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Its acquisition procedure is a separate obligation.","$\\bar F_u$, $\\bar q$ & An update and a query decoder operating on the retained state.","$T$, $\\tau$ & Translation of representations and the corresponding operation names.","$V\\subseteq\\M$ & Current set of candidate models, not a probability distribution.","$C_{e,o}(V)$ & Candidates retained after experiment $e$ returns observation $o$.","$\\phi$, $\\eta$ & Translation of candidate models and translation of observations.","$\\TV(P,Q)$ & $\\tfrac12\\sum_x|P(x)-Q(x)|$ on a finite outcome space.","\\section{A small instrument with a hidden register}\\label{sec:instrument}\nConsider a mathematical instrument with three registers\n\\[\nx=(a,b,c)\\in\\Ffive^3.\n\\]\nEach register contains one of $0,1,2,3,4$. All arithmetic in this example is modulo five. The display reports\n\\[\nq(a,b,c)=a+b.\n\\]\nThere are $125$ possible complete states, but only five possible readings. The registers and their operations are fully specified analytical objects. They are not measurements from an agent experiment.","Initially, two operations are available:\n\\begin{equation}\\label{eq:instrument-ops}\nH(a,b,c)=(a+b,b,c),\\qquad W(a,b,c)=(b,a,c).\n\\end{equation}\nThe first adds the second register into the first. The second exchanges the first two registers. The third register is preserved by both and does not affect the display. The operation order is public. A prediction concerns what the display would show after a named finite sequence of these operations.","\\subsection{Knowing the current display is insufficient}\nA record containing only $s=a+b$ answers the current question exactly. It cannot, in general, predict the display after $H$. The two complete states\n\\[\nx=(0,0,0),\\qquad y=(1,4,0)\n\\]\nboth display zero. After $H$, their displays are zero and four. Any predictor receiving only the common old display must return the same answer for both, so it cannot be exact in both cases.","A sufficient retained state for the original operation family is\n\\[\nz=(s,b)=(a+b,b).\n\\]\nIts updates are\n\\begin{equation}\\label{eq:display-coordinates}\n\\bar H(s,b)=(s+b,b),\\qquad \\bar W(s,b)=(s,s-b),\n\\end{equation}\nand its displayed output is its first coordinate. These equations retain the information needed for every finite $H/W$ continuation without retaining $c$.","The representation is also minimal in its number of states. The present reading $s$ and the reading $s_H$ after a hypothetical $H$ determine\n\\[\nb=s_H-s,\\qquad a=2s-s_H.\n\\]\nTwo different pairs $(a,b)$ therefore differ on at least one of these questions. Every exact reusable representation for this contract needs at least $25$ distinguishable values. Section~\\ref{sec:quotients} gives the general argument.","\\subsection{A new operation changes what must be known}\nNow admit\n\\begin{equation}\\label{eq:new-operation}\nJ(a,b,c)=(a+c,b,c).\n\\end{equation}\nThe old representation remains correct for $H$ and $W$. It is no longer sufficient for the enlarged contract. States with the same $(a,b)$ but different $c$ now produce different readings after $J$.","The additional reading $s_J$ yields\n\\[\nc=s_J-s.\n\\]\nTogether with the earlier distinctions for $a$ and $b$, the enlarged question family distinguishes all $125$ complete states. The useful state counts are therefore five for the current display, $25$ for reusable $H/W$ prediction, and $125$ after $J$ is admitted.","\\subsection{Correction requires evidence, not a renamed variable}\nA predictor that has retained only $(s,b)$ cannot recover $c$ merely by adding a third field to its data structure. Conditional on a fixed $(a,b)$ and a uniformly distributed $c$, every possible display after $J$ is equally likely. Any guess based only on the old record succeeds with probability $1/5$. Randomizing the guess does not improve that probability.","An actual execution of $J$, followed by a new reading, supplies the missing distinction. Because $J$ changes the instrument, the recovered state is the state \\emph{after} that operation:\n\\begin{equation}\\label{eq:recover-after-J}\n(a',b',c')=(s_J-b,\\ b,\\ s_J-s).\n\\end{equation}\nThe old state can then be reconstructed because this particular operation is invertible. Neither reversibility nor a free observation should be assumed for other instruments.","\\section{A contract for reusable understanding}\\label{sec:contract}\nLet $\\X$ and $\\Y$ be finite nonempty sets. Let $\\U$ be a finite operation alphabet, possibly empty, and let $\\Q$ be a finite nonempty family of queries $q:\\X\\to\\Y$. Each $u\\in\\U$ has an implemented total transition $F_u:\\X\\to\\X$. The deterministic mathematical contract is\n\\[\n\\C=(\\X,\\U,(F_u)_{u\\in\\U},\\Q,\\Y).\n\\]\nFor a word $w=u_1\\cdots u_n$, define $F_w=F_{u_n}\\circ\\cdots\\circ F_{u_1}$. The empty word acts as the identity. Thus $q(F_w(x))$ asks a question after a stated continuation.","t for reusable understanding}\\label{sec:contract}\nLet $\\X$ and $\\Y$ be finite nonempty sets. Let $\\U$ be a finite operation alphabet, possibly empty, and let $\\Q$ be a finite nonempty family of queries $q:\\X\\to\\Y$. Each $u\\in\\U$ has an implemented total transition $F_u:\\X\\to\\X$. The deterministic mathematical contract is\n\\[\n\\C=(\\X,\\U,(F_u)_{u\\in\\U},\\Q,\\Y).\n\\]\nFor a word $w=u_1\\cdots u_n$, define $F_w=F_{u_n}\\circ\\cdots\\circ F_{u_1}$. The empty word acts as the identity. Thus $q(F_w(x))$ asks a question after a stated continuation.","empty sets. Let $\\U$ be a finite operation alphabet, possibly empty, and let $\\Q$ be a finite nonempty family of queries $q:\\X\\to\\Y$. Each $u\\in\\U$ has an implemented total transition $F_u:\\X\\to\\X$. The deterministic mathematical contract is\n\\[\n\\C=(\\X,\\U,(F_u)_{u\\in\\U},\\Q,\\Y).\n\\]\nFor a word $w=u_1\\cdots u_n$, define $F_w=F_{u_n}\\circ\\cdots\\circ F_{u_1}$. The empty word acts as the identity. Thus $q(F_w(x))$ asks a question after a stated continuation.","\\begin{definition}[Exact reusable representation]\\label{def:representation}\nAn exact reusable representation for $\\C$ is a surjection $r:\\X\\to\\Z$, updates $\\bar F_u:\\Z\\to\\Z$, and decoders $\\bar q:\\Z\\to\\Y$ such that, for all $x,u,q$,\n\\begin{equation}\\label{eq:representation-contract}\nr(F_u(x))=\\bar F_u(r(x)),\\qquad q(x)=\\bar q(r(x)).\n\\end{equation}\n\\end{definition}\nInduction immediately gives $q(F_w(x))=\\bar q(\\bar F_w(r(x)))$ for every finite word. The same initial record supports many future questions. It is not allowed to inspect the selected future query and replace itself with that query's answer before the challenge begins.","\\subsection{A representation of the problem is not yet a representation in the solver}\nThe map $r$ specifies what information would suffice. It does not establish that a particular agent can acquire $r(x)$ from its sensors or that the agent uses this state. Those claims require an acquisition procedure, a realized computational path, and admissible interventions on that path.","For the instrument, an omniscient mathematical encoder can read $(a,b,c)$. A device restricted to the display cannot do so. It can obtain $b$ through a controlled $H$ observation, but that procedure costs an operation and changes the state. Any account of its understanding must include those effects rather than identify mathematical existence with operational access.","\\begin{counterexample}[A separate answer for each question is not one reusable record]\\label{ex:joint}\nLet $x=(a,b)\\in\\{0,1\\}^2$ and let the two queries return $a$ and $b$. If the encoder knows which query will be asked, one bit suffices: retain that answer. If one record must be fixed before the query is selected and must answer both exactly, it needs four distinct values. A one-bit record has only two values and must identify two different answer profiles.\n\\end{counterexample}\nThe order of quantifiers is decisive. A different adequate representation for every question is not one adequate representation for their union. This is why a claimed method should be assessed with its encoder, updater, and decoder fixed across the family it promises to support.","\\section{Behavioral compression and the price of new questions}\\label{sec:quotients}\nDefine\n\\begin{equation}\\label{eq:equivalence}\nx\\Eq y\\quad\\Longleftrightarrow\\quad q(F_w(x))=q(F_w(y))\\ \\text{for all }q\\in\\Q,\\ w\\in\\U^*.\n\\end{equation}\nThis is equality of all declared future answer profiles. It is an equivalence relation. It describes behavioral sufficiency under the contract, not equality of hidden mechanisms or identity outside that contract.","\\begin{theorem}[The minimal exact reusable state]\\label{thm:quotient}\nThe quotient $\\X/{\\Eq}$ carries an exact reusable representation for $\\C$. Every other exact reusable representation $r:\\X\\to\\Z$ refines this quotient: if $r(x)=r(y)$, then $x\\Eq y$. There is a unique map $h:\\Z\\to\\X/{\\Eq}$ with $h(r(x))=[x]$. In particular,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","ntation for $\\C$. Every other exact reusable representation $r:\\X\\to\\Z$ refines this quotient: if $r(x)=r(y)$, then $x\\Eq y$. There is a unique map $h:\\Z\\to\\X/{\\Eq}$ with $h(r(x))=[x]$. In particular,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","xact reusable representation $r:\\X\\to\\Z$ refines this quotient: if $r(x)=r(y)$, then $x\\Eq y$. There is a unique map $h:\\Z\\to\\X/{\\Eq}$ with $h(r(x))=[x]$. In particular,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","$ refines this quotient: if $r(x)=r(y)$, then $x\\Eq y$. There is a unique map $h:\\Z\\to\\X/{\\Eq}$ with $h(r(x))=[x]$. In particular,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","q y$. There is a unique map $h:\\Z\\to\\X/{\\Eq}$ with $h(r(x))=[x]$. In particular,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","(r(x))=[x]$. In particular,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","r,\n\\[\n|\\Z|\\geq|\\X/{\\Eq}|.\n\\]\nA fixed-length binary state needs at least $\\lceil\\log_2|\\X/{\\Eq}|\\rceil$ bits. The quotient can be computed by finite partition refinement.\n\\end{theorem}\n\\begin{proof}\nIf $x\\Eq y$, the empty word gives $q(x)=q(y)$ for every query. Moreover, $F_u(x)\\Eq F_u(y)$ because any continuation after $u$ is a continuation in the original contract. Thus $[x]\\mapsto[F_u(x)]$ and $[x]\\mapsto q(x)$ are well-defined updates and decoders.","For any exact representation, the equations in~\\eqref{eq:representation-contract} imply equality of all future decoded answers whenever $r(x)=r(y)$. This proves refinement. Surjectivity of $r$ gives existence and uniqueness of $h$ and the cardinality bound. The binary bound follows by counting available codewords.","For computation, let $P_0$ group states with identical current query vectors. Given a partition $P_k$, refine each block by the block of $F_u(x)$ for every $u$. Induction shows that $P_k$ identifies exactly the states agreeing on words of length at most $k$. Every strict refinement increases the block count. At most $|\\X|-1$ strict refinements are possible. At a fixed point the partition is closed under all operations, hence under every finite word, and is exactly~\\eqref{eq:equivalence}.\n\\end{proof}","Even uniqueness of the behavioral quotient does not supply a uniquely intelligible notation. The coordinates $(a,b)$ and $(s,b)$ describe the same $25$ behavioral classes in the original instrument. One makes the physical register update obvious. The other makes the readout immediate. Their difference is computational and explanatory rather than a difference in information content.","\\subsection{The cost of expanding the contract}\nLet $\\C'$ have the same complete state space and contain every operation and query of $\\C$, while possibly adding more. Write $\\equiv_{\\C'}$ for its finer behavioral equivalence. Take any representation $r$ adequate for $\\C$. For each $z$, let $k_z$ be the number of $\\equiv_{\\C'}$ classes that meet the fibre $r^{-1}(z)$.","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the low","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainme","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign di","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within eac","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label.","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ t","\\begin{proposition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded beha","osition}[Contract expansion and supplementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder s…","plementary information]\\label{prop:expansion}\nThe relation $\\equiv_{\\C'}$ refines $\\equiv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, …","iv_{\\C}$. Any supplementary code $s:\\X\\to\\mathcal S$ for which $(r,s)$ supports exact reusable behavior under $\\C'$ must satisfy\n\\[\n|\\mathcal S|\\geq\\max_z k_z.\n\\]\nIf every $\\equiv_{\\C'}$ class lies within a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its…","a single fibre of $r$, this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","this bound is attained by an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","an exact reusable pair $(r,s)$. This condition holds when $r$ is the old minimal behavioral quotient. A supplementary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","ntary value computed solely from $r(x)$ cannot repair a fibre meeting more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","more than one expanded class.\n\\end{proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","proposition}\n\\begin{proof}\nAgreement on the larger continuation family implies agreement on the smaller one. Within a fixed fibre of $r$, an adequate pair $(r,s)$ must separate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","rate the $k_z$ new classes, giving the lower bound. For attainment under the additional condition, assign distinct labels $1,\\ldots,k_z$ to the expanded classes within each old fibre and let $s(x)$ be its class label. The pair $(r(x),s(x))$ then identifies exactly the expanded behavioral class. The quotient update and query decoder supply its exact dynamics. Finally, any value computed only from $r(x)$ is constant on that fibre and separates none of its states.\n\\end{proof}","The attainment condition matters. An old representation can retain irrelevant extra coordinates that cease to be dynamically closed under new operations. For example, start with two bits $(a,b)$, no operations, a constant-zero query, and the unnecessarily detailed record $r(a,b)=a$. Add the swap operation while keeping the query constant. There is still only one expanded behavioral class, so $k_z=1$ in each old fibre. Yet updating the retained coordinate $a$ after the swap requires $b$. The information lower bound for answers is not sufficient to preserve arbitrary redundant coordinates. The minimal old quotient has no such problem.","retain irrelevant extra coordinates that cease to be dynamically closed under new operations. For example, start with two bits $(a,b)$, no operations, a constant-zero query, and the unnecessarily detailed record $r(a,b)=a$. Add the swap operation while keeping the query constant. There is still only one expanded behavioral class, so $k_z=1$ in each old fibre. Yet updating the retained coordinate $a$ after the swap requires $b$. The information lower bound for answers is not sufficient to preserve arbitrary redundant coordinates. The minimal old quotient has no such problem.","dinates that cease to be dynamically closed under new operations. For example, start with two bits $(a,b)$, no operations, a constant-zero query, and the unnecessarily detailed record $r(a,b)=a$. Add the swap operation while keeping the query constant. There is still only one expanded behavioral class, so $k_z=1$ in each old fibre. Yet updating the retained coordinate $a$ after the swap requires $b$. The information lower bound for answers is not sufficient to preserve arbitrary redundant coordinates. The minimal old quotient has no such problem.","In the instrument, each old $(a,b)$ class contains five expanded classes. An additional five-valued symbol is sufficient and necessary. A separately stored fixed-length symbol therefore needs three bits. This should not be confused with subtracting the rounded bounds for a joint $25$-state and $125$-state encoding, which are five and seven bits respectively. Separate buffers and joint recoding are different coding contracts.","\\begin{counterexample}[A finite continuation test leaves an unbounded alternative]\\label{ex:horizon}\nFor any fixed $h\\geq0$, a one-operation machine can return zero through the first $h$ steps and return one at step $h+1$. A constant-zero machine agrees with it on all those earlier observations. Without a state bound or another restriction on the hypothesis family, success through horizon $h$ cannot establish agreement at every future horizon.\n\\end{counterexample}\nThis does not invalidate finite certification of a supplied finite model. It distinguishes exhaustive checking of a known transition system from identifying an unrestricted unknown system by a finite record.","\\subsection{A carrier can be readable and irrelevant}\nConsider a solver state $m=(a,b,c,z)$, with $z=a+b$. Suppose its output circuit reads $a$ and $b$ directly and ignores $z$. A decoder applied to $z$ predicts the output perfectly. Replacing $z$ while keeping the actual circuit inputs unchanged has no effect. The code $z$ is available to an observer but is not the operative route used by this solver.","Now alter the implementation so that the output reads $z$, while a repair circuit recomputes $z$ from $a,b$ before every output. An attempted replacement may again have no observable effect, but for a different reason: the affected carrier is restored. A null intervention cannot identify which of these implementations is present unless its timing and the remaining information routes are controlled.","\\begin{proposition}[The identification limit of a transcript]\\label{prop:identification}\nSuppose two proposed implementations induce complete transcript laws $P_0$ and $P_1$ on a finite transcript alphabet under a specified permitted experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differenc","\\begin{proposition}[The identification limit of a transcript]\\label{prop:identification}\nSuppose two proposed implementations induce complete transcript laws $P_0$ and $P_1$ on a finite transcript alphabet under a specified permitted experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to","gin{proposition}[The identification limit of a transcript]\\label{prop:identification}\nSuppose two proposed implementations induce complete transcript laws $P_0$ and $P_1$ on a finite transcript alphabet under a specified permitted experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to total variation. This choice attains the bound.