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Examples include a finite real signal, a complex field on a finite grid, or an array of categorical labels. Its type includes shape, scalar field or palette, and boundary convention. A zero-extended finite sequence and a periodic sequence are different objects even when their displayed samples agree.","A **predictive state** is a summary $z=\\gamma(x)$, or more generally a summary of an observation history. It is sufficient only relative to a specified family of future quantities, actions and dynamics. There is no implication from exact storage of $x$ to sufficiency of $\\gamma(x)$.","Let $X$ and $C_a$ be finite-dimensional Hilbert spaces. A linear atlas consists of analysis and synthesis maps","$$\nA_a:X\\longrightarrow C_a,\\qquad S_a:C_a\\longrightarrow X,\n\\qquad S_aA_a=I_X.\n$$","The coefficients representing actual signals form $\\mathcal C_a=\\operatorname{im}A_a$. This distinction matters for redundant representations, where $\\mathcal C_a$ is a proper subspace of the ambient coefficient space.","**Proposition 1. Exact realization and consistency.** The analysis map is injective, $\\Pi_a=A_aS_a$ is an idempotent with image $\\mathcal C_a$, and $\\Pi_ac=c$ exactly when $c$ is a realizable coefficient vector. Moreover,","$$\n\\|A_ax\\|\\geq \\frac{\\|x\\|}{\\|S_a\\|},\\qquad\n\\|S_a(c+e)-S_ac\\|\\leq\\|S_a\\|\\,\\|e\\|.\n$$","The first inequality is understood for nonzero $X$. No assertion of orthogonality follows from a left inverse alone.","**Proof.** If $A_ax=0$, then $x=S_aA_ax=0$. Also $\\Pi_a^2=A_a(S_aA_a)S_a=\\Pi_a$. Its image is contained in $\\mathcal C_a$ and it fixes every $A_ax$, so its image is exactly $\\mathcal C_a$. Both norm inequalities follow from the operator-norm inequality applied to $S_a$. $\\square$","For a unitary atlas $U_a$, the synthesis is $U_a^*$ and the condition number is one. For a learned invertible atlas, exact invertibility is not enough: the synthesis norm controls amplification of coefficient errors. An arbitrarily ill-conditioned exact transform is not a stable representation.","For unitary $U_a$ and positive weights $w_a$, put","$$\n\\mathcal Ax=(w_aU_ax)_a,\\qquad\n\\mathcal S(c_a)_a=\\frac{1}{c}\\sum_aw_aU_a^*c_a,\n\\qquad c=\\sum_aw_a^2.\n$$","Then $\\mathcal S\\mathcal A=I$ and $\\|\\mathcal Ax\\|^2=c\\|x\\|^2$, as proved in [K1]. Here $\\mathcal A\\mathcal S=c^{-1}\\mathcal A\\mathcal A^*$ is the orthogonal projection onto the consistent coefficient subspace.","For a nonempty atlas family, define transport on realizable coefficients from atlas $a$ to atlas $b$ by","$$\nT_{ba}=A_bS_a:\\mathcal C_a\\longrightarrow\\mathcal C_b.\n$$","For an endomorphism $F:X\\to X$, define its coordinate realization by $F_a=A_aFS_a$ on $\\mathcal C_a$. In this paragraph $F$ may be nonlinear. For a partial map, every formula is restricted to the transported admissible domain. Maps between different signal types use separate input and output atlases.","$$\nT_{aa}=I,\\qquad T_{cb}T_{ba}=T_{ca},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n$$","Conversely, suppose maps $G_a:\\mathcal C_a\\to\\mathcal C_a$ satisfy $G_bT_{ba}=T_{ba}G_a$ for every pair. Then there is a unique map $F:X\\to X$ with $G_a=A_aFS_a$ for every $a$.","**Proof.** The transport identities follow by inserting $S_bA_b=I$. The operator identity follows because either path is $A_bFS_a$. For the converse, choose a reference atlas $r$ and put $F=S_rG_rA_r$. Compatibility gives $G_aA_a=A_aF$. Every consistent coefficient is $A_ax$, which proves the asserted realization. Applying $S_a$ proves uniqueness. $\\square$","$$\nd_{ba}(x)=G_b(A_bx)-T_{ba}G_a(A_ax).\n$$","$$\nS_bd_{ba}(x)=S_bG_bA_bx-S_aG_aA_ax.\n$$","Thus $\\|S_bd_{ba}(x)\\|\\leq\\|S_b\\|\\,\\|d_{ba}(x)\\|$. Zero discrepancy establishes that the chart descriptions agree, not that their common prediction is correct. All charts can agree on the same wrong operator.","Redundant coefficient outputs must also be tested for consistency. Synthesizing an arbitrary coefficient vector always produces some signal, but reanalysis returns $\\Pi_ac$, not necessarily the original $c$. Reconstruction and coefficient consistency are different checks.","Within one orthogonal atlas, let $E_i$ be pairwise orthogonal coordinate selectors with $\\sum_iE_i=I$. Pull them back by a unitary map:","$$\nP_i=U^*E_iU,\\qquad P_iP_j=\\delta_{ij}P_i,\n\\qquad P_i^*=P_i,\\qquad\\sum_iP_i=I_X.\n$$","A packet is the component $P_ix$ together with its projector, domain, atlas and resolution. A bare coefficient block without this information is not a fully specified packet.","**Proposition 3. Complete refinement preserves reconstruction.** Suppose a finite rooted tree has root projector $I$, and every internal projector is the sum of pairwise orthogonal child projectors. Every complete antichain $\\Lambda$, meeting each root-to-leaf path exactly once, satisfies","$$\n\\sum_{v\\in\\Lambda}P_v=I,\\qquad\nx=\\sum_{v\\in\\Lambda}P_vx,\\qquad\n\\|x\\|^2=\\sum_{v\\in\\Lambda}\\|P_vx\\|^2.\n$$","**Proof.** Replace an internal node by the sum of its children. Repeated replacement reaches any complete antichain and preserves the total projector. Different selected nodes have disjoint descendant branches, hence orthogonal ranges. The norm identity is Pythagoras. $\\square$","This is a conservation law for information under complete refinement. Deleting a child, storing only its energy, or imposing a coefficient threshold changes the contract. For a retained subset $J$ of an orthogonal decomposition, the lost signal is exactly $\\sum_{i\\notin J}P_ix$, and its squared norm is the sum of the omitted packet energies.","$$\nV(v)=\\min\\left(c(v),\\sum_{u\\in\\operatorname{ch}(v)}V(u)\\right)\n$$","Let $L:X\\to Y$ be linear, with orthogonal input projectors $P_i$ and output projectors $Q_j$. Its transfer blocks are $L_{ji}=Q_jLP_i$.","$$\nL=\\sum_{j,i}Q_jLP_i.\n$$","If $P_i=\\sum_rP_{i,r}$ and $Q_j=\\sum_sQ_{j,s}$ are complete refinements, then","$$\nQ_jLP_i=\\sum_{s,r}Q_{j,s}LP_{i,r}.\n$$","For another linear map $M:Y\\to Z$ and output projectors $R_k$,","$$\nR_kMLP_i=\\sum_j(R_kMQ_j)(Q_jLP_i).\n$$","**Proof.** Insert the appropriate identity decomposition on the input, output or intermediate space. Use $Q_j^2=Q_j$ in the composition formula. $\\square$","For a nonlinear map, $F(\\sum_iP_ix)$ generally differs from $\\sum_iF(P_ix)$. Applying a linear transfer formula to such a map silently loses interactions.","$$\nF(x)=b+\\sum_{m=1}^{p}B_m(x,\\ldots,x),\n$$","where each $B_m:X^m\\to Y$ is multilinear. Then","$$\nQ_jF(x)=Q_jb+\n\\sum_{m=1}^{p}\\ \\sum_{i_1,\\ldots,i_m}\nQ_jB_m(P_{i_1}x,\\ldots,P_{i_m}x).\n$$","**Proof.** Substitute $x=\\sum_iP_ix$ in every argument and distribute by multilinearity. $\\square$","For example, $(u+v)^2=u^2+2uv+v^2$. A grammar containing only independent packet squares misses $2uv$. For nonpolynomial maps, one needs direct nonlinear semantics or a local expansion with a stated remainder bound. The word algebraic does not remove that obligation.","Let $\\gamma:X\\to Z=\\gamma(X)$ be a summary and let $F_u:X\\to X$ describe a deterministic transition for action $u$. The reduced description closes when there is $\\bar F_u:Z\\to Z$ such that","$$\n\\gamma F_u=\\bar F_u\\gamma.\n$$","$$\n\\gamma(x)=\\gamma(x')\\ \\Longrightarrow\\\n\\gamma(F_u(x))=\\gamma(F_u(x'))\n$$","for every admissible $x,x'$ and every action. When it exists it is unique on $\\gamma(X)$.","**Proof.** Necessity follows by applying $\\bar F_u$ to equal summaries. For sufficiency, define $\\bar F_u(z)=\\gamma(F_u(x))$ for any representative with $\\gamma(x)=z$. The stated condition makes this well-defined and also proves uniqueness. $\\square$","This criterion also applies when $X$ is a latent physical state and $\\gamma$ is an observation map. Two physically different states can produce identical images and different futures. Exact image reconstruction does not resolve that hidden-state ambiguity. A history, a belief state or additional observations may be needed.","Suppose $X$ is a Hilbert space, $P$ is an orthogonal projection, $Q=I-P$ and $x_{t+1}=Lx_t$. The retained state is $z_t=Px_t$. Then","$$\nz_{t+1}=PLPz_t+PLQx_t.\n$$","**Proposition 7. Closure defect and compositional defect.** The retained state closes for every initial signal if and only if $PLQ=0$. If it does not close, the error in using only $PLP$ is","$$\ne_t=PLQx_t,\\qquad\n\\|e_t\\|\\leq\\|PLQ\\|\\,\\|Qx_t\\|.\n$$","For two linear operators $L$ and $M$,","$$\n\\boxed{PLMP-(PLP)(PMP)=PLQMP.}\n$$","**Proof.** If $PLQ=0$, the displayed recurrence depends only on $z_t$. Conversely, apply Proposition 6 to $x$ and $Px$, which have the same retained state, to obtain $PLQx=0$ for every $x$. The norm bound is submultiplicativity. The compositional identity follows by inserting $I=P+Q$ between $L$ and $M$. $\\square$","There is a constructive finite solution when the dynamics are a known finite family of linear maps $L_u$ on a $d$-dimensional space. It identifies which additional observables are required, rather than merely requesting a larger state vector.","Let the initial summary be $C_0x$. Regard its rows as linear observables. Starting from $W_0=\\operatorname{row}(C_0)$, define","$$\nW_{n+1}=W_n+\\sum_uW_nL_u,\n\\qquad W_nL_u=\\{wL_u:w\\in W_n\\}.\n$$","**Proposition 8. Minimal linear predictive completion.** The sequence stabilizes after at most $d-\\dim W_0$ strict enlargements. Its limit $W_*$ is the smallest row space containing $W_0$ and invariant under right multiplication by every $L_u$. If $R$ is any full-row-rank matrix whose rows form a basis of $W_*$, then","$$\nC_0=HR,\\qquad RL_u=B_uR\n$$","for suitable matrices $H$ and $B_u$. Therefore $z=Rx$ retains the original summary and has exact reduced dynamics $z_{t+1}=B_uz_t$. Every linear summary with those two properties has row space containing $W_*$.","**Proof.** The spaces are nested and have dimension at most $d$. Equality at one step means invariance under every $L_u$, so it is a fixed point. Containment in any invariant row space containing $W_0$ follows by induction. The matrix identities express inclusion of the relevant rows in the basis $R$. Conversely, if $C_0=HR$ and $RL_u=B_uR$, the row space of $R$ is invariant and contains $W_0$, so it contains every $W_n$. $\\square$","s are nested and have dimension at most $d$. Equality at one step means invariance under every $L_u$, so it is a fixed point. Containment in any invariant row space containing $W_0$ follows by induction. The matrix identities express inclusion of the relevant rows in the basis $R$. Conversely, if $C_0=HR$ and $RL_u=B_uR$, the row space of $R$ is invariant and contains $W_0$, so it contains every $W_n$. $\\square$","ested and have dimension at most $d$. Equality at one step means invariance under every $L_u$, so it is a fixed point. Containment in any invariant row space containing $W_0$ follows by induction. The matrix identities express inclusion of the relevant rows in the basis $R$. Conversely, if $C_0=HR$ and $RL_u=B_uR$, the row space of $R$ is invariant and contains $W_0$, so it contains every $W_n$. $\\square$","$$\nC_0,\\ C_0L,\\ \\ldots,\\ C_0L^{d-1}.\n$$","Its annihilator is the largest $L$-invariant subspace inside $\\ker C_0$, the standard unobservable-subspace construction [R2]. With several actions, observables pulled back through action words are required, but the linear span still stabilizes in finite dimension.","This gives a precise role for obstruction-driven refinement: add observable directions created by the dynamics until the chosen prediction variables close. It does not guarantee that the completion is small. In some systems $W_*=X^*$ and exact memoryless closure needs the full state.","The theorem assumes the operators and relevant state are available. Applying it to estimated operators constructs closure for those estimates. It is not a proof that the real system has the same closure. If only $C_0x$ is measured and the full $x$ is unavailable, the augmented state $Rx$ is not automatically accessible.","Let $\\Omega$ be a finite state set. Let $K_u(i,j)$ be known row-stochastic transition matrices. Let $\\ell_y(i)>0$ be a finite family of observation likelihoods. A partition $\\pi$ has aggregation matrix","$$\nC_\\pi(a,i)=\\mathbf 1_{i\\in B_a}.\n$$","Distributions are column vectors. Prediction is $p\\mapsto K_u^\\top p$, and a Bayesian update is","$$\n\\mathcal B_{\\ell_y}(p)\n=\\frac{p\\odot\\ell_y}{\\ell_y^\\top p}.\n$$","$$\nC_\\pi K_u^\\top=\\bar K_u^\\top C_\\pi,\n\\qquad\nC_\\pi\\mathcal B_{\\ell_y}(p)\n=\\mathcal B_{\\bar\\ell_y}(C_\\pi p).\n$$","Fix a mandatory initial partition $\\pi_0$ with at least two blocks. This preserves distinctions that the intended application requires and excludes the trivial one-state quotient.","**Proposition 9. Coarsest joint finite closure.** First split $\\pi_0$ by the likelihood vectors $(\\ell_y(i))_y$. Then repeatedly split each current block by the signatures","$$\n\\left(\\sum_{j\\in B}K_u(i,j)\\right)_{u,B},\n$$","where $B$ ranges over the current partition. The process terminates after at most $|\\Omega|-|\\pi_{\\mathrm{init}}|$ strict refinement rounds. The final partition is the unique coarsest refinement of $\\pi_0$ on which all declared predictions and Bayesian updates close for every prior.","**Proof.** At a stable partition, states in a common block have the same transition probability into every coarse block. Those probabilities define $\\bar K_u$ and give the first commutation identity, by checking it on point-mass priors. Likelihood constancy on each block defines $\\bar\\ell_y$ and gives the second identity by direct summation.","For minimality, every admissible partition refining $\\pi_0$ has at least two blocks. By Proposition 12 below, exact updating for all priors requires each fine likelihood to be constant within each block. Thus it refines the first likelihood split. Suppose it refines a current partition. Each current block is a union of its blocks. Exact prediction makes the probability of reaching each such union constant within any admissible block, so it also refines the next split. Induction proves it refines the final partition. Each strict refinement increases the number of blocks, which proves termination. $\\square$","Thus it refines the first likelihood split. Suppose it refines a current partition. Each current block is a union of its blocks. Exact prediction makes the probability of reaching each such union constant within any admissible block, so it also refines the next split. Induction proves it refines the final partition. Each strict refinement increases the number of blocks, which proves termination. $\\square$","$$\nz_t=Px_t,\\qquad q_t=Qx_t,\n$$","$$\nA=PLP,\\quad B=PLQ,\\quad C=QLP,\\quad D=QLQ.\n$$","Thus $z_{t+1}=Az_t+Bq_t$ and $q_{t+1}=Cz_t+Dq_t$.","**Proposition 10. Exact projected memory and a conditional tail bound.** For every $t\\geq0$,","$$\n\\boxed{\nz_{t+1}=Az_t+\n\\sum_{k=0}^{t-1}BD^kCz_{t-1-k}+BD^tq_0.\n}\n$$","If $\\|D\\|\\leq\\rho<1$ and $\\|z_s\\|\\leq Z$ over the relevant history, omitting terms with $k\\geq H$ changes the one-step right-hand side by at most","$$\n\\|B\\|\\,\\|C\\|Z\\frac{\\rho^H}{1-\\rho}.\n$$","Dropping the initial unresolved-state term adds at most $\\|B\\|\\rho^t\\|q_0\\|$.","$$\nq_t=D^tq_0+\\sum_{s=0}^{t-1}D^{t-1-s}Cz_s.\n$$","Substitute this into the retained recurrence and change indices. Bound the omitted terms by the geometric series. $\\square$","The memory kernel is $BD^kC$. It records a path from retained variables into unresolved variables, evolution there, and return. The first kernel $BC$ is precisely the return path appearing in the two-step composition defect. The term $BD^tq_0$ is deterministic when $q_0$ is specified. It becomes uncertain only relative to incomplete information or a probability model for $q_0$.","The tail bound compares right-hand sides using the same resolved history. Replacing that history by recursively generated predictions introduces additional rollout error. Stability of that recursion needs its own bound. Also, a spectral-radius condition alone does not imply $\\|D\\|\\leq\\rho$ in the norm already being used, especially for nonnormal operators.","**Proposition 11. Ideal statistical benefit of refinement.** Let a target $Y$ take values in a finite-dimensional Hilbert space with $\\mathbb E\\|Y\\|^2<\\infty$. Let $\\mathcal F\\subseteq\\mathcal G$ be information sets and define $m_{\\mathcal F}=\\mathbb E[Y\\mid\\mathcal F]$ and $m_{\\mathcal G}=\\mathbb E[Y\\mid\\mathcal G]$. Then","$$\n\\mathbb E\\|Y-m_{\\mathcal F}\\|^2\n=\n\\mathbb E\\|Y-m_{\\mathcal G}\\|^2\n+\\mathbb E\\|m_{\\mathcal G}-m_{\\mathcal F}\\|^2.\n$$","**Proof.** Expand $Y-m_{\\mathcal F}=(Y-m_{\\mathcal G})+(m_{\\mathcal G}-m_{\\mathcal F})$. The cross term has zero conditional expectation given $\\mathcal G$. $\\square$","This section imports a specific existing Komorebi foundation rather than inventing another inference heuristic. Let $p_i$ be a prior on a finite fine state set and let $B_a$ be coarse blocks. Define","$$\nP_a=\\sum_{i\\in B_a}p_i,\\qquad\nr_a(p)=\\frac{\\sum_{i\\in B_a}p_i\\ell_i}{P_a}\n\\quad(P_a>0).\n$$","On zero-mass blocks the value of $r_a$ is immaterial. Updating finely and then aggregating is exactly updating $P$ with the effective likelihood $r(p)$. The effective likelihood generally depends on the fine prior inside each coarse block, information that $P$ alone does not retain.","$$\n\\ell_i=c\\bar\\ell_a\\quad\\text{for every }i\\in B_a\n$$","with one common $c>0$. For a fixed prior, put $\\delta=r(p)-\\bar\\ell$, $\\bar b=\\mathcal B_{\\bar\\ell}(P)$ and $Z_r=r(p)^\\top P$. Then","$$\n\\boxed{\nC_\\pi\\mathcal B_{\\ell}(p)-\\bar b\n=\n\\frac{1}{Z_r}\\left(\\operatorname{diag}P-\\bar bP^\\top\\right)\\delta.\n}\n$$","global factor gives commutation by summing the fine posterior. Conversely, compare priors supported on one state in each of two distinct coarse blocks. Equality of coarse posterior odds forces the ratio of those states' likelihoods to equal the ratio of the coarse likelihoods. Varying the representatives proves constancy within each block and the common factor across blocks. For the defect, use $Z_r=\\bar\\ell^\\top P+P^\\top\\delta$ and subtract the two normalized vectors. $\\square$","compare priors supported on one state in each of two distinct coarse blocks. Equality of coarse posterior odds forces the ratio of those states' likelihoods to equal the ratio of the coarse likelihoods. Varying the representatives proves constancy within each block and the common factor across blocks. For the defect, use $Z_r=\\bar\\ell^\\top P+P^\\top\\delta$ and subtract the two normalized vectors. $\\square$","The matrix in the defect formula annihilates $\\bar\\ell$, recording the irrelevant common-scale direction. It does not make every likelihood discrepancy irrelevant. The construction in Proposition 9 refines until both the likelihood and transition defects vanish for a specified finite family.","$$\n\\operatorname{dom}(g\\circ f)\n=\\{x\\in\\operatorname{dom}f:f(x)\\in\\operatorname{dom}g\\}.\n$$","**Proposition 13. Strong fixed-horizon refinement limit.** Let $X$ be a Hilbert space, $L:X\\to X$ bounded, and $P_n$ orthogonal projections converging strongly to $I$. Put $L_n=P_nLP_n$. For every fixed $x\\in X$ and every fixed nonnegative integer $k$,","$$\nL_nx\\longrightarrow Lx,\\qquad\nL_n^kx\\longrightarrow L^kx.\n$$","**Proof.** Since $\\|P_n\\|\\leq1$,","$$\n\\|(L_n-L)x\\|\n\\leq\\|L\\|\\,\\|(P_n-I)x\\|+\\|(P_n-I)Lx\\|\n\\longrightarrow0.\n$$","Also $\\|L_n\\|\\leq\\|L\\|$. Induct on $k$, using","$$\n(L_n^{k+1}-L^{k+1})x\n=L_n(L_n^k-L^k)x+(L_n-L)L^kx.\n$$","The first term tends to zero by the uniform norm bound and induction, and the second by strong convergence on the fixed vector $L^kx$. $\\square$","If estimated operators $\\widehat L_n$ are uniformly bounded and $(\\widehat L_n-L_n)x\\to0$ for every fixed $x$, the same argument gives fixed-horizon convergence for $\\widehat L_n$. These are explicit assumptions, not conclusions of the finite learning procedure.","This result is strong convergence, not operator-norm convergence and not a bound uniform in time. On $\\ell^2$, finite coordinate projections converge strongly to $I$ while $\\|I-P_n\\|=1$ for every $n$. Unbounded operators require domain control. Adaptive data-dependent partitions and fitted coefficients require additional consistency arguments. These remain separate mathematical tasks.","$$\nx_{t+1}=Lx_t,\\qquad\nL=\\begin{pmatrix}\\tfrac12&1\\\\[2pt]\\tfrac14&\\tfrac13\\end{pmatrix},\n\\qquad x_0=\\begin{pmatrix}1\\\\0\\end{pmatrix}.\n$$","The spatial archive stores $x_t$. A second atlas uses the orthogonal Haar matrix","$$\nU=\\frac{1}{\\sqrt2}\\begin{pmatrix}1&1\\\\1&-1\\end{pmatrix}.