three fresh reports all equal to one, the exact target probability is $\\Pr(s=1\\mid111)=41/56$.","The learner retains the four masses using $N=65\\,536$ and the stated remainder rule after each update. There are 68 literal probability-field bits because each of the four counts includes values through $65\\,536$. The initial masses are exactly representable, so $e_0=0$. With $\\beta=1/4$, $\\varepsilon=0$, and three observations, (11) gives the sharper bound\n\\[\n e_3\\le\\frac{1+4+16}{32\\,768}=\\frac{21}{32\\,768}.\n\\]\nThis uses the actual initialization, not the larger generic bound in (12).","The later response is produced by three applications of one fixed candidate matrix,\n\\[\n T_n=\\begin{pmatrix}1/2&(-1)^n/64\\\\1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, an","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor s","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\x","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterio","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-r","1/8&1/2\\end{pmatrix},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied","},\n \\qquad (v_0,h_0)=(s,1).\n\\tag{17}\n\\]\nThe predictor for cell $(s,\\theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit…","theta)$ uses $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","es $\\bar v_t=2^{-t}s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","s$ and therefore reports $\\bar y_i=s/8$. The nuisance bit $n$ affects the physical response but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","but remains unidentified by the reference observations. With $g=d=1/2$, $b=1/64$, $c=1/8$, and $V_0=H_0=1$, the comparison determinant is $127/512>0$. Directly evaluating (14) at three prediction steps yields\n\\[\n d_3=\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","\\frac{481}{32\\,768}.\n\\]\nChoose between numerical predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","predictions $a=0$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","$ and $a=1/8$, with loss $L(y,a)=\\min\\{1,|y-a|\\}$. It is bounded and 1-Lipschitz. The common radius is consequently\n\\[\n \\xi=\\frac{21+481}{32\\,768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.",",768}=\\frac{251}{16\\,384}.\n\\]\nThe exact companion computes the actual retained masses, approximate risks, both true risks, and their comparison. It verifies that the retained predictor selects $a=1/8$, that its approximate advantage exceeds $2\\xi$, and that $\\widehat R(1/8)+\\xi<3/50$. Thus the same finite construction supports a unique posterior-optimal decision and a posterior-risk threshold of $0.06$ under the supplied model. The complete prior, sensor laws, candidate matrices, bit record, and loss function all belong to this one example.","midrule\n$a=0$ & $23991/262144\\ \\approx0.09152$ & $173737/1835008\\ \\approx0.09468$","$a=1/8$ & $8777/262144\\ \\approx0.03348$ & $81105/1835008\\ \\approx0.04420$","bottomrule\n\\end{tabularx}\n\\end{center}\nThe certified upper risk for the selected decision is $12793/262144\\approx0.04880$, below $3/50$. The actual cell-mass error is $1/114688$, smaller than its uniform allowance. The risk certificate uses the proved allowance, not a substituted post hoc exact error. Exact enumeration is available here only because the example is fully supplied and finite.","The probability coordinates do not encode $n$, and the dynamical allowance does not pretend they do. The result remains approximate precisely where the evidence leaves the physical response unresolved. Reducing the denominator, admitting an observation that depends on $n$, or increasing the hidden return coupling changes a different premise. These revisions require separate responses rather than an undifferentiated instruction to trust the old certificate.","\\subsection{What changes when the evidence contract changes}\nA new likelihood can split an old probability cell. A new candidate operator can shrink $I_Q$. A new action can propagate a previously invisible discrepancy into the output. A new initial-code restriction can fail to initialize a quotient that still exists mathematically. Each revision therefore needs a named compatibility check. The smallest representation for an old contract is not an entitlement to make new claims without acquiring the missing distinctions.","\\subsection{A revision changes specific claims, not every claim at once}\nThe sensor-to-predictor construction provides one ledger for revision. Its probability cells are indexed by $(s,\\theta)$, while the hidden candidate index $n$ is represented only through the response allowance. Each row below describes a change to that same example, not a newly selected favorable model.","midrule\nChange only $N$ & The likelihood-profile partition and physical model. & Rounded history, memory error, probability-field size, and the resulting decision margin.","Let the reference depend on $n$ & Old predictions under the old reference. & New profile cells and their masses. The old four masses cannot universally supply a split.","\\clearpage\n\\section{Small exact illustrations and reproduction}\\label{app:checks}\nThree small exact suites check the declared constructions with integers, sets, rational arithmetic, and $\\mathbb F_2$. The retained observable suite compares paired row-space closure with direct word enumeration, tests common-quotient and initialization criteria, and compares likelihood-profile updates with raw-world Bayes updates. It includes explicit switching, hidden-return, zero-probability, and memory-migration controls. The two retained suites are unchanged. The first-edition suite additionally checks quantized probability records, matched adaptive updates, block-error recurrences, and the joint sensor decision. These are bounded implementation checks, not universal proofs or agent experiments."]}