\n\\end{proof}\nThe experiment may include interventions. The proposition says only that the \\emph{permitted complete transcript} bounds identification. Adding a new sensor or an admissib…","implementations induce complete transcript laws $P_0$ and $P_1$ on a finite transcript alphabet under a specified permitted experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to total variation. This choice attains the bound.\n\\end{proof}\nThe experiment may include interventions. The proposition says only that the \\emph{permitted complete transcript} bounds identification. Adding a new sensor or an admissible internal intervention changes that transcript law and can change the limit.","laws $P_0$ and $P_1$ on a finite transcript alphabet under a specified permitted experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to total variation. This choice attains the bound.\n\\end{proof}\nThe experiment may include interventions. The proposition says only that the \\emph{permitted complete transcript} bounds identification. Adding a new sensor or an admissible internal intervention changes that transcript law and can change the limit.","d experiment, including any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to total variation. This choice attains the bound.\n\\end{proof}\nThe experiment may include interventions. The proposition says only that the \\emph{permitted complete transcript} bounds identification. Adding a new sensor or an admissible internal intervention changes that transcript law and can change the limit.","uding any adaptive choices made by its evaluator. With equal prior probability, the greatest identification accuracy available from that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to total variation. This choice attains the bound.\n\\end{proof}\nThe experiment may include interventions. The proposition says only that the \\emph{permitted complete transcript} bounds identification. Adding a new sensor or an admissible internal intervention changes that transcript law and can change the limit.","that transcript is\n\\[\n\\frac12\\bigl(1+\\TV(P_0,P_1)\\bigr).\n\\]\nIn particular, identical transcript laws permit no accuracy above $1/2$.\n\\end{proposition}\n\\begin{proof}\nLet $d(o)\\in[0,1]$ be the probability of guessing implementation 0 after transcript $o$. Accuracy is\n\\[\n\\frac12\\sum_o\\bigl(P_0(o)d(o)+P_1(o)(1-d(o))\\bigr)\n=\\frac12+\\frac12\\sum_o(P_0(o)-P_1(o))d(o).\n\\]\nThe sum is maximized by setting $d(o)=1$ on positive differences and zero on negative ones. The positive differences sum to total variation. This choice attains the bound.\n\\end{proof}\nThe experiment may include interventions. The proposition says only that the \\emph{permitted complete transcript} bounds identification. Adding a new sensor or an admissible internal intervention changes that transcript law and can change the limit.","\\subsection{The interpreter must not secretly do all the work}\nTake any bijection $p:\\X\\to\\Z$ of a finite state space. Define\n\\[\n\\bar F_u=p\\circ F_u\\circ p^{-1},\\qquad \\bar q=q\\circ p^{-1}.\n\\]\nAll exact representation equations hold. The labels in $\\Z$ may nevertheless be arbitrary, and the translated updates may require a complete lookup table. Exact representability alone does not explain why the representation is easier to acquire, use, or communicate.","\\subsection{A bounded stochastic extension}\nThe distinction between a current answer and a reusable state also survives stochastic dynamics. Let $K_u(x,x')$ be finite transition kernels and let $r:\\X\\to\\Z$ be surjective. Assume the current query values are constant on fibres of $r$.","\\begin{proposition}[Uniform controlled stochastic reuse]\\label{prop:lumpability}\nA single reduced kernel $\\bar K_u$ reproduces every projected one-step law from every complete state if and only if, whenever $r(x)=r(y)$,\n\\[\n\\sum_{v:r(v)=z'}K_u(x,v)=\\sum_{v:r(v)=z'}K_u(y,v)\n\\quad\\text{for every }u,z'.\n\\]\nWith matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history.\n\\end{proposition}\n\\begin{proof}\nA common projected row is necessary by considering point-mass initial states. If the rows agree, use their shared value to define $\\bar K_u(r(x),z')$. Conditional on any reduced history, the complete state has a distribution supported on the current fibre. Averaging equal projected rows gives","\\begin{proposition}[Uniform controlled stochastic reuse]\\label{prop:lumpability}\nA single reduced kernel $\\bar K_u$ reproduces every projected one-step law from every complete state if and only if, whenever $r(x)=r(y)$,\n\\[\n\\sum_{v:r(v)=z'}K_u(x,v)=\\sum_{v:r(v)=z'}K_u(y,v)\n\\quad\\text{for every }u,z'.\n\\]\nWith matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history.\n\\end{proposition}\n\\begin{proof}\nA common projected row is necessary by considering point-mass initial states. If the rows agree, use their shared value to define $\\bar K_u(r(x),z')$. Conditional on any reduced history, the complete state has a distribution supported on the current fibre. Averaging equal projected rows gives the same reduced transition law. Induction yields equality of the finite history laws, including the de","\\begin{proposition}[Uniform controlled stochastic reuse]\\label{prop:lumpability}\nA single reduced kernel $\\bar K_u$ reproduces every projected one-step law from every complete state if and only if, whenever $r(x)=r(y)$,\n\\[\n\\sum_{v:r(v)=z'}K_u(x,v)=\\sum_{v:r(v)=z'}K_u(y,v)\n\\quad\\text{for every }u,z'.\n\\]\nWith matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history.\n\\end{proposition}\n\\begin{proof}\nA common projected row is necessary by considering point-mass initial states. If the rows agree, use their shared value to define $\\bar K_u(r(x),z')$. Conditional on any reduced history, the complete state has a distribution supported on the current fibre. Averaging equal projected rows gives the same reduced transition law. Induction yields equality of the finite history laws, including the deterministic query values. A common randomized history policy preserves the same argu","m_{v:r(v)=z'}K_u(y,v)\n\\quad\\text{for every }u,z'.\n\\]\nWith matched initial projected laws, the reduced model preserves the law of every finite sequence of reduced states and queries under any policy depending only on the reduced history.\n\\end{proposition}\n\\begin{proof}\nA common projected row is necessary by considering point-mass initial states. If the rows agree, use their shared value to define $\\bar K_u(r(x),z')$. Conditional on any reduced history, the complete state has a distribution supported on the current fibre. Averaging equal projected rows gives the same reduced transition law. Induction yields equality of the finite history laws, including the deterministic query values. A common randomized history policy preserves the same argument after conditioning on its selected action.\n\\end{proof}\nThis is the controlled lumpability condition underlying stochastic state aggregation~\\cite{givan}. The policy qualification is essential. A controller that secretly observes a discarded variable can choose different actions in states that the reduced description identifies. Small predictio…","Let a source representation have state space $Z$, updates $f_u$, and query decoders $q$. Let a target have state space $W$, updates $g_v$, and decoders $q'$. A map $\\tau$ translates operation names. Query pairs $(q,q')$ and a representation map $T:Z\\to W$ are fixed in advance. The operation names alone do not establish that the target implements the intended source operation.","\\begin{theorem}[Exact transport of reusable operations]\\label{thm:transport}\nSuppose, for every source operation and matched query,\n\\begin{equation}\\label{eq:transport-exact}\nT f_u=g_{\\tau(u)}T,\\qquad q'T=q.\n\\end{equation}\nThen, for every source state and every finite source word $w$,\n\\[\nT f_w=g_{\\tau(w)}T,\\qquad q'(g_{\\tau(w)}(Tz))=q(f_w(z)).\n\\]\nTwo such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity.","\\begin{theorem}[Exact transport of reusable operations]\\label{thm:transport}\nSuppose, for every source operation and matched query,\n\\begin{equation}\\label{eq:transport-exact}\nT f_u=g_{\\tau(u)}T,\\qquad q'T=q.\n\\end{equation}\nThen, for every source state and every finite source word $w$,\n\\[\nT f_w=g_{\\tau(w)}T,\\qquad q'(g_{\\tau(w)}(Tz))=q(f_w(z)).\n\\]\nTwo such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intert","\\begin{theorem}[Exact transport of reusable operations]\\label{thm:transport}\nSuppose, for every source operation and matched query,\n\\begin{equation}\\label{eq:transport-exact}\nT f_u=g_{\\tau(u)}T,\\qquad q'T=q.\n\\end{equation}\nThen, for every source state and every finite source word $w$,\n\\[\nT f_w=g_{\\tau(w)}T,\\qquad q'(g_{\\tau(w)}(Tz))=q(f_w(z)).\n\\]\nTwo such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for intervention","rt}\nSuppose, for every source operation and matched query,\n\\begin{equation}\\label{eq:transport-exact}\nT f_u=g_{\\tau(u)}T,\\qquad q'T=q.\n\\end{equation}\nThen, for every source state and every finite source word $w$,\n\\[\nT f_w=g_{\\tau(w)}T,\\qquad q'(g_{\\tau(w)}(Tz))=q(f_w(z)).\n\\]\nTwo such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward","ad q'T=q.\n\\end{equation}\nThen, for every source state and every finite source word $w$,\n\\[\nT f_w=g_{\\tau(w)}T,\\qquad q'(g_{\\tau(w)}(Tz))=q(f_w(z)).\n\\]\nTwo such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward answers only from states in the image of $T$ and for translated source operations. It does not guarantee coverage of all tar…","end{equation}\nThen, for every source state and every finite source word $w$,\n\\[\nT f_w=g_{\\tau(w)}T,\\qquad q'(g_{\\tau(w)}(Tz))=q(f_w(z)).\n\\]\nTwo such transports compose, with their operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward answers only from states in the image of $T$ and for translated source operations. It does not guarantee coverage of all target states …","r operation and query correspondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward answers only from states in the image of $T$ and for translated source operations. It does not guarantee coverage of all target states or all target operations. Surjectivity, reverse translation, and recovery of discarded source distinctions are separate requirements.","orrespondences composed as well. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward answers only from states in the image of $T$ and for translated source operations. It does not guarantee coverage of all target states or all target operations. Surjectivity, reverse translation, and recovery of discarded source distinctions are separate requirements.","ll. No injectivity assumption on $T$ is needed for this forward guarantee.\n\\end{theorem}\n\\begin{proof}\nThe empty word is immediate. If the identity holds for $w$, then\n\\[\nT f_u f_w=g_{\\tau(u)}T f_w\n=g_{\\tau(u)}g_{\\tau(w)}T.\n\\]\nInduction proves the word identity, and applying the matched decoder proves the answer identity. For a second transport $S$ with operation map $\\sigma$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward answers only from states in the image of $T$ and for translated source operations. It does not guarantee coverage of all target states or all target operations. Surjectivity, reverse translation, and recovery of discarded source distinctions are separate requirements.","$, substitution gives $STf_u=h_{\\sigma(\\tau(u))}ST$ and the corresponding decoder identity. This is the same condition for the composite.\n\\end{proof}\nThis is an intertwining argument. Related composition questions for interventional causal models are studied in greater generality by Rischel and Weichwald~\\cite{rischel}. Here the theorem guarantees forward answers only from states in the image of $T$ and for translated source operations. It does not guarantee coverage of all target states or all target operations. Surjectivity, reverse translation, and recovery of discarded source distinctions are separate requirements.","\\subsection{The instrument in display coordinates}\nFor the original instrument quotient, let the source state be $(a,b)$ and set\n\\[\nT(a,b)=(a+b,b).\n\\]\nThe target updates in~\\eqref{eq:display-coordinates} satisfy~\\eqref{eq:transport-exact}. The source decoder $a+b$ becomes the target's first coordinate. Thus a method can be carried between a register description and a display-centered description without relearning the meaning of every operation sequence.","\\begin{figure}[htbp]\n\\centering\\includegraphics[width=.96\\linewidth]{figures/transport_square.pdf}\n\\caption{One verified operation square for the instrument. Translating and then applying the target update produces the same state as applying the source update and then translating. The equality concerns the declared $H$ operation, not arbitrary interventions.}\\label{fig:transport}\n\\end{figure}","\\begin{counterexample}[Correct current answers with a wrong transported operation]\\label{ex:badtransport}\nRetain the correct map $T(a,b)=(a+b,b)$ and the correct display decoder. Now incorrectly define the target swap by $\\widetilde W(s,b)=(s,b)$. The current display and the display immediately after a swap are still correct. Starting from $(a,b)=(1,3)$, however, the source sequence $W$ then $H$ ends with display zero. The incorrect target sequence ends with display two. Matching a restricted collection of outputs did not preserve the method.\n\\end{counterexample}\nThe error becomes visible only when the altered internal state is used by a later operation. That is precisely why reusable understanding requires continuations rather than a list of isolated answers.","\\subsection{Approximate transfer accumulates error}\nExact equalities are not always available. For this result let the target state space carry a metric $d$. Suppose each target update is $L$-Lipschitz for a common $L\\geq0$, and suppose\n\\[\nd(Tf_u(z),g_{\\tau(u)}(Tz))\\leq\\epsilon\n\\]\nuniformly on every state and operation that the contract permits. Let a target query be $M$-Lipschitz with real-valued output and satisfy $|q'(Tz)-q(z)|\\leq\\delta$.","te transfer accumulates error}\nExact equalities are not always available. For this result let the target state space carry a metric $d$. Suppose each target update is $L$-Lipschitz for a common $L\\geq0$, and suppose\n\\[\nd(Tf_u(z),g_{\\tau(u)}(Tz))\\leq\\epsilon\n\\]\nuniformly on every state and operation that the contract permits. Let a target query be $M$-Lipschitz with real-valued output and satisfy $|q'(Tz)-q(z)|\\leq\\delta$.","\\begin{proposition}[Finite-horizon transfer budget]\\label{prop:error}\nFor matched source and target operation sequences, start from $z_0,w_0$ with $d(Tz_0,w_0)\\leq e_0$. After $n$ operations,\n\\begin{align*}\nd(Tz_n,w_n)&\\leq L^n e_0+\\epsilon\\sum_{j=0}^{n-1}L^j,","\\begin{proposition}[Finite-horizon transfer budget]\\label{prop:error}\nFor matched source and target operation sequences, start from $z_0,w_0$ with $d(Tz_0,w_0)\\leq e_0$. After $n$ operations,\n\\begin{align*}\nd(Tz_n,w_n)&\\leq L^n e_0+\\epsilon\\sum_{j=0}^{n-1}L^j,\\\\\n|q(z_n)-q'(w_n)|&\\leq\\delta+M\\left(L^n e_0+\\epsilon\\sum_{j=0}^{n-1}L^j\\right).\n\\end{align*}\nThe empty sum is zero and $L^0=1$, including when $L=0$. At $L=1$ the state bound is $e_0+n\\epsilon$.\n\\end{proposition}\n\\begin{proof}\nWriting $e_n=d(Tz_n,w_n)$, insert $g_{\\tau(u)}(Tz_n)$ and use the triangle inequality and the Lipschitz bound to obtain $e_{n+1}\\leq\\epsilon+Le_n$. Iterating gives the stated expression. Insert $q'(Tz_n)$ between the two outputs and use the decoder mismatch and its Lipschitz constant.\n\\end{proof}\nThe bound concerns the matched operation sequence. It does not automatically cover two controllers that select different operations after slightly different observations. Adaptive choices require a matching policy contract or a separate stability argument","|q(z_n)-q'(w_n)|&\\leq\\delta+M\\left(L^n e_0+\\epsilon\\sum_{j=0}^{n-1}L^j\\right).\n\\end{align*}\nThe empty sum is zero and $L^0=1$, including when $L=0$. At $L=1$ the state bound is $e_0+n\\epsilon$.\n\\end{proposition}\n\\begin{proof}\nWriting $e_n=d(Tz_n,w_n)$, insert $g_{\\tau(u)}(Tz_n)$ and use the triangle inequality and the Lipschitz bound to obtain $e_{n+1}\\leq\\epsilon+Le_n$. Iterating gives the stated expression. Insert $q'(Tz_n)$ between the two outputs and use the decoder mismatch and its Lipschitz constant.\n\\end{proof}\nThe bound concerns the matched operation sequence. It does not automatically cover two controllers that select different operations after slightly different observations. Adaptive choices require a matching policy contract or a separate stability argument for the controller.","\\delta+M\\left(L^n e_0+\\epsilon\\sum_{j=0}^{n-1}L^j\\right).\n\\end{align*}\nThe empty sum is zero and $L^0=1$, including when $L=0$. At $L=1$ the state bound is $e_0+n\\epsilon$.\n\\end{proposition}\n\\begin{proof}\nWriting $e_n=d(Tz_n,w_n)$, insert $g_{\\tau(u)}(Tz_n)$ and use the triangle inequality and the Lipschitz bound to obtain $e_{n+1}\\leq\\epsilon+Le_n$. Iterating gives the stated expression. Insert $q'(Tz_n)$ between the two outputs and use the decoder mismatch and its Lipschitz constant.\n\\end{proof}\nThe bound concerns the matched operation sequence. It does not automatically cover two controllers that select different operations after slightly different observations. Adaptive choices require a matching policy contract or a separate stability argument for the controller.","\\begin{figure}[htbp]\n\\centering\\includegraphics[width=.94\\linewidth]{figures/error_accumulation.pdf}\n\\caption{Exact values of the recurrence $e_{n+1}=Le_n+1/100$, with $e_0=0$, illustrate the transfer bound. The curves are analytical constructions, not measured model errors. A small local defect can remain bounded, accumulate linearly, or be amplified.