\n$$","Its coefficients $c_t=Ux_t$ reconstruct as $x_t=U^*c_t$. The same dynamics in that atlas are $ULU^*$, not a separate physical law. Thus coordinate transport preserves the full dynamics.","Now retain only the first spatial coordinate. With $P=\\operatorname{diag}(1,0)$, the retained scalar $z_t$ and discarded scalar $q_t$ obey","$$\nz_{t+1}=\\tfrac12z_t+q_t,\\qquad\nq_{t+1}=\\tfrac14z_t+\\tfrac13q_t.\n$$","$$\nz_0=1,\\qquad z_1=\\tfrac12,\\qquad z_2=\\tfrac12,\n\\qquad z_3=\\tfrac{11}{24}.\n$$","The compressed one-step matrix predicts $z_2=1/4$ instead of $1/2$. The lost return path contributes exactly","$$\nBC=1\\cdot\\tfrac14=\\tfrac14.\n$$","There are two exact repairs. One is state refinement: start with $C_0=(1,0)$ and add $C_0L=(1/2,1)$. The resulting row space is the full two-dimensional space, so it closes. The other is memory:","$$\nz_{t+1}=\\tfrac12z_t+\n\\tfrac14\\sum_{k=0}^{t-1}\\left(\\tfrac13\\right)^kz_{t-1-k},\n$$","since $q_0=0$. This is the exact same system expressed without retaining $q_t$ as a separate current-state variable. At $t=2$, it yields","$$\nz_3=\\tfrac12\\cdot\\tfrac12\n+\\tfrac14\\cdot\\tfrac12\n+\\tfrac1{12}\\cdot1\n=\\tfrac{11}{24}.\n$$","To see a nontrivial reduced closure, adjoin a third independent coordinate with multiplier $5/6$. The smallest closed refinement of the first coordinate then contains the first two coordinates, not the third. Exact prediction of the chosen summary does not require reconstructing every component of the enlarged signal.","$$\nL'=\\begin{pmatrix}0&1000\\\\0&0\\end{pmatrix},\n\\qquad x=(0,1/1000)^\\top,\n$$","omitting the second coordinate loses norm $1/1000$, but changes the next first coordinate by exactly one. The correct certificate is the product of omitted magnitude and transfer amplification.","Let a finite-dimensional autonomous linear process have hidden state $x_t\\in\\mathbb R^d$ and observed output $y_t\\in\\mathbb R^p$:","$$\nx_{t+1}=Lx_t,\\qquad y_t=Cx_t.\n$$","The observation map need not be injective. For a positive integer $h$, define the observation map","$$\nO_h=\\begin{pmatrix}C\\\\CL\\\\\\vdots\\\\CL^{h-1}\\end{pmatrix},\n\\qquad V_h=\\operatorname{im}O_h.\n$$","The coordinates of $O_hx_s$ are future observations relative to time $s$. They are not available at time $s$ merely because this matrix is written down. At time $t\\geq h-1$, however, the same vector is the observed history","$$\nz_t=\\begin{pmatrix}y_{t-h+1}\\\\\\vdots\\\\y_t\\end{pmatrix}\n=O_hx_{t-h+1}.\n$$","1. There is a linear map $D_h:V_h\\to V_h$ satisfying $O_hL=D_hO_h$.\n2. $\\ker O_h\\subseteq\\ker(O_hL)$.\n3. $\\operatorname{rank}O_{h+1}=\\operatorname{rank}O_h$.","When these statements hold, $D_h$ is unique on $V_h$. The history state obeys $z_{t+1}=D_hz_t$, and its last output block is $y_t$. Its iterates therefore determine every subsequent observed output. If a positive integer $d$ bounds the hidden dimension, $h=d$ is sufficient for closure, though a smaller positive $h$ may suffice.","**Proof.** The equivalence of the first two statements is the linear factorization criterion: define $D_h(O_hx)=O_hLx$ and require independence from the representative $x$. The rows of $O_hL$ are $CL,\\ldots,CL^h$. All except the final block already occur in $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","f.** The equivalence of the first two statements is the linear factorization criterion: define $D_h(O_hx)=O_hLx$ and require independence from the representative $x$. The rows of $O_hL$ are $CL,\\ldots,CL^h$. All except the final block already occur in $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","statements is the linear factorization criterion: define $D_h(O_hx)=O_hLx$ and require independence from the representative $x$. The rows of $O_hL$ are $CL,\\ldots,CL^h$. All except the final block already occur in $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","linear factorization criterion: define $D_h(O_hx)=O_hLx$ and require independence from the representative $x$. The rows of $O_hL$ are $CL,\\ldots,CL^h$. All except the final block already occur in $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","=O_hLx$ and require independence from the representative $x$. The rows of $O_hL$ are $CL,\\ldots,CL^h$. All except the final block already occur in $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$",",CL^h$. All except the final block already occur in $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","n $O_h$. The kernel condition is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","is therefore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","efore equivalent to $CL^h$ being in the row space of $O_h$, which is exactly the rank equality. It also implies $O_hLx\\in V_h$ because this vector is $O_h$ applied to $Lx$. Uniqueness follows because every element of $V_h$ has a representative. The history recurrence follows directly. Cayley-Hamilton expresses $L^d$ as a linear combination of lower powers, so $CL^d$ is in the row space of $O_d$. $\\square$","This does not recover hidden variables in $\\ker O_h$. It identifies the quotient needed for the declared outputs. An independent hidden component can remain unobservable forever without preventing exact output prediction.","For several actions $L_u$, a fixed delay vector does not automatically have one autonomous transition matrix. Products of the actual action matrices determine its observation map, which can change with the action history. One needs an action-aware state, a common closed observable family, or a separately justified belief-state construction. Proposition 14 is not silently generalized to arbitrary controlled or stochastic dynamics.","Suppose candidate states and their successors have been recorded as columns of matrices $Z_-,Z_+\\in\\mathbb R^{m\\times N}$. Their construction may use output histories. No hidden-state coordinates need be supplied to the fitting procedure.","Write $W=\\operatorname{im}Z_-$. Exact interpolation of these columns is a question about a linear map on $W$, not yet about every physically realizable state.","**Proposition 15. Data-span identification and its boundary.** A linear map $B$ on the ambient space with $BZ_-=Z_+$ exists if and only if","$$\n\\ker Z_-\\subseteq\\ker Z_+.\n$$","Its restriction to $W$ is unique. It is an endomorphism of $W$ if and only if $\\operatorname{im}Z_+\\subseteq W$. A particular ambient extension is","$$\nB_0=Z_+Z_-^\\dagger,\n$$","where $\\dagger$ denotes the Moore-Penrose inverse. Every other fitted extension agrees with $B_0$ on $W$. If $W$ is a proper subspace of the true realizable state space $V$, the data alone do not identify the operator on $V\\setminus W$.","**Proof.** Necessity follows by applying $B$ to every linear dependence among the input columns. Under the kernel condition, define $B(Z_-a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","essity follows by applying $B$ to every linear dependence among the input columns. Under the kernel condition, define $B(Z_-a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","dence among the input columns. Under the kernel condition, define $B(Z_-a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","he input columns. Under the kernel condition, define $B(Z_-a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","Under the kernel condition, define $B(Z_-a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","condition, define $B(Z_-a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","a)=Z_+a$. Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$",". Two choices of $a$ differ by a kernel vector and give the same answer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","nswer. This defines and uniquely determines the map on $W$. It maps $W$ into $W$ exactly under the image condition. The pseudoinverse formula gives that map because $a-Z_-^\\dagger Z_-a\\in\\ker Z_-$. If $W\\subsetneq V$, choose a nonzero linear functional $\\varphi$ on $V$ vanishing on $W$ and a nonzero $v\\in V$. The operators $B$ and $B+v\\varphi$ agree on every observed column and differ elsewhere. $\\square$","The three checks are different. The kernel test checks consistency of a deterministic linear fit. The image test checks whether the observed span is closed under the fitted transitions. Equality $W=V$ is a *coverage condition*. Neither of the first two checks establishes it.","For a known fixed feature space of dimension $r$, coverage is witnessed by $r$ independent input states. For an unknown larger feature space, the same rank only gives a lower bound on dimension. A model card must say which interpretation is being used.","If several action-specific operators act on a common declared $r$-dimensional state, each action's paired data must determine that operator on the required span. Pooling all actions into one high-rank matrix does not identify an action that has never been tried on some direction.","Work in a fixed $r$-dimensional coordinate system, where $Z_-\\in\\mathbb R^{r\\times N}$ has full row rank. Let","$$\nZ_+=BZ_-+R,\\qquad\n\\widetilde Z_-=Z_-+E,\\qquad\n\\widetilde Z_+=Z_++F.\n$$","The matrix $R$ records deviations from the declared linear recurrence on these samples. The matrices $E$ and $F$ record input and successor perturbations. All norms in the next proposition are spectral norms.","**Proposition 16. Fixed-space identification bound.** Put $s=\\sigma_{\\min}(Z_-)>0$. If $\\|E\\|<s$, then $\\widetilde Z_-$ has full row rank, and","$$\n\\widehat B=\\widetilde Z_+\\widetilde Z_-^\\dagger\n$$","$$\n\\begin{aligned}\n\\widehat B-B\n&=(R+F-BE)\\widetilde Z_-^\\dagger,\\\\\n\\|\\widehat B-B\\|\n&\\leq\\frac{\\|R\\|+\\|F\\|+\\|B\\|\\,\\|E\\|}\n{s-\\|E\\|}.\n\\end{aligned}\n$$","**Proof.** For every unit $v\\in\\mathbb R^r$,\n$\\|\\widetilde Z_-^\\top v\\|\\geq\\|Z_-^\\top v\\|-\\|E^\\top v\\|\\geq s-\\|E\\|$.\nThus the perturbed matrix has full row rank and its pseudoinverse has norm at most $(s-\\|E\\|)^{-1}$. Since $\\widetilde Z_-\\widetilde Z_-^\\dagger=I_r$, subtract $B$ from the fitted expression and collect terms. The bound follows by submultiplicativity and the triangle inequality. $\\square$","This is a bound for a fixed state space and a fixed coordinate basis. If the basis or its dimension is also estimated from noisy data, subspace-estimation error must be added. No such error is hidden inside $R$ without a derivation. Correlated and adversarial perturbations are allowed by the displayed deterministic inequality, but there is no sample-complexity claim until their magnitudes are bounded under a sampling model.","The denominator explains why numerical rank alone is insufficient. Full rank can coexist with severe noise amplification. The reference calculations include examples attaining the $1/s$ dependence and a rank-collapse example at the excluded boundary $\\|E\\|=s$.","**Proposition 17. Finite-prefix indistinguishability.** For every integer $T\\geq0$, there are two finite-dimensional autonomous rational linear systems with identical scalar observations at times $0,\\ldots,T$ and different observations at time $T+1$. Both transition matrices can be chosen nilpotent.","**Proof.** One system has scalar output identically zero. For the other, take dimension $T+2$, initial state $x_0=e_{T+2}$, observation row $C=e_1^\\top$, and shift $Le_j=e_{j-1}$ for $j\\geq2$, $Le_1=0$. Then $CL^tx_0=0$ for $0\\leq t\\leq T$ and $CL^{T+1}x_0=1$. The shift is rational and nilpotent. $\\square$","$$\nL=\\begin{pmatrix}\n1/2&1&0\\\\\n1/4&1/3&0\\\\\n0&0&5/6\n\\end{pmatrix},\n\\qquad C=\\begin{pmatrix}1&0&0\\end{pmatrix}.\n$$","$$\nO_2=\\begin{pmatrix}1&0&0\\\\1/2&1&0\\end{pmatrix}\n$$","has rank two and $O_3$ has the same rank. Proposition 14 gives an exact two-dimensional delay state. Choose $x_0=(1,0,q)^\\top$ for any $q$. The first four observations are","$$\ny_0=1,\\quad y_1=1/2,\\quad y_2=1/2,\\quad y_3=11/24.\n$$","A learner receives those four outputs and the declared delay length $h=2$. It does **not** receive $L$, $C$, $O_2$ or $q$. From consecutive observed windows it forms","$$\nZ_-=\\begin{pmatrix}1&1/2\\\\1/2&1/2\\end{pmatrix},\n\\qquad\nZ_+=\\begin{pmatrix}1/2&1/2\\\\1/2&11/24\\end{pmatrix}.\n$$","Since $\\det Z_-=1/4$, these windows span the declared two-dimensional state. Exact arithmetic gives","$$\n\\boxed{\nD_2=Z_+Z_-^{-1}\n=\\begin{pmatrix}0&1\\\\1/12&5/6\\end{pmatrix}.\n}\n$$","$$\n\\boxed{y_{t+1}=\\frac56y_t+\\frac1{12}y_{t-1},\\qquad t\\geq1.}\n$$","It predicts $y_4=61/144$, which was not used to identify the coefficients. Proposition 14 and the verified rank condition on the true observation maps establish why this recurrence continues for every initial state of this particular system. The four samples alone establish the fitted recurrence only within the declared closed two-state model class. Proposition 17 forbids confusing those two conclusions.","Allow the representation to change with time. Let $\\gamma_t:X\\to Z_t$ be the declared representation at step $t$, with a norm on each $Z_t$. Let $x_{t+1}=F_t(x_t)$ be the true transition and let $\\widehat f_t:Z_t\\to Z_{t+1}$ be the implemented step. Its defect is","$$\nr_t(x)=\\gamma_{t+1}(F_t(x))-\\widehat f_t(\\gamma_t(x)).\n$$","**Proposition 18. Compositional defect propagation.** Suppose $\\widehat f_t$ is $\\lambda_t$-Lipschitz on a domain containing both $\\gamma_t(x_t)$ and the predicted state $\\widehat z_t$. Suppose\n$\\|r_t(x_t)\\|\\leq\\delta_t$ and\n$\\widehat z_{t+1}=\\widehat f_t(\\widehat z_t)$.\nThen, with $e_t=\\|\\gamma_t(x_t)-\\widehat z_t\\|$,","$$\n\\boxed{\ne_T\\leq\n\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\n\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n}\n$$","Empty products are one. If the output $h_t(x)$ has a decoder $H_t$ with Lipschitz constant $\\kappa_t$ and\n$\\|h_t(x_t)-H_t(\\gamma_t(x_t))\\|\\leq\\eta_t$, then","$$\n\\|h_t(x_t)-H_t(\\widehat z_t)\\|\n\\leq\\eta_t+\\kappa_te_t.\n$$","**Proof.** Add and subtract $\\widehat f_t(\\gamma_t(x_t))$ to obtain\n$e_{t+1}\\leq\\delta_t+\\lambda_te_t$.\nRepeated substitution yields the product-sum formula. Add and subtract $H_t(\\gamma_t(x_t))$ for the output inequality. $\\square$","For a fixed linear representation $R$ and an estimated transition $\\widehat B_u$, the defect is $(RL_u-\\widehat B_uR)x$. For a reference reduced operator $B_u$, it splits exactly as","$$\nRL_u-\\widehat B_uR\n=(RL_u-B_uR)+(B_u-\\widehat B_u)R.\n$$","If $\\lambda_t\\leq\\rho<1$ and $\\delta_t\\leq\\delta$, then","$$\ne_T\\leq\\rho^Te_0+\\delta\\frac{1-\\rho^T}{1-\\rho}.\n$$","For linear candidate operators, one sufficient way to obtain a common contraction norm is to exhibit a positive-definite matrix $M$ and $0\\leq\\rho<1$ with","$$\n\\widehat B_u^\\top M\\widehat B_u\\preceq\\rho^2M\n\\quad\\text{for every declared action }u.\n$$","Then $\\|z\\|_M=(z^\\top Mz)^{1/2}$ gives\n$\\|\\widehat B_uz\\|_M\\leq\\rho\\|z\\|_M$ for all actions, directly from the displayed inequality. Proposition 18 therefore applies under arbitrary switching among them. This is the common quadratic Lyapunov construction from control theory [R8], used here as an explicit certificate rather than inferred from a list of eigenvalues.","$$\nA=\\begin{pmatrix}0&2\\\\0&0\\end{pmatrix},\\qquad\nB=\\begin{pmatrix}0&0\\\\2&0\\end{pmatrix}\n$$","are both nilpotent, but $BA=\\operatorname{diag}(0,4)$. Alternating them amplifies an appropriate state by four every two steps. Even a single stable nonnormal matrix can produce large transient amplification in a specified norm.","$$\nz_{t+1}=Az_t+\\sum_{k=0}^{t-1}K_kz_{t-1-k}+f_t,\n\\qquad K_k=BD^kC,\\quad f_t=BD^tq_0.\n$$","For negative indices set $z_s=0$, solely as a convention for missing prehistory in this initialized formula. A length-$H$ approximate model, with $H\\geq0$, is","$$\n\\widehat z_{t+1}=\\widehat A\\widehat z_t+\n\\sum_{k=0}^{H-1}\\widehat K_k\\widehat z_{t-1-k}.\n$$","An empty sum is zero. Let $d_t$ be the exact right-hand side minus this truncated right-hand side evaluated on the **true** history.","$$\na=\\|\\widehat A\\|+\\sum_{k=0}^{H-1}\\|\\widehat K_k\\|<1,\n\\qquad\\sup_{t\\geq0}\\|d_t\\|\\leq\\Delta.\n$$","Let $E_0$ bound the initialization errors at indices $-H,\\ldots,0$. Then the recursively generated predictions satisfy","$$\n\\boxed{\n\\sup_{t\\geq0}\\|z_t-\\widehat z_t\\|\n\\leq\\max\\left(E_0,\\frac{\\Delta}{1-a}\\right).\n}\n$$","If $\\|z_t\\|\\leq Z$ for $t\\geq0$ and $\\|D\\|\\leq\\rho<1$, a valid choice of defect bound is","$$\n\\begin{aligned}\n\\Delta={}&Z\\left(\\|A-\\widehat A\\|\n+\\sum_{k=0}^{H-1}\\|K_k-\\widehat K_k\\|\\right)\\\\\n&+\\|B\\|\\,\\|C\\|Z\\frac{\\rho^H}{1-\\rho}\n+\\|B\\|\\,\\|q_0\\|.\n\\end{aligned}\n$$","The last term can be removed if the exact initialized forcing $f_t$ is retained in the approximate recurrence.","**Proof.** Subtract the two recurrences and bound every retained history error by the largest of the errors at indices $t-H,\\ldots,t$. This gives\n$e_{t+1}\\leq a\\max_{t-H\\leq s\\leq t}e_s+\\Delta$.\nThe ball of radius $E=\\max(E_0,\\Delta/(1-a))$ is invariant because $aE+\\Delta\\leq E$. Induction proves the bound. The estimate for $\\Delta$ separates coefficient errors, the geometric kernel tail and the initialized forcing, using Proposition 10. $\\square$","bound every retained history error by the largest of the errors at indices $t-H,\\ldots,t$. This gives\n$e_{t+1}\\leq a\\max_{t-H\\leq s\\leq t}e_s+\\Delta$.\nThe ball of radius $E=\\max(E_0,\\Delta/(1-a))$ is invariant because $aE+\\Delta\\leq E$. Induction proves the bound. The estimate for $\\Delta$ separates coefficient errors, the geometric kernel tail and the initialized forcing, using Proposition 10. $\\square$","For the scalar example with $q_0=0$, take exact retained coefficients and $z_0=1$. All outputs lie in $[0,1]$: the full nonnegative memory coefficients sum to at most $7/8$, so induction bounds every new output by the previous maximum. Thus $Z=1$ is valid. With","$$\nA=1/2,\\quad K_k=\\frac14\\left(\\frac13\\right)^k,\n$$","$$\na_H=\\frac78-\\frac38\\,3^{-H},\\qquad\n\\Delta_H=\\frac38\\,3^{-H},\\qquad\n\\boxed{\\sup_t|z_t-\\widehat z_t|\\leq\n\\frac{3\\,3^{-H}}{1+3\\,3^{-H}}.}\n$$","Let $P_i$ be an orthogonal resolution of a finite-dimensional real Hilbert space. Suppose","with bounded multilinear maps. For each ordered tuple $\\boldsymbol i=(i_1,\\ldots,i_m)$, let $\\beta_{m,\\boldsymbol i}$ bound the norm of $B_m$ restricted to\n$\\operatorname{im}P_{i_1}\\times\\cdots\\times\\operatorname{im}P_{i_m}$.\nLet $\\mathcal I_m$ be the retained tuple set, and let $F_{\\mathcal I}$ retain the constant term and exactly those interactions.","**Proposition 20. Packet-interaction truncation bound.** Put $a_i=\\|P_ix\\|$. Then","$$\n\\boxed{\n\\|F(x)-F_{\\mathcal I}(x)\\|\n\\leq\\sum_{m=1}^{p}\\ \\sum_{\\boldsymbol i\\notin\\mathcal I_m}\n\\beta_{m,\\boldsymbol i}\\prod_{j=1}^{m}a_{i_j}.\n}\n$$","On the ball $\\|x\\|\\leq R$, a valid Lipschitz bound for $F_{\\mathcal I}$ is","$$\n\\operatorname{Lip}(F_{\\mathcal I})\n\\leq\\sum_{m=1}^{p}\\ \\sum_{\\boldsymbol i\\in\\mathcal I_m}\nm\\beta_{m,\\boldsymbol i}R^{m-1}.\n$$","**Proof.** Subtract the retained terms from the exact multilinear expansion. Bound each omitted term by its restricted multilinear norm and sum. For the Lipschitz estimate, telescope the difference between two $m$-fold evaluations by replacing one argument at a time. Each changed argument has norm at most $\\|x-y\\|$ after projection, and each unchanged argument has norm at most $R$. There are $m$ terms for each retained tuple. $\\square$","d terms from the exact multilinear expansion. Bound each omitted term by its restricted multilinear norm and sum. For the Lipschitz estimate, telescope the difference between two $m$-fold evaluations by replacing one argument at a time. Each changed argument has norm at most $\\|x-y\\|$ after projection, and each unchanged argument has norm at most $R$. There are $m$ terms for each retained tuple. $\\square$","If the declared polynomial itself approximates a nonpolynomial map with remainder at most $\\varepsilon_R$ on the ball, add $\\varepsilon_R$ to the defect bound. Without such a remainder estimate, the result is only a theorem for the polynomial, not for the unexpanded physical law.","For a retained state $z=Px$ and omitted state $q=(I-P)x$, dropping every term that uses $q$ gives the coarser majorant","$$\n\\|F(z+q)-F(z)\\|\n\\leq\\sum_{m=1}^{p}\\|B_m\\|\n\\left((\\|z\\|+\\|q\\|)^m-\\|z\\|^m\\right).\n$$","A prediction guarantee follows by inserting these defect and Lipschitz bounds into Proposition 18, provided the true and predicted trajectories remain in the declared domain. That domain condition cannot be inferred from a small initial error. One possible construction is to assume that the true map preserves a closed convex set $K$ and to project every approximate output back onto $K$. Euclidean projection is nonexpansive and fixes the true output, so it does not increase the one-step approximation bound. This is a declared modification of the predictor, not an identity for an unmodified recurrence.","For the scalar quadratic $(u+v)^2$, retaining only $u^2+v^2$ omits $2uv$. The packet bound is $2|u||v|$ and is exact in magnitude. Small energy in one packet matters in proportion to both its amplitude and the cross-interaction coefficient.","Fix a finite-dimensional represented space $X$. It may be the observed signal space or a predictive quotient, but it must be the *same* space for all atlases in this section. Let $A_a:X\\to C_a$ and $S_a:C_a\\to X$ satisfy $S_aA_a=I$. Let $G_a:C_a\\to C_a$ be linear fitted maps. Pull them back to object space:","$$\nL_a=S_aG_aA_a.