}\\label{fig:error}\n\\end{figure}","\\begin{definition}[Recipient-relative explanatory adequacy]\\label{def:explanation}\nFix a recipient procedure $L$, background information $B$, resource allowance, and continuation family. An artifact $E$ is adequate for a target model $m$ when $L(B,E)$ yields a representation and operations that answer the required continuations for $m$ within the declared error and resource limits. The artifact's inputs, the recipient's prior access, and any target-specific background are included in the contract.\n\\end{definition}\nThis is an operational property of an artifact and a recipient. It does not equate instruction with human comprehension, and it does not require natural language. A diagram, proof, worked example, executable transformation, or combination can play the role of $E$.","ts. The artifact's inputs, the recipient's prior access, and any target-specific background are included in the contract.\n\\end{definition}\nThis is an operational property of an artifact and a recipient. It does not equate instruction with human comprehension, and it does not require natural language. A diagram, proof, worked example, executable transformation, or combination can play the role of $E$.","\\subsection{An exact communication lower bound}\nLet $\\Theta$ be a finite family of target models. Relative to the required continuation questions, divide these models into $N$ classes with distinct answer profiles. Use one fixed recipient background and one fixed deterministic recipient. The teacher can select a message but cannot change the recipient's background separately for each target without that change being counted as communication.","\\begin{proposition}[The information an exact explanation must carry]\\label{prop:explanation}\nIf a fixed-length $b$-bit artifact lets the recipient answer the declared questions exactly for every target in $\\Theta$, then $2^b\\geq N$. With target-dependent background $B_\\theta$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe b","\\begin{proposition}[The information an exact explanation must carry]\\label{prop:explanation}\nIf a fixed-length $b$-bit artifact lets the recipient answer the declared questions exactly for every target in $\\Theta$, then $2^b\\geq N$. With target-dependent background $B_\\theta$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe bound is about information, not explanatory elegance. It does not establish that any shortest messag","\\begin{proposition}[The information an exact explanation must carry]\\label{prop:explanation}\nIf a fixed-length $b$-bit artifact lets the recipient answer the declared questions exactly for every target in $\\Theta$, then $2^b\\geq N$. With target-dependent background $B_\\theta$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe bound is about information, not explanatory elegance. It does not establish that any shortest message is easy to under","\\begin{proposition}[The information an exact explanation must carry]\\label{prop:explanation}\nIf a fixed-length $b$-bit artifact lets the recipient answer the declared questions exactly for every target in $\\Theta$, then $2^b\\geq N$. With target-dependent background $B_\\theta$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe bound is about information, not explanatory elegance. It does not establish that any shortest message is easy to understand or that the recipient can decode it eff","\\begin{proposition}[The information an exact explanation must carry]\\label{prop:explanation}\nIf a fixed-length $b$-bit artifact lets the recipient answer the declared questions exactly for every target in $\\Theta$, then $2^b\\geq N$. With target-dependent background $B_\\theta$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe bound is about information, not explanatory elegance. It does not establish that any shortest message is easy to understand or that the recipient can decode it efficiently. A compressed index into an enormous shared","\\begin{proposition}[The information an exact explanation must carry]\\label{prop:explanation}\nIf a fixed-length $b$-bit artifact lets the recipient answer the declared questions exactly for every target in $\\Theta$, then $2^b\\geq N$. With target-dependent background $B_\\theta$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe bound is about information, not explanatory elegance. It does not establish that any shortest message is easy to understand or that the recipient can decode it efficiently. A compressed index into an enormous shared table can me","a$, the same argument gives\n\\[\n2^b\\geq\\max_{B}\\ N_B,\n\\]\nwhere $N_B$ counts distinct answer profiles among targets sharing that exact background value.\n\\end{proposition}\n\\begin{proof}\nTwo different required answer profiles cannot produce the same background and message. The deterministic recipient would then make the same responses for both and fail on a distinguishing question. There are at most $2^b$ fixed-length messages. Apply this counting argument to the whole family, or separately within each background fibre.\n\\end{proof}\nThe bound is about information, not explanatory elegance. It does not establish that any shortest message is easy to understand or that the recipient can decode it efficiently. A compressed index into an enormous shared table can meet the bound while transferring little new method. The table's construction and availability must be charged to the background account.","\\subsection{What the instrument explanation transfers}\nFor the original instrument, an adequate artifact can specify arithmetic modulo five, the representation $(s,b)$, the two update rules in~\\eqref{eq:display-coordinates}, the output decoder, and the domain restriction excluding $J$. A recipient supplied with an initial $(s,b)$ can then answer every $H/W$ continuation by applying these operations.","That establishes a subject explanation for the specified recipient. It does not establish that a neural system that produced the artifact used $(s,b)$ internally. A separate process explanation might show that the system's actual intermediate states realize these coordinates and respond to corresponding interventions. These two certificates can support one another, but they answer different questions.","An especially useful explanation includes the failure boundary. The line ``the third register is irrelevant'' is too broad. The line ``the third register does not affect any display reachable through the current $H/W$ operation family'' is exact. The latter statement tells the recipient what should be reconsidered when a new operation arrives.","\\subsection{An accessible sender and a family of recipients}\nLet $\\M$ be a finite nonempty set of possible cases. The sender observes only $\\alpha(m)$, where $\\alpha:\\M\\to S$ is a declared access map. Recipient $i\\in\\{1,\\ldots,k\\}$ receives background $\\beta_i(m)$ and must recover $d_i(m)$. A demand can be a single answer, a reusable state, or the complete finite answer profile for a declared task. The maps and recipients are fixed before the case is selected.","One encoder $E:S\\to W$ produces a common artifact. A decoder $D_i$ for each recipient must satisfy\n\\begin{equation}\\label{eq:audience-contract}\nD_i(E(\\alpha(m)),\\beta_i(m))=d_i(m)\\qquad\\text{for every }m,i.\n\\end{equation}\nThe encoder is not allowed to consult a recipient-specific hidden state. Recipients do not receive an uncharged transcript or a separate description of the selected case. Here decoders may be designed once for the declared finite contract and then remain fixed. If an existing interpreter is prescribed, the criterion below is only a necessary information condition unless that interpreter can implement the constructed decoder.","Build a graph $G_\\alpha$ on $\\alpha(\\M)$. Join $s$ and $t$ when some cases $m,n$ and recipient $i$ satisfy\n\\[\n\\alpha(m)=s,\\quad\\alpha(n)=t,\\quad\n\\beta_i(m)=\\beta_i(n),\\quad d_i(m)\\ne d_i(n).\n\\]\nAn edge can be a loop when $s=t$. It then records a distinction that neither the sender's access nor that recipient's background can resolve.","\\begin{theorem}[A common explanatory artifact under sender and recipient constraints]\\label{thm:audience}\nAn exact artifact satisfying~\\eqref{eq:audience-contract} with at most $K\\geq1$ symbols exists if and only if $G_\\alpha$ has no loop and admits a proper coloring with at most $K$ colors. In the loop-free case, the minimum alphabet size is $\\chi(G_\\alpha)$, and the minimum fixed binary length with designable decoders is $\\lceil\\log_2\\chi(G_\\alpha)\\rceil$. A loop makes the exact contract infeasible for every alphabet size.\n\\end{theorem}\n\\begin{proof}\nAn encoder assigns a symbol to every accessible sender value. If adjacent vertices receive the same symbol, their witnessing recipient receives the same artifact and background in two cases requiring different outputs. No deterministic decoder can be correct in both. Thus every valid encoder is a proper coloring, and a loop","\\begin{theorem}[A common explanatory artifact under sender and recipient constraints]\\label{thm:audience}\nAn exact artifact satisfying~\\eqref{eq:audience-contract} with at most $K\\geq1$ symbols exists if and only if $G_\\alpha$ has no loop and admits a proper coloring with at most $K$ colors. In the loop-free case, the minimum alphabet size is $\\chi(G_\\alpha)$, and the minimum fixed binary length with designable decoders is $\\lceil\\log_2\\chi(G_\\alpha)\\rceil$. A loop makes the exact contract infeasible for every alphabet size.\n\\end{theorem}\n\\begin{proof}\nAn encoder assigns a symbol to every accessible sender value. If adjacent vertices receive the same symbol, their witnessing recipient receives the same artifact and background in two cases requiring different outputs. No deterministic decoder can be correct in both. Thus every valid encoder is a proper coloring, and a loop is impossible.","lanatory artifact under sender and recipient constraints]\\label{thm:audience}\nAn exact artifact satisfying~\\eqref{eq:audience-contract} with at most $K\\geq1$ symbols exists if and only if $G_\\alpha$ has no loop and admits a proper coloring with at most $K$ colors. In the loop-free case, the minimum alphabet size is $\\chi(G_\\alpha)$, and the minimum fixed binary length with designable decoders is $\\lceil\\log_2\\chi(G_\\alpha)\\rceil$. A loop makes the exact contract infeasible for every alphabet size.\n\\end{theorem}\n\\begin{proof}\nAn encoder assigns a symbol to every accessible sender value. If adjacent vertices receive the same symbol, their witnessing recipient receives the same artifact and background in two cases requiring different outputs. No deterministic decoder can be correct in both. Thus every valid encoder is a proper coloring, and a loop is impossible.","\\subsection{Complementary knowledge can reduce communication}\nFor two bits $(a,b)$, let the first recipient know $a$ and need $b$, while the second knows $b$ and needs $a$. A sender with access only to $a\\oplus b$ can broadcast that one bit. Both recipients recover their missing bit. The sender need not reconstruct the complete pair. An inference from successful explanation to possession of the recipient's full state would therefore be invalid.","\\begin{counterexample}[Individually binary explanations can require three common symbols]\\label{ex:audience-cycle}\nTake five cases labeled $0,\\ldots,4$ and five recipients. Recipient $i$ cannot distinguish cases $i$ and $i+1$ modulo five from its background, but distinguishes every other case individually. It demands zero on case $i$ and one on case $i+1$, with arbitrary fixed demands elsewhere. The sender knows the case. Each recipient separately needs only two artifact symbols. The common confusability graph is a five-cycle. It is not two-colorable, but the sequence of colors $A,B,A,B,C$ is proper. One common artifact therefore needs three symbols, or two fixed binary bits.\n\\end{counterexample}","cannot distinguish cases $i$ and $i+1$ modulo five from its background, but distinguishes every other case individually. It demands zero on case $i$ and one on case $i+1$, with arbitrary fixed demands elsewhere. The sender knows the case. Each recipient separately needs only two artifact symbols. The common confusability graph is a five-cycle. It is not two-colorable, but the sequence of colors $A,B,A,B,C$ is proper. One common artifact therefore needs three symbols, or two fixed binary bits.\n\\end{counterexample}","\\subsection{A finite model of informative correction}\nLet $\\M$ be a finite nonempty set of deterministic candidate models. Let $\\mathcal E$ be a finite set of permitted experiments. Each experiment $e$ has an outcome map $f_e:\\M\\to\\mathcal O_e$. The current information is a candidate set $V\\subseteq\\M$. On observing $o$ from $e$, update it by\n\\begin{equation}\\label{eq:filter}\nC_{e,o}(V)=\\{m\\in V:f_e(m)=o\\}.\n\\end{equation}\nThis is consistency filtering, not Bayesian inference and not a general model of scientific reasoning. An empty set is meaningful: the observation contradicts every currently retained candidate. It must not be replaced by an invented best answer.","An experiment is informative for $V$ when its outcome map is not constant on $V$. Models indistinguishable under every permitted experiment form an observational equivalence class. The aim can be to identify such a class rather than a uniquely named model.","\\begin{proposition}[Finite correction under separating evidence]\\label{prop:queries}\nSuppose the true model $m_*$ belongs to the initial set $V_0$, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty","\\begin{proposition}[Finite correction under separating evidence]\\label{prop:queries}\nSuppose the true model $m_*$ belongs to the initial set $V_0$, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true o","\\begin{proposition}[Finite correction under separating evidence]\\label{prop:queries}\nSuppose the true model $m_*$ belongs to the initial set $V_0$, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies an","\\begin{proposition}[Finite correction under separating evidence]\\label{prop:queries}\nSuppose the true model $m_*$ belongs to the initial set $V_0$, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive d","under separating evidence]\\label{prop:queries}\nSuppose the true model $m_*$ belongs to the initial set $V_0$, every observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and bran…","observation is exact, and the allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis …","allowed experiment family separates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis elementary result shares the q…","rates any two distinct remaining observational classes. Repeatedly selecting an informative experiment preserves $m_*$ and reaches a single observational class after at most $|V_0|-1$ informative observations. If there are $N$ distinguishable target classes, every adaptive procedure identifying all of them using experiments with at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis elementary result shares the query-and-counterexample perspe…","ith at most $b\\geq2$ outcomes and at most $h$ observations must satisfy $b^h\\geq N$.\n\\end{proposition}\n\\begin{proof}\nThe true model predicts each exact observation, so filtering never removes it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis elementary result shares the query-and-counterexample perspective of classical active learning~\\cite{angluin}. It does not give an efficient method for discovering the candidate family, finding an inexpensive separating experiment, or dealing with inaccurate observations. Its upper bound is a count of informative observations, not a bound on wall-clock work or physical cost.","moves it. An informative experiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis elementary result shares the query-and-counterexample perspective of classical active learning~\\cite{angluin}. It does not give an efficient method for discovering the candidate family, finding an inexpensive separating experiment, or dealing with inaccurate observations. Its upper bound is a count of informative observations, not a bound on wall-clock work or physical cost.","xperiment splits the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis elementary result shares the query-and-counterexample perspective of classical active learning~\\cite{angluin}. It does not give an efficient method for discovering the candidate family, finding an inexpensive separating experiment, or dealing with inaccurate observations. Its upper bound is a count of informative observations, not a bound on wall-clock work or physical cost.","the current set into at least two nonempty outcome classes. Whatever true outcome occurs, at least one candidate is removed. The nonempty set can strictly shrink at most $|V_0|-1$ times. If two observational classes remain, the separating assumption supplies another informative experiment. For the second claim, an adaptive decision tree of depth at most $h$ and branching at most $b$ has at most $b^h$ leaves. Distinct required classes must reach different leaves.\n\\end{proof}\nThis elementary result shares the query-and-counterexample perspective of classical active learning~\\cite{angluin}. It does not give an efficient method for discovering the candidate family, finding an inexpensive separating experiment, or dealing with inaccurate observations. Its upper bound is a count of informative observations, not a bound on wall-clock work or physical cost.","\\subsection{Repairing the running example}\nBefore $J$ is admitted, all five possible values of $c$ are equivalent under the base contract. Treating them as one class is correct, not ignorance that the old task required the predictor to resolve. After $J$ is admitted, these classes split. The system can report that its old state no longer determines the answer.","The current record supplies $(s,b)$. Executing $J$ and observing $s_J$ determines $c=s_J-s$, while~\\eqref{eq:recover-after-J} updates the affected first register. The corrected extended coordinates are $(s_J,b,c)$. Their operations are\n\\[\n\\bar H(s,b,c)=(s+b,b,c),\\quad\n\\bar W(s,b,c)=(s,s-b,c),\\quad\n\\bar J(s,b,c)=(s+c,b,c).\n\\]\nThe base method survives as the restriction of the enlarged method. Only the newly relevant distinction is acquired.","A different kind of failure would occur if the instrument secretly implemented $H(a,b,c)=(a+2b,b,c)$. That would contradict the old transition model rather than merely extend the operation catalogue. A correction record should say which assumption changed. Otherwise a sequence of patched answers can conceal a permanently incorrect account of the mechanism.","\\subsection{Experiments change both information and state}\nLet $X$ be a finite nonempty state space. Each experiment $e$ has an admissibility set $A_e\\subseteq X$, a transition $T_e:X\\to X$, an observation $o_e:X\\to O_e$, and a known cost $c_e\\geq0$. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example $o_e=q\\circ T_e$. At an information state $V\\subseteq X$, the experiment is permitted only when $V\\subseteq A_e$. After outcome $o$, the new support is\n\\begin{equation}\\label{eq:acquisition-update}\nV_{e,o}=\\{T_e(x):x\\in V,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are possible. This update tracks the world after the intervention, not just a narrowed list of old states.","xperiments change both information and state}\nLet $X$ be a finite nonempty state space. Each experiment $e$ has an admissibility set $A_e\\subseteq X$, a transition $T_e:X\\to X$, an observation $o_e:X\\to O_e$, and a known cost $c_e\\geq0$. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example $o_e=q\\circ T_e$. At an information state $V\\subseteq X$, the experiment is permitted only when $V\\subseteq A_e$. After outcome $o$, the new support is\n\\begin{equation}\\label{eq:acquisition-update}\nV_{e,o}=\\{T_e(x):x\\in V,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are possible. This update tracks the world after the intervention, not just a narrowed list of old states.","be a finite nonempty state space. Each experiment $e$ has an admissibility set $A_e\\subseteq X$, a transition $T_e:X\\to X$, an observation $o_e:X\\to O_e$, and a known cost $c_e\\geq0$. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example $o_e=q\\circ T_e$. At an information state $V\\subseteq X$, the experiment is permitted only when $V\\subseteq A_e$. After outcome $o$, the new support is\n\\begin{equation}\\label{eq:acquisition-update}\nV_{e,o}=\\{T_e(x):x\\in V,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are possible. This update tracks the world after the intervention, not just a narrowed list of old states.",". Each experiment $e$ has an admissibility set $A_e\\subseteq X$, a transition $T_e:X\\to X$, an observation $o_e:X\\to O_e$, and a known cost $c_e\\geq0$. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example $o_e=q\\circ T_e$. At an information state $V\\subseteq X$, the experiment is permitted only when $V\\subseteq A_e$. After outcome $o$, the new support is\n\\begin{equation}\\label{eq:acquisition-update}\nV_{e,o}=\\{T_e(x):x\\in V,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are possible. This update tracks the world after the intervention, not just a narrowed list of old states.","ity set $A_e\\subseteq X$, a transition $T_e:X\\to X$, an observation $o_e:X\\to O_e$, and a known cost $c_e\\geq0$. The observation is indexed by the state before the experiment, but can describe a reading taken afterward: for example $o_e=q\\circ T_e$. At an information state $V\\subseteq X$, the experiment is permitted only when $V\\subseteq A_e$. After outcome $o$, the new support is\n\\begin{equation}\\label{eq:acquisition-update}\nV_{e,o}=\\{T_e(x):x\\in V,\\ o_e(x)=o\\}.\n\\end{equation}\nOnly nonempty branches are possible. This update tracks the world after the intervention, not just a narrowed list of old states.","Let $r:X\\to Z$ be the target record. A support is terminal when $r$ is constant on it. Let $h$ be the number of experiments still available. Write $\\mathcal V_h(V)$ for the least worst-case sum of experiment costs among policies that terminate with the target known within $h$ steps. Infeasibility is denoted $+\\infty$. Then define\n\\begin{equation}\\label{eq:acquisition-bellman}\n\\begin{aligned}\n\\mathcal V_h(V)&=0 &&\\text{if }|r(V)|=1,","Let $r:X\\to Z$ be the target record. A support is terminal when $r$ is constant on it. Let $h$ be the number of experiments still available. Write $\\mathcal V_h(V)$ for the least worst-case sum of experiment costs among policies that terminate with the target known within $h$ steps. Infeasibility is denoted $+\\infty$. Then define\n\\begin{equation}\\label{eq:acquisition-bellman}\n\\begin{aligned}\n\\mathcal V_h(V)&=0 &&\\text{if }|r(V)|=1,\\\\\n\\mathcal V_0(V)&=+\\infty &&\\text{otherwise},\\\\\n\\mathcal V_h(V)&=\\min_{e:V\\subseteq A_e}\n\\left(c_e+\\max_{o:V_{e,o}\\ne\\varnothing}\\mathcal V_{h-1}(V_{e,o})\\right)\n&&\\substack{\\text{otherwise,}\\\\h>0.}\n\\end{aligned}\n\\end{equation}\nAn empty minimum is $+\\infty$. The terminal case permits stopping immediately, so an unnecessary additional experiment is never forced.","h>0.}\n\\end{aligned}\n\\end{equation}\nAn empty minimum is $+\\infty$. The terminal case permits stopping immediately, so an unnecessary additional experiment is never forced.","\\begin{theorem}[Exact finite acquisition with state-changing observations]\\label{thm:acquisition}\nFor every nonempty support $V$ and finite horizon $h$, recurrence~\\eqref{eq:acquisition-bellman} equals the minimum worst-case acquisition cost among deterministic adaptive policies with that horizon, access boundary, and current-state target. Every finite value has an attaining policy obtained by choosing a minimizing experiment and its recursively attaining branches. The value is infinite exactly when no such policy exists.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, a target can be returned without further evidence precisely when it is constant over the current support. This gives the base case. At a nonterminal support with $h>0$, the first experiment must be admissible at every possible state. After its observed o","\\begin{theorem}[Exact finite acquisition with state-changing observations]\\label{thm:acquisition}\nFor every nonempty support $V$ and finite horizon $h$, recurrence~\\eqref{eq:acquisition-bellman} equals the minimum worst-case acquisition cost among deterministic adaptive policies with that horizon, access boundary, and current-state target. Every finite value has an attaining policy obtained by choosing a minimizing experiment and its recursively attaining branches. The value is infinite exactly when no such policy exists.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, a target can be returned without further evidence precisely when it is constant over the current support. This gives the base case. At a nonterminal support with $h>0$, the first experiment must be admissible at every possible state. After its observed outcome the possible cur","rget. Every finite value has an attaining policy obtained by choosing a minimizing experiment and its recursively attaining branches. The value is infinite exactly when no such policy exists.\n\\end{theorem}\n\\begin{proof}\nAt horizon zero, a target can be returned without further evidence precisely when it is constant over the current support. This gives the base case. At a nonterminal support with $h>0$, the first experiment must be admissible at every possible state. After its observed outcome the possible current states are exactly~\\eqref{eq:acquisition-update}. By induction, the best possible remaining worst-case cost on that branch is $\\mathcal V_{h-1}$. Every nonempty branch can occur, giving the maximum over branches. Minimizing over the finite experiment set gives a lower bound for every adaptive policy. Choosing a minimizing experiment and attaining branch policies achieves that bound whenever it is finite. If every choice has an impossible branch, no guaranteed policy exists.\n\\end{proof}","can be returned without further evidence precisely when it is constant over the current support. This gives the base case. At a nonterminal support with $h>0$, the first experiment must be admissible at every possible state. After its observed outcome the possible current states are exactly~\\eqref{eq:acquisition-update}. By induction, the best possible remaining worst-case cost on that branch is $\\mathcal V_{h-1}$. Every nonempty branch can occur, giving the maximum over branches. Minimizing over the finite experiment set gives a lower bound for every adaptive policy. Choosing a minimizing experiment and attaining branch policies achieves that bound whenever it is finite. If every choice has an impossible branch, no guaranteed policy exists.\n\\end{proof}","The state search can be large: there are up to $2^{|X|}-1$ nonempty supports, before accounting for additional memory variables. The theorem is an exact finite characterization, not a polynomial-time learning guarantee. The delivered reference implementation memoizes supports and remaining horizons. Its companion check compares the result with a separately implemented enumeration of complete observation-history policies on small systems.","\\subsection{The diagnostic operation is a paid transformation}\nIn the instrument, take a known base record $(s,b)$ and let the five possible values of $c$ define the initial support. Set the cost of $J$ to two units, and the cost of $H$ and $W$ to one unit each. Each experiment returns the exact display after its operation. The target is the enlarged current record $(s,b,c)$.","A $J$ observation resolves all five possibilities and costs two. An $H/W$ sequence cannot reveal $c$, so no policy costing less than two can meet the target. The optimum is therefore two, attained by executing $J$ and retaining the post-operation record. This example fixes its sensor and operation catalogue. It does not claim that every physical diagnostic with five outcomes has the same cost.","\\begin{figure}[htbp]\n\\centering\\includegraphics[width=.98\\linewidth]{figures/acquisition_branches.pdf}\n\\caption{A single $J$ diagnostic from base record $(s,b)=(4,3)$ has five possible exact reports. Each report determines a different \\emph{post-operation} record. The cost is charged to the operation, and the hidden value is acquired rather than supplied to the decoder.}\\label{fig:acquisition}\n\\end{figure}","To require recovery of a past distinction, replace the state by $(x_0,x_t)$, let every transition preserve $x_0$, and make the terminal target include the required part of $x_0$. The same recurrence then treats historical reconstruction honestly. In the reset example, the two origins remain indistinguishable at every horizon. If both current knowledge and preserved history matter, use their joint target rather than claim that either implies the other.","For the instrument, the diagnostic $J$ is invertible once $c$ is known, so the exact observation can support both records. That property comes from the specified operation. It cannot be transferred to an arbitrary experiment. Likewise, a planner that chooses an informative experiment is not thereby credited with discovering its transition law or building its sensor.","The horizon is a contractual limit, not an arbitrary truncation of a failed computation. A negative result at horizon $h$ says that the declared task cannot be guaranteed within that many permitted experiments. Increasing the horizon changes the problem. It does not retroactively validate a promise made at the smaller horizon.","Let $\\M$ and $\\mathcal N$ be finite model families, with model translation $\\phi:\\M\\to\\mathcal N$. For one source experiment $e$, let $f:\\M\\to\\mathcal O$ give its outcomes. A corresponding target experiment has outcome map $g:\\mathcal N\\to\\mathcal O'$. Let $\\eta:\\mathcal O\\to\\mathcal O'$ translate evidence. Assume the basic observation square commutes:\n\\begin{equation}\\label{eq:evidence-square}\ng(\\phi(m))=\\eta(f(m))\\quad\\text{for all }m\\in\\M.\n\\end{equation}\nThis condition says that a model's predicted evidence translates correctly. It permits $\\eta$ to merge source outcomes.","O$ give its outcomes. A corresponding target experiment has outcome map $g:\\mathcal N\\to\\mathcal O'$. Let $\\eta:\\mathcal O\\to\\mathcal O'$ translate evidence. Assume the basic observation square commutes:\n\\begin{equation}\\label{eq:evidence-square}\ng(\\phi(m))=\\eta(f(m))\\quad\\text{for all }m\\in\\M.\n\\end{equation}\nThis condition says that a model's predicted evidence translates correctly. It permits $\\eta$ to merge source outcomes.","Write $\\phi[V]=\\{\\phi(m):m\\in V\\}$. Let $C_o(V)=\\{m\\in V:f(m)=o\\}$ and $C'_{o'}(W)=\\{n\\in W:g(n)=o'\\}$. Both are ordinary consistency filters.","\\begin{theorem}[Correction transport and evidence reflection]\\label{thm:correction}\nUnder~\\eqref{eq:evidence-square}, for every $V\\subseteq\\M$ and every attainable source outcome $o\\in f(\\M)$,\n\\begin{equation}\\label{eq:correction-inclusion}\n\\phi[C_o(V)]\\subseteq C'_{\\eta(o)}(\\phi[V]).\n\\end{equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","theorem}[Correction transport and evidence reflection]\\label{thm:correction}\nUnder~\\eqref{eq:evidence-square}, for every $V\\subseteq\\M$ and every attainable source outcome $o\\in f(\\M)$,\n\\begin{equation}\\label{eq:correction-inclusion}\n\\phi[C_o(V)]\\subseteq C'_{\\eta(o)}(\\phi[V]).\n\\end{equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","$o\\in f(\\M)$,\n\\begin{equation}\\label{eq:correction-inclusion}\n\\phi[C_o(V)]\\subseteq C'_{\\eta(o)}(\\phi[V]).\n\\end{equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","tion}\\label{eq:correction-inclusion}\n\\phi[C_o(V)]\\subseteq C'_{\\eta(o)}(\\phi[V]).\n\\end{equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","tion-inclusion}\n\\phi[C_o(V)]\\subseteq C'_{\\eta(o)}(\\phi[V]).\n\\end{equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","}\n\\phi[C_o(V)]\\subseteq C'_{\\eta(o)}(\\phi[V]).\n\\end{equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","equation}\nEquality holds for all such $V$ and $o$ if and only if $\\eta$ is injective on the attainable outcome set $f(\\M)$. The claim includes candidate sets contradicted by the observation, so that preservation of an empty update is part of the contract.\n\\end{theorem}\n\\begin{proof}\nIf $n\\in\\phi[C_o(V)]$, then $n=\\phi(m)$ for some $m\\in V$ with $f(m)=o$. Equation~\\eqref{eq:evidence-square} gives $g(n)=\\eta(o)$, proving inclusion.","If $\\eta$ is injective on $f(\\M)$ and $n\\in C'_{\\eta(o)}(\\phi[V])$, choose $m\\in V$ with $\\phi(m)=n$. Then $\\eta(f(m))=g(n)=\\eta(o)$, so $f(m)=o$. Thus $n\\in\\phi[C_o(V)]$, proving equality.","Conversely, suppose two distinct attainable outcomes $a,b$ have $\\eta(a)=\\eta(b)$. Choose $m$ with $f(m)=b$, let $V=\\{m\\}$, and consider source observation $a$. The source update is empty, whereas the target update retains $\\phi(m)$ by~\\eqref{eq:evidence-square}. Uniform equality therefore fails.\n\\end{proof}\nThis is an exact set-theoretic result about direct images and filtering. The institutional or cognitive interpretation does not make it a new general theorem of set theory. Its value is to expose a condition that output translation alone omits: the translated evidence must preserve the distinctions by which a candidate is excluded.","\\begin{counterexample}[A translated observation preserves both mistaken candidates]\\label{ex:correction}\nTake $\\M=\\{m_0,m_1\\}$ with $f(m_0)=0$ and $f(m_1)=1$. Translate models bijectively to $\\{n_0,n_1\\}$, but translate both observations to a single symbol $*$. Let $g(n_0)=g(n_1)=*$. The observation square commutes. Starting from both candidates, source observation $1$ leaves only $m_1$, whose image is $\\{n_1\\}$. Translating first and then filtering by $*$ leaves both $n_0$ and $n_1$.\n\\end{counterexample}","counterexample}[A translated observation preserves both mistaken candidates]\\label{ex:correction}\nTake $\\M=\\{m_0,m_1\\}$ with $f(m_0)=0$ and $f(m_1)=1$. Translate models bijectively to $\\{n_0,n_1\\}$, but translate both observations to a single symbol $*$. Let $g(n_0)=g(n_1)=*$. The observation square commutes. Starting from both candidates, source observation $1$ leaves only $m_1$, whose image is $\\{n_1\\}$. Translating first and then filtering by $*$ leaves both $n_0$ and $n_1$.\n\\end{counterexample}","both mistaken candidates]\\label{ex:correction}\nTake $\\M=\\{m_0,m_1\\}$ with $f(m_0)=0$ and $f(m_1)=1$. Translate models bijectively to $\\{n_0,n_1\\}$, but translate both observations to a single symbol $*$. Let $g(n_0)=g(n_1)=*$. The observation square commutes. Starting from both candidates, source observation $1$ leaves only $m_1$, whose image is $\\{n_1\\}$. Translating first and then filtering by $*$ leaves both $n_0$ and $n_1$.\n\\end{counterexample}","ndidates]\\label{ex:correction}\nTake $\\M=\\{m_0,m_1\\}$ with $f(m_0)=0$ and $f(m_1)=1$. Translate models bijectively to $\\{n_0,n_1\\}$, but translate both observations to a single symbol $*$. Let $g(n_0)=g(n_1)=*$. The observation square commutes. Starting from both candidates, source observation $1$ leaves only $m_1$, whose image is $\\{n_1\\}$. Translating first and then filtering by $*$ leaves both $n_0$ and $n_1$.\n\\end{counterexample}","label{ex:correction}\nTake $\\M=\\{m_0,m_1\\}$ with $f(m_0)=0$ and $f(m_1)=1$. Translate models bijectively to $\\{n_0,n_1\\}$, but translate both observations to a single symbol $*$. Let $g(n_0)=g(n_1)=*$. The observation square commutes. Starting from both candidates, source observation $1$ leaves only $m_1$, whose image is $\\{n_1\\}$. Translating first and then filtering by $*$ leaves both $n_0$ and $n_1$.\n\\end{counterexample}","The failure is not an inaccurate target answer in its coarse language. Both models really do predict $*$ there. The failure is the loss of a previously meaningful ability to discriminate. This is why a representation that can communicate conclusions may still be an inadequate carrier of scientific revision.","\\subsection{The instrument explains the boundary}\nFor base experiments built only from $H$, $W$, and the display, the translation from $(a,b,c)$ to $(a,b)$ can preserve evidence without collapsing outcome symbols. Different values of $c$ produce the same permitted observations, so their identification loses no base correction power.","For $J$, that translation cannot preserve the full outcome map at all: two states with identical $(a,b)$ can have different readings after $J$. One could force a commuting square by translating all new readings into one coarse symbol. That would preserve the trivial coarse prediction, but it would destroy the ability to discover $c$. Enlarging the representation is a different response from weakening the question until the old representation appears adequate.","Let $M\\in\\M$ be a fixed unknown model in a finite family, with prior $p(m)$. An experiment produces $O\\in\\mathcal O$ with likelihood $L_m(o)$. A common reporting channel $B(o,z)$ produces $Z\\in\\mathcal Z$. Each likelihood row and each channel row is a probability distribution. The complete joint model is\n\\begin{equation}\\label{eq:bayes-joint}\n\\Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).\n\\end{equation}\nIn particular, the channel does not secretly inspect $M$. Define\n\\[\nQ_m(z)=\\sum_o L_m(o)B(o,z),\\qquad\np^o(m)=\\frac{p(m)L_m(o)}{\\sum_n p(n)L_n(o)},\\qquad\np^z(m)=\\frac{p(m)Q_m(z)}{\\sum_n p(n)Q_n(z)}.\n\\]\nA posterior is defined only when its denominator is positive. An event assigned probability zero by the entire curr","Let $M\\in\\M$ be a fixed unknown model in a finite family, with prior $p(m)$. An experiment produces $O\\in\\mathcal O$ with likelihood $L_m(o)$. A common reporting channel $B(o,z)$ produces $Z\\in\\mathcal Z$. Each likelihood row and each channel row is a probability distribution. The complete joint model is\n\\begin{equation}\\label{eq:bayes-joint}\n\\Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).\n\\end{equation}\nIn particular, the channel does not secretly inspect $M$. Define\n\\[\nQ_m(z)=\\sum_o L_m(o)B(o,z),\\qquad\np^o(m)=\\frac{p(m)L_m(o)}{\\sum_n p(n)L_n(o)},\\qquad\np^z(m)=\\frac{p(m)Q_m(z)}{\\sum_n p(n)Q_n(z)}.\n\\]\nA posterior is defined only when its denominator is positive. An event assigned probability zero by the entire current model has no posterior under that model. It is a model-fail","Let $M\\in\\M$ be a fixed unknown model in a finite family, with prior $p(m)$. An experiment produces $O\\in\\mathcal O$ with likelihood $L_m(o)$. A common reporting channel $B(o,z)$ produces $Z\\in\\mathcal Z$. Each likelihood row and each channel row is a probability distribution. The complete joint model is\n\\begin{equation}\\label{eq:bayes-joint}\n\\Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).\n\\end{equation}\nIn particular, the channel does not secretly inspect $M$. Define\n\\[\nQ_m(z)=\\sum_o L_m(o)B(o,z),\\qquad\np^o(m)=\\frac{p(m)L_m(o)}{\\sum_n p(n)L_n(o)},\\qquad\np^z(m)=\\frac{p(m)Q_m(z)}{\\sum_n p(n)Q_n(z)}.\n\\]\nA posterior is defined only when its denominator is positive. An event assigned probability zero by the entire current model has no posterior under that model. It is a model-failure signal, not permission to invent a pr","Let $M\\in\\M$ be a fixed unknown model in a finite family, with prior $p(m)$. An experiment produces $O\\in\\mathcal O$ with likelihood $L_m(o)$. A common reporting channel $B(o,z)$ produces $Z\\in\\mathcal Z$. Each likelihood row and each channel row is a probability distribution. The complete joint model is\n\\begin{equation}\\label{eq:bayes-joint}\n\\Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).\n\\end{equation}\nIn particular, the channel does not secretly inspect $M$. Define\n\\[\nQ_m(z)=\\sum_o L_m(o)B(o,z),\\qquad\np^o(m)=\\frac{p(m)L_m(o)}{\\sum_n p(n)L_n(o)},\\qquad\np^z(m)=\\frac{p(m)Q_m(z)}{\\sum_n p(n)Q_n(z)}.\n\\]\nA posterior is defined only when its denominator is positive. An event assigned probability zero by the entire current model has no posterior under that model. It is a model-failure signal, not permission to invent a probability distribution.","family, with prior $p(m)$. An experiment produces $O\\in\\mathcal O$ with likelihood $L_m(o)$. A common reporting channel $B(o,z)$ produces $Z\\in\\mathcal Z$. Each likelihood row and each channel row is a probability distribution. The complete joint model is\n\\begin{equation}\\label{eq:bayes-joint}\n\\Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).\n\\end{equation}\nIn particular, the channel does not secretly inspect $M$. Define\n\\[\nQ_m(z)=\\sum_o L_m(o)B(o,z),\\qquad\np^o(m)=\\frac{p(m)L_m(o)}{\\sum_n p(n)L_n(o)},\\qquad\np^z(m)=\\frac{p(m)Q_m(z)}{\\sum_n p(n)Q_n(z)}.\n\\]\nA posterior is defined only when its denominator is positive. An event assigned probability zero by the entire current model has no posterior under that model. It is a model-failure signal, not permission to invent a probability distribution.","prior $p(m)$. An experiment produces $O\\in\\mathcal O$ with likelihood $L_m(o)$. A common reporting channel $B(o,z)$ produces $Z\\in\\mathcal Z$. Each likelihood row and each channel row is a probability distribution. The complete joint model is\n\\begin{equation}\\label{eq:bayes-joint}\n\\Pr(M=m,O=o,Z=z)=p(m)L_m(o)B(o,z).\n\\end{equation}\nIn particular, the channel does not secretly inspect $M$. Define\n\\[\nQ_m(z)=\\sum_o L_m(o)B(o,z),\\qquad\np^o(m)=\\frac{p(m)L_m(o)}{\\sum_n p(n)L_n(o)},\\qquad\np^z(m)=\\frac{p(m)Q_m(z)}{\\sum_n p(n)Q_n(z)}.\n\\]\nA posterior is defined only when its denominator is positive. An event assigned probability zero by the entire current model has no posterior under that model. It is a model-failure signal, not permission to invent a probability distribution.","\\subsection{When a coarser report preserves every update}\nThe reporting model implies $M\\to O\\to Z$. Therefore, on every positive-probability pair $(o,z)$, the full posterior after observing both records equals $p^o$. The coarse posterior is a mixture of these fine posteriors:\n\\begin{equation}\\label{eq:posterior-mixture}\np^z=\\sum_o\\Pr(O=o\\mid Z=z)\\,p^o.\n\\end{equation}\nAveraging is not generally the same as retaining each possible update. The distinction is classical statistical sufficiency, whose factorization foundations include Halmos and Savage~\\cite{halmos}. We state and prove the finite channel criterion used here.","\\begin{theorem}[Exact Bayesian correction through a reporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition i","\\begin{theorem}[Exact Bayesian correction through a reporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose","\\begin{theorem}[Exact Bayesian correction through a reporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\","\\begin{theorem}[Exact Bayesian correction through a reporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z","\\begin{theorem}[Exact Bayesian correction through a reporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positiv","act Bayesian correction through a reporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n…","eporting channel]\\label{thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel…","thm:bayes}\nFor fixed finite likelihoods $L$ and reporting channel $B$, the following are equivalent:\n\\begin{enumerate}[label=(\\roman*)]\n\\item For every prior $p$, $p^o=p^z$ on every positive-probability pair $(o,z)$.\n\\item This equality holds for one prior assigning positive mass to every model.\n\\item There exists a row-stochastic kernel $\\rho(o\\mid z)$ such that, for all $m,o,z$,\n\\begin{equation}\\label{eq:bayes-factorization}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If…","on}\nL_m(o)B(o,z)=Q_m(z)\\rho(o\\mid z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","d z).\n\\end{equation}\n\\end{enumerate}\nThus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","Thus a report preserves every posterior update exactly when the discarded observation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","rvation can be conditionally reconstructed from that report without knowing the model, in the actual joint experiment.\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.",".\n\\end{theorem}\n\\begin{proof}\nThe first condition implies the second. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","d. Suppose the second holds under a fully supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","lly supported prior. For each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","or each $z$ of positive probability set $\\rho(o\\mid z)=\\Pr(O=o\\mid Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","id Z=z)$. If $(o,z)$ has positive probability, posterior equality and Bayes' formula give\n\\[\np(m)L_m(o)B(o,z)\n=\\Pr(O=o,Z=z)p^o(m)\n=\\rho(o\\mid z)\\Pr(Z=z)p^z(m)\n=p(m)Q_m(z)\\rho(o\\mid z).\n\\]\nCancel the positive $p(m)$. If the pair has probability zero, full prior support makes $L_m(o)B(o,z)=0$ for every model, and $\\rho(o\\mid z)=0$. If $z$ is impossible under every model, choose any probability row $\\rho(\\cdot\\mid z)$ since both sides are zero. This proves the third condition.","Conversely, the factorization makes the likelihood of $(o,z)$ a model-independent multiple of the likelihood of $z$. Normalization under any prior with positive pair probability therefore gives $p(M\\mid o,z)=p^z$. The common reporting channel gives $p(M\\mid o,z)=p^o$. This proves the first condition.\n\\end{proof}","For deterministic reporting $z=\\eta(o)$, the criterion says that the nonzero vectors $(L_m(o))_{m\\in\\M}$ within each report fibre must be proportional. An ancillary symbol may be discarded even when the report is not injective. In the deterministic evidence model, distinct attainable outcomes have nonzero likelihood vectors with disjoint supports. Those vectors cannot be proportional, recovering the injectivity requirement for that special case. Noise therefore changes the correct preservation test, not merely the notation used for filtering.","The kernel $\\rho$ is conditional reconstruction for the specified coupling $(M,O,Z)$. The existence of some unrelated procedure simulating an experiment from another record is a different assertion. Neither assertion shows that the real sensor has the stated likelihood or that a learner inferred it correctly.","\\subsection{An exact measure of the lost update}\nA coarser report may fail the criterion while still retaining much of the useful evidence. There is a natural average loss, though it should not be mistaken for a worst-case bound on every possible observation. For finite distributions use natural logarithms and the convention $0\\log 0=0$.","\\begin{proposition}[Expected posterior loss is conditional information]\\label{prop:posterior-information}\nUnder~\\eqref{eq:bayes-joint},\n\\begin{equation}\\label{eq:posterior-kl}\n\\mathbb E_{O,Z}\\!\\left[D_{\\mathrm{KL}}(p^O\\Vert p^Z)\\right]\n=I(M;O\\mid Z)=I(M;O)-I(M;Z).\n\\end{equation}\nThe common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~\\ref{thm:bayes}.\n\\end{proposition}\n\\begin{proof}\nOn the support of the joint law, $p(m\\mid o,z)=p^o(m)$. Substituting this into the finite definition of conditional mutual information gives the expected divergence. Expanding its logarithm as $\\log(p^o(m)/p(m))-\\log(p^z(m)/p(m))$ and summing gives the difference of mutual informations. Terms with zero prior mass carry no weight. Every positive-weight numerator has a positive denominator, so the expression i",",Z}\\!\\left[D_{\\mathrm{KL}}(p^O\\Vert p^Z)\\right]\n=I(M;O\\mid Z)=I(M;O)-I(M;Z).\n\\end{equation}\nThe common value is nonnegative. It is zero exactly when the two posteriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~\\ref{thm:bayes}.\n\\end{proposition}\n\\begin{proof}\nOn the support of the joint law, $p(m\\mid o,z)=p^o(m)$. Substituting this into the finite definition of conditional mutual information gives the expected divergence. Expanding its logarithm as $\\log(p^o(m)/p(m))-\\log(p^z(m)/p(m))$ and summing gives the difference of mutual informations. Terms with zero prior mass carry no weight. Every positive-weight numerator has a positive denominator, so the expression is finite.","teriors agree on every positive-probability observation pair for that prior. For a fully supported prior, zero is equivalent to the factorization in Theorem~\\ref{thm:bayes}.\n\\end{proposition}\n\\begin{proof}\nOn the support of the joint law, $p(m\\mid o,z)=p^o(m)$. Substituting this into the finite definition of conditional mutual information gives the expected divergence. Expanding its logarithm as $\\log(p^o(m)/p(m))-\\log(p^z(m)/p(m))$ and summing gives the difference of mutual informations. Terms with zero prior mass carry no weight. Every positive-weight numerator has a positive denominator, so the expression is finite.","For probability distributions, $D_{\\mathrm{KL}}(a\\Vert b)\\geq0$, with equality only at $a=b$. This follows from $\\log t\\leq t-1$ on the common support, with missing support only increasing the divergence. A finite weighted sum of nonnegative terms is zero exactly when every positive-weight term is zero. The final assertion follows from Theorem~\\ref{thm:bayes}.\n\\end{proof}","\\subsection{The noisy instrument and a harmless discarded flag}\nReturn to the instrument immediately after a known $J$ intervention. The earlier $(s,b)$ is known, while $c$ is initially uniform on five values. A read-only sensor reports the shifted value $Y=\\text{reported display}-s$ modulo five. It returns the correct $c$ with probability $3/4$ and each wrong value with probability $1/16$. The physical state remains fixed during repeated read-only observations.","Suppose the raw report also contains a binary flag with probabilities $1/3$ and $2/3$, independent of $c$ and of $Y$. Removing that flag satisfies~\\eqref{eq:bayes-factorization} exactly. Both recipients retain the same posterior over $c$ and hence the same distribution over the post-$J$ state $(s+c-b,b,c)$. The flag does not carry corrective information for this family.","\\begin{counterexample}[A compact report can destroy the useful noisy distinction]\\label{ex:noisy-parity}\nIn the same instrument, replace $Y\\in\\{0,1,2,3,4\\}$ by its ordinary integer parity, not a field operation. On observing the full symbol $Y=2$, the posterior puts $3/4$ on $c=2$ and $1/16$ on each alternative. On observing only even parity, it puts $7/24$ on each even candidate and $1/16$ on each odd candidate. The report is correct as a parity statement but fails to preserve the posterior needed to distinguish the three even values.\n\\end{counterexample}","For $n$ conditionally independent identical reports $Y=2$, the posterior of $c=2$ is\n\\[\n\\frac{12^n}{12^n+4}.\n\\]\nFor $n$ reports retained only as even parity, the posterior of that same candidate is\n\\[\n\\frac{14^n}{3\\cdot14^n+2\\cdot3^n}.\n\\]\nAt $n=0$ both equal the prior $1/5$. The full-report value at $n=3$ is $432/433$, whereas the parity value cannot exceed $1/3$. Repeatedly receiving the same coarse fact does not restore distinctions the reporting language never records.","\\begin{figure}[htbp]\n\\centering\\includegraphics[width=.96\\linewidth]{figures/noisy_correction.pdf}\n\\caption{Posterior mass on $c=2$ along the specified repeated-report histories. Removing the ancillary flag gives exactly the full-report curve. Retaining only parity leaves three even candidates indistinguishable. The curves are analytic conditional calculations, not measured learning trajectories or probabilities of obtaining these histories.}\\label{fig:noisy}\n\\end{figure}","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its to","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total mass","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$,","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_","\\begin{proposition}[A conditional stability bound with an evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the …","evidence denominator]\\label{prop:conditioning}\nLet $P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","$P$ and $Q$ be finite joint laws of a model and an observation, with $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","ith $\\TV(P,Q)\\leq\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","q\\delta$. Let $E$ be an observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","observation event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","on event with $a=P(E)>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}",")>0$ and $b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","$b=Q(E)>0$. Then\n\\begin{equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","equation}\\label{eq:conditional-bound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","ound}\n\\TV\\bigl(P(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","(M\\mid E),Q(M\\mid E)\\bigr)\n\\leq\\min\\left\\{1,\\frac{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","{\\delta}{\\max(a,b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","b)}\\right\\}.\n\\end{equation}\nIn particular, a justified lower bound $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","nd $P(E)\\geq\\gamma>0$ supplies the weaker usable bound $\\min\\{1,\\delta/\\gamma\\}$.\n\\end{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","nd{proposition}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","ion}\n\\begin{proof}\nWrite $u_m=P(M=m,E)$, $v_m=Q(M=m,E)$, and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","and let $p=u/a$, $q=v/b$. Exchanging $P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","P$ and $Q$ if necessary, assume $a\\leq b$. The variation inside $E$ is at least $\\tfrac12\\sum_m|u_m-v_m|$. The variation outside $E$ is at least $|a-b|/2$, since its total masses differ by that amount. Hence\n\\[\n\\delta\\geq \\tfrac12\\sum_m|u_m-v_m|+\\tfrac12(b-a)\n=b-\\sum_m\\min(u_m,v_m).\n\\]\nSince $a\\leq b$, $\\min(ap_m,bq_m)\\leq b\\min(p_m,q_m)$. Substitution gives $\\delta\\geq b\\,\\TV(p,q)$. Division by $b=\\max(a,b)$ and the universal upper bound one prove the claim.\n\\end{proof}","The bound is sharp for general joint laws: put the same mass $a$ on $E$ in both laws, choose conditional laws at distance $t$, and make their complements identical. Then the joint distance is $at$. The result neither estimates $\\delta$ from data nor proves that the evidence model is calibrated. Those are separate inputs.","\\begin{counterexample}[Rare evidence can reverse a posterior despite small joint error]\\label{ex:rare-evidence}\nFor $0<\\epsilon<1/2$, take a binary model with equal prior probabilities. Under $P$, let $P(M=0,E)=\\epsilon$ and $P(M=1,E)=0$. Under $Q$, interchange these two masses. Choose the complement masses so each model retains prior probability $1/2$. Both laws give $E$ probability $\\epsilon$, their joint total variation is $2\\epsilon$, and their posteriors on $E$ are opposite point masses. Joint error can tend to zero while conditional error remains one.\n\\end{counterexample}","dence can reverse a posterior despite small joint error]\\label{ex:rare-evidence}\nFor $0<\\epsilon<1/2$, take a binary model with equal prior probabilities. Under $P$, let $P(M=0,E)=\\epsilon$ and $P(M=1,E)=0$. Under $Q$, interchange these two masses. Choose the complement masses so each model retains prior probability $1/2$. Both laws give $E$ probability $\\epsilon$, their joint total variation is $2\\epsilon$, and their posteriors on $E$ are opposite point masses. Joint error can tend to zero while conditional error remains one.