\n$$","$$\n\\overline L=\\sum_aw_aL_a,\n\\qquad\\widetilde G_a=A_a\\overline LS_a\n\\quad\\text{on }\\mathcal C_a=\\operatorname{im}A_a.\n$$","**Proposition 21. Coherent reconciliation and coordinate-stable transport.** The family $\\widetilde G_a$ is consistent and satisfies every atlas intertwining identity. The operator $\\overline L$ is the unique minimizer of","$$\n\\sum_aw_a\\|L-L_a\\|_F^2\n$$","over linear maps on $X$, where $\\|\\cdot\\|_F$ is the Frobenius norm. For any target operator $L_*$ and any chosen operator norm,","$$\n\\|\\overline L-L_*\\|\\leq\\sum_aw_a\\|L_a-L_*\\|.\n$$","Further, the norm $\\|c\\|_a=\\|S_ac\\|$ on $\\mathcal C_a$ makes every exact transport $T_{ba}=A_bS_a$ an isometry. A represented transition $A_bLS_a$ has exactly the same induced Lipschitz constant in these norms as $L$ has in object space.","**Proof.** Coherence follows from Proposition 2 because every chart realizes the same $\\overline L$. Expanding the squares gives","$$\n\\sum_aw_a\\|L-L_a\\|_F^2\n=\\|L-\\overline L\\|_F^2+\n\\sum_aw_a\\|L_a-\\overline L\\|_F^2.\n$$","The weighted cross term vanishes, proving the unique minimizer. The target bound follows from the triangle inequality. The pullback expression is a norm because $S_a$ is inverse to $A_a$ on $\\mathcal C_a$. Finally,\n$\\|T_{ba}c\\|_b=\\|S_bA_bS_ac\\|=\\|S_ac\\|$.\nThe same identity for a transition, and the bijection between $X$ and $\\mathcal C_a$, establish equality of Lipschitz constants. $\\square$","his reconciliation is ordinary weighted least squares in a common semantic space. It does not prove that consensus is correct, reduce statistical error under arbitrary dependence, or turn deterministic recodings into independent evidence. The output must also be reanalyzed into a realizable coefficient vector. An independently fitted redundant output can have an additional inconsistency residual $(I-A_aS_a)G_aA_ax$ before reconciliation.","For Euclidean coefficient errors, conditioning remains important. If $A_a$ has smallest singular value at least $m_a>0$ and largest singular value at most $M_a$, its canonical synthesis $A_a^\\dagger$ has norm at most $1/m_a$, and\n$\\|T_{ba}\\|\\leq M_b/m_a$.\nAn arbitrary left inverse can behave much worse away from the consistent coefficient space, so the canonical-synthesis qualification matters.","**Proposition 22. The upward-lift obstruction.** Let $\\gamma:X\\to Z$ be a coarse representation and $\\eta:X\\to W$ a proposed refined representation. There is a deterministic map $J:\\gamma(X)\\to\\eta(X)$ satisfying $\\eta=J\\gamma$ if and only if","$$\n\\gamma(x)=\\gamma(x')\\ \\Longrightarrow\\ \\eta(x)=\\eta(x')\n\\quad\\text{for every }x,x'.\n$$","For linear representations, this is equivalent to $\\ker\\gamma\\subseteq\\ker\\eta$. If two states share a coarse representation but their refined representations differ, any predicted refined value $w$ has worst-case error at least","$$\n\\max\\bigl(\\|w-\\eta(x)\\|,\\|w-\\eta(x')\\|\\bigr)\n\\geq\\tfrac12\\|\\eta(x)-\\eta(x')\\|.\n$$","**Proof.** A lift must be constant on each coarse fiber. Conversely, the stated condition defines it unambiguously on every fiber. The linear version is the kernel factorization criterion. The lower bound follows from the triangle inequality between the two refined values through $w$. $\\square$","For a strict refinement, the coarse state is recoverable from the refined one but not conversely. In the linear case this typically means $\\ker\\eta\\subsetneq\\ker\\gamma$, exactly the opposite of the condition required for an upward lift.","For example, $\\gamma(x_1,x_2)=x_1$ and $\\eta(x_1,x_2)=(x_1,x_2)$. The states $(0,1)$ and $(0,-1)$ have the same coarse state, but any common point initialization has worst-case error at least one in Euclidean norm. Padding a vector with zeros is a choice, not information recovery.","For adaptive prediction, an attempted lift $J_t$ contributes its own initialization defect","$$\n\\eta_t(x_t)-J_t\\gamma_t(x_t).\n$$","Let $\\mathcal G$ be a countable declared program set, with one fixed probability mass function $\\pi(g)>0$ satisfying $\\sum_g\\pi(g)=1$. Every program specifies a likelihood $L_g(D)$ for the **same** observed data event or density relative to a common base measure. Assume","$$\n0<Z(D)=\\sum_{g\\in\\mathcal G}\\pi(g)L_g(D)<\\infty.\n$$","Let finite active sets $\\mathcal G_n$ increase to $\\mathcal G$. Their unnormalized evidence is $Z_n=\\sum_{g\\in\\mathcal G_n}\\pi(g)L_g(D)$. Only stages with $Z_n>0$ are considered.","**Proposition 23. Exact accounting for an enumerated grammar.** Define the full posterior $P(g\\mid D)=\\pi(g)L_g(D)/Z$ and extend the active posterior $P_n(g\\mid D)=\\pi(g)L_g(D)/Z_n$ by zero outside $\\mathcal G_n$. Then","$$\n\\boxed{\n\\|P-P_n\\|_{\\mathrm{TV}}=\\tau_n,\n\\qquad \\tau_n=1-\\frac{Z_n}{Z}.\n}\n$$","For any common family of candidate predictive probability kernels, the corresponding predictive mixtures differ in total variation by at most $\\tau_n$. If the omitted unnormalized evidence has a bound","$$\n\\sum_{g\\notin\\mathcal G_n}\\pi(g)L_g(D)\\leq b_n,\n$$","$$\n\\tau_n\\leq\\frac{b_n}{Z_n+b_n}.\n$$","**Proof.** On the active set, $P=(Z_n/Z)P_n$, so the total mass of the difference there is $\\tau_n$. The mass outside is also $\\tau_n$. Half the sum of absolute differences is therefore $\\tau_n$. Applying the same probability kernel to two mixing measures cannot increase total variation: integrate any event probability, a function taking values in $[0,1]$, against their difference. For the final bound, write $Z=Z_n+W_n$ with $0\\leq W_n\\leq b_n$ and use that $W_n/(Z_n+W_n)$ increases with $W_n$. $\\square$","the active set, $P=(Z_n/Z)P_n$, so the total mass of the difference there is $\\tau_n$. The mass outside is also $\\tau_n$. Half the sum of absolute differences is therefore $\\tau_n$. Applying the same probability kernel to two mixing measures cannot increase total variation: integrate any event probability, a function taking values in $[0,1]$, against their difference. For the final bound, write $Z=Z_n+W_n$ with $0\\leq W_n\\leq b_n$ and use that $W_n/(Z_n+W_n)$ increases with $W_n$. $\\square$","$P=(Z_n/Z)P_n$, so the total mass of the difference there is $\\tau_n$. The mass outside is also $\\tau_n$. Half the sum of absolute differences is therefore $\\tau_n$. Applying the same probability kernel to two mixing measures cannot increase total variation: integrate any event probability, a function taking values in $[0,1]$, against their difference. For the final bound, write $Z=Z_n+W_n$ with $0\\leq W_n\\leq b_n$ and use that $W_n/(Z_n+W_n)$ increases with $W_n$. $\\square$","of the difference there is $\\tau_n$. The mass outside is also $\\tau_n$. Half the sum of absolute differences is therefore $\\tau_n$. Applying the same probability kernel to two mixing measures cannot increase total variation: integrate any event probability, a function taking values in $[0,1]$, against their difference. For the final bound, write $Z=Z_n+W_n$ with $0\\leq W_n\\leq b_n$ and use that $W_n/(Z_n+W_n)$ increases with $W_n$. $\\square$","au_n$. The mass outside is also $\\tau_n$. Half the sum of absolute differences is therefore $\\tau_n$. Applying the same probability kernel to two mixing measures cannot increase total variation: integrate any event probability, a function taking values in $[0,1]$, against their difference. For the final bound, write $Z=Z_n+W_n$ with $0\\leq W_n\\leq b_n$ and use that $W_n/(Z_n+W_n)$ increases with $W_n$. $\\square$","The mass outside is also $\\tau_n$. Half the sum of absolute differences is therefore $\\tau_n$. Applying the same probability kernel to two mixing measures cannot increase total variation: integrate any event probability, a function taking values in $[0,1]$, against their difference. For the final bound, write $Z=Z_n+W_n$ with $0\\leq W_n\\leq b_n$ and use that $W_n/(Z_n+W_n)$ increases with $W_n$. $\\square$","Since the active sets exhaust the grammar and the evidence sum is finite, $Z_n\\uparrow Z$ and $\\tau_n\\downarrow0$. This is posterior approximation *inside the declared grammar*. It does not establish that the real signal-generating mechanism belongs to it.","For a discrete data sequence, a probability likelihood is at most one, so the omitted prior mass is a valid $b_n$. A continuous-data density need not be at most one. It requires a density envelope or another justified bound on omitted evidence. A small unnormalized tail can also carry substantial posterior mass when the active evidence is very small.","The fixed prior need not be uniform. For an explicit enumeration $g_1,g_2,\\ldots$, $\\pi(g_k)=2^{-k}$ is a normalized example, with omitted prior mass $2^{-n}$ after $n$ programs. It expresses a particular enumeration bias, not a canonical prior on mathematical meaning.","**Proposition 24. Prior-preserving program refinement.** Suppose a parent program $g$ is replaced by children $g_j$ with weights $\\alpha_j\\geq0$, $\\sum_j\\alpha_j=1$, and suppose its likelihood is the mixture","$$\nL_g(D)=\\sum_j\\alpha_jL_{g_j}(D).\n$$","Give child $j$ prior mass $\\pi(g)\\alpha_j$ and leave all other programs unchanged. Then total evidence is unchanged and the sum of the children's posterior masses equals the former parent's posterior mass. Posterior predictive distributions are also unchanged when the same mixture relation holds for the joint likelihood of $(D,y)$ for every predicted event $y$.","**Proof.** The children's total unnormalized evidence is\n$\\sum_j\\pi(g)\\alpha_jL_{g_j}(D)=\\pi(g)L_g(D)$.\nThus both the total normalizer and every other program's posterior are unchanged. The same sum for joint evidence gives the predictive statement after normalization by the unchanged evidence of $D$. $\\square$","1. The data come from an autonomous finite-dimensional linear process with declared outputs $y_t=Cx_t$.\n2. A delay length $h$ satisfies $\\operatorname{rank}O_{h+1}=\\operatorname{rank}O_h$.\n3. Exact observed input windows in $Z_-$ span $V_h=\\operatorname{im}O_h$, and $Z_+$ records their true one-step successor windows.","Then an algorithm using only $(Z_-,Z_+)$ recovers the unique transition on $V_h$. Initialized with a fully observed length-$h$ window, it predicts every later output exactly. Any lossless atlas of $V_h$, any complete packet decomposition of that space, and every coherently transported realization of the learned transition preserve the same predictions. Hidden variables outside the observable quotient need not be recovered.","**Proof.** Proposition 14 gives the unique true transition $D_h$ on $V_h$. Because the observed windows span $V_h$, Proposition 15 identifies exactly its restriction, which is the entire required operator. Equality of the learned and true states follows by induction after initialization. The current output is the last block of the history state. Propositions 1--4 preserve state reconstruction, transport and complete decomposition, so they preserve that output sequence. The perturbation statement is the direct composition of Propositions 16 and 18 with the declared output decoder. $\\square$","equired operator. Equality of the learned and true states follows by induction after initialization. The current output is the last block of the history state. Propositions 1--4 preserve state reconstruction, transport and complete decomposition, so they preserve that output sequence. The perturbation statement is the direct composition of Propositions 16 and 18 with the declared output decoder. $\\square$","Let $\\mathcal A$ and $\\mathcal O$ be finite nonempty action and observation sets. A symbol $s=(a,o)$ records an action followed by an observation. A word $w=s_1\\cdots s_k$ has temporal order from left to right. The empty word is $\\epsilon$.","$$\nf(w)=\\Pr(o_1,\\ldots,o_k\\mid \\operatorname{do}(a_1,\\ldots,a_k)).\n$$","$$\nf(\\epsilon)=1,\\qquad f(w)\\geq0,\\qquad\n\\sum_{o\\in\\mathcal O}f(w(a,o))=f(w)\n\\quad\\text{for every }w,a.\n$$","A concrete realization has a belief $p\\in\\mathbb R^n_+$ with $\\mathbf1^\\top p=1$. For each $s=(a,o)$, let $M_s\\in\\mathbb R^{n\\times n}$ be entrywise nonnegative, with","$$\n\\mathbf1^\\top\\sum_oM_{a,o}=\\mathbf1^\\top.\n$$","The matrix entry $M_{a,o}(j,i)$ is the probability of next hidden state $j$ and observation $o$, given current hidden state $i$ and chosen action $a$. This is a discrete probability instrument. It includes an action-specific transition followed by an observation, but does not require that factorization.","Put $M_w=M_{s_k}\\cdots M_{s_1}$. Then","$$\nf(w)=\\mathbf1^\\top M_wp_0,\\qquad\np^+=\\frac{M_sp}{\\mathbf1^\\top M_sp}\n$$","**Proposition 26. Finite instrument closure.** Starting with $W_0=\\operatorname{span}\\{\\mathbf1^\\top\\}$, define","$$\nW_{j+1}=W_j+\\sum_{s\\in\\mathcal A\\times\\mathcal O}W_jM_s.\n$$","This stabilizes after at most $n-1$ strict enlargements at","$$\nW_* = \\operatorname{span}\\{\\mathbf1^\\top M_w:w\\text{ a finite word}\\}.\n$$","Choose a full row rank matrix $R$ with row space $W_*$ and first row $\\mathbf1^\\top$. There are unique matrices $B_s$ with","$$\nRM_s=B_sR,\\qquad \\lambda=e_1^\\top,\\qquad \\lambda R=\\mathbf1^\\top.\n$$","For $z=Rp$, both the next-observation likelihood and the posterior predictive state close:","$$\n\\Pr(o\\mid z,a)=\\lambda B_{a,o}z,\\qquad\nz^+=\\frac{B_{a,o}z}{\\lambda B_{a,o}z}.\n$$","Every linear summary determining all word probabilities by linear readouts for every initial belief has row space containing $W_*$.","**Proof.** The nested-space argument from Proposition 8 applies to the finite family $M_s$. Multiplication of a test row by $M_s$ prepends the symbol $s$ to that test, so the stable span is exactly the displayed word span. Invariance supplies the matrices $B_s$, and full row rank makes them unique. Apply $R$ to the unnormalized and normalized hidden-state updates. If all test probabilities are linear functions of another summary, then each row $\\mathbf1^\\top M_w$ lies in its row space. Equality on all beliefs implies equality on their linear span, which is $\\mathbb R^n$. $\\square$","tion 8 applies to the finite family $M_s$. Multiplication of a test row by $M_s$ prepends the symbol $s$ to that test, so the stable span is exactly the displayed word span. Invariance supplies the matrices $B_s$, and full row rank makes them unique. Apply $R$ to the unnormalized and normalized hidden-state updates. If all test probabilities are linear functions of another summary, then each row $\\mathbf1^\\top M_w$ lies in its row space. Equality on all beliefs implies equality on their linear span, which is $\\mathbb R^n$. $\\square$","st, so the stable span is exactly the displayed word span. Invariance supplies the matrices $B_s$, and full row rank makes them unique. Apply $R$ to the unnormalized and normalized hidden-state updates. If all test probabilities are linear functions of another summary, then each row $\\mathbf1^\\top M_w$ lies in its row space. Equality on all beliefs implies equality on their linear span, which is $\\mathbb R^n$. $\\square$","le span is exactly the displayed word span. Invariance supplies the matrices $B_s$, and full row rank makes them unique. Apply $R$ to the unnormalized and normalized hidden-state updates. If all test probabilities are linear functions of another summary, then each row $\\mathbf1^\\top M_w$ lies in its row space. Equality on all beliefs implies equality on their linear span, which is $\\mathbb R^n$. $\\square$","This proposition uses a known hidden-state model. It is not yet an observable-data learner. It also concerns all initial beliefs, whereas a model initialized at one fixed $p_0$ can have a smaller reachable predictive dimension. The next construction handles that distinction without requiring recovery of hidden states.","For a history $h$, define its unnormalized future-behavior column","$$\nv_h(t)=f(ht),\n$$","where $t$ ranges over every finite test word. Form the infinite history/test matrix","$$\n\\mathscr H_{t,h}=f(ht).\n$$","**Proposition 27. Minimal finite behavior realization.** If $\\operatorname{rank}\\mathscr H=r<\\infty$, the vector space $V=\\operatorname{span}\\{v_h\\}$ has dimension $r$. There are well-defined linear maps $B_s:V\\to V$, an initial vector $v_\\epsilon$, and a linear mass functional $\\lambda(v)=v(\\epsilon)$ satisfying","$$\nB_sv_h=v_{hs},\\qquad\nf(s_1\\cdots s_k)=\\lambda B_{s_k}\\cdots B_{s_1}v_\\epsilon.\n$$","Every linear realization of $f$ has dimension at least $r$. Any two reachable and observable realizations of dimension $r$ differ by an invertible change of coordinates. For $f(h)>0$, the normalized state $v_h/f(h)$ determines every future test probability conditional on $h$ and the declared future interventions.","**Proof.** If $\\sum_hc_hv_h=0$, then for every test $t$,","$$\n\\sum_hc_hv_{hs}(t)=\\sum_hc_hf(hst)=0,\n$$","because $st$ is another test. Thus shifting columns preserves all linear relations and defines $B_s$. The probability formula follows by induction. For any $d$-dimensional realization, the history/test matrix factors into test rows and reachable state columns, so its rank is at most $d$.","ector after each history in one minimal realization to the corresponding vector in the other. If a linear combination vanishes in the first, every future test evaluates to zero on it. Those same evaluations vanish in the second, where observability makes the combination zero. The map is therefore well-defined and bijective, and intertwines every symbol operator and both boundary vectors. Finally $f(ht)/f(h)$ is exactly the conditional experiment probability. $\\square$","sponding vector in the other. If a linear combination vanishes in the first, every future test evaluates to zero on it. Those same evaluations vanish in the second, where observability makes the combination zero. The map is therefore well-defined and bijective, and intertwines every symbol operator and both boundary vectors. Finally $f(ht)/f(h)$ is exactly the conditional experiment probability. $\\square$","The coordinate dimension $r$ includes probability mass. Normalized states lie in the affine hyperplane $\\lambda z=1$, so their affine dimension is at most $r-1$. A finite hidden-state model of size $n$ has behavior rank at most $n$. The theorem itself requires finite behavior rank, not a preferred hidden-state ontology.","Select $r$ tests $t_1,\\ldots,t_r$ and $r$ histories $h_1,\\ldots,h_r$, with $t_1=\\epsilon$, for which the square matrix","$$\nH_{ij}=f(h_jt_i)\n$$","For each symbol $s$, also obtain the shifted table and initial test vector","$$\n(G_s)_{ij}=f(h_jst_i),\\qquad (b_0)_i=f(t_i).\n$$","**Proposition 28. Observable table reconstruction.** Assume the full behavior rank is at most $r$ and the displayed $r\\times r$ matrix $H$ is invertible. Then the full rank is exactly $r$, and","$$\n\\boxed{B_s=G_sH^{-1},\\qquad \\lambda=e_1^\\top,\\qquad b_0=(f(t_i))_i}\n$$","is a minimal realization of every word probability. Its state after history $h$ is","$$\nb(h)=(f(ht_i))_i=B_hb_0.\n$$","The normalized state is $b(h)/f(h)$ when $f(h)>0$. Its update uses the formula in Proposition 26, with no hidden-state coordinates supplied to the reconstruction algorithm.","**Proof.** Invertibility gives rank at least $r$, so the assumed upper bound is attained. Evaluation on the selected tests is an isomorphism $V\\to\\mathbb R^r$, because it maps the chosen basis of history columns to the invertible matrix $H$. The shifted images of those basis columns are precisely the columns of $G_s$, so $B_sH=G_s$. The first selected test evaluates the mass. Initial evaluation gives $b_0$. Proposition 27 supplies correctness for every history, not only the selected columns. $\\square$","of.** Invertibility gives rank at least $r$, so the assumed upper bound is attained. Evaluation on the selected tests is an isomorphism $V\\to\\mathbb R^r$, because it maps the chosen basis of history columns to the invertible matrix $H$. The shifted images of those basis columns are precisely the columns of $G_s$, so $B_sH=G_s$. The first selected test evaluates the mass. Initial evaluation gives $b_0$. Proposition 27 supplies correctness for every history, not only the selected columns. $\\square$","on on the selected tests is an isomorphism $V\\to\\mathbb R^r$, because it maps the chosen basis of history columns to the invertible matrix $H$. The shifted images of those basis columns are precisely the columns of $G_s$, so $B_sH=G_s$. The first selected test evaluates the mass. Initial evaluation gives $b_0$. Proposition 27 supplies correctness for every history, not only the selected columns. $\\square$","There are two ways to justify coverage in this statement. One can know that the selected row and column families span the true behavior space. Alternatively, a justified upper bound $r$ on the process rank, together with the invertible $r\\times r$ block, proves that coverage. The observed invertibility by itself only gives a lower bound. An unknown higher-rank process can agree on a finite table and differ later.","In the canonical behavior space $V$, define","$$\n\\mathcal C=\\{v\\in V:v(t)\\geq0\\text{ for every finite test }t\\}.