\n\\end{counterexample}","small joint error]\\label{ex:rare-evidence}\nFor $0<\\epsilon<1/2$, take a binary model with equal prior probabilities. Under $P$, let $P(M=0,E)=\\epsilon$ and $P(M=1,E)=0$. Under $Q$, interchange these two masses. Choose the complement masses so each model retains prior probability $1/2$. Both laws give $E$ probability $\\epsilon$, their joint total variation is $2\\epsilon$, and their posteriors on $E$ are opposite point masses. Joint error can tend to zero while conditional error remains one.\n\\end{counterexample}","\\begin{figure}[htbp]\n\\centering\\includegraphics[width=.95\\linewidth]{figures/conditioning_bound.pdf}\n\\caption{The certified upper bound $\\min(1,\\delta/\\gamma)$ for $\\delta=1/100$. A small joint-law error does not give a useful guarantee when the evidence probability is comparably small. The plot is a bound, not an empirical estimate of posterior error.}\\label{fig:conditioning}\n\\end{figure}","\\subsection{From approximate beliefs to a recipient's decisions}\nNot every posterior distinction matters equally to an action. To link correction with use, fix one finite nonempty action set $A$ and losses $\\ell(m,a)\\in[0,L_{\\max}]$. A receiver evaluates actions from the distribution its actual interpreter reconstructs. The target loss function and available actions must be the same in the comparison.","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two dis","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions dif","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ b","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$.","\\begin{proposition}[A joint posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive b…","int posterior and recipient decision certificate]\\label{prop:decision}\nLet $p$ be the reference posterior, $\\widehat p$ an approximate or transmitted posterior, and $r_i$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$…","$ recipient $i$'s reconstructed posterior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ …","terior. Suppose\n\\[\n\\TV(p,\\widehat p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the stric…","at p)\\leq\\epsilon,\\qquad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","uad\n\\TV(\\widehat p,r_i)\\leq\\eta_i.\n\\]\nIf $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","f $a_i$ minimizes expected loss under $r_i$, then, with $e_i=\\min\\{1,\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}",",\\epsilon+\\eta_i\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","\\}$,\n\\begin{equation}\\label{eq:recipient-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","t-regret}\n\\mathbb E_p\\ell(M,a_i)-\\min_{a\\in A}\\mathbb E_p\\ell(M,a)\n\\leq\\min\\{L_{\\max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","max},2L_{\\max}e_i\\}.\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}",".\n\\end{equation}\nIf the expected-loss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","ss gap between $a_i$ and every other action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","ther action under $r_i$ is strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","strictly greater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","reater than $2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","$2L_{\\max}e_i$, then $a_i$ is also the unique optimum under $p$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","$.\n\\end{proposition}\n\\begin{proof}\nThe triangle inequality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","quality bounds $\\TV(p,r_i)$ by $e_i$. For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","For a function taking values in $[0,L_{\\max}]$, its expectations under these two distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","o distributions differ by at most $L_{\\max}e_i$. Let $a_*$ minimize expected loss under $p$. Insert the two risks under $r_i$ between the risks of $a_i$ and $a_*$ under $p$. The middle difference is nonpositive by optimality under $r_i$, and the two outside differences are at most $L_{\\max}e_i$. Regret is also at most $L_{\\max}$. Applying the same two expectation bounds to the difference between $a_i$ and any competitor proves the strict-gap assertion.\n\\end{proof}","The two errors concern one actual chain of records. One cannot use the best transmission for one recipient and the best posterior approximation for another while reporting a common certificate. When $\\widehat p=Q(M\\mid E)$, Proposition~\\ref{prop:conditioning} can supply $\\epsilon$. A separate information-loss estimate or an explicitly verified decoder may instead supply the bound. The result is relative to the stipulated posterior and loss, not a guarantee that a real decision is wise or that a protected real-world outcome can be represented by a bounded scalar loss.","Let the unknown model $M$ remain fixed. At time $t$, a known policy selects experiment $e_t$ using only the prior reported history and recorded experiment names. Randomization is permitted when its conditional law is independent of $M$ given that history. Given $M=m$ and $e_t=e$, a fresh raw observation has law $L_{e,m}$, and its reported form is drawn through $B_e$. Conditional on the model and selected experiment, these draws do not depend on earlier hidden observations. Write $Q_{e,m}=L_{e,m}B_e$.","t $e_t$ using only the prior reported history and recorded experiment names. Randomization is permitted when its conditional law is independent of $M$ given that history. Given $M=m$ and $e_t=e$, a fresh raw observation has law $L_{e,m}$, and its reported form is drawn through $B_e$. Conditional on the model and selected experiment, these draws do not depend on earlier hidden observations. Write $Q_{e,m}=L_{e,m}B_e$.","\\begin{theorem}[Sequential preservation of noisy correction under adaptive experiments]\\label{thm:adaptive-bayes}\nSuppose each permitted experiment admits a kernel $\\rho_e$ with\n\\[\nL_{e,m}(o)B_e(o,z)=Q_{e,m}(z)\\rho_e(o\\mid z)\n\\quad\\text{for all }m,o,z.\n\\]\nFor every finite positive-probability history under the declared policy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization","\\begin{theorem}[Sequential preservation of noisy correction under adaptive experiments]\\label{thm:adaptive-bayes}\nSuppose each permitted experiment admits a kernel $\\rho_e$ with\n\\[\nL_{e,m}(o)B_e(o,z)=Q_{e,m}(z)\\rho_e(o\\mid z)\n\\quad\\text{for all }m,o,z.\n\\]\nFor every finite positive-probability history under the declared policy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization gives a product of the terms $Q_{e_t,m}(z_t)$ multiplied by the model-independent",".\n\\]\nFor every finite positive-probability history under the declared policy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization gives a product of the terms $Q_{e_t,m}(z_t)$ multiplied by the model-independent factors $\\rho_{e_t}(o_t\\mid z_t)$ and the same policy factors. Posterior normalization cancels those factors. The remaining product is exactly the likelihood for the reported history. A bounded stopping decision determined by that history adds no model-dependent factor, so the same calculation applies.\n\\end{proof}","cy, the posterior conditioned on the complete raw and reported record equals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization gives a product of the terms $Q_{e_t,m}(z_t)$ multiplied by the model-independent factors $\\rho_{e_t}(o_t\\mid z_t)$ and the same policy factors. Posterior normalization cancels those factors. The remaining product is exactly the likelihood for the reported history. A bounded stopping decision determined by that history adds no model-dependent factor, so the same calculation applies.\n\\end{proof}","uals the posterior conditioned only on the reported outcomes and experiment names. The claim also holds at a bounded stopping time determined by that reported history.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization gives a product of the terms $Q_{e_t,m}(z_t)$ multiplied by the model-independent factors $\\rho_{e_t}(o_t\\mid z_t)$ and the same policy factors. Posterior normalization cancels those factors. The remaining product is exactly the likelihood for the reported history. A bounded stopping decision determined by that history adds no model-dependent factor, so the same calculation applies.\n\\end{proof}","at a bounded stopping time determined by that reported history.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization gives a product of the terms $Q_{e_t,m}(z_t)$ multiplied by the model-independent factors $\\rho_{e_t}(o_t\\mid z_t)$ and the same policy factors. Posterior normalization cancels those factors. The remaining product is exactly the likelihood for the reported history. A bounded stopping decision determined by that history adds no model-dependent factor, so the same calculation applies.\n\\end{proof}","y.\n\\end{theorem}\n\\begin{proof}\nFor a fixed realized history, its likelihood under model $m$ is the product of the selected-experiment probabilities and the terms $L_{e_t,m}(o_t)B_{e_t}(o_t,z_t)$. The policy factors are independent of $m$ once the reported history is fixed. Applying the assumed factorization gives a product of the terms $Q_{e_t,m}(z_t)$ multiplied by the model-independent factors $\\rho_{e_t}(o_t\\mid z_t)$ and the same policy factors. Posterior normalization cancels those factors. The remaining product is exactly the likelihood for the reported history. A bounded stopping decision determined by that history adds no model-dependent factor, so the same calculation applies.\n\\end{proof}","\\begin{counterexample}[Copying a record is not repeating an experiment]\\label{ex:duplicate-record}\nLet a binary model have equal prior probabilities, and suppose one observed positive report has likelihood $4/5$ under model 1 and $1/5$ under model 0. Its posterior probability for model 1 is $4/5$. Sending two copies of that same report does not add evidence, so the posterior remains $4/5$. Treating the copies as conditionally independent instead gives $16/17$. Both calculations are arithmetically valid for their respective likelihoods, but only the first describes duplication of one record.\n\\end{counterexample}","an experiment]\\label{ex:duplicate-record}\nLet a binary model have equal prior probabilities, and suppose one observed positive report has likelihood $4/5$ under model 1 and $1/5$ under model 0. Its posterior probability for model 1 is $4/5$. Sending two copies of that same report does not add evidence, so the posterior remains $4/5$. Treating the copies as conditionally independent instead gives $16/17$. Both calculations are arithmetically valid for their respective likelihoods, but only the first describes duplication of one record.\n\\end{counterexample}","\\subsection{The complete instrument witness}\nFor the original contract, the instrument witness contains $r(a,b,c)=(a+b,b)$, updates~\\eqref{eq:display-coordinates}, the first-coordinate decoder, and a statement that $c$ is outside the relevant future distinctions. It also names how an observer acquires the initial record. A recipient with that record and the equations can reproduce all declared predictions.","After $J$ is admitted, the witness changes. The old representation does not predict the new reading. An explicit $J$ observation supplies $c$ and updates the first coordinate. The enlarged state has the exact updates in Section~\\ref{sec:revision}. The original $H/W$ explanations remain valid on their unchanged domain. New predictions require the acquired evidence, and any translated explanation must preserve that evidence's discriminating meaning.","A simple amortization comparison illustrates the issue. Suppose a representation costs $D$ to construct, each later use costs $c$, and the matched baseline costs $b>c$ per use. The representation has lower total cost after $N$ uses exactly when\n\\[\nD+Nc<Nb,\\quad\\text{or equivalently}\\quad N>\\frac{D}{b-c}.\n\\]\nThis accounting does not prove that the representation can be discovered for cost $D$, that the uses are comparable, or that all target instances share it. It locates where such evidence would enter.","\\subsection{The graph belongs to the acquired transcript}\nFix a finite set $X_0$ of possible initial states, a finite deterministic experiment catalogue as in Section~\\ref{sec:acquisition}, and a finite horizon $h$. A deterministic experimental policy is an observation-history tree. Its choices cannot depend on the hidden initial state. Only uniformly admissible choices are permitted at each reachable node.","For a policy tree $T$ and initial state $x$, write $s_T(x)$ for its terminal sender transcript and $x_T(x)$ for its final physical state. The tree has finitely many reachable leaves. Specify recipient backgrounds $\\beta_i^T(x)$ and demands $d_i^T(x)$ by a declared rule. A demand may concern $x_T(x)$, the original $x$, or a stated pair of past and present values. These are different contracts.","The recipient sees only the common artifact and its specified background. If an experiment name, its timing, or another transcript is visible to the recipient, that information belongs in $\\beta_i^T$. It must not be used as an uncharged auxiliary message. Let $C(T)$ be the maximum total experimental cost across initial states. Build $G_T$ on the reachable sender transcripts using the graph construction of Section~\\ref{sec:audience}, with $\\alpha=s_T$.","and its specified background. If an experiment name, its timing, or another transcript is visible to the recipient, that information belongs in $\\beta_i^T$. It must not be used as an uncharged auxiliary message. Let $C(T)$ be the maximum total experimental cost across initial states. Build $G_T$ on the reachable sender transcripts using the graph construction of Section~\\ref{sec:audience}, with $\\alpha=s_T$.","\\begin{theorem}[A finite joint certificate for acquisition and explanation]\\label{thm:joint-acquisition}\nFor finite $X_0$, a finite experiment catalogue, horizon $h$, cost budget $C$, and artifact alphabet bound $K\\geq1$, there exists a uniformly admissible deterministic acquisition policy and one common artifact satisfying every recipient demand exactly if and only if some attainable policy tree $T$ satisfies\n\\[\nC(T)\\leq C,\\qquad G_T\\text{ has no loop},\\qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system i","\\begin{theorem}[A finite joint certificate for acquisition and explanation]\\label{thm:joint-acquisition}\nFor finite $X_0$, a finite experiment catalogue, horizon $h$, cost budget $C$, and artifact alphabet bound $K\\geq1$, there exists a uniformly admissible deterministic acquisition policy and one common artifact satisfying every recipient demand exactly if and only if some attainable policy tree $T$ satisfies\n\\[\nC(T)\\leq C,\\qquad G_T\\text{ has no loop},\\qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, b","\\begin{theorem}[A finite joint certificate for acquisition and explanation]\\label{thm:joint-acquisition}\nFor finite $X_0$, a finite experiment catalogue, horizon $h$, cost budget $C$, and artifact alphabet bound $K\\geq1$, there exists a uniformly admissible deterministic acquisition policy and one common artifact satisfying every recipient demand exactly if and only if some attainable policy tree $T$ satisfies\n\\[\nC(T)\\leq C,\\qquad G_T\\text{ has no loop},\\qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, because its experi","\\begin{theorem}[A finite joint certificate for acquisition and explanation]\\label{thm:joint-acquisition}\nFor finite $X_0$, a finite experiment catalogue, horizon $h$, cost budget $C$, and artifact alphabet bound $K\\geq1$, there exists a uniformly admissible deterministic acquisition policy and one common artifact satisfying every recipient demand exactly if and only if some attainable policy tree $T$ satisfies\n\\[\nC(T)\\leq C,\\qquad G_T\\text{ has no loop},\\qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, because its experiment choices depend only on its availa","begin{theorem}[A finite joint certificate for acquisition and explanation]\\label{thm:joint-acquisition}\nFor finite $X_0$, a finite experiment catalogue, horizon $h$, cost budget $C$, and artifact alphabet bound $K\\geq1$, there exists a uniformly admissible deterministic acquisition policy and one common artifact satisfying every recipient demand exactly if and only if some attainable policy tree $T$ satisfies\n\\[\nC(T)\\leq C,\\qquad G_T\\text{ has no loop},\\qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, because its experiment choices depend only on its available history. The sender's final artifact is a function of its terminal transcript. Applying Theorem~\\ref{thm:audience} to those transcripts proves the graph conditions. The worst-cas…","A finite joint certificate for acquisition and explanation]\\label{thm:joint-acquisition}\nFor finite $X_0$, a finite experiment catalogue, horizon $h$, cost budget $C$, and artifact alphabet bound $K\\geq1$, there exists a uniformly admissible deterministic acquisition policy and one common artifact satisfying every recipient demand exactly if and only if some attainable policy tree $T$ satisfies\n\\[\nC(T)\\leq C,\\qquad G_T\\text{ has no loop},\\qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, because its experiment choices depend only on its available history. The sender's final artifact is a function of its terminal transcript. Applying Theorem~\\ref{thm:audience} to those transcripts proves the graph conditions. The worst-case experimental …","qquad \\chi(G_T)\\leq K.\n\\]\nThis is a finite decision problem. An accepting witness consists of the policy tree, a proper coloring, and the induced fixed decoders. Among feasible trees, a minimum experimental cost is attained. The statement does not assert that the search or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, because its experiment choices depend only on its available history. The sender's final artifact is a function of its terminal transcript. Applying Theorem~\\ref{thm:audience} to those transcripts proves the graph conditions. The worst-case experimental cost is $C(T)$, so it satisfies the budget.","arch or the resulting decoders are efficient.\n\\end{theorem}\n\\begin{proof}\nAny feasible system induces a policy tree of depth at most $h$, because its experiment choices depend only on its available history. The sender's final artifact is a function of its terminal transcript. Applying Theorem~\\ref{thm:audience} to those transcripts proves the graph conditions. The worst-case experimental cost is $C(T)$, so it satisfies the budget.","\\begin{counterexample}[A useful explanation need not identify the state at its source]\\label{ex:joint-source}\nLet the hidden state be two bits $(a,b)$. The sender's only experiment reads $a\\oplus b$ without changing the state. Repeating it never identifies the complete pair. One recipient already knows $a$ and needs $b$, while another knows $b$ and needs $a$. The single parity reading and its one-bit broadcast satisfy both demands. Before that reading, the sender has only one transcript and the explanation graph has a loop. After the reading, the graph is one edge and the one-bit message is exact.\n\\end{counterexample}","In the noisy instrument, every readout has positive likelihood under every candidate value of $c$. With a fully supported prior, any finite sequence of those readouts leaves every candidate with positive posterior probability. Repeated reports can make one value overwhelmingly likely, but they cannot give zero-error identification within a fixed finite number of such observations. A decision-loss tolerance, a posterior threshold, and an exact reconstruction target are different terminal conditions.","First, the reasoner retains the base coordinates and the exact scope of $H/W$ reuse. Admitting $J$ splits old behavioral classes. The exact diagnostic plan uses a real $J$ operation and preserves its effect on the first register. If only the noisy sensor is available, the retained object is a posterior over possible post-$J$ states, not an invented exact value of $c$.","Second, the evidence description identifies which readouts are conditionally independent and which tokens are copied or ancillary. Removing the independent flag is correction preserving. Replacing all five possible readings by parity is not. Three identical full readings give the conditional posterior $432/433$ for the corresponding candidate, but that number remains conditional on the declared likelihood model and prior.","\\subsection{Task-dependent interpreter changes are information}\nLet $\\Theta$ be a finite nonempty family of tasks and let each task have a required answer profile over a fixed query family. A fixed interpreter $U$ receives background $B_\\theta$, an installed program string $p_\\theta$, an artifact $e_\\theta$, and a query. It must return the required answer. The interpreter is fixed across tasks. The strings may depend on the task, but their lengths are charged. Let $N_B$ be the number of distinct required answer profiles among tasks having the same background value $B$.","Theta$ be a finite nonempty family of tasks and let each task have a required answer profile over a fixed query family. A fixed interpreter $U$ receives background $B_\\theta$, an installed program string $p_\\theta$, an artifact $e_\\theta$, and a query. It must return the required answer. The interpreter is fixed across tasks. The strings may depend on the task, but their lengths are charged. Let $N_B$ be the number of distinct required answer profiles among tasks having the same background value $B$.","xed query family. A fixed interpreter $U$ receives background $B_\\theta$, an installed program string $p_\\theta$, an artifact $e_\\theta$, and a query. It must return the required answer. The interpreter is fixed across tasks. The strings may depend on the task, but their lengths are charged. Let $N_B$ be the number of distinct required answer profiles among tasks having the same background value $B$.","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exact","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tas","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. Ther","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinc","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms prov","\\begin{proposition}[Charging a task-dependent interpreter]\\label{prop:advice}\nSuppose $p_\\theta$ and $e_\\theta$ have fixed lengths $d$ and $b$ bits, respectively, and the deterministic interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms proves the second assertion.\n\\end{proof}","c interpreter answers every required query exactly. Then\n\\[\n2^{d+b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms proves the second assertion.\n\\end{proof}","b}\\geq\\max_B N_B.\n\\]\nIn particular, representing every Boolean function on $n$ input bits under one fixed background requires $d+b\\geq2^n$. This counts task-dependent information, not the running time of the interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms proves the second assertion.\n\\end{proof}","interpreter or the information needed for any one specially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms proves the second assertion.\n\\end{proof}","ecially chosen function.\n\\end{proposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms proves the second assertion.\n\\end{proof}","roposition}\n\\begin{proof}\nFix a background $B$. If two distinct required answer profiles used the same pair $(p_\\theta,e_\\theta)$, the fixed interpreter would receive identical inputs for a distinguishing query and would have to give the same answer. Exactness would fail for one of the two tasks. There are at most $2^{d+b}$ pairs, proving the first bound. There are $2^{2^n}$ Boolean functions on $n$ input bits, with distinct complete answer profiles. Taking base-two logarithms proves the second assertion.\n\\end{proof}","This is the counting argument of Proposition~\\ref{prop:explanation} applied to the complete task-dependent package. It is not a new general description-complexity theorem. The point is to prevent a moving interpreter from becoming an unreported communication channel. Variable-length descriptions require a declared framing or prefix convention. Merely saying that a message has length at most $b$ is not the same fixed-length model, because the length itself can carry information.","\\subsection{Amortization without disappearing costs}\nSuppose a method has setup cost $S$ and per-use cost $r$, while a comparator costs $c$ on each use. In the same declared cost units and with no other differences, the method costs less over $N$ uses exactly when $S+Nr<Nc$. When $c>r$, this is $N>S/(c-r)$. At equality the costs tie. If $c\\leq r$ and $S\\geq0$, setup cannot create a strict cost saving in this model.","Let $X$ be a finite nonempty set of cases, $\\alpha:X\\to S$ the sender's access map, and $\\mathcal A$ a finite nonempty artifact catalogue. Recipient $i$ has a background map $\\beta_i:X\\to B_i$, a demanded output $d_i:X\\to D_i$, and a fixed procedure $L_i$. Evaluate $L_i(\\beta_i(x),a)$ under a declared instruction and memory bound. A timeout, an invalid output, or a rejected artifact counts as failure. Since the inputs and execution bounds are finite, the resulting acceptance table is finite even when $L_i$ is written in a general programming language.","te nonempty artifact catalogue. Recipient $i$ has a background map $\\beta_i:X\\to B_i$, a demanded output $d_i:X\\to D_i$, and a fixed procedure $L_i$. Evaluate $L_i(\\beta_i(x),a)$ under a declared instruction and memory bound. A timeout, an invalid output, or a rejected artifact counts as failure. Since the inputs and execution bounds are finite, the resulting acceptance table is finite even when $L_i$ is written in a general programming language.","Define\n\\[\n\\mathcal A(x)=\\{a\\in\\mathcal A:\\text{every recipient returns }d_i(x)\n\\text{ within its declared limits}\\},\n\\]\nand, for each reachable sender value $s$, define\n\\begin{equation}\\label{eq:decoder-intersection}\n\\mathcal A_s=\\bigcap_{x:\\alpha(x)=s}\\mathcal A(x).\n\\end{equation}\nAn artifact selected from $\\mathcal A_s$ must work for all cases the sender still considers possible. Choosing a different acceptable artifact after secretly learning which hidden case is true is not allowed.","\\begin{theorem}[The fixed-recipient covering criterion]\\label{thm:fixed-decoder}\nUnder this finite contract, an exact encoder $E:S\\to\\mathcal A$ exists on the reachable sender values if and only if every $\\mathcal A_s$ is nonempty. Its smallest possible image size is\n\\begin{equation}\\label{eq:decoder-cover}\n\\min\\{|C|:C\\subseteq\\mathcal A,\\ C\\cap\\mathcal A_s\\ne\\varnothing\n\\text{ for every reachable }s\\},\n\\end{equation}\nwhen the feasible family is nonempty. Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must","\\begin{theorem}[The fixed-recipient covering criterion]\\label{thm:fixed-decoder}\nUnder this finite contract, an exact encoder $E:S\\to\\mathcal A$ exists on the reachable sender values if and only if every $\\mathcal A_s$ is nonempty. Its smallest possible image size is\n\\begin{equation}\\label{eq:decoder-cover}\n\\min\\{|C|:C\\subseteq\\mathcal A,\\ C\\cap\\mathcal A_s\\ne\\varnothing\n\\text{ for every reachable }s\\},\n\\end{equation}\nwhen the feasible family is nonempty. Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for eve","\\begin{theorem}[The fixed-recipient covering criterion]\\label{thm:fixed-decoder}\nUnder this finite contract, an exact encoder $E:S\\to\\mathcal A$ exists on the reachable sender values if and only if every $\\mathcal A_s$ is nonempty. Its smallest possible image size is\n\\begin{equation}\\label{eq:decoder-cover}\n\\min\\{|C|:C\\subseteq\\mathcal A,\\ C\\cap\\mathcal A_s\\ne\\varnothing\n\\text{ for every reachable }s\\},\n\\end{equation}\nwhen the feasible family is nonempty. Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finit…","his finite contract, an exact encoder $E:S\\to\\mathcal A$ exists on the reachable sender values if and only if every $\\mathcal A_s$ is nonempty. Its smallest possible image size is\n\\begin{equation}\\label{eq:decoder-cover}\n\\min\\{|C|:C\\subseteq\\mathcal A,\\ C\\cap\\mathcal A_s\\ne\\varnothing\n\\text{ for every reachable }s\\},\n\\end{equation}\nwhen the feasible family is nonempty. Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","exists on the reachable sender values if and only if every $\\mathcal A_s$ is nonempty. Its smallest possible image size is\n\\begin{equation}\\label{eq:decoder-cover}\n\\min\\{|C|:C\\subseteq\\mathcal A,\\ C\\cap\\mathcal A_s\\ne\\varnothing\n\\text{ for every reachable }s\\},\n\\end{equation}\nwhen the feasible family is nonempty. Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","ble family is nonempty. Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","Thus a codebook of at most $K$ artifacts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","facts is possible exactly when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","when a set of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","t of at most $K$ artifacts meets all these requirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","quirements. Every valid encoder induces a proper coloring of the sender-access graph, so its image size is at least that graph's chromatic number. The converse graph condition need not hold for the fixed procedures.\n\\end{theorem}\n\\begin{proof}\nA sender seeing $s$ must emit one value $E(s)$, and that value must lie in $\\mathcal A(x)$ for every $x$ with $\\alpha(x)=s$. This is exactly membership in $\\mathcal A_s$. Hence a valid encoder implies nonempty intersections. Conversely, selecting one value from each nonempty finite intersection constructs a valid encoder.","The image of any such encoder intersects every $\\mathcal A_s$. If a set $C$ intersects every requirement, selecting $E(s)$ from $C\\cap\\mathcal A_s$ gives an encoder whose image is contained in $C$. This proves the minimum formula by finite minimization. If two adjacent sender values received the same artifact, a recipient with the same background would receive identical decoder inputs while requiring distinct outputs. Its fixed procedure could not satisfy both. This proves the graph lower bound.\n\\end{proof}","\\begin{counterexample}[Information suffices but the installed decoder needs a larger codebook]\\label{ex:fixed-codebook}\nTake three cases $x_0,x_1,x_2$ fully visible to the sender. One recipient already has background $i$ in case $x_i$ and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts $A,B,C$. It returns zero on $A,B$ at background 0, on $B,C$ at background 1, and on $A,C$ at background 2. It fails on the remaining pairs. Every two of the three acceptable-artifact sets intersect, but their common intersection is empty. Two catalogue entries suffice, while one does not. The graph lower bound is one and the fixed-decoder optimum is two.\n\\end{counterexample}","nformation suffices but the installed decoder needs a larger codebook]\\label{ex:fixed-codebook}\nTake three cases $x_0,x_1,x_2$ fully visible to the sender. One recipient already has background $i$ in case $x_i$ and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts $A,B,C$. It returns zero on $A,B$ at background 0, on $B,C$ at background 1, and on $A,C$ at background 2. It fails on the remaining pairs. Every two of the three acceptable-artifact sets intersect, but their common intersection is empty. Two catalogue entries suffice, while one does not. The graph lower bound is one and the fixed-decoder optimum is two.\n\\end{counterexample}","installed decoder needs a larger codebook]\\label{ex:fixed-codebook}\nTake three cases $x_0,x_1,x_2$ fully visible to the sender. One recipient already has background $i$ in case $x_i$ and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts $A,B,C$. It returns zero on $A,B$ at background 0, on $B,C$ at background 1, and on $A,C$ at background 2. It fails on the remaining pairs. Every two of the three acceptable-artifact sets intersect, but their common intersection is empty. Two catalogue entries suffice, while one does not. The graph lower bound is one and the fixed-decoder optimum is two.\n\\end{counterexample}","larger codebook]\\label{ex:fixed-codebook}\nTake three cases $x_0,x_1,x_2$ fully visible to the sender. One recipient already has background $i$ in case $x_i$ and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts $A,B,C$. It returns zero on $A,B$ at background 0, on $B,C$ at background 1, and on $A,C$ at background 2. It fails on the remaining pairs. Every two of the three acceptable-artifact sets intersect, but their common intersection is empty. Two catalogue entries suffice, while one does not. The graph lower bound is one and the fixed-decoder optimum is two.\n\\end{counterexample}","d-codebook}\nTake three cases $x_0,x_1,x_2$ fully visible to the sender. One recipient already has background $i$ in case $x_i$ and always needs output zero. The sender-access graph has no edges, so a designable decoder could ignore a single common message and return zero. Now fix an installed decoder on artifacts $A,B,C$. It returns zero on $A,B$ at background 0, on $B,C$ at background 1, and on $A,C$ at background 2. It fails on the remaining pairs. Every two of the three acceptable-artifact sets intersect, but their common intersection is empty. Two catalogue entries suffice, while one does not. The graph lower bound is one and the fixed-decoder optimum is two.\n\\end{counterexample}","\\subsection{Sender uncertainty is a different obstruction}\nNonempty casewise sets $\\mathcal A(x)$ do not suffice. If the sender cannot distinguish two cases whose acceptable-artifact sets are disjoint, their intersection in~\\eqref{eq:decoder-intersection} is empty. More recipient compute cannot repair a distinction missing from both its background and the transmitted artifact. A new observation may repair the sender's access. An added recipient background may repair the receiving side. Neither can be credited without being part of the actual workflow.","\\subsection{The explicit finite implementation model}\nUse a finite augmented state space $X$. Include every origin variable, history flag, recipient background, and permission needed to make the declared transition and terminal tables valid. Each experiment $e$ has the transition, observation, and admissibility semantics of Section~\\ref{sec:acquisition}, together with a fixed nonnegative integer cost vector $c_e\\in\\mathbb N^r$. Its coordinates have declared units. The charge includes the specified source-side interface work for that experiment. We do not assume that acquiring information leaves the physical state unchanged.","The explicit finite implementation model}\nUse a finite augmented state space $X$. Include every origin variable, history flag, recipient background, and permission needed to make the declared transition and terminal tables valid. Each experiment $e$ has the transition, observation, and admissibility semantics of Section~\\ref{sec:acquisition}, together with a fixed nonnegative integer cost vector $c_e\\in\\mathbb N^r$. Its coordinates have declared units. The charge includes the specified source-side interface work for that experiment. We do not assume that acquiring information leaves the physical state unchanged.","For each state $x$ and artifact $a$ accepted by all fixed recipient procedures, let $\\gamma(x,a)\\in\\mathbb N^r$ be the terminal charge. It includes transmission and the declared source and recipient execution work. Invalid or out-of-budget decoder executions have no terminal entry. At support $V$, a stop using $a$ is allowed only if it works for every $x\\in V$, with worst-case vector\n\\[\ng_a(V)=\\max_{x\\in V}\\gamma(x,a),\n\\]\nwhere maxima and comparisons of vectors are componentwise. Different coordinates may attain their maxima on different cases. A coordinatewise budget nevertheless has to cover every one of them.","A controller is an explicit observation tree of depth at most $h$. Each decision node or stop node occupies one fixed-size record, with fields for the operation or artifact and the bounded number of successor pointers. The count $n(T)$ includes all stored branches, not just those visited in one execution. The interpreter, recipient programs, and state tables are installed separately and remain fixed throughout the optimization. Their storage is added to the recorded controller size. Sharing equal subtrees in a directed acyclic graph is a different storage model.","For a tree $T$, let $v(T)$ be its componentwise worst-case charged online cost. Let $\\mathcal P_h(V)$ be the attainable pairs $(v(T),n(T))$. The stop pairs are\n\\begin{equation}\\label{eq:resource-stop}\n\\bigl(g_a(V),1\\bigr)\\quad\\text{for each artifact accepted throughout }V.\n\\end{equation}\nFor $h>0$, each uniformly admissible experiment gives the additional pairs\n\\begin{equation}\\label{eq:resource-branch}\n\\left(c_e+\\max_o v_o,\\ 1+\\sum_o n_o\\right),\n\\quad (v_o,n_o)\\in\\mathcal P_{h-1}(V_{e,o}),\n\\end{equation}\nwith one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At $h=0$ only stops are allowed. Sets are finite because t","For a tree $T$, let $v(T)$ be its componentwise worst-case charged online cost. Let $\\mathcal P_h(V)$ be the attainable pairs $(v(T),n(T))$. The stop pairs are\n\\begin{equation}\\label{eq:resource-stop}\n\\bigl(g_a(V),1\\bigr)\\quad\\text{for each artifact accepted throughout }V.\n\\end{equation}\nFor $h>0$, each uniformly admissible experiment gives the additional pairs\n\\begin{equation}\\label{eq:resource-branch}\n\\left(c_e+\\max_o v_o,\\ 1+\\sum_o n_o\\right),\n\\quad (v_o,n_o)\\in\\mathcal P_{h-1}(V_{e,o}),\n\\end{equation}\nwith one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At $h=0$ only stops are allowed. Sets are finite because the catalogue","For a tree $T$, let $v(T)$ be its componentwise worst-case charged online cost. Let $\\mathcal P_h(V)$ be the attainable pairs $(v(T),n(T))$. The stop pairs are\n\\begin{equation}\\label{eq:resource-stop}\n\\bigl(g_a(V),1\\bigr)\\quad\\text{for each artifact accepted throughout }V.\n\\end{equation}\nFor $h>0$, each uniformly admissible experiment gives the additional pairs\n\\begin{equation}\\label{eq:resource-branch}\n\\left(c_e+\\max_o v_o,\\ 1+\\sum_o n_o\\right),\n\\quad (v_o,n_o)\\in\\mathcal P_{h-1}(V_{e,o}),\n\\end{equation}\nwith one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At $h=0$ only stops are allowed. Sets are finite because the catalogue and horizon are finite.","throughout }V.\n\\end{equation}\nFor $h>0$, each uniformly admissible experiment gives the additional pairs\n\\begin{equation}\\label{eq:resource-branch}\n\\left(c_e+\\max_o v_o,\\ 1+\\sum_o n_o\\right),\n\\quad (v_o,n_o)\\in\\mathcal P_{h-1}(V_{e,o}),\n\\end{equation}\nwith one child choice for every nonempty observation branch. An experiment contributes no pair when even one possible branch has no completion. At $h=0$ only stops are allowed. Sets are finite because the catalogue and horizon are finite.","\\begin{theorem}[A joint resource certificate for fixed recipients]\\label{thm:resource-synthesis}\nRecurrences~\\eqref{eq:resource-stop}--\\eqref{eq:resource-branch} enumerate exactly the cost profiles of the permitted explicit trees. A workflow within online budget $b\\in\\mathbb N^r$ and controller-record budget $N$ exists if and only if some $(v,n)\\in\\mathcal P_h(V)$ satisfies $v\\leq b$ and $n\\leq N$. Every such pair has a finite executable tree witness. Keeping only componentwise nondominated pairs at each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each ch","\\begin{theorem}[A joint resource certificate for fixed recipients]\\label{thm:resource-synthesis}\nRecurrences~\\eqref{eq:resource-stop}--\\eqref{eq:resource-branch} enumerate exactly the cost profiles of the permitted explicit trees. A workflow within online budget $b\\in\\mathbb N^r$ and controller-record budget $N$ exists if and only if some $(v,n)\\in\\mathcal P_h(V)$ satisfies $v\\leq b$ and $n\\leq N$. Every such pair has a finite executable tree witness. Keeping only componentwise nondominated pairs at each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable","\\begin{theorem}[A joint resource certificate for fixed recipients]\\label{thm:resource-synthesis}\nRecurrences~\\eqref{eq:resource-stop}--\\eqref{eq:resource-branch} enumerate exactly the cost profiles of the permitted explicit trees. A workflow within online budget $b\\in\\mathbb N^r$ and controller-record budget $N$ exists if and only if some $(v,n)\\in\\mathcal P_h(V)$ satisfies $v\\leq b$ and $n\\leq N$. Every such pair has a finite executable tree witness. Keeping only componentwise nondominated pairs at each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge","\\begin{theorem}[A joint resource certificate for fixed recipients]\\label{thm:resource-synthesis}\nRecurrences~\\eqref{eq:resource-stop}--\\eqref{eq:resource-branch} enumerate exactly the cost profiles of the permitted explicit trees. A workflow within online budget $b\\in\\mathbb N^r$ and controller-record budget $N$ exists if and only if some $(v,n)\\in\\mathcal P_h(V)$ satisfies $v\\leq b$ and $n\\leq N$. Every such pair has a finite executable tree witness. Keeping only componentwise nondominated pairs at each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge is $c_e$ plus the co","\\begin{theorem}[A joint resource certificate for fixed recipients]\\label{thm:resource-synthesis}\nRecurrences~\\eqref{eq:resource-stop}--\\eqref{eq:resource-branch} enumerate exactly the cost profiles of the permitted explicit trees. A workflow within online budget $b\\in\\mathbb N^r$ and controller-record budget $N$ exists if and only if some $(v,n)\\in\\mathcal P_h(V)$ satisfies $v\\leq b$ and $n\\leq N$. Every such pair has a finite executable tree witness. Keeping only componentwise nondominated pairs at each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge is $c_e$ plus the componentwise ma","actly the cost profiles of the permitted explicit trees. A workflow within online budget $b\\in\\mathbb N^r$ and controller-record budget $N$ exists if and only if some $(v,n)\\in\\mathcal P_h(V)$ satisfies $v\\leq b$ and $n\\leq N$. Every such pair has a finite executable tree witness. Keeping only componentwise nondominated pairs at each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge is $c_e$ plus the componentwise maximum of the child charges. The installed explicit tree contains the root and every child record, giving the sum in~\\eqref{eq:resource-branch}. Conversely, combining an admiss…","each recursive step preserves feasibility for every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge is $c_e$ plus the componentwise maximum of the child charges. The installed explicit tree contains the root and every child record, giving the sum in~\\eqref{eq:resource-branch}. Conversely, combining an admissible root with the witnessing child trees realizes that pair. This proves exact enumeration and the budget equivalence.","r every pair of budgets $(b,N)$.\n\\end{theorem}\n\\begin{proof}\nAt depth zero, an admissible workflow must stop with one artifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge is $c_e$ plus the componentwise maximum of the child charges. The installed explicit tree contains the root and every child record, giving the sum in~\\eqref{eq:resource-branch}. Conversely, combining an admissible root with the witnessing child trees realizes that pair. This proves exact enumeration and the budget equivalence.","tifact that works for every possible state. Its worst-case terminal charge and its one stored record give exactly~\\eqref{eq:resource-stop}. For the inductive step, any nonterminal tree first executes an experiment available throughout $V$. Its possible next supports are exactly $V_{e,o}$. By induction each child tree supplies one attainable pair. Every branch is possible, so the online charge is $c_e$ plus the componentwise maximum of the child charges. The installed explicit tree contains the root and every child record, giving the sum in~\\eqref{eq:resource-branch}. Conversely, combining an admissible root with the witnessing child trees realizes that pair. This proves exact enumeration and the budget equivalence.","\\begin{counterexample}[Separate resource optima do not define a feasible common workflow]\\label{ex:resource-frontier}\nLet a hidden origin take four values and remain fixed under every experiment. A direct experiment reveals its value at charge four units. Its explicit controller has one decision node and four stop nodes, giving profile $(4,5)$. Three other experiments ask whether the origin equals 0, 1, or 2, respectively, each at charge one unit. A decision chain of these equality tests distinguishes the four origins with worst-case charge three and seven stored records, giving profile $(3,7)$. The fixed recipient decoder reads a two-bit origin artifact at a leaf. Its work and the common message length are held fixed and are not included in the one online coordinate shown here.","A direct experiment reveals its value at charge four units. Its explicit controller has one decision node and four stop nodes, giving profile $(4,5)$. Three other experiments ask whether the origin equals 0, 1, or 2, respectively, each at charge one unit. A decision chain of these equality tests distinguishes the four origins with worst-case charge three and seven stored records, giving profile $(3,7)$. The fixed recipient decoder reads a two-bit origin artifact at a leaf. Its work and the common message length are held fixed and are not included in the one online coordinate shown here.","Within horizon three the nondominated profiles are $(4,5)$ and $(3,7)$. A tree with charge at most three cannot use the direct experiment. A binary equality tree distinguishing all four origins needs at least four leaves and hence at least seven records. Conversely, any tree distinguishing four origins needs four distinct stopping artifacts and at least one decision node, so five is the smallest possible record count. Thus the separate minima are three online units and five records, but the combined budget $(3,5)$ admits no tree.\n\\end{counterexample}","se the direct experiment. A binary equality tree distinguishing all four origins needs at least four leaves and hence at least seven records. Conversely, any tree distinguishing four origins needs four distinct stopping artifacts and at least one decision node, so five is the smallest possible record count. Thus the separate minima are three online units and five records, but the combined budget $(3,5)$ admits no tree.\n\\end{counterexample}","\\subsection{Which implementation costs remain outside the recurrence}\nOffline search and proof checking do not disappear because the controller is compiled. Their time, software, and input tables are recorded separately. The online interpreter must be specified, and its dispatch costs can be included in $c_e$ and $\\gamma$. A record count becomes a storage guarantee only after the fixed record layout and installed tables are included. The number of source-file bytes is an audit quantity, not a theorem of minimal description length.","\\section{A costed first-edition instrument certificate}\\label{sec:first-ledger}\nThe common realization can now be described without a designable recipient or an omniscient sender. The instrument is still $(a,b,c)\\in\\mathbb F_5^3$. Before the diagnostic, the sender knows only $(s,b)=(a+b,b)$. Recipient A has the old register pair $(a,b)$. Recipient B has the old display pair $(s,b)$. Neither recipient has $c$. All three parties know the operation laws and the fixed diagnostic protocol before the particular hidden state is selected.","A costed first-edition instrument certificate}\\label{sec:first-ledger}\nThe common realization can now be described without a designable recipient or an omniscient sender. The instrument is still $(a,b,c)\\in\\mathbb F_5^3$. Before the diagnostic, the sender knows only $(s,b)=(a+b,b)$. Recipient A has the old register pair $(a,b)$. Recipient B has the old display pair $(s,b)$. Neither recipient has $c$. All three parties know the operation laws and the fixed diagnostic protocol before the particular hidden state is selected.","The sole diagnostic in this version applies $J$ exactly once and returns its exact post-operation display $s_J$. The sender computes $m=s_J-s=c$ and broadcasts that one field element. The report $s_J$ and the sender's private controller address are not transmitted separately. The recipients are told by the installed protocol that their demanded record is the state after this one diagnostic, not an arbitrary later state.","The two frozen decoders are\n\\begin{equation}\\label{eq:fixed-instrument-decoders}\nL_A((a,b),m)=(a+m,b,m),\\qquad\nL_B((s,b),m)=(s+m,b,m).\n\\end{equation}\nRecipient A obtains the post-$J$ register description. Recipient B obtains the post-$J$ display-centered description. Both compute one modular addition. Their installed update procedures then support all $H/W/J$ continuations in their respective coordinates. These procedures are supplied, not learned by the diagnostic.","For a fixed old $(s,b)$, five possible values of $c$ remain. Before an observation, no one message can supply both exact demanded records for every one of those cases. After the diagnostic, the five report values lead to five stop nodes and each selects its one valid message. The explicit tree has six records. A fixed three-bit field encodes the five messages. Two bits do not suffice. Three additional unused binary patterns are rejected, not assigned secret meanings.","midrule\nPhysical diagnostic & 1 & One $J$ operation with one returned exact display. Sensor construction is supplied.","Recipient A modular arithmetic & 1 & Compute $a+m$ in its fixed decoder.","Recipient B modular arithmetic & 1 & Compute $s+m$ in its fixed decoder.","Installed method & Recorded separately & Both decoder programs, transition semantics, framing, and prior background. Not charged as newly acquired $c$.","\\subsection{Reuse has an explicit implementation}\nIn B's coordinates $(s,b,c)$, each $H$, $W$, or $J$ update uses one modular addition or subtraction and the query reads the first coordinate. Thus a length-$n$ continuation uses exactly $n$ modular arithmetic operations after initialization, plus its recorded dispatch, assignment, and output operations. In A's coordinates the current display requires the additional addition $a+b$, while swap can be a coordinate permutation. The programs are different but agree on the declared answers through the exact translation theorem.","licit implementation}\nIn B's coordinates $(s,b,c)$, each $H$, $W$, or $J$ update uses one modular addition or subtraction and the query reads the first coordinate. Thus a length-$n$ continuation uses exactly $n$ modular arithmetic operations after initialization, plus its recorded dispatch, assignment, and output operations. In A's coordinates the current display requires the additional addition $a+b$, while swap can be a coordinate permutation. The programs are different but agree on the declared answers through the exact translation theorem.","\\subsection{An uncertain reading requires a different artifact}\nThe deterministic certificate does not apply to the noisy sensor merely because the same physical register is present. Under the noisy contract, the sender carries a posterior over $c$, not a certified exact value. A recipient can use a transmitted likelihood history or a specified posterior representation together with the existing correction and decision guarantees. Its message size and arithmetic work must then be charged again. They are not inherited from the three-bit exact-state message.","\\clearpage\\appendix\n\\section{A complete trace of the instrument}\\label{app:trace}\nStart from $(a,b,c)=(1,3,2)$. The first display is four. The base record is $(s,b)=(4,3)$. Apply $H$, then $W$, then the newly admitted $J$.","midrule\nInitial & $(1,3,2)$ & $4$ & $(4,3)$ & $H/W$ contract adequate","After $H$ & $(4,3,2)$ & $2$ & $(2,3)$ & Exact base update","After $W$ & $(3,4,2)$ & $2$ & $(2,4)$ & Exact base update","Before predicting $J$ & $(3,4,2)$ & $2$ & $(2,4)$ & Missing $c$ distinction","After $J$ & $(0,4,2)$ & $4$ & $(4,4)$ & New reading acquired","The last base record does not by itself retain all newly relevant information. The previous display was two and the new display is four, so $c=4-2=2$. The post-operation first register is $4-4=0$. The enlarged display-centered record is therefore $(s,b,c)=(4,4,2)$.","A subsequent $H$ gives $(s,b,c)=(3,4,2)$ and display three. A subsequent $W$ leaves the display at three but changes the second coordinate to $3-4=4$ modulo five. A subsequent $J$ changes the display to zero. Each assertion follows from the enlarged update equations, not from extrapolating a list of displayed numbers.","\\subsection{What was actually learned}\nThe original prediction method already encoded the effect of $H$ and $W$. The $J$ observation supplied a new state distinction. It did not establish the $J$ transition law from scratch, because that law was part of the expanded analytical contract. Learning that law would require a candidate family and experiments capable of distinguishing it from alternatives.","\\subsection{A correction record for the trace}\nThe record should retain the old scope, the new operation, the failed sufficiency condition, the acquired observation, and the exact post-intervention state reconstruction. It should also preserve the unchanged base equations. A concise corrected explanation would say that the third coordinate is irrelevant to $H/W$ continuations but relevant to any contract admitting $J$.","ction record for the trace}\nThe record should retain the old scope, the new operation, the failed sufficiency condition, the acquired observation, and the exact post-intervention state reconstruction. It should also preserve the unchanged base equations. A concise corrected explanation would say that the third coordinate is irrelevant to $H/W$ continuations but relevant to any contract admitting $J$.","\\subsection{The same trace with an uncertain readout}\nIn the preceding trace, the state immediately before $J$ is $(3,4,2)$ and its exact display is two. Suppose the reasoner retains only the base record $(2,4)$ and models $c$ as uniform. After $J$, the physical state is $(0,4,2)$. Under the noisy sensor of Section~\\ref{sec:bayesian}, a reported display of four gives the shifted report two. The resulting belief is over states $(c-2,4,c)$, not over unchanged pre-intervention states.","an uncertain readout}\nIn the preceding trace, the state immediately before $J$ is $(3,4,2)$ and its exact display is two. Suppose the reasoner retains only the base record $(2,4)$ and models $c$ as uniform. After $J$, the physical state is $(0,4,2)$. Under the noisy sensor of Section~\\ref{sec:bayesian}, a reported display of four gives the shifted report two. The resulting belief is over states $(c-2,4,c)$, not over unchanged pre-intervention states.","midrule\nFull shifted report $2$ & $3/4$ & $36/37$ & $432/433$","Only even parity & $7/24$ & $98/303$ & $1372/4143$","bottomrule\n\\end{tabular}\n\\end{center}\nThe entries are posterior mass on $c=2$ along the named histories. Their probabilities of occurrence are not the entries in the table. Every other candidate remains possible after finitely many noisy reports. Duplicate copies of one reading count as one observation. The executable demonstration reproduces the fractions, checks the harmless ancillary flag, and distinguishes the full-state target from the recipient-relative communication target."]}