\n$$","**Proposition 29. Canonical probability cone.** The set $\\mathcal C$ is a closed convex pointed cone that spans $V$. Every $B_s$ preserves it. The mass functional is strictly positive on its nonzero elements, and","$$\n\\sum_o\\lambda B_{a,o}=\\lambda.\n$$","$$\n\\mathcal K=\\{v\\in\\mathcal C:\\lambda v=1\\}\n$$","is a compact convex probability-valid state space. Every positive-probability branch maps $\\mathcal K$ into itself by normalized updating. In the core-test coordinates of Proposition 28, $\\mathcal K\\subseteq\\{1\\}\\times[0,1]^{r-1}$.","**Proof.** Each test evaluation is a continuous linear functional on finite-dimensional $V$, so the cone is closed and convex. If both $v$ and $-v$ belong to it, every test value is zero and $v=0$. It spans $V$ because it contains every column $v_h$. Also $(B_sv)(t)=v(st)$, proving invariance.","The consistency law $\\sum_ov(t(a,o))=v(t)$ holds for each history column and hence throughout $V$. Iterating it along any fixed action schedule expresses $v(\\epsilon)$ as a sum of nonnegative values for all observation words on that schedule. If $v(\\epsilon)=0$, each summand is zero, for every schedule and length, so $v=0$. The same sums show $0\\leq v(t)\\leq1$ when $\\lambda v=1$. They also give mass conservation. The base is closed and bounded in the finite core coordinates and thus compact. Invariance and positive normalization prove the posterior assertion. $\\square$","as a sum of nonnegative values for all observation words on that schedule. If $v(\\epsilon)=0$, each summand is zero, for every schedule and length, so $v=0$. The same sums show $0\\leq v(t)\\leq1$ when $\\lambda v=1$. They also give mass conservation. The base is closed and bounded in the finite core coordinates and thus compact. Invariance and positive normalization prove the posterior assertion. $\\square$","A practical sufficient certificate can instead use a known closed cone $C_0$ with nonempty normalized base, a strictly positive mass functional, $b_0\\in C_0$, $B_sC_0\\subseteq C_0$, and mass conservation. A nonnegative hidden-state realization supplies one after a coordinate change. Such a certificate may use information unavailable to an observable-only learner and must be labeled accordingly.","Negative entries of $B_s$ do not, by themselves, prove invalidity. Coordinates can be signed while the represented state remains in the correct cone. Conversely, clipping negative coefficients can destroy the operator identities. Validity belongs to the pair consisting of the operators and their admissible cone, not to a coordinatewise heuristic.","Let $P_1,\\ldots,P_m$ be a complete family of linear packet maps on a predictive space, with $\\sum_iP_i=I$. Algebraic reconstruction gives $z=\\sum_iP_iz$. It does not imply that $P_iz$ is an unnormalized probability state.","**Proposition 30. Positive packet decomposition and its boundary.** Suppose a cone $C$ has a strictly positive mass functional $\\lambda$. If every packet map preserves $C$, then each normalized $z\\in C$ decomposes as a genuine mixture","$$\nz=\\sum_{i:w_i>0} w_i z_i,\\qquad\nw_i=\\lambda P_iz\\geq0,\\qquad\nz_i=P_iz/w_i,\\qquad\\sum_iw_i=1.\n$$","A zero weight has $P_iz=0$. In the standard cone $\\mathbb R^n_+$, however, a complete family of cone-preserving idempotents must consist of diagonal coordinate-selection matrices. In particular, a nontrivial Fourier or Haar packet family cannot generally be reinterpreted as a mixture of probability states merely because it is orthogonal and complete.","**Proof.** Cone preservation and strict positivity give the weights and the zero-weight claim. Apply $\\lambda$ to the complete decomposition to obtain their sum. Normalization of each nonzero term gives the mixture.","On $\\mathbb R^n_+$, cone preservation is entrywise nonnegativity. Since the matrices sum to $I$, every off-diagonal entry in every matrix is zero. Idempotence forces each diagonal entry to be zero or one, and completeness makes the selections disjoint and exhaustive. $\\square$","For a two-state belief $p=(3/4,1/4)^\\top$, the Haar detail projector","$$\nP_{\\mathrm d}=\\frac12\\begin{pmatrix}1&-1\\\\-1&1\\end{pmatrix}\n$$","produces $(1/4,-1/4)^\\top$. It is a valid signed detail packet, but not a subprobability distribution. Its total mass is zero despite being nonzero, which cannot happen inside a cone with a strictly positive mass functional.","**Proposition 31. Missing-intervention nonidentifiability.** There are finite controlled stochastic models that agree on every possible observation record generated using only action $a$, but disagree under another action $b$. Therefore even infinitely many passive records confined to $a$ cannot identify the unrestricted controlled model.","**Proof.** Use one hidden state and two observations. In both models, action $a$ gives probabilities $(1/2,1/2)$. Under $b$, let the first model give $(1,0)$ and the second $(0,1)$. Every length-$k$ record under $a$ has probability $2^{-k}$ in both models. Their first observation under $b$ differs with probability one. $\\square$","This is a coverage obstruction, not an optimization problem. It remains even when the hidden dimension is known exactly. Richer transforms of the same action-$a$ observations do not create data about $b$.","Action selection based on past observations requires a further distinction. Under a known policy $\\pi$, an action/observation word has joint probability","$$\n\\Pr_\\pi(w)=f(w)\\prod_{j=1}^{|w|}\\pi(a_j\\mid h_{j-1}),\n$$","Let $X\\in\\mathbb R^{m\\times n}$ have rank $r\\geq1$, let $s=\\sigma_r(X)>0$, and observe $Y=X+E$ with $\\|E\\|_2\\leq\\varepsilon<s/2$. Let $P$ project onto $\\operatorname{im}X$, and let $\\widehat P$ project onto the leading $r$ left singular vectors of $Y$.","$$\n\\sigma_r(Y)\\geq s-\\varepsilon,\\qquad\n\\sigma_{r+1}(Y)\\leq\\varepsilon,\\qquad\n\\|P-\\widehat P\\|_2\\leq\\frac{2\\varepsilon}{s}<1.\n$$","The $(r+1)$st singular value is interpreted as zero when $r$ is the maximum possible rank. If $U$ and $\\widehat U$ are orthonormal bases of the two ranges, then","$$\n\\sigma_{\\min}(\\widehat U^\\top U)\n\\geq\\sqrt{1-(2\\varepsilon/s)^2}.\n$$","Thus $x\\mapsto\\widehat U^\\top x$ is injective on the true column space. Its inverse on that space is","$$\ny\\longmapsto U(\\widehat U^\\top U)^{-1}y,\n$$","with norm at most $[1-(2\\varepsilon/s)^2]^{-1/2}$.","**Proof.** The min-max characterization of singular values and the triangle inequality imply that adding $E$ changes each singular value by at most $\\varepsilon$. The leading-$r$ singular subspace gives $\\|(I-\\widehat P)Y\\|_2=\\sigma_{r+1}(Y)\\leq\\varepsilon$. Hence","$$\n\\|(I-\\widehat P)X\\|_2\\leq2\\varepsilon,\n\\qquad\n\\|(I-\\widehat P)P\\|_2\n=\\|(I-\\widehat P)XX^\\dagger\\|_2\\leq2\\varepsilon/s.\n$$","For equal-rank orthogonal projections, the latter norm equals $\\|P-\\widehat P\\|_2$. To see the equality, diagonalize $U^\\top\\widehat U$ by singular value decomposition. Each nontrivial paired principal plane contributes a sine to both norms, while common directions contribute zero. On the same planes the singular values of $\\widehat U^\\top U$ are the corresponding cosines. This proves the last bound and the inverse formula. $\\square$","jections, the latter norm equals $\\|P-\\widehat P\\|_2$. To see the equality, diagonalize $U^\\top\\widehat U$ by singular value decomposition. Each nontrivial paired principal plane contributes a sine to both norms, while common directions contribute zero. On the same planes the singular values of $\\widehat U^\\top U$ are the corresponding cosines. This proves the last bound and the inverse formula. $\\square$","A threshold $\\tau$ with $\\varepsilon<\\tau<s-\\varepsilon$ recovers the rank of this finite matrix. The unknown true gap must be justified or replaced by a valid confidence interval. Rank thresholds do not exclude modes absent from the selected tests or unexcited by the selected histories.","The inverse formula contains the unknown true basis $U$. It proves identifiability and conditioning relative to that space, not that a learner already knows a perfect decoder. When using the estimated subspace as coordinates for a realization, all boundary vectors, symbol operators and probability constraints must be transformed together. Signed singular-vector coordinates are not new independent observations.","Let $G_s=B_sH$ with $s_H=\\sigma_{\\min}(H)>0$. Observe","$$\n\\widetilde H=H+E,\\qquad\n\\widetilde G_s=G_s+F_s,\\qquad\n\\|E\\|_2<s_H.\n$$","**Proposition 33. Noisy core-block realization.** The estimate $\\widehat B_s=\\widetilde G_s\\widetilde H^{-1}$ satisfies","$$\n\\widehat B_s-B_s=(F_s-B_sE)\\widetilde H^{-1},\n$$","$$\n\\boxed{\n\\|\\widehat B_s-B_s\\|_2\n\\leq\\frac{\\|F_s\\|_2+\\|B_s\\|_2\\|E\\|_2}{s_H-\\|E\\|_2}.\n}\n$$","$$\n\\|\\widehat B_s-B_s\\|_2\n\\leq\n\\frac{\\|F_s\\|_2+\\|\\widehat B_s\\|_2\\|E\\|_2}{s_H}.\n$$","If $\\widetilde s=\\sigma_{\\min}(\\widetilde H)$ is known and $\\|E\\|_2\\leq e<\\widetilde s$, the last denominator may be replaced by $\\widetilde s-e$.","**Proof.** Since $\\widetilde H$ is invertible, subtract $B_s$ from its fitted expression and collect the terms. The least singular value of $\\widetilde H$ is at least $s_H-\\|E\\|_2$. For the second identity, use","$$\n(\\widehat B_s-B_s)H=F_s-\\widehat B_sE\n$$","and multiply by $H^{-1}$. Finally $s_H\\geq\\widetilde s-e$ by singular-value perturbation. $\\square$","When the first test is empty, the mass row $\\lambda=e_1^\\top$ is fixed rather than estimated. Set the first component of the initial vector to one as required by $f(\\epsilon)=1$. Errors in other initial test probabilities remain initial-state error. This convention removes an avoidable normalizer-estimation ambiguity but does not enforce cone preservation or mass conservation for the fitted symbol family.","Let $C$ be a probability cone with compact normalized base $K$ and mass row $\\lambda$. Work in a declared norm and its dual. Suppose true normalized states have norm at most $R$. Let a true branch have","$$\nu=B_sz,\\qquad p=\\lambda u\\geq\\eta>0,\\qquad z^+=u/p\\in K.\n$$","For a predicted state $\\widehat z$, form $\\widehat u=\\widehat B_s\\widehat z$, $\\widehat p=\\lambda\\widehat u$, and define $d=\\widehat u-u$. Write $e=\\|\\widehat z-z\\|$, $\\|\\widehat B_s\\|\\leq L$, and $\\|\\widehat B_s-B_s\\|\\leq\\epsilon_B$. Then","$$\n\\|d\\|\\leq Le+\\epsilon_BR=:D.\n$$","**Proposition 34. Normalization-aware filtering stability.** If $\\|\\lambda\\|_*D<\\eta$, then $\\widehat p>0$ and","$$\n\\widehat z^+-z^+\n=\\frac{(I-z^+\\lambda)d}{\\widehat p},\n$$","$$\n\\boxed{\n\\|\\widehat z^+-z^+\\|\n\\leq\\frac{(1+R\\|\\lambda\\|_*)D}{\\eta-\\|\\lambda\\|_*D}.\n}\n$$","Alternatively, if each realized fitted denominator is checked to satisfy $\\widehat p_t\\geq\\widehat\\eta_t>0$, the pathwise recursion obeys","$$\ne_{t+1}\\leq\\alpha_te_t+\\beta_t,\\qquad\n\\alpha_t=\\frac{(1+R\\|\\lambda\\|_*)L_t}{\\widehat\\eta_t},\\qquad\n\\beta_t=\\frac{(1+R\\|\\lambda\\|_*)\\epsilon_{B,t}R}{\\widehat\\eta_t}.\n$$","**Proof.** The branch likelihood differs by $\\lambda d$, giving the lower bound on $\\widehat p$. Subtract $u/p$ from $(u+d)/(p+\\lambda d)$ and use $u=pz^+$. The induced norm of $I-z^+\\lambda$ is at most $1+R\\|\\lambda\\|_*$. Applying that identity at each step gives the recursion. $\\square$","Rare evidence can create a large posterior change from a small absolute error. For example, take a true unnormalized vector $(\\eta,0)$ and a perturbed vector $(\\eta,\\eta)$. Their $\\ell^1$ difference is $\\eta$, while their normalized vectors differ by one in $\\ell^1$. Both arise as possible branch outputs of valid probability instruments when $0<\\eta\\leq1/2$, after assigning the remaining mass to another branch. There is no uniform posterior Lipschitz constant independent of a probability margin.","Fix positive integers $M$ and $N$ and choose $M$ finite words before estimation. For each word, reset independently to the same initial distribution, execute its prescribed action sequence, and record the indicator that all observations match. Obtain $N$ independent trials per word. Dependencies between different words' estimates are allowed, but the $N$ trials used for each individual mean must be independent with the declared success probability.","**Proposition 35. Reset-experiment probability certificate.** Let $\\widehat f(w)$ be the sample mean for each selected word. For $0<\\delta<1$, with probability at least $1-\\delta$, all $M$ estimates satisfy","$$\n|\\widehat f(w)-f(w)|\\leq\n\\epsilon_N,\\qquad\n\\epsilon_N=\\sqrt{\\frac{\\log(2M/\\delta)}{2N}}.\n$$","If the selected words cover every entry of $H$, every $G_s$, and $b_0$, then on the same event","$$\n\\|\\widetilde H-H\\|_2\\leq r\\epsilon_N,\\qquad\n\\|\\widetilde G_s-G_s\\|_2\\leq r\\epsilon_N,\\qquad\n\\|\\widetilde b_0-b_0\\|_2\\leq\\sqrt r\\epsilon_N.\n$$","**Proof.** For a centered Bernoulli variable $X-p$, write $g(t)=\\log\\mathbb E\\exp(t(X-p))$. Then $g(0)=g'(0)=0$, and $g''(t)$ is the variance of a Bernoulli under exponential tilting, at most $1/4$. Integrating twice gives $g(t)\\leq t^2/8$. Independence and the exponential Markov inequality give an upper-tail bound $\\exp(-2N\\epsilon^2)$ after optimizing $t=4\\epsilon$. The lower tail has the same bound. A union bound over the $M$ words proves the simultaneous result. The spectral norm is bounded by the Frobenius norm, yielding the matrix and vector estimates. $\\square$","oulli variable $X-p$, write $g(t)=\\log\\mathbb E\\exp(t(X-p))$. Then $g(0)=g'(0)=0$, and $g''(t)$ is the variance of a Bernoulli under exponential tilting, at most $1/4$. Integrating twice gives $g(t)\\leq t^2/8$. Independence and the exponential Markov inequality give an upper-tail bound $\\exp(-2N\\epsilon^2)$ after optimizing $t=4\\epsilon$. The lower tail has the same bound. A union bound over the $M$ words proves the simultaneous result. The spectral norm is bounded by the Frobenius norm, yielding the matrix and vector estimates. $\\square$","onential tilting, at most $1/4$. Integrating twice gives $g(t)\\leq t^2/8$. Independence and the exponential Markov inequality give an upper-tail bound $\\exp(-2N\\epsilon^2)$ after optimizing $t=4\\epsilon$. The lower tail has the same bound. A union bound over the $M$ words proves the simultaneous result. The spectral norm is bounded by the Frobenius norm, yielding the matrix and vector estimates. $\\square$","The relation to the realization calculus is particularly direct before normalization. Let $R$ be a linear summary of a hidden probability space, let $\\lambda R=\\mathbf1^\\top$, and let $B_s$ be a proposed reduced branch operator. For a true belief $p$, put","$$\nz=Rp,\\qquad\n\\delta_s=(RM_s-B_sR)p.\n$$","The defect $\\delta_s$ is a vector in the reduced representation, not a probability of model failure. It records the failure of the unnormalized instrument to commute with the summary.","$$\na=\\mathbf1^\\top M_sp>0,\\qquad\nb=\\lambda B_sz>0\n$$","are positive, and let $\\widehat z^+=B_sz/b$. Then","$$\n\\boxed{\nR\\left(\\frac{M_sp}{a}\\right)-\\widehat z^+\n=\\frac{(I-\\widehat z^+\\lambda)\\delta_s}{a},\n\\qquad a-b=\\lambda\\delta_s.\n}\n$$","If $M_s\\neq0$ and both likelihoods and normalized updates agree on every strictly positive belief, then $RM_s=B_sR$. Agreement of normalized updates alone need not force this equality.","**Proof.** The true unnormalized summary is $B_sz+\\delta_s$, with mass $a=b+\\lambda\\delta_s$. Subtract the two normalized vectors and collect terms. Since a nonzero nonnegative branch has positive likelihood on every strictly positive belief, agreement permits multiplication by the common mass. This gives equality of unnormalized summaries on the interior of the probability simplex, whose linear span is the whole hidden vector space. For the last claim, a one-state model can assign different positive branch probabilities while its normalized state always remains the scalar one. $\\square$","s positive likelihood on every strictly positive belief, agreement permits multiplication by the common mass. This gives equality of unnormalized summaries on the interior of the probability simplex, whose linear span is the whole hidden vector space. For the last claim, a one-state model can assign different positive branch probabilities while its normalized state always remains the scalar one. $\\square$","For an orthogonal retained projector $P$ preserving the mass row, put $Q=I-P$, $R=P$ on its image and $B_s=PM_sP$. Then","$$\n\\delta_s=PM_sQp.\n$$","This is the unresolved-to-retained transfer block from Part I. Its posterior effect is not simply its norm: normalization applies the oblique map $I-\\widehat z^+\\lambda$ and divides by the true branch probability. A component proportional to $\\widehat z^+$ changes only the branch mass and disappears from the normalized defect, while still changing likelihood-based inference over mechanisms.","A certificate should therefore retain both $\\lambda\\delta_s$ and the posterior defect. A posterior that looks correct can hide a wrong evidence model. This is the stochastic counterpart of the distinction between reconstruction and predictive closure.","$$\nT_0=\\begin{pmatrix}3/4&1/4\\\\1/4&3/4\\end{pmatrix},\\qquad\nT_1=\\begin{pmatrix}1/4&2/3\\\\3/4&1/3\\end{pmatrix},\\qquad\np_0=\\begin{pmatrix}2/3\\\\1/3\\end{pmatrix}.\n$$","$$\nO_0=\\operatorname{diag}(1/4,3/4),\\qquad\nO_1=\\operatorname{diag}(3/4,1/4),\\qquad\nM_{a,o}=O_oT_a.\n$$","Choose tests $(\\epsilon,(0,1))$ and histories $(\\epsilon,(0,1))$. The observable tables give","$$\nH=\\begin{pmatrix}1&13/24\\\\13/24&5/16\\end{pmatrix},\\qquad\n\\det H=11/576,\\qquad\nb_0=\\begin{pmatrix}1\\\\13/24\\end{pmatrix}.\n$$","The selected coordinate $q$ is the probability of seeing observation one after choosing action zero. In the hidden realization,","$$\nR=\\begin{pmatrix}1&1\\\\5/8&3/8\\end{pmatrix},\\qquad\nz=Rp=(1,q)^\\top.\n$$","The observable reconstruction $B_s=G_sH^{-1}$ yields","$$\n\\begin{aligned}\nB_{0,0}&=\\begin{pmatrix}1&-1\\\\11/32&-1/4\\end{pmatrix},&\nB_{0,1}&=\\begin{pmatrix}0&1\\\\-3/32&3/4\\end{pmatrix},\\\\[4pt]\nB_{1,0}&=\\begin{pmatrix}5/48&5/6\\\\23/192&5/24\\end{pmatrix},&\nB_{1,1}&=\\begin{pmatrix}43/48&-5/6\\\\37/64&-5/8\\end{pmatrix}.\n\\end{aligned}\n$$","The coefficients include negatives. Nevertheless, every operator preserves the cone $R\\mathbb R^2_+$, because $B_sR=RM_s$. A sufficient normalized base is","$$\nK_0=\\{(1,q)^\\top:3/8\\leq q\\leq5/8\\}.\n$$","This cone certificate uses the known generating model and is checked separately. It is not inferred merely from the finite probability table, and $K_0$ is not claimed to be the maximal canonical base of Proposition 29.","$$\np(1\\mid z,0)=q,\\qquad\nq^+=\\frac{-3/32+(3/4)q}{q}\n=\\frac34-\\frac{3}{32q}.\n$$","Starting at $q_0=13/24$, this gives probability $13/24$ and $q_1=15/26$. If action one then produces observation zero, its probability is","$$\n\\frac5{48}+\\frac56\\frac{15}{26}=\\frac{365}{624},\n$$","$$\nq_2=\\frac{23/192+(5/24)(15/26)}{365/624}=\\frac{599}{1460}.\n$$","The joint probability of this two-step action/observation word is $365/1152$. The reference checks compare these values and every word up to a declared finite test length against the hidden generator in exact rational arithmetic. The proposition, under its rank and coverage assumptions, is what extends the result to all word lengths.","**Proposition 37. Conditional observation-based stochastic completion.** Suppose a causal finite-alphabet signal process has behavior rank at most $r$, and a selected $r\\times r$ history/test block containing the empty test is invertible. Exact values of that block, every one-symbol shifted block, and the initial core-test vector determine a minimal realization of every future experiment probability. Its normalized state is recursively accessible after each positive-probability history, without reconstructing the hidden physical state.","$$\nz'=Tz,\\qquad B_s'=TB_sT^{-1},\\qquad\n\\lambda'=\\lambda T^{-1},\\qquad C'=TC.\n$$","**Proof.** Proposition 28 reconstructs the unique minimal behavior realization, and Proposition 27 gives its conditional interpretation. Proposition 29 supplies probability validity for the exact realization. Substituting the transformed quantities cancels every $T^{-1}T$, including in the scalar normalization. Full packet synthesis is the identity, whereas Proposition 30 characterizes the additional probability-mixture condition. The perturbation conclusions are the stated chain of propositions with their separate assumptions retained. No missing gap, coverage or positivity assumption follows from matrix inversion alone. $\\square$","formed quantities cancels every $T^{-1}T$, including in the scalar normalization. Full packet synthesis is the identity, whereas Proposition 30 characterizes the additional probability-mixture condition. The perturbation conclusions are the stated chain of propositions with their separate assumptions retained. No missing gap, coverage or positivity assumption follows from matrix inversion alone. $\\square$","Fix a real predictive space $E=\\mathbb R^r$, a nonzero row $\\lambda$, and a finite family $B_s$, indexed by $s=(a,o)$. The action and observation alphabets are finite and nonempty. Every declared observation branch is present, including branches represented by the zero operator. Temporal words act from right to left:","$$\nB_{s_1\\cdots s_k}=B_{s_k}\\cdots B_{s_1},\\qquad B_\\epsilon=I.\n$$","$$\n\\lambda\\sum_o B_{a,o}=\\lambda \\quad\\text{for every action }a.\n$$","An initial state has $\\lambda z_0=1$. The proposed probability of a word is $\\lambda B_wz_0$. An invertible coordinate change must transform $z_0$, $B_s$, $\\lambda$ and the admissible cone together, as in Proposition 37.","A **probability-domain certificate** specifies a closed, pointed, generating cone $C\\subset E$, with $\\lambda$ strictly positive on $C\\setminus\\{0\\}$, such that","$$\nz_0\\in C,\\qquad B_sC\\subseteq C.\n$$","Its normalized base is $K=\\{z\\in C:\\lambda z=1\\}$. These conditions imply nonnegative word probabilities and their per-action normalization at every depth. Positive-probability updates remain in $K$. Zero-probability branches have no conditional posterior and must not be assigned one by denominator flooring.","$$\nC_h=\\{z:\\lambda B_wz\\geq0\\text{ for every }|w|\\leq h\\},\n\\qquad C_0=\\{z:\\lambda z\\geq0\\}.\n$$","Each $C_h$ is polyhedral because it is an intersection of finitely many homogeneous half-spaces. It need not be a pointed cone or have a bounded normalized base.","$$\nC_{h+1}=C_h\\cap\\bigcap_s B_s^{-1}C_h,\n\\qquad C_{h+1}\\subseteq C_h.\n$$","Their intersection $C_\\infty$ is the largest cone contained in $C_0$ and invariant under all $B_s$. If an exact equality $C_{h+1}=C_h$ occurs, then $C_h=C_\\infty$ and the finite inequalities certify all future word probabilities. If the representation is observable, meaning that the rows $\\lambda B_w$ span $E^*$, then $C_\\infty$ is pointed and $\\lambda$ is strictly positive on its nonzero elements. If reachable vectors $B_wz_0\\in C_\\infty$ span $E$, the cone is also generating.","nfty$ is the largest cone contained in $C_0$ and invariant under all $B_s$. If an exact equality $C_{h+1}=C_h$ occurs, then $C_h=C_\\infty$ and the finite inequalities certify all future word probabilities. If the representation is observable, meaning that the rows $\\lambda B_w$ span $E^*$, then $C_\\infty$ is pointed and $\\lambda$ is strictly positive on its nonzero elements. If reachable vectors $B_wz_0\\in C_\\infty$ span $E$, the cone is also generating.","contained in $C_0$ and invariant under all $B_s$. If an exact equality $C_{h+1}=C_h$ occurs, then $C_h=C_\\infty$ and the finite inequalities certify all future word probabilities. If the representation is observable, meaning that the rows $\\lambda B_w$ span $E^*$, then $C_\\infty$ is pointed and $\\lambda$ is strictly positive on its nonzero elements. If reachable vectors $B_wz_0\\in C_\\infty$ span $E$, the cone is also generating.","**Proof.** For a first symbol $s$ followed by a word $v$, $B_{sv}=B_vB_s$. The inequalities for all words of length at most $h+1$ are therefore exactly the displayed recurrence. If $z\\in C_\\infty$, every test on $B_sz$ is another test on $z$, so $C_\\infty$ is invariant. Conversely, invariance of a cone contained in $C_0$ makes every successive test nonnegative, hence places it in $C_\\infty$. Equality at one finite step establishes invariance immediately and therefore gives the intersection.","If both $z$ and $-z$ lie in $C_\\infty$, every test row annihilates $z$, so observability gives $z=0$. For $z\\in C_\\infty$ with $\\lambda z=0$, sum word probabilities along any fixed action sequence. Mass conservation makes the sum zero. Every term is nonnegative, so every word test vanishes and again $z=0$. Spanning by reachable cone elements gives the final assertion. $\\square$","Let $V=[v_1\\ \\cdots\\ v_m]\\in\\mathbb R^{r\\times m}$ have full row rank and satisfy","$$\n\\lambda V=\\mathbf1_m^\\top.\n$$","Its columns generate $C=V\\mathbb R_+^m$ and its normalized base is $K=\\operatorname{conv}\\{v_1,\\ldots,v_m\\}$. Finite generation makes the cone closed. Its normalized generators and full row rank make it pointed and generating, with a strictly positive mass functional.","**Proposition 39. Finite positive-lift certificate.** This cone is invariant under $B_s$ if and only if there is an entrywise nonnegative matrix $A_s\\in\\mathbb R^{m\\times m}$ such that","$$\n\\boxed{B_sV=VA_s,\\qquad A_s\\geq0.}\n$$","If these identities hold for every symbol and $z_0=V\\alpha_0$ for $\\alpha_0\\geq0$ with $\\mathbf1^\\top\\alpha_0=1$, then","$$\n\\mathbf1^\\top\\sum_o A_{a,o}=\\mathbf1^\\top,\n\\qquad\n\\lambda B_wz_0=\\mathbf1^\\top A_w\\alpha_0.\n$$","Thus the finite witness defines a nonnegative probability instrument whose image under $V$ is the predictive model. It certifies all word probabilities without enumerating them.","**Proof.** Invariance places each $B_sv_j$ in the conical hull of the columns of $V$. Choose its nonnegative coefficients as column $j$ of $A_s$. Conversely, $B_sV\\alpha=VA_s\\alpha\\in C$ for every $\\alpha\\geq0$. Multiplying the intertwining identity by $\\lambda$ and summing over observations gives","$$\n\\mathbf1^\\top\\sum_oA_{a,o}\n=\\lambda\\sum_oB_{a,o}V\n=\\lambda V=\\mathbf1^\\top.\n$$","Iterating $B_sV=VA_s$ gives $B_wV=VA_w$. Applying the mass row and the initial coefficients proves the word identity. Strict positivity of $\\lambda$ follows because a nonzero conical combination has positive total coefficient mass. $\\square$","The $m$ nonnegative coordinates need not be physical hidden states, and $m$ need not equal the minimal predictive dimension $r$. They are one finite realization witnessing validity. When $V$ has dependent columns, the lift can be nonunique. Those redundant coordinates do not change the predictive state after synthesis.","There is a dual finite test. If a complete inequality description of the same cone is $C=\\{z:Gz\\geq0\\}$, then $GB_sv_j\\geq0$ for every generator $j$ is equivalent to invariance. One may also certify $GB_s=H_sG$ with $H_s\\geq0$. The latter certificate is sufficient by direct multiplication and necessary by the polyhedral cone form of Farkas' lemma [R13]. The reference calculation uses the direct generator test, not an unreported numerical optimization certificate.","Let $\\mathcal U$ be the convex hull of finitely many **complete operator families** $\\{B_s^{(j)}\\}_s$. Every family must satisfy the exact per-action mass identities. The convex coefficient is shared across that family's symbols. A product of independent uncertainty sets can instead be handled by its joint vertices.","**Proposition 40. Robust finite-domain certificate.** For a fixed normalized generator matrix $V$, suppose every vertex family has witnesses","$$\nB_s^{(j)}V=VA_s^{(j)},\\qquad A_s^{(j)}\\geq0.\n$$","Then every family in $\\mathcal U$ preserves $C=V\\mathbb R_+^m$ and is mass conserving. If its initial state lies in $K$, it defines valid finite observation-tree probabilities at every horizon. This remains true when a family in $\\mathcal U$ is selected separately at each history.","For a complete facet description $C=\\{z:Gz\\geq0\\}$, let $g_i^\\top$ be its rows. A sufficient norm-ball certificate around $\\widehat B_s$ is","$$\n\\boxed{\n g_i^\\top\\widehat B_sv_j\n \\geq\\epsilon_s\\|g_i\\|_2\\|v_j\\|_2\n \\quad\\text{for all }s,i,j.\n}\n$$","It covers all perturbations with $\\|B_s-\\widehat B_s\\|_2\\leq\\epsilon_s$ that also satisfy the exact mass constraints.","**Proof.** For convex weights $\\theta_j$, take $A_s=\\sum_j\\theta_jA_s^{(j)}$. The equality and nonnegativity survive convex combination, as does mass conservation. At each history, every selected family maps the current cone state back into the cone. Applying the mass identity at that node proves recursive normalization, regardless of how the family was selected. For the norm ball,","$$\n|g_i^\\top(B_s-\\widehat B_s)v_j|\n\\leq\\epsilon_s\\|g_i\\|_2\\|v_j\\|_2.\n$$","The margin makes every perturbed generator image satisfy all facets. Proposition 39, or the direct generator test, then supplies invariance. $\\square$","Once $V$ is fixed, certificate-constrained fitting is a convex problem. Unknowns include the entries of $\\widetilde B_s$, $A_s$, $\\widetilde z_0$ and $\\alpha_0$. The constraints","$$\n\\begin{gathered}\n\\widetilde B_sV=VA_s,\\quad A_s\\geq0,\n\\quad \\lambda\\sum_o\\widetilde B_{a,o}=\\lambda,\\\\\n\\widetilde z_0=V\\alpha_0,\\quad\\alpha_0\\geq0,\n\\quad\\mathbf1^\\top\\alpha_0=1\n\\end{gathered}\n$$","are linear. Entrywise confidence intervals and fixed-input regression tolerances add linear inequalities. Norm balls add convex constraints. Jointly optimizing $V$ and $A_s$ is different: the product $VA_s$ is then bilinear. No general convexity is claimed for learning the domain and dynamics together.","**Proposition 41. Certified correction and its error budget.** For fixed $V$, every feasible solution of the displayed constraints is an internally valid model. Suppose a center estimate $\\widehat B_s$ and a feasible corrected estimate $\\widetilde B_s$ satisfy","$$\n\\|B_s^*-\\widehat B_s\\|\\leq\\epsilon_s,\n\\qquad\n\\|\\widetilde B_s-\\widehat B_s\\|\\leq\\epsilon_s.\n$$","Then $\\|\\widetilde B_s-B_s^*\\|\\leq2\\epsilon_s$. The same conclusion holds for initial states in their specified norm. If the true model lies in these confidence sets and itself admits the fixed-domain certificate, the convex constraints are feasible. Feasibility alone does not establish that the true model preserves this domain.","tion 39. The error bound is the triangle inequality through the center estimate. Under the final assumptions, insert the true operators, true initial state and their nonnegative witnesses into the constraints. They provide a feasible point. The converse fails because existence of one valid point does not say that every point in a confidence set, or its true member, has the same invariant domain. $\\square$","When $V$ is square and invertible, put $A_s=V^{-1}\\widehat B_sV$. Assume mass conservation has already been imposed. Let $N_s$ be an entrywise positive reference instrument with the same per-action mass identities. For $0\\leq\\eta\\leq1$, the mixture","$$\n\\widetilde A_s=(1-\\eta)A_s+\\eta N_s\n$$","$$\n\\eta\\geq\\max_{s,i,j:(A_s)_{ij}<0}\n\\frac{-(A_s)_{ij}}{(N_s)_{ij}-(A_s)_{ij}}.\n$$","An empty maximum is zero. If every negative entry has magnitude at most $d$ and every reference entry is at least $m_0>0$, the simpler sufficient choice is $\\eta=d/(m_0+d)$. These statements follow by solving each scalar inequality. Mass conservation is unchanged by convex mixing.","This is an explicit model modification, not coefficient clipping or an exact coordinate change. Relative to a true instrument on the same cone, the one-step error is at most $(1-\\eta)\\epsilon+2\\eta$ in the summed base-norm metric defined next, when the original error was at most $\\epsilon$. The factor two follows because two valid instruments each distribute unit total mass. The reference example retains the repair amount as part of its approximation record.","Assume $C$ is a closed, pointed, generating cone and $\\lambda$ is strictly positive on its nonzero elements. Its base $K$ is compact. Define the base norm","$$\n\\|x\\|_C=\\inf\\{\\lambda u+\\lambda v:x=u-v,\\ u,v\\in C\\}.\n$$","The infimum is finite because the cone is generating. Compactness of $K$ makes this a genuine norm. Its unit ball is $\\operatorname{conv}(K\\cup(-K))$. In particular $\\|u\\|_C=\\lambda u$ on $C$, and $|\\lambda x|\\leq\\|x\\|_C$. On the ordinary probability orthant this is the $\\ell^1$ norm. These are standard ordered-space constructions [R15], used here to avoid mixing Euclidean coefficient magnitude with probability mass.","**Proposition 42. Instrument contraction before normalization.** If $\\{B_{a,o}\\}_o$ is cone preserving and mass conserving for a fixed action, then","$$\n\\boxed{\\sum_o\\|B_{a,o}x\\|_C\\leq\\|x\\|_C.}\n$$","For a polyhedral base $K=\\operatorname{conv}\\{v_j\\}$, define the discrepancy between two instruments by","$$\n\\epsilon_a=\\sup_{z\\in K}\\sum_o\n\\|(B_{a,o}-\\widetilde B_{a,o})z\\|_C.\n$$","The supremum equals the maximum of the same expression at the finitely many generators $v_j$. When both instruments are valid on $C$, $\\epsilon_a\\leq2$.","**Proof.** For any $x=u-v$ with $u,v\\in C$, cone preservation gives","$$\n\\sum_o\\|B_{a,o}x\\|_C\n\\leq\\sum_o\\lambda B_{a,o}(u+v)\n=\\lambda u+\\lambda v.\n$$","Take the infimum. For the generator statement, the discrepancy is convex in $z$. At a convex combination its value is at most the same combination of generator values, hence at most their maximum. The generators themselves belong to $K$, proving equality. Finally, for $z\\in K$, the triangle inequality bounds the sum by the total masses of both instruments, which are one each. $\\square$","$$\n\\|x\\|_C=\\min_{a,b\\geq0}\n\\{\\mathbf1^\\top(a+b):V(a-b)=x\\}.\n$$","If $V$ is square and invertible, this reduces to $\\|V^{-1}x\\|_1$. The exact worked interval certificate is of this type. With dependent generators, choosing one pseudoinverse does not in general compute the base norm.","Let two models share $(C,\\lambda)$ and have initial states $z_0,\\widetilde z_0\\in K$. At each time use the same policy, a specified function or randomization depending only on the observed history. Thus on any common history the action probabilities are identical in the two comparisons. This is not a comparison of two policies optimized separately for the two models.","**Proposition 43. Observation-tree total-variation bound.** If both instruments preserve $C$ and conserve mass, and if every action at time $t$ has the discrepancy in Proposition 42 bounded by $\\epsilon_t$, then the distributions $P_T,\\widetilde P_T$ of length-$T$ observation records under the common policy satisfy","$$\n\\boxed{\n\\|P_T-\\widetilde P_T\\|_{\\mathrm{TV}}\n\\leq\\min\\left(1,\\frac12\\|z_0-\\widetilde z_0\\|_C\n+\\frac12\\sum_{t=0}^{T-1}\\epsilon_t\\right).\n}\n$$","**Proof.** Let $u_h,\\widetilde u_h\\in C$ be the unnormalized states on each history at depth $t$, including the common policy factors. Their masses sum to one in each model. Define $D_t=\\sum_h\\|u_h-\\widetilde u_h\\|_C$. At a history with selected action $a$, write","$$\nB_{a,o}u_h-\\widetilde B_{a,o}\\widetilde u_h\n=B_{a,o}(u_h-\\widetilde u_h)\n+(B_{a,o}-\\widetilde B_{a,o})\\widetilde u_h.\n$$","Sum the first term over observations and use Proposition 42. When $\\lambda\\widetilde u_h>0$, normalize only to apply the definition of $\\epsilon_t$ to the second term. Its sum is bounded by $\\epsilon_t\\lambda\\widetilde u_h$. If the mass is zero, strict positivity implies $\\widetilde u_h=0$. Summing histories yields $D_{t+1}\\leq D_t+\\epsilon_t$. Averaging over common policy probabilities gives the same inequality for randomized actions. Finally each word-probability difference is bounded by its state difference because $|\\lambda x|\\leq\\|x\\|_C$. Total variation is half the sum of those absolute differences. $\\square$","nition of $\\epsilon_t$ to the second term. Its sum is bounded by $\\epsilon_t\\lambda\\widetilde u_h$. If the mass is zero, strict positivity implies $\\widetilde u_h=0$. Summing histories yields $D_{t+1}\\leq D_t+\\epsilon_t$. Averaging over common policy probabilities gives the same inequality for randomized actions. Finally each word-probability difference is bounded by its state difference because $|\\lambda x|\\leq\\|x\\|_C$. Total variation is half the sum of those absolute differences. $\\square$","ilde u_h$. If the mass is zero, strict positivity implies $\\widetilde u_h=0$. Summing histories yields $D_{t+1}\\leq D_t+\\epsilon_t$. Averaging over common policy probabilities gives the same inequality for randomized actions. Finally each word-probability difference is bounded by its state difference because $|\\lambda x|\\leq\\|x\\|_C$. Total variation is half the sum of those absolute differences. $\\square$","Let $P,Q$ be probability distributions on a finite record space. A record event $H$ might be a particular observed prefix. Conditional continuation laws are obtained by restriction to $H$ and normalization.","**Proposition 44. Sharp event-probability dependence of conditioning.** If $\\|P-Q\\|_{\\mathrm{TV}}\\leq\\delta$, $p=P(H)>0$ and $q=Q(H)>0$, then","$$\n\\boxed{\\|P(\\cdot\\mid H)-Q(\\cdot\\mid H)\\|_{\\mathrm{TV}}\n\\leq\\min(1,\\delta/p).}\n$$","When $p>\\delta$, positivity of $q$ follows automatically. The factor $1/p$ cannot be removed in general.","**Proof.** Write the subprobability restrictions on $H$ as $u,v$, of masses $p,q$. On the complementary event, the sum of absolute probability differences is at least $|p-q|$. Therefore","$$\n\\|u-v\\|_1+|p-q|\\leq2\\delta.\n$$","Adding and subtracting $v/p$ gives","$$\n\\frac12\\|u/p-v/q\\|_1\n\\leq\\frac{\\|u-v\\|_1+|p-q|}{2p}\n\\leq\\delta/p.\n$$","Also $|p-q|\\leq\\delta$, establishing the positivity condition. For sharpness, put total mass $p$ on an event with two outcomes. Let $P$ assign all of it to the first, and let $Q$ move mass $\\delta\\leq p$ to the second, leaving the complement unchanged. The unconditional total variation is $\\delta$ and the conditional distance is $\\delta/p$. $\\square$","**Proposition 45. Rational finite-certificate search.** Suppose $\\lambda$, $B_s$ and $z_0$ are rational and satisfy the exact mass identities. There is an enumeration procedure that returns a valid finite certificate whenever a certificate of Proposition 39 exists with a rational normalized full-row-rank generator matrix $V$. The procedure is sound whenever it returns. No termination guarantee is asserted when no such rational certificate exists.","**Proof.** Enumerate all finite rational matrices $V$ in an order that eventually includes every size and every bounded numerator and denominator range. Retain those of full row rank with $\\lambda V=\\mathbf1^\\top$. For each retained matrix, the requirements $B_sV=VA_s$, $A_s\\geq0$, $V\\alpha_0=z_0$, $\\alpha_0\\geq0$ and $\\mathbf1^\\top\\alpha_0=1$ are rational linear feasibility problems. A nonempty rational polyhedron has a rational feasible point, as follows from elimination of rational linear inequalities. These problems can therefore be decided by exact rational linear programming, or witnesses can be enumerated in a dovetailed rational search. Recheck every returned equality and sign exactly. Proposition 39 proves soundness. If a rational ge","**Proof.** Enumerate all finite rational matrices $V$ in an order that eventually includes every size and every bounded numerator and denominator range. Retain those of full row rank with $\\lambda V=\\mathbf1^\\top$. For each retained matrix, the requirements $B_sV=VA_s$, $A_s\\geq0$, $V\\alpha_0=z_0$, $\\alpha_0\\geq0$ and $\\mathbf1^\\top\\alpha_0=1$ are rational linear feasibility problems. A nonempty rational polyhedron has a rational feasible point, as follows from elimination of rational linear inequalities. These problems can therefore be decided by exact rational linear programming, or witnesses can be enumerated in a dovetailed rational search. Recheck every returned equality and sign exactly. Proposition 39 proves soundness. If a rational generator certificate exists, its finite representation is eventually considered and its feasible witnesses are found. $\\square$","ination of rational linear inequalities. These problems can therefore be decided by exact rational linear programming, or witnesses can be enumerated in a dovetailed rational search. Recheck every returned equality and sign exactly. Proposition 39 proves soundness. If a rational generator certificate exists, its finite representation is eventually considered and its feasible witnesses are found. $\\square$","The observable tables of Section 38 determine the two-dimensional operators $B_s$ and initial vector $(1,13/24)^\\top$. The new reference procedure receives only these fitted operators and that vector. It enumerates intervals with rational endpoints in $[0,1]$ of reduced denominator at most eight, ordered by increasing width and then by endpoints. The coordinate restriction $[0,1]$ comes from its declared test-probability meaning.","$$\n\\boxed{K=\\{(1,q)^\\top:2/5\\leq q\\leq3/5\\}.}\n$$","$$\nV=\\begin{pmatrix}1&1\\\\2/5&3/5\\end{pmatrix},\n\\qquad \\alpha_0=\\begin{pmatrix}7/24\\\\17/24\\end{pmatrix},\n\\qquad A_s=V^{-1}B_sV.\n$$","Every entry of every $A_s$ is at least $1/64$. The full per-action column sums are exactly one. These rational identities certify all word probabilities of the fitted model, not only the words used in the fit. The certificate was not supplied to the search, and no hidden generator is imported by the standalone example.","The positive slack gives a robust statement: any perturbation of the lifted entries by at most $1/64$, coupled to retain the exact per-action column sums, remains nonnegative. Interior perturbations of size $1/128$ are used in the worked comparison. Repeated composition is valid for all horizons, while Proposition 43 bounds the changed observation law for each fixed horizon.","Let $z=(t,x,y)^\\top$, $\\lambda=(1,0,0)$, $z_0=(1,1,0)^\\top$, and","$$\nR=\\begin{pmatrix}3/5&-4/5\\\\4/5&3/5\\end{pmatrix},\n\\qquad T=\\operatorname{diag}(1,R).\n$$","$$\nB_{\\mathrm r,0}=T,\\qquad B_{\\mathrm r,1}=0.\n$$","$$\nB_{\\mathrm p,0}=z_0(1/2,1/2,0),\\qquad\nB_{\\mathrm p,1}=z_0(1/2,-1/2,0).\n$$","The rotate outcome zero is certain. The probe has probabilities $(t+x)/2$ and $(t-x)/2$ and resets every nonzero branch to $z_0$ after normalization. Both actions conserve the mass $t$.","$$\nC_{\\mathrm L}=\\{(t,x,y):t\\geq\\sqrt{x^2+y^2}\\}\n$$","is probability valid for these operators. It is the canonical all-test cone. Every closed convex invariant cone containing $z_0$ and admitting the nonnegative mass interpretation equals $C_{\\mathrm L}$. In particular no finite normalized generator certificate exists for this exact model. Its reachable and observable predictive dimension is three.","**Proof.** Rotation preserves $t$ and $x^2+y^2$. On $C_{\\mathrm L}$ both probe effects are nonnegative and each probe image is a nonnegative multiple of $z_0$. Thus the displayed cone is invariant and valid.","Write the rotation angle as $\\theta$. It is not a rational multiple of $2\\pi$. Otherwise $\\zeta=e^{i\\theta}$ would be a root of unity, so $\\zeta+\\zeta^{-1}=6/5$ would be a rational algebraic integer. Every rational algebraic integer is an integer, by the rational-root argument for a monic integer polynomial, contradicting $6/5\\notin\\mathbb Z$. The forward powers of an irrational circle rotation are dense. Compactness supplies powers approaching the identity, so their closure contains the inverse rotation and is a closed subgroup. An infinite closed subgroup has arbitrarily small positive angles. Integer multiples of such angles approximate every point on the circle, so that subgroup is the entire circle.","A state passing every test must pass a probe after any number $k$ of rotations. Hence","$$\nt\\pm e_1^\\top R^k(x,y)^\\top\\geq0\n\\quad\\text{for every }k\\geq0.\n$$","Density gives $t\\geq\\sqrt{x^2+y^2}$. The reverse inclusion was already proved, so the canonical cone is exactly $C_{\\mathrm L}$.","Any invariant cone with nonnegative mass lies inside this canonical cone, by Proposition 38. If it is closed, convex and contains $z_0$, it contains the orbit $T^kz_0$ and the closed convex hull of that orbit. On $t=1$, this hull is the full unit disk. Conical scaling gives all of $C_{\\mathrm L}$. The unit disk has infinitely many extreme points, so it is not the convex hull of finitely many points.","Finally the reachable columns $z_0,Tz_0,T^2z_0$ are independent. The mass row and probe rows before zero and one rotations span the dual space because the rotation has nonzero sine. Reachability and observability establish predictive dimension three. $\\square$","Let $P_n$ be the regular $n$-gon inscribed in the unit disk, with a vertex at $(1,0)$ and $n\\geq3$. Replace the rotate branch by","$$\nT_\\alpha=\\operatorname{diag}(1,\\alpha R),\\qquad 0\\leq\\alpha<1,\n$$","$$\n\\alpha\\leq\\cos(\\pi/n),\n$$","then the cone over $P_n$ is invariant under all modified operators and contains $z_0$. Consequently the modified model admits a finite positive lift. The original and modified models are both valid on the larger cone $C_{\\mathrm L}$, and for every common policy their length-$T$ laws satisfy","$$\n\\boxed{\\|P_T-P_T^{(\\alpha)}\\|_{\\mathrm{TV}}\n\\leq\\min(1,T(1-\\alpha)/2).}\n$$","The angle-independent sufficient condition can be improved for a fixed polygon and rotation. For the rational diamond $|x|+|y|\\leq t$, the present rotation is invariant after attenuation whenever $\\alpha\\leq5/7$. Thus $\\alpha=2/3$ gives an entirely rational four-generator certificate.","**Proof.** The regular polygon contains the disk of radius $\\cos(\\pi/n)$ and lies in the unit disk. Every $\\alpha Rv$ with $v\\in P_n$ has Euclidean norm at most $\\alpha$, hence lies in $P_n$ under the stated condition. The probe effects remain nonnegative because $P_n$ lies in the disk, and their normalized images are the retained vertex $(1,0)$.","The base norm of $C_{\\mathrm L}$ is","$$\n\\|(t,x,y)\\|_{C_{\\mathrm L}}=\\max(|t|,\\sqrt{x^2+y^2}).\n$$","Indeed the convex hull of the unit disks in the planes $t=1$ and $t=-1$ is the cylinder with this unit ball. The rotate-branch discrepancy on a normalized state is at most $1-\\alpha$ in this norm. Probe discrepancies are zero. Proposition 43 gives the law bound. For the diamond, the induced $\\ell^1$ norm of $R$ is $7/5$, so $\\alpha\\|R\\|_1\\leq1$ proves invariance. $\\square$","For a target attenuation loss $0<\\eta\\leq1$, choosing $\\alpha=1-\\eta$ and sufficiently many polygon vertices satisfies the certificate. The inequality $1-\\cos u\\leq u^2/2$ shows that $n\\geq\\pi/\\sqrt{2\\eta}$ is sufficient, with an integer ceiling and $n\\geq3$. This is a constructive upper bound on certificate size, not a lower bound or a proof of optimality.","The distinction is exact. At $\\alpha=1$ there is no finite polyhedral certificate for this model. At $\\alpha<1$ a sufficiently fine polygon works, at the cost of a declared change in the observation law. The reference figures show that tradeoff rather than labeling the approximation an identical model.","**Proposition 48. Sequential confidence accounting for certified proposals.** At stage $n$, let the proposal be measurable from previous data, and let a verification procedure return a set $\\mathcal U_n$ for the true model in the declared identifiable coordinate system. Assume conditional coverage","$$\n\\Pr(\\text{true model}\\notin\\mathcal U_n\\mid\\text{previous data})\n\\leq\\delta_n,\n\\qquad\\sum_{n\\geq1}\\delta_n\\leq\\delta.\n$$","Then with probability at least $1-\\delta$, every returned uncertainty set covers its true model. On this event, any data-selected domain with an exactly verified robust certificate for **all** members of its corresponding $\\mathcal U_n$ is also valid for the true model, provided the true initial state is covered by the certified base. The conclusion remains valid at any stopping time chosen from the recorded results.","**Proof.** Take expectations of the conditional failure bounds, then apply the countable union bound. Off the union of failure events, all coverages hold simultaneously. Robust invariance over an entire set then applies to its true member by deterministic implication, independently of how its domain was chosen. A stopped stage is one of the already covered stages. $\\square$","For example, $\\delta_n=\\delta 2^{-n}$ spends a finite total confidence budget. Fresh reset trials for a proposal fixed by past data can use Proposition 35 conditionally, with this stage budget. An exact witness needs no statistical correction for how many candidate cones were tried. The coverage statement does need correction for repeatedly constructing confidence sets or inspecting post-selected data.","**Proposition 49. Conditional learning with an all-horizon probability certificate.** Suppose the observable identification assumptions of Proposition 37 hold for a declared finite-rank controlled process. Suppose a justified confidence procedure produces an operator and initial-state set containing the true realization on an event of probability at least $1-\\delta$. Assume the per-action mass constraints are exact.","If a fitted realization admits a checked finite witness of Proposition 39, it is an internally valid stochastic model at every horizon. If the witness is robust over the whole confidence set and covers its initial states, then on the coverage event the same probability domain is valid for both the fitted and true models. When their summed base-norm instrument discrepancies are at most $\\epsilon_t$ and their initial discrepancy is at most $e_0$, every common policy has the bound","nite witness of Proposition 39, it is an internally valid stochastic model at every horizon. If the witness is robust over the whole confidence set and covers its initial states, then on the coverage event the same probability domain is valid for both the fitted and true models. When their summed base-norm instrument discrepancies are at most $\\epsilon_t$ and their initial discrepancy is at most $e_0$, every common policy has the bound","$$\n\\|P_T-\\widehat P_T\\|_{\\mathrm{TV}}\n\\leq\\min\\left(1,\\frac{e_0+\\sum_{t<T}\\epsilon_t}{2}\\right).\n$$","nd the confidence procedure supplies the stated event, not an inferred event of its own. Proposition 39 supplies internal validity for the fitted model. Proposition 40 extends the same domain to every member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","nfidence procedure supplies the stated event, not an inferred event of its own. Proposition 39 supplies internal validity for the fitted model. Proposition 40 extends the same domain to every member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","procedure supplies the stated event, not an inferred event of its own. Proposition 39 supplies internal validity for the fitted model. Proposition 40 extends the same domain to every member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","stated event, not an inferred event of its own. Proposition 39 supplies internal validity for the fitted model. Proposition 40 extends the same domain to every member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","vent of its own. Proposition 39 supplies internal validity for the fitted model. Proposition 40 extends the same domain to every member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","oposition 39 supplies internal validity for the fitted model. Proposition 40 extends the same domain to every member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","y member of the uncertainty set, hence to its true member on that event. Propositions 42 and 43 give the total-variation bound, and Proposition 44 gives the conditional bound. For a coordinate change $T$, use $V'=TV$, $B_s'=TB_sT^{-1}$ and $\\lambda'=\\lambda T^{-1}$. Then $B_s'V'=V'A_s$ and $\\lambda'V'=\\mathbf1^\\top$ with the unchanged nonnegative witnesses. Full packet synthesis is the identity. $\\square$","Let $E=\\mathbb R^r$ and let $\\mathbb S^d$ be the real symmetric $d\\times d$ matrices. Fix an injective linear map","$$\n\\mathcal J:E\\longrightarrow\\mathbb S^d,\n\\qquad \\lambda z=\\operatorname{tr}\\mathcal J(z).\n$$","Suppose some $e\\in E$ satisfies $\\mathcal J(e)\\succ0$. Define","$$\nC_{\\mathcal J}=\\{z:\\mathcal J(z)\\succeq0\\},\n\\qquad K_{\\mathcal J}=\\{z\\in C_{\\mathcal J}:\\lambda z=1\\}.\n$$","The map $\\mathcal J$ is an embedding, not a many-to-one decoder of a larger positive coordinate space. The cone is an inverse image of the positive semidefinite cone. This is different from the image cone $V\\mathbb R_+^m$ used in Proposition 39.","**Proposition 50. Matrix-embedding probability certificate.** The cone $C_{\\mathcal J}$ is closed, pointed and generating, and $\\lambda$ is strictly positive on its nonzero elements. Its normalized base is compact. Suppose a finite-alphabet predictive model has $z_0\\in K_{\\mathcal J}$, exact mass conservation and identities","$$\n\\boxed{\\mathcal J B_s=\\Phi_s\\mathcal J,}\n\\qquad\n\\Phi_s(X)=\\sum_{j=1}^{m_s} w_{sj}K_{sj}XK_{sj}^{\\mathsf T},\n\\quad w_{sj}\\geq0.\n$$","Then every $B_s$ preserves $C_{\\mathcal J}$, and every finite observation word has a nonnegative, normalized probability. Zero maps may use an empty sum. An invertible change of predictive coordinates transports the certificate by","$$\nB'_s=TB_sT^{-1},\\qquad\n\\mathcal J'=\\mathcal JT^{-1},\\qquad\n\\lambda'=\\lambda T^{-1}.\n$$","**Proof.** Closedness follows from continuity. If both $z$ and $-z$ are in the cone, $\\mathcal J(z)$ is both positive and negative semidefinite, hence zero. Injectivity gives $z=0$. A sufficiently small neighborhood of $e$ maps into positive definite matrices, so the cone has an interior point and is generating. A nonzero positive semidefinite matrix has positive trace. The trace-one positive semidefinite matrices form a compact set. The image of $\\mathcal J$ is a closed finite-dimensional subspace and its inverse on that image is continuous, so $K_{\\mathcal J}$ is compact.","th $z$ and $-z$ are in the cone, $\\mathcal J(z)$ is both positive and negative semidefinite, hence zero. Injectivity gives $z=0$. A sufficiently small neighborhood of $e$ maps into positive definite matrices, so the cone has an interior point and is generating. A nonzero positive semidefinite matrix has positive trace. The trace-one positive semidefinite matrices form a compact set. The image of $\\mathcal J$ is a closed finite-dimensional subspace and its inverse on that image is continuous, so $K_{\\mathcal J}$ is compact.","ro. Injectivity gives $z=0$. A sufficiently small neighborhood of $e$ maps into positive definite matrices, so the cone has an interior point and is generating. A nonzero positive semidefinite matrix has positive trace. The trace-one positive semidefinite matrices form a compact set. The image of $\\mathcal J$ is a closed finite-dimensional subspace and its inverse on that image is continuous, so $K_{\\mathcal J}$ is compact.","Every term $wKXK^{\\mathsf T}$ is positive semidefinite when $X$ is. The intertwining identity therefore proves invariance. Iteration gives invariance of every branch state, and the mass identity normalizes every finite observation tree. Substituting the transported quantities proves the final statement. $\\square$","$$\n\\|\\mathcal J(z)\\|_1\\leq\\|z\\|_{C_{\\mathcal J}},\n$$","because every positive decomposition in $E$ gives one in the matrix space. Equality holds when $\\mathcal J$ maps onto all of $\\mathbb S^d$, by the positive and negative spectral parts. For a proper matrix subspace, those spectral parts may leave the subspace. Norm transport therefore needs its own stated hypothesis.","$$\n\\mathcal J(m,x,y)=\\frac12\n\\begin{pmatrix}m+x&y\\\\y&m-x\\end{pmatrix}.\n$$","Then $\\operatorname{tr}\\mathcal J=m$ and","$$\n\\det\\mathcal J=\\frac{m^2-x^2-y^2}{4}.\n$$","Its eigenvalues are $(m\\pm\\sqrt{x^2+y^2})/2$, so positivity is exactly the Lorentz-cone condition. The normalized base is the unit disk.","**Proposition 51. Exact rational matrix witness for the rotation model.** Let $T=\\operatorname{diag}(1,R)$ with the unchanged rotation","$$\nR=\\begin{pmatrix}3/5&-4/5\\\\4/5&3/5\\end{pmatrix},\n\\qquad W=\\begin{pmatrix}2&-1\\\\1&2\\end{pmatrix}.\n$$","$$\n\\begin{aligned}\n\\mathcal J(Tz)&=\\frac15W\\mathcal J(z)W^{\\mathsf T},\\\\\n\\mathcal J(B_{\\mathrm p,0}z)&=E_{00}\\mathcal J(z)E_{00}^{\\mathsf T},\\\\\n\\mathcal J(B_{\\mathrm p,1}z)&=E_{01}\\mathcal J(z)E_{01}^{\\mathsf T},\n\\end{aligned}\n$$","where $E_{ij}$ is the corresponding $2\\times2$ matrix unit. The other rotate branch is zero. Thus this exact model has a finite rational matrix certificate without attenuation or a polygonal replacement.","**Proof.** Direct multiplication gives the rotate identity. The two matrix-unit congruences retain, respectively, the upper and lower diagonal entries of $\\mathcal J(z)$ and place them in its upper diagonal entry. These are $(m+x)/2$ and $(m-x)/2$, exactly the two probe masses resetting to $(1,1,0)$. Also $W^{\\mathsf T}W=5I$ and","$$\nE_{00}^{\\mathsf T}E_{00}+E_{01}^{\\mathsf T}E_{01}=I.\n$$","The actions conserve trace. The initial state embeds as $\\operatorname{diag}(1,0)$, and $(1,0,0)$ is a strictly positive interior witness. Proposition 50 applies. $\\square$","Let $(Y,\\mathcal Y)$ be a standard Borel observation space and let $\\mu$ be a fixed probability reference measure. Keep a finite action set. For each action $a$, let $B_a(y):E\\to E$ be measurable and integrable in operator norm. Let $(C,\\lambda)$ be a proper probability cone and mass functional as in Section 46. Assume that, outside one $\\mu$-null set for each action,","$$\nB_a(y)C\\subseteq C,\n\\qquad\n\\lambda\\int_Y B_a(y)\\,\\mu(dy)=\\lambda.\n$$","**Proposition 52. Continuous-observation instrument semantics.** For $z\\in K=\\{z\\in C:\\lambda z=1\\}$,","$$\n\\boxed{\nP_a(A\\mid z)=\\int_A\\lambda B_a(y)z\\,\\mu(dy),\n\\qquad\nz^+(y)=\\frac{B_a(y)z}{\\lambda B_a(y)z}\n}\n$$","define a probability kernel and, for $\\mu$-almost every observation with positive denominator, a conditional state. They generate consistent finite record laws under any common measurable history-dependent policy. For prescribed actions, the joint observation density is","$$\n\\lambda B_{a_T}(y_T)\\cdots B_{a_1}(y_1)z_0\n$$","with respect to $\\mu^{\\otimes T}$. Replacing the reference measure by $\\mu'=r\\mu$, with $r>0$ almost everywhere and $\\int r\\,d\\mu=1$, requires $B'_a(y)=B_a(y)/r(y)$. Event probabilities and normalized states then remain unchanged.","**Proof.** Cone positivity makes the density nonnegative and the integrated mass identity makes its integral one. A positive-density update lies in $K$. Assign any fixed state of $K$ on the common exceptional null set and at zero-density points to obtain a measurable version of the update. This arbitrary choice has no effect on the probability kernel. Repeated integration, or conditioning one step at a time, proves normalization and consistency of the finite laws. Measurable policy factors are nonnegative and sum to one at each history. The reference-measure transformation cancels $r$ in each integral and cancels $1/r$ in the normalized update. $\\square$","common exceptional null set and at zero-density points to obtain a measurable version of the update. This arbitrary choice has no effect on the probability kernel. Repeated integration, or conditioning one step at a time, proves normalization and consistency of the finite laws. Measurable policy factors are nonnegative and sum to one at each history. The reference-measure transformation cancels $r$ in each integral and cancels $1/r$ in the normalized update. $\\square$","at zero-density points to obtain a measurable version of the update. This arbitrary choice has no effect on the probability kernel. Repeated integration, or conditioning one step at a time, proves normalization and consistency of the finite laws. Measurable policy factors are nonnegative and sum to one at each history. The reference-measure transformation cancels $r$ in each integral and cancels $1/r$ in the normalized update. $\\square$","a measurable version of the update. This arbitrary choice has no effect on the probability kernel. Repeated integration, or conditioning one step at a time, proves normalization and consistency of the finite laws. Measurable policy factors are nonnegative and sum to one at each history. The reference-measure transformation cancels $r$ in each integral and cancels $1/r$ in the normalized update. $\\square$","When $\\mu$ has no atoms, an individual point has probability zero even when its density is positive. A density may exceed one. Neither fact is a numerical pathology. Flooring a vanishing denominator changes the model and is not part of this theorem.","Specialize to $Y=[-1,1]$ and $\\mu(du)=du/2$. Suppose $B_a(u)$ is a matrix polynomial. A finite congruence certificate can prove positivity for every $u$, not merely for values on a grid.","For a real $d^2\\times d^2$ symmetric matrix $F$, define","$$\n[\\Phi_F(X)]_{ij}=\\sum_{p,q=1}^d F_{(i,p),(j,q)}X_{pq}.\n$$","$$\nF=\\sum_\\nu w_\\nu\\operatorname{vec}(K_\\nu)\n\\operatorname{vec}(K_\\nu)^{\\mathsf T},\\qquad w_\\nu\\geq0,\n$$","then $\\Phi_F(X)=\\sum_\\nu w_\\nu K_\\nu X K_\\nu^{\\mathsf T}$. Every positive semidefinite $F$ has a real factorization of this kind. The matrix is the usual Choi/Gram representation, with an explicit index convention [R16].","**Proposition 53. Polynomial Gram certificate for interval observations.** Fix the injective embedding $\\mathcal J$ and its strict interior witness from Proposition 50. Set","$$\nZ_q(u)=\\begin{pmatrix}I_{d^2}\\\\uI_{d^2}\\\\\\vdots\\\\u^qI_{d^2}\\end{pmatrix}.\n$$","Suppose each action has constant positive semidefinite matrices $G_{a,0},G_{a,1}$, with the second term omitted when $q=0$, such that","$$\n\\begin{aligned}\nF_a(u)&=Z_q(u)^{\\mathsf T}G_{a,0}Z_q(u)\\\\\n&\\quad +(1-u^2)Z_{q-1}(u)^{\\mathsf T}G_{a,1}Z_{q-1}(u),\\\\\n\\mathcal J B_a(u)&=\\Phi_{F_a(u)}\\mathcal J.\n\\end{aligned}\n$$","Assume also $z_0\\in K_{\\mathcal J}$ and the exact integrated mass identity. Then the model is internally probability-valid for every observation value in $[-1,1]$ and every finite horizon. The intertwining condition is finitely many coefficient equalities. The normalization condition uses only","$$\n\\int_{-1}^1u^k\\,\\frac{du}{2}\n=\\begin{cases}0&k\\text{ odd},\\\\1/(k+1)&k\\text{ even}.\\end{cases}\n$$","**Proof.** Both Gram expressions are positive semidefinite on the interval because $1-u^2\\geq0$ there. Factoring $F_a(u)$ at a fixed $u$ gives a positive congruence map, whether or not its individual factors depend polynomially on $u$. The intertwining identity therefore proves $B_a(u)C_{\\mathcal J}\\subseteq C_{\\mathcal J}$. Polynomials on a compact interval are integrable. The mass identity and Proposition 52 prove the assertion. Equality of two finite matrix polynomials is equivalent to equality of every coefficient. The moment values follow by elementary integration. $\\square$","or not its individual factors depend polynomially on $u$. The intertwining identity therefore proves $B_a(u)C_{\\mathcal J}\\subseteq C_{\\mathcal J}$. Polynomials on a compact interval are integrable. The mass identity and Proposition 52 prove the assertion. Equality of two finite matrix polynomials is equivalent to equality of every coefficient. The moment values follow by elementary integration. $\\square$","For fixed $\\mathcal J$ and fixed degree bounds, unknown entries of $B_a$ and the Gram matrices obey linear equalities and positive semidefinite constraints. A convex data-fit objective therefore gives a semidefinite feasibility or optimization problem in this fixed representation. Jointly choosing $\\mathcal J$ changes the problem: the intertwining equation then contains products of unknown coefficients.","$$\n\\Phi_u(X)=\\sum_\\nu w_\\nu g_\\nu(u)K_\\nu(u)XK_\\nu(u)^{\\mathsf T},\n\\qquad g_\\nu(u)\\in\\{1,1-u^2\\}.\n$$","It checks the polynomial identities, weights, support, embedding, initial state and normalization exactly. It does not run a general semidefinite optimizer. The operator identity holds on the image of $\\mathcal J$; trace preservation of an arbitrary extension outside that image is not required.","$$\nB_a(u)=\\sum_{k=0}^D u^k C_{a,k},\n\\qquad\nM_{a,j}=\\int_{-1}^1 u^j B_a(u)\\,\\frac{du}{2}.\n$$","Define the moment Gram matrix $W_D$ by","$$\n(W_D)_{jk}=\\int_{-1}^1u^{j+k}\\,\\frac{du}{2},\n\\qquad 0\\leq j,k\\leq D.\n$$","A history is now a prescribed action sequence with observations constrained to measurable events of positive or zero probability. It produces an unnormalized predictive vector $h_j$. A future event experiment produces a row functional $t_i$. Choose $r$ of each and form","$$\nH_{ij}=t_i h_j,\\qquad\n(G_{a,j})_{i\\ell}=t_i M_{a,j}h_\\ell.\n$$","The index $j$ in $G_{a,j}$ is a moment order. Odd-order entries can be signed expectations. They are not branch probabilities and are never normalized as if they were.","**Proposition 54. Conditional finite-moment identification.** Suppose the true predictive dimension is $r$, the degree bound $D$ is correct, and the selected $r\\times r$ core $H$ is invertible. Exact population values of $H$, every $G_{a,j}$ and the initial test vector determine the complete polynomial operator family in test coordinates:","$$\n\\boxed{\n\\widehat M_{a,j}=G_{a,j}H^{-1},\n\\qquad\n\\widehat C_{a,k}=\\sum_{j=0}^D(W_D^{-1})_{kj}\\widehat M_{a,j}.\n}\n$$","If every moment-operator error is at most $\\eta$ in one fixed induced norm, then coefficient reconstruction alone obeys","$$\n\\|\\widetilde C_{a,k}-\\widehat C_{a,k}\\|\n\\leq\\eta\\sum_j|(W_D^{-1})_{kj}|.\n$$","**Proof.** A nonzero polynomial has strictly positive squared integral, so its coefficient vector $c$ satisfies $c^{\\mathsf T}W_Dc>0$. Thus $W_D$ is invertible. Let $T$ have rows $t_i$ and $S$ have columns $h_j$. Since $H=TS$ is an invertible square core in an $r$-dimensional space, both $T$ and $S$ are invertible. Therefore $G_{a,j}H^{-1}=TM_{a,j}T^{-1}$. The relation $M_{a,j}=\\sum_k(W_D)_{jk}C_{a,k}$ is exact integration of the polynomial expansion. Invert it and conjugate by $T$ to obtain the displayed reconstruction. The error bound is the triangle inequality for that same fixed linear inversion. $\\square$","icient vector $c$ satisfies $c^{\\mathsf T}W_Dc>0$. Thus $W_D$ is invertible. Let $T$ have rows $t_i$ and $S$ have columns $h_j$. Since $H=TS$ is an invertible square core in an $r$-dimensional space, both $T$ and $S$ are invertible. Therefore $G_{a,j}H^{-1}=TM_{a,j}T^{-1}$. The relation $M_{a,j}=\\sum_k(W_D)_{jk}C_{a,k}$ is exact integration of the polynomial expansion. Invert it and conjugate by $T$ to obtain the displayed reconstruction. The error bound is the triangle inequality for that same fixed linear inversion. $\\square$","j$. Since $H=TS$ is an invertible square core in an $r$-dimensional space, both $T$ and $S$ are invertible. Therefore $G_{a,j}H^{-1}=TM_{a,j}T^{-1}$. The relation $M_{a,j}=\\sum_k(W_D)_{jk}C_{a,k}$ is exact integration of the polynomial expansion. Invert it and conjugate by $T$ to obtain the displayed reconstruction. The error bound is the triangle inequality for that same fixed linear inversion. $\\square$","The rank and degree assumptions are not learned from the determinant of $H$. The determinant establishes coverage only after the ambient predictive dimension has been justified. The moment Gram can become poorly conditioned at higher degree. An orthogonal polynomial basis can improve this representation without removing an incorrect degree assumption.","The entries have an operational population meaning. A reset experiment for an event word, with one observation weighted by $u^j$, returns a bounded random variable in $[-1,1]$. Its expectation is the corresponding entry of $G_{a,j}$. With $N$ independent trials for each of $M$ predeclared entries, the same exponential-moment argument as Proposition 35 gives the simultaneous bound","$$\n\\epsilon_N=\\sqrt{\\frac{2\\log(2M/\\delta)}{N}}\n$$","with failure probability at most $\\delta$. The factor differs from the Bernoulli bound because the value range is two. Dependent trajectories, unknown resets and post-selection are not covered by that observation protocol. This edition uses exact constructed population tables, not a new sampling experiment.","$$\nL_3(u)=\\frac{5u^3-3u}{2},\\qquad\np_\\pm(u)=1\\pm\\epsilon L_3(u),\\qquad 0<\\epsilon<1.\n$$","Relative to $du/2$, these are distinct strictly positive normalized densities with the same moments of orders zero, one and two. Their total-variation distance is","$$\n\\boxed{\\|p_+\\mu-p_-\\mu\\|_{\\mathrm{TV}}=\\frac{13\\epsilon}{40}.}\n$$","**Proof.** The endpoints and stationary points of $L_3$ show $|L_3(u)|\\leq1$ on the interval. Direct integration gives $\\int u^jL_3(u)\\,du/2=0$ for $j=0,1,2$. This proves normalization and moment agreement, while strict positivity follows from $\\epsilon<1$. Its nonnegative root other than zero is $\\sqrt{3/5}$. Integrating $|L_3|$ on the two subintervals of $[0,1]$ and using symmetry gives $\\int|L_3|\\,du/2=13/40$. The density difference is $2\\epsilon L_3$, so its half-integral is the stated distance. Independence makes every product moment factor into the identical one-step moments. $\\square$","ts of $L_3$ show $|L_3(u)|\\leq1$ on the interval. Direct integration gives $\\int u^jL_3(u)\\,du/2=0$ for $j=0,1,2$. This proves normalization and moment agreement, while strict positivity follows from $\\epsilon<1$. Its nonnegative root other than zero is $\\sqrt{3/5}$. Integrating $|L_3|$ on the two subintervals of $[0,1]$ and using symmetry gives $\\int|L_3|\\,du/2=13/40$. The density difference is $2\\epsilon L_3$, so its half-integral is the stated distance. Independence makes every product moment factor into the identical one-step moments. $\\square$","oment agreement, while strict positivity follows from $\\epsilon<1$. Its nonnegative root other than zero is $\\sqrt{3/5}$. Integrating $|L_3|$ on the two subintervals of $[0,1]$ and using symmetry gives $\\int|L_3|\\,du/2=13/40$. The density difference is $2\\epsilon L_3$, so its half-integral is the stated distance. Independence makes every product moment factor into the identical one-step moments. $\\square$","Let $A_1,\\ldots,A_n$ be a measurable partition of $Y$. Define the cell operators","$$\nB_{a,j}=\\int_{A_j}B_a(y)\\,\\mu(dy).\n$$","**Proposition 56. Exact bin coarsening and its posterior.** Every $B_{a,j}$ preserves the probability cone, and $\\lambda\\sum_jB_{a,j}=\\lambda$. For a policy measurable with respect to bin histories, the finite model $\\{B_{a,j}\\}$ reproduces exactly the binned observation law of the continuous model at every finite horizon. On an observed cell with $p_j=\\lambda B_{a,j}z>0$,","$$\n\\frac{B_{a,j}z}{p_j}\n=\\frac1{p_j}\\int_{A_j}z^+(y)\\lambda B_a(y)z\\,\\mu(dy).\n$$","cone remain in the cone. Summing over a partition gives the original mass identity. For a fixed bin history, its policy factors are constant across the corresponding cells. Successive integration of the operator product therefore gives the product of the cell operators, in the original temporal order. This establishes every finite binned record probability. In the posterior identity, substitute $z^+(y)\\lambda B_a(y)z=B_a(y)z$ on the positive-density set. The remaining outputs are zero because they lie in the cone and have zero mass. $\\square$","across the corresponding cells. Successive integration of the operator product therefore gives the product of the cell operators, in the original temporal order. This establishes every finite binned record probability. In the posterior identity, substitute $z^+(y)\\lambda B_a(y)z=B_a(y)z$ on the positive-density set. The remaining outputs are zero because they lie in the cone and have zero mass. $\\square$","A common discretization replaces a family of matrices $K(y)$ by its mean and then applies $X\\mapsto KXK^{\\mathsf T}$. This is generally the wrong mean operation.","**Proposition 57. Exact covariance correction for an averaged congruence.** Let $\\nu$ be a probability measure and suppose $K(y)$ has square-integrable entries. Put $\\bar K=\\int K(y)\\,\\nu(dy)$. For every symmetric $X$,","$$\n\\boxed{\n\\int K(y)XK(y)^{\\mathsf T}\\,\\nu(dy)\n=\\bar KX\\bar K^{\\mathsf T}\n+\\int (K(y)-\\bar K)X(K(y)-\\bar K)^{\\mathsf T}\\,\\nu(dy).\n}\n$$","The correction is positive semidefinite when $X\\succeq0$. For an affine matrix function $K(u)=K_0+uK_1$ with scalar $u$ of finite second moment, it equals $\\operatorname{Var}_\\nu(u)K_1XK_1^{\\mathsf T}$. On a uniform interval of width $h$, the variance is $h^2/12$.","**Proof.** Substitute $K=\\bar K+(K-\\bar K)$ and expand. The two cross terms integrate to zero. The remaining integrand is a positive congruence when $X$ is positive. For an affine matrix function the centered term is $(u-\\mathbb E_\\nu u)K_1$, giving the variance formula. Integrating the square of a centered uniform variable gives $h^2/12$. $\\square$","$$\n\\boxed{\\frac{3}{4n^2}}\n$$","when there are $n$ bins. With one bin, midpoint evaluation retains only one quarter of the required mass. Exact integration or the covariance correction restores the missing contribution. Dividing an inaccurate approximation by a guessed normalizer would define another model rather than establish this identity.","$$\nB_a^\\Pi(y)=\\frac{B_{a,j}}{\\mu(A_j)}\\quad(y\\in A_j).\n$$","**Proposition 58. Positive cell-average approximation.** The family $B_a^\\Pi$ is cone preserving and mass conserving. It has exactly the same cell operators as $B_a$. Let $Y$ have a metric and suppose","$$\n\\|B_a(y)-B_a(v)\\|_{C\\to C}\\leq L\\,d(y,v)\n$$","in the induced base norm, with every cell of diameter at most $h$. Then","$$\n\\epsilon_\\Pi:=\\sup_{a,z\\in K}\\int\n\\|(B_a(y)-B_a^\\Pi(y))z\\|_C\\,\\mu(dy)\n\\leq Lh.\n$$","For equal intervals of width $h$ under the uniform measure on $[-1,1]$, the sharper bound is $\\epsilon_\\Pi\\leq Lh/3$.","**Proof.** Each cell average is a positive scalar multiple of a positive cell operator. Its integral over that cell is $B_{a,j}$, proving both normalization and exact cell agreement. For $y\\in A_j$, subtract the cell average and use","$$\n\\|(B_a(y)-B_a^\\Pi(y))z\\|_C\n\\leq\\frac{L}{\\mu(A_j)}\\int_{A_j}d(y,v)\\,\\mu(dv),\n$$","because $\\|z\\|_C=1$. Integrating over $y$ proves the diameter bound. Two independent uniform points in an interval of width $h$ have mean absolute separation $h/3$, obtained by integrating over its two triangles. Applying this inside every equal interval gives the sharper bound. $\\square$","In the full $2\\times2$ symmetric-matrix example, the pullback base norm is","$$\n\\|(m,x,y)\\|_C=\\max\\{|m|,\\sqrt{x^2+y^2}\\}=\\|\\mathcal J(m,x,y)\\|_1.\n$$","There $\\|K(u)\\|_2\\leq2$ and $\\|K(u)-K(v)\\|_2\\leq(3/2)|u-v|$. The identity","$$\nK(u)XK(u)^{\\mathsf T}-K(v)XK(v)^{\\mathsf T}\n=(K(u)-K(v))XK(u)^{\\mathsf T}\n+K(v)X(K(u)-K(v))^{\\mathsf T}\n$$","and $\\|AXB\\|_1\\leq\\|A\\|_2\\|X\\|_1\\|B\\|_2$ give $L=6$. The rotated action uses an orthogonal matrix before this map and has the same bound. Equal bins therefore give $\\epsilon_\\Pi\\leq4/n$, capped by two for two valid instruments. The inequality is sufficient, not asserted sharp.","**Proposition 59. Continuous-record total-variation bound.** Let $B_a(y)$ and $\\widetilde B_a(y)$ satisfy Proposition 52 on a common $(C,\\lambda,\\mu)$, and let the two models use the same measurable policy on every common observation history. Suppose their initial states satisfy $\\|z_0-\\widetilde z_0\\|_C\\leq e_0$ and their time-$t$ instruments satisfy","$$\n\\sup_{a,z\\in K}\\int\\|(B_{a,t}(y)-\\widetilde B_{a,t}(y))z\\|_C\\,\\mu(dy)\n\\leq\\epsilon_t.\n$$","The length-$T$ record laws, including actions when the policy is randomized, obey","$$\n\\boxed{\n\\|P_T-\\widetilde P_T\\|_{\\mathrm{TV}}\n\\leq\\min\\left\\{1,\\frac{e_0+\\sum_{t=0}^{T-1}\\epsilon_t}{2}\\right\\}.\n}\n$$","No lower bound on each observation density is required. For identical initial states and the equal-bin approximation in Section 64, the example has the sufficient bound $\\min(1,2T/n)$.","**Proof.** For any positive decomposition $x=u-v$, cone preservation and mass conservation give","$$\n\\int\\|B_a(y)x\\|_C\\,\\mu(dy)\n\\leq\\int\\lambda B_a(y)(u+v)\\,\\mu(dy)\n=\\lambda u+\\lambda v.\n$$","Taking the infimum proves integrated instrument contraction. At each continuous history let $v_h,\\widetilde v_h$ be the unnormalized state densities, including the policy factors. Split the next difference as","$$\nB_a(y)(v_h-\\widetilde v_h)\n+(B_a(y)-\\widetilde B_a(y))\\widetilde v_h.\n$$","Integrate the first term using contraction. The second is bounded by $\\epsilon_t\\lambda\\widetilde v_h$ after normalizing only its nonzero positive mass. Its integral over histories is at most $\\epsilon_t$. Tonelli's theorem applies to these nonnegative norm bounds, and operator integrability supplies well-defined finite-dimensional integrals. The integrated state discrepancy therefore satisfies $D_{t+1}\\leq D_t+\\epsilon_t$. Finally $|\\lambda x|\\leq\\|x\\|_C$ bounds the record-density difference, and total variation is half its absolute integral. $\\square$","ction. The second is bounded by $\\epsilon_t\\lambda\\widetilde v_h$ after normalizing only its nonzero positive mass. Its integral over histories is at most $\\epsilon_t$. Tonelli's theorem applies to these nonnegative norm bounds, and operator integrability supplies well-defined finite-dimensional integrals. The integrated state discrepancy therefore satisfies $D_{t+1}\\leq D_t+\\epsilon_t$. Finally $|\\lambda x|\\leq\\|x\\|_C$ bounds the record-density difference, and total variation is half its absolute integral. $\\square$",". Its integral over histories is at most $\\epsilon_t$. Tonelli's theorem applies to these nonnegative norm bounds, and operator integrability supplies well-defined finite-dimensional integrals. The integrated state discrepancy therefore satisfies $D_{t+1}\\leq D_t+\\epsilon_t$. Finally $|\\lambda x|\\leq\\|x\\|_C$ bounds the record-density difference, and total variation is half its absolute integral. $\\square$","For a measurable event $H$ of probability $p=P_T(H)>0$, the restriction-and-normalization argument of Proposition 44 works for measures as well as finite vectors. When both conditional laws exist, their total-variation distance is at most $\\min(1,\\delta/p)$ if the full-law distance is at most $\\delta$. The event probability is invariant under a reference-measure change. A density value at one point is not a substitute for that event probability.","The full-law comparison between $B$ and $B^\\Pi$ permits any common measurable policy because both are continuous-output models on the same space. The exact equivalence to the finite bin model in Proposition 56 instead requires a bin-measurable policy. These are different conclusions with different observation information.","$$\nK(u)=\\frac12\\begin{pmatrix}1+3u&0\\\\0&1-3u\\end{pmatrix},\n\\qquad -1\\leq u\\leq1,\n\\qquad \\mu(du)=du/2.\n$$","The direct action uses $X\\mapsto K(u)XK(u)^{\\mathsf T}$. The rotated action first applies the unchanged rotation congruence from Section 57, giving","$$\nX\\longmapsto\\frac15 K(u)WXW^{\\mathsf T}K(u)^{\\mathsf T}.\n$$","Both are exactly normalized because $\\int K(u)^{\\mathsf T}K(u)\\,du/2=I$ and $W^{\\mathsf T}W=5I$. In predictive coordinates, write","$$\n\\begin{gathered}\na(u)=\\frac{1+9u^2}{4},\\qquad\nb(u)=\\frac{3u}{2},\\qquad\nc(u)=\\frac{1-9u^2}{4},\\\\[3pt]\nB_0(u)=\\begin{pmatrix}a(u)&b(u)&0\\\\b(u)&a(u)&0\\\\0&0&c(u)\\end{pmatrix},\n\\qquad B_1(u)=B_0(u)T.\n\\end{gathered}\n$$","Although $c(u)$ is negative for part of the interval, both actions preserve the disk cone. Entrywise sign is not the positivity criterion in these coordinates. The initial state is $z_0=(1,0,0)^{\\mathsf T}$, or the matrix $I/2$.","$$\nQ=\\begin{pmatrix}\n1&0&0\\\\\n1/2&3/8&0\\\\\n1/2&9/40&-3/10\n\\end{pmatrix}.\n$$","$$\nS=\\begin{pmatrix}\n1&1/2&107/320\\\\\n0&3/8&3/10\\\\\n0&0&-3/40\n\\end{pmatrix}.\n$$","$$\n\\boxed{\nH=\\begin{pmatrix}\n1&1/2&107/320\\\\\n1/2&25/64&179/640\\\\\n1/2&107/320&823/3200\n\\end{pmatrix},\\qquad\\det H=\\frac{81}{25600}.\n}\n$$","$$\nW_2=\\begin{pmatrix}1&0&1/3\\\\0&1/3&0\\\\1/3&0&1/5\\end{pmatrix},\n\\qquad\nW_2^{-1}=\\begin{pmatrix}9/4&0&-15/4\\\\0&3&0\\\\-15/4&0&45/4\\end{pmatrix},\n\\qquad\\det W_2=\\frac4{135}.\n$$","Proposition 54 reconstructs $QB_a(u)Q^{-1}$ from $H$ and the six inserted-moment tables. The code that performs this reconstruction imports no generating matrices. The separate generator produces the tables and is used only in the verification suite. These are population values from an exact example, not learned sample estimates.","For a direct observation value $u=1/2$, the density and updated state are","$$\np_1=\\frac{13}{16},\\qquad z_1=(1,12/13,0)^{\\mathsf T}.\n$$","If the rotated action next gives $u=-1/2$, they are","$$\np_2=\\frac{413}{1040},\\qquad\nz_2=(1,-312/413,-240/413)^{\\mathsf T}.\n$$","Their product $413/1280$ is a joint density with respect to $(du/2)^{\\otimes2}$, not the probability of those two exact values. The two point values each have zero probability in this nonatomic model.","By contrast, the event $u_1\\geq0$ under action zero followed by $u_2\\geq0$ under action one has actual probability","$$\n\\boxed{\\frac{107}{320},}\n$$","and its conditional state is $(1,96/107,-24/107)^{\\mathsf T}$. Exact bin operators produce this value without quadrature.","$$\n\\det\\mathcal J(z_2)=\\frac{15625}{682276}>0.\n$$","On an interval with midpoint $v$ and width $h$, the average matrix instrument is","$$\n\\mathbb E[K(u)XK(u)^{\\mathsf T}]\n=K(v)XK(v)^{\\mathsf T}\n+\\frac{h^2}{12}K_1XK_1^{\\mathsf T},\n\\qquad K_1=\\frac32\\operatorname{diag}(1,-1).\n$$","The correction is itself a positive congruence. The cell operator multiplies this mean by the cell mass $h/2$. Equal bins have width $2/n$, so omitting the covariance loses trace $3/(4n^2)$ over the full partition. The reference calculation checks both the exact matrix identity and this mass defect for several rational partitions.","Suppose a structural library supplies a certified template with polynomial coefficients $C_{a,k}$, mass row $\\lambda$ and embedding $\\mathcal J$. Suppose observed population tables produce coefficients $\\widehat C_{a,k}$ and mass row $\\widehat\\lambda$ of the same dimension.","$$\n\\boxed{\n\\widehat C_{a,k}T=TC_{a,k},\\qquad\n\\widehat\\lambda T=\\lambda\n}\n$$","are linear in $T$. Every invertible solution transports the template certificate to the fitted coordinates by $\\widehat{\\mathcal J}=\\mathcal JT^{-1}$. If the fitted coefficient family is observable, meaning the rows $\\widehat\\lambda\\widehat C_w$ span the dual space, the solution is unique. The fitted initial state must still be tested in the transported normalized cone.","**Proof.** All coefficient matrices are fixed, so the displayed constraints are linear. An invertible solution yields $\\widehat B_a(u)=TB_a(u)T^{-1}$ coefficient by coefficient. Substitution in the template intertwining identity proves the transported certificate. To prove uniqueness, let $D$ be the difference of two solutions. It satisfies $\\widehat C_{a,k}D=DC_{a,k}$ and $\\widehat\\lambda D=0$. Therefore $\\widehat\\lambda\\widehat C_wD=0$ for every coefficient word. Observability makes every column of $D$ zero. An initial state outside the transported cone is not repaired by this operator identity, which explains the separate check. $\\square$","All coefficient matrices are fixed, so the displayed constraints are linear. An invertible solution yields $\\widehat B_a(u)=TB_a(u)T^{-1}$ coefficient by coefficient. Substitution in the template intertwining identity proves the transported certificate. To prove uniqueness, let $D$ be the difference of two solutions. It satisfies $\\widehat C_{a,k}D=DC_{a,k}$ and $\\widehat\\lambda D=0$. Therefore $\\widehat\\lambda\\widehat C_wD=0$ for every coefficient word. Observability makes every column of $D$ zero. An initial state outside the transported cone is not repaired by this operator identity, which explains the separate check. $\\square$","ion yields $\\widehat B_a(u)=TB_a(u)T^{-1}$ coefficient by coefficient. Substitution in the template intertwining identity proves the transported certificate. To prove uniqueness, let $D$ be the difference of two solutions. It satisfies $\\widehat C_{a,k}D=DC_{a,k}$ and $\\widehat\\lambda D=0$. Therefore $\\widehat\\lambda\\widehat C_wD=0$ for every coefficient word. Observability makes every column of $D$ zero. An initial state outside the transported cone is not repaired by this operator identity, which explains the separate check. $\\square$","ransported certificate. To prove uniqueness, let $D$ be the difference of two solutions. It satisfies $\\widehat C_{a,k}D=DC_{a,k}$ and $\\widehat\\lambda D=0$. Therefore $\\widehat\\lambda\\widehat C_wD=0$ for every coefficient word. Observability makes every column of $D$ zero. An initial state outside the transported cone is not repaired by this operator identity, which explains the separate check. $\\square$","For the worked example, the linear system has the unique solution $T=Q$ from Section 66. The learner receives the population tables and a fixed structural template. It reconstructs the operators, solves the linear match, transports the embedding and exports a certificate. Independent replay then needs only that certificate. It does not refit, search, or import the template or generator.","**Proposition 61. Conditional continuous realization with a finite domain certificate.** Assume a controlled continuous-observation process has the declared finite predictive dimension and polynomial operator-density degree, a fixed uniform reference measure on $[-1,1]$, and an invertible complete event-test core. Exact moment tables identify its operator density in test coordinates by Proposition 54.","the exact integral mass identity supplies normalization through Proposition 52. Proposition 60 supplies a transported witness in the template case. Coordinate cancellation and full packet synthesis are exact identities. Proposition 56 proves the cell-model statement. Propositions 58 and 59 provide the two approximation conclusions with their common-domain and common-policy assumptions unchanged. $\\square$","Work in mass-first coordinates $z=(m,x)\\in\\mathbb R\\times\\mathbb R^d$, with $\\lambda=(1,0,\\ldots,0)$. For a center $c\\in\\mathbb R^d$ and a symmetric positive definite matrix $H$, define","$$\nC_{c,H}=\\{(m,x):m\\geq0,\\ (x-mc)^{\\mathsf T}H(x-mc)\\leq m^2\\}.\n$$","$$\nS_c=\\begin{pmatrix}1&0\\\\-c&I\\end{pmatrix},\\qquad\nQ=S_c^{\\mathsf T}\\begin{pmatrix}1&0\\\\0&-H\\end{pmatrix}S_c.\n$$","Then $C_{c,H}=\\{z:m\\geq0,\\ z^{\\mathsf T}Qz\\geq0\\}$. The cone is proper and generating, and $m$ is strictly positive on every nonzero cone element. These facts follow by an invertible linear transformation to the Lorentz cone. In particular, an output with zero mass in the cone is the zero vector.","Let $B_a(u)$ be polynomial operator densities on $[-1,1]$, relative to $du/2$. Write","$$\n\\lambda B_a(u)=(b_{a,0}(u),b_a(u)^{\\mathsf T}),\\qquad\n\\beta_a(u)=b_{a,0}(u)+b_a(u)^{\\mathsf T}c.\n$$","**Proposition 62. Oriented quadratic probability certificate.** Suppose that for each action and every $u\\in[-1,1]$ there are $\\tau_a(u)\\geq0$ and $R_a(u)\\succeq0$ satisfying","$$\n\\boxed{B_a(u)^{\\mathsf T}QB_a(u)-\\tau_a(u)Q=R_a(u),}\n$$","$$\n\\boxed{\\beta_a(u)\\geq0,\\qquad\n\\beta_a(u)^2-b_a(u)^{\\mathsf T}H^{-1}b_a(u)\\geq0.}\n$$","If $z_0\\in C_{c,H}$ has mass one and $\\lambda\\int B_a(u)\\,du/2=\\lambda$, then every operator preserves $C_{c,H}$. The supplied model has valid probabilities at every finite horizon. Polynomial sum-of-squares witnesses for the scalar inequalities and for $R_a$ make this a finite certificate.","**Proof.** For $z\\in C_{c,H}$, the quadratic identity gives","$$\n(B_a z)^{\\mathsf T}Q(B_a z)\n=\\tau_a z^{\\mathsf T}Qz+z^{\\mathsf T}R_a z\\geq0.\n$$","For a nonzero $z$, write $x=m(c+w)$ with $w^{\\mathsf T}Hw\\leq1$. Its output mass is $m(\\beta_a+b_a^{\\mathsf T}w)$. Cauchy--Schwarz in the $H$ inner product bounds this below by","$$\nm\\left(\\beta_a-\\sqrt{b_a^{\\mathsf T}H^{-1}b_a}\\right)\\geq0.\n$$","Thus both inequalities defining the desired sheet of the quadratic cone hold. The zero input causes no difficulty. Integrated mass conservation gives normalization, and induction gives the finite-horizon conclusion. For polynomial data, identities between coefficients and nonnegative sums of polynomial squares establish the displayed hypotheses throughout the interval. Terms multiplied by $1-u^2$ are also nonnegative on that interval. $\\square$","red sheet of the quadratic cone hold. The zero input causes no difficulty. Integrated mass conservation gives normalization, and induction gives the finite-horizon conclusion. For polynomial data, identities between coefficients and nonnegative sums of polynomial squares establish the displayed hypotheses throughout the interval. Terms multiplied by $1-u^2$ are also nonnegative on that interval. $\\square$","The orientation test is essential. The matrix $-I$ preserves every quadratic form, but maps positive mass to negative mass. A quadratic inequality without its mass orientation can certify the wrong sheet. Likewise, testing finitely many values of $u$ is not a substitute for a polynomial identity and its positivity witness.","This is a sufficient language, not a completeness theorem for all cones. The reference checker accepts rational scalar and matrix sum-of-squares witnesses. Its discovery routine uses the narrower conformal case $R_a=0$, but the checker does not confuse that search restriction with the theorem's scope.","An operator identity can reveal a useful coordinate geometry without specifying that geometry in advance. Suppose two fitted operators evaluated at an anchor $u_*$ give an invertible ratio","$$\nF=B_b(u_*)B_a(u_*)^{-1},\\qquad \\lambda F=\\lambda.\n$$","$$\nF=\\begin{pmatrix}1&0\\\\b&A\\end{pmatrix}.\n$$","If $I-A$ is invertible, its fixed center is $c=(I-A)^{-1}b$. The centered spatial action is $A$. A candidate invariant metric obeys","$$\nA^{\\mathsf T}HA=H.\n$$","With $A$ fixed, this is a linear equation in the independent entries of symmetric $H$. No coordinate template appears in that equation.","**Proposition 63. Conditional operator-only ellipsoid discovery.** Suppose the symmetric solution space of $A^{\\mathsf T}HA=H$ is one-dimensional and contains a positive definite matrix $H_0$. Let a rank-one fitted branch at a declared observation have a nonzero-mass image, normalized to $v=(1,x_v)$. If $x_v\\neq c$, define","$$\nH=\\frac{H_0}{(x_v-c)^{\\mathsf T}H_0(x_v-c)}.\n$$","This gives a unique positive definite metric in that solution space for which $v$ is on the normalized boundary. If this proposed geometry passes every hypothesis of Proposition 62 for the complete fitted family and its initial state, it is an internally valid probability-domain certificate. The proposal requires no supplied template. Failure at any step withholds this certificate and does not establish model invalidity.","**Proof.** Invertibility of $I-A$ gives the unique center. One-dimensionality makes every symmetric invariant form a scalar multiple of $H_0$. The nonzero displacement has strictly positive squared $H_0$ length, so the normalization is well-defined, positive and independent of the initially chosen positive scale. The final acceptance test is exactly Proposition 62. Neither invariance under the ratio nor the choice of a boundary image alone is asserted to certify the full family. $\\square$","very symmetric invariant form a scalar multiple of $H_0$. The nonzero displacement has strictly positive squared $H_0$ length, so the normalization is well-defined, positive and independent of the initially chosen positive scale. The final acceptance test is exactly Proposition 62. Neither invariance under the ratio nor the choice of a boundary image alone is asserted to certify the full family. $\\square$","At $u_*=0$, the operator ratio gives the required invariant-form equation. A rational rank-one observation supplies a boundary image. The recovered center and metric are","$$\n\\boxed{\nc=\\begin{pmatrix}1/2\\\\1/2\\end{pmatrix},\\qquad\nH=\\begin{pmatrix}100/9&-20/3\\\\-20/3&100/9\\end{pmatrix}.\n}\n$$","The two metric eigenvalues are $40/9$ and $160/9$. For both actions the exact certificate is","$$\n\\begin{aligned}\nB_a(u)^{\\mathsf T}QB_a(u)&=\\frac{(1-9u^2)^2}{16}Q,\\\\\n\\beta_a(u)&=\\frac{1+9u^2}{4},\\\\\n\\beta_a(u)^2-b_a(u)^{\\mathsf T}H^{-1}b_a(u)\n&=\\frac{(1-9u^2)^2}{16}.\n\\end{aligned}\n$$","Every scalar on the right has an exact rational sum-of-squares representation. Integrated mass is unchanged. The fitted initial state is the ellipsoid center, and the two densities $13/16$ and $413/1040$ from Section 66 are unchanged. This is a new certificate construction for the same model, not a change to its predictions.","In this section observations lie in $[0,1]$ with Lebesgue reference measure. This is an explicit change from the preceding $[-1,1]$ convention. Under $v=(u+1)/2$, the probability measure $du/2$ becomes $dv$. A Lipschitz constant with respect to $u$ becomes twice as large with respect to $v$.","$$\nb_{n,k}(x)=\\binom nk x^k(1-x)^{n-k},\\qquad\nD_n(x,v)=(n+1)\\sum_{k=0}^n b_{n,k}(x)b_{n,k}(v).\n$$","Both variables range over $[0,1]$. The kernel is nonnegative and integrates to one in either variable. Let $B_a(v)$ be any operator-norm integrable density preserving a common probability cone $C$, with mass functional $\\lambda$ and normalized base $K$. Define","$$\nB_a^{[n]}(x)=\\int_0^1D_n(x,v)B_a(v)\\,dv\n=\\sum_{k=0}^n b_{n,k}(x)C_{a,n,k},\n$$","$$\nC_{a,n,k}=(n+1)\\int_0^1b_{n,k}(v)B_a(v)\\,dv.\n$$","**Proposition 64. Valid positive polynomial approximation.** Every $B_a^{[n]}$ preserves $C$, has degree at most $n$, and satisfies exact integrated mass conservation. If","$$\n\\|B_a(v)-B_a(x)\\|_{C\\to C}\\leq L|v-x|\n$$","$$\n\\boxed{\n\\sup_{a,z\\in K}\\int_0^1\n\\|(B_a^{[n]}(x)-B_a(x))z\\|_C\\,dx\n\\leq\\frac{L}{\\sqrt{3(n+2)}}.\n}\n$$","With identical initial states, the corresponding length-$T$ laws under a common policy differ in total variation by at most","$$\n\\boxed{\\min\\left(1,\\frac{TL}{2\\sqrt{3(n+2)}}\\right).}\n$$","**Proof.** Nonnegative integration preserves the closed convex cone. Since $\\int_0^1b_{n,k}=1/(n+1)$ and $\\sum_k b_{n,k}=1$, the kernel is doubly stochastic. Its integral in $x$ gives $\\int B_a^{[n]}=\\int B_a$, proving exact mass conservation.","For the error, subtract $B_a(x)$ inside the kernel integral, apply the Lipschitz inequality and use $\\|z\\|_C=1$. The integrated discrepancy is at most $L\\mathbb E|V-X|$ under the joint probability density $D_n(x,v)$. This joint law can be constructed by first choosing $J$ uniformly from $\\{0,\\ldots,n\\}$, and then drawing $X,V$ independently, conditionally on $J=k$, with density $(n+1)b_{n,k}$.","$$\n\\operatorname{Var}(X\\mid J=k)\n=\\frac{(k+1)(n-k+1)}{(n+2)^2(n+3)}.\n$$","The average of $(k+1)(n-k+1)$ over $k=0,\\ldots,n$ is $(n+2)(n+3)/6$, obtained by summing the first two integer powers. Hence","$$\n\\mathbb E(V-X)^2\n=2\\mathbb E\\operatorname{Var}(X\\mid J)\n=\\frac1{3(n+2)}.\n$$","Cauchy--Schwarz gives the claimed integrated bound. The whole-law conclusion is Proposition 59 applied to the two valid instruments on the same domain. $\\square$","Operationally, it replaces the raw observation $v$ by a noisy observation $x$ drawn from $D_n(x,v)$. It preserves total mass and underlying unconditioned evolution, not the entire original observation law. That changed law is what the bound measures. A prediction policy must be compared as the same function of the reported history.","$$\np(v)=\\frac12+2\\left|v-\\frac12\\right|.\n$$","It integrates to one, takes values between $1/2$ and $3/2$, and has Lipschitz constant two. It is not a single polynomial on $[0,1]$. Proposition 64 nevertheless supplies normalized positive polynomial approximations at every degree, with an explicit approximation budget.","An ordinary orthogonal truncation need not preserve that validity. The valid density $p(v)=3v^2$ has degree-one $L^2$ projection","$$\nq(v)=3v-\\frac12,\n$$","$$\np^{[1]}(v)=v+\\frac12.\n$$","**Proposition 65. Certified coefficient error and approximation bias.** Suppose every fitted coefficient $\\widetilde C_{a,n,k}$ preserves the same cone $C$, and","$$\n\\lambda\\frac1{n+1}\\sum_{k=0}^n\\widetilde C_{a,n,k}=\\lambda.\n$$","Then $\\widetilde B_a^{[n]}(x)=\\sum_k b_{n,k}(x)\\widetilde C_{a,n,k}$ is internally valid. If","$$\n\\sup_{z\\in K}\\|(C_{a,n,k}-\\widetilde C_{a,n,k})z\\|_C\n\\leq\\eta_{a,k},\n$$","$$\n\\boxed{\n\\frac{L}{\\sqrt{3(n+2)}}+\n\\max_a\\frac1{n+1}\\sum_{k=0}^n\\eta_{a,k}.\n}\n$$","**Proof.** Nonnegative Bernstein weights preserve cone positivity, and their integrals prove normalization. For any $z\\in K$, the triangle inequality gives","$$\n\\int\\|\\textstyle\\sum_k b_{n,k}(x)(C_{a,n,k}-\\widetilde C_{a,n,k})z\\|_C\\,dx\n\\leq\\frac1{n+1}\\sum_k\\eta_{a,k}.\n$$","Add the approximation error from Proposition 64. A separate correction is handled by another application of the triangle inequality. $\\square$","This is a conditional approximation-and-estimation theorem, not a claim that more coefficients always improve a finite-data estimator. Increasing $n$ reduces the stated bias bound while changing the number and conditioning of the quantities to be estimated. A cone witness proves that the resulting object is a probability model. It does not prove that the coefficient error budgets cover the generating process.","The moments needed for $C_{a,n,k}$ are bounded Bernstein-weighted experiments. Their interpretation as full operator coefficients still requires suitable predictive-state coverage or a justified observable core. The next section provides sequential evidence for conditional experiments, not an automatic replacement for those identification assumptions.","Let $\\mathcal F_{t-1}$ denote the available information immediately before observing the next measurement, including the chosen action, current model and any test selected using past data. A predictable test is a measurable function $\\phi_{t,j}$ of the upcoming measurement, chosen from this information, with values in $[0,1]$. Put","$$\nX_{t,j}=\\phi_{t,j}(Y_t),\\quad\nm_{t,j}=\\mathbb E[X_{t,j}\\mid\\mathcal F_{t-1}],\\quad\np_{t,j}=\\text{the model's prediction for that test}.\n$$","The model prediction is also predictable. It may change as the model learns. The true conditional mean $m_{t,j}$ may depend on the entire past. No independence, stationarity or known reset state is assumed in this paragraph.","Choose budgets $\\alpha_j>0$ for a finite or countable prospectively specified family of test rules, with $\\sum_j\\alpha_j\\leq\\delta$. Define","$$\nR_{t,j}=\\sum_{s=1}^t(X_{s,j}-p_{s,j}),\\qquad\nD_{t,j}=\\sum_{s=1}^t(m_{s,j}-p_{s,j}).\n$$","**Proposition 66. Time-uniform predictable-residual certificate.** With probability at least $1-\\delta$, for every time $t\\geq1$ and every registered rule $j$,","$$\n\\boxed{\n|R_{t,j}-D_{t,j}|\\leq\nb_{t,j},\\qquad\nb_{t,j}=\\sqrt{\\frac t2\\log\\frac{2t(t+1)}{\\alpha_j}}.\n}\n$$","For variables taking values in an interval of width $c$, multiply the boundary by $c$. The simultaneous guarantee remains valid when the observation time is chosen from the data.","**Proof.** For a random variable $X\\in[0,1]$ with conditional mean $m$, its conditional centered logarithmic moment-generating function has value and first derivative zero at the origin. Its second derivative is the variance under exponential tilting. A variable supported on $[0,1]$ has variance at most $1/4$, because its squared distance from $1/2$ is at most $1/4$. Integrating twice gives","$$\n\\mathbb E[e^{\\theta(X-m)}\\mid\\mathcal F]\\leq e^{\\theta^2/8}.\n$$","Iterating this conditional inequality gives the same exponential bound for the martingale sum $\\sum_{s\\leq t}(X_{s,j}-m_{s,j})$. Optimizing the exponential Markov inequality gives the two-sided bound $2\\exp(-2b^2/t)$. At the displayed $b_{t,j}$ this is $\\alpha_j/[t(t+1)]$. Sum over $t$, using $\\sum_{t\\geq1}1/[t(t+1)]=1$, and then over $j$. The resulting event holds at every time, so it also holds at a data-selected time. Scaling the variable proves the range version. $\\square$","r the martingale sum $\\sum_{s\\leq t}(X_{s,j}-m_{s,j})$. Optimizing the exponential Markov inequality gives the two-sided bound $2\\exp(-2b^2/t)$. At the displayed $b_{t,j}$ this is $\\alpha_j/[t(t+1)]$. Sum over $t$, using $\\sum_{t\\geq1}1/[t(t+1)]=1$, and then over $j$. The resulting event holds at every time, so it also holds at a data-selected time. Scaling the variable proves the range version. $\\square$","For intervention evidence use the information $\\mathcal H_{t-1}$ just before choosing the action. Let the action be randomized from a known history-dependent policy $\\pi_t$. The true next-observation kernel under action $a$ is $P_t(\\cdot\\mid\\mathcal H_{t-1},a)$. Randomization uses no unrecorded future information. For a predictable test $\\phi_t\\in[0,1]$, define","$$\n\\mu_t^a=\\int\\phi_t(y)P_t(dy\\mid\\mathcal H_{t-1},a),\\qquad\nW_t^a=\\frac{\\mathbf1_{\\{A_t=a\\}}\\phi_t(Y_t)}{\\pi_t(a\\mid\\mathcal H_{t-1})}.\n$$","**Proposition 67. Covered adaptive-action evidence.** If the target action has $\\pi_t(a\\mid\\mathcal H_{t-1})\\geq\\pi_{\\min}>0$ at every time, then","$$\n\\mathbb E[W_t^a\\mid\\mathcal H_{t-1}]=\\mu_t^a,\n\\qquad 0\\leq W_t^a\\leq1/\\pi_{\\min}.\n$$","Consequently Proposition 66 gives a time-uniform certificate for $\\sum_{s\\leq t}\\mu_s^a$ from $\\sum_{s\\leq t}W_s^a$, with its boundary multiplied by $1/\\pi_{\\min}$. The target is the average action-conditioned quantity along the encountered contexts, not a reset-population table or the state distribution under a different long-term policy.","**Proof.** Condition first on the action and then on its observation. Only action $a$ contributes, with probability $\\pi_t(a\\mid\\mathcal H_{t-1})$, cancelling the denominator. The range bound follows directly. Apply the scaled conditional moment-generating-function argument to the resulting martingale differences. $\\square$","This simple bound is conservative. Variance-aware weighting can improve it, but cannot remove the need for information about the action of interest. When $\\pi_{\\min}=0$, two models can give exactly the same law under the observed policy and disagree on a never-selected action. For example, both can emit a Bernoulli observation with parameter $1/2$ under action zero, while their action-one parameters are $1/10$ and $9/10$. Repeating action zero cannot distinguish them.","ple bound is conservative. Variance-aware weighting can improve it, but cannot remove the need for information about the action of interest. When $\\pi_{\\min}=0$, two models can give exactly the same law under the observed policy and disagree on a never-selected action. For example, both can emit a Bernoulli observation with parameter $1/2$ under action zero, while their action-one parameters are $1/10$ and $9/10$. Repeating action zero cannot distinguish them.","s conservative. Variance-aware weighting can improve it, but cannot remove the need for information about the action of interest. When $\\pi_{\\min}=0$, two models can give exactly the same law under the observed policy and disagree on a never-selected action. For example, both can emit a Bernoulli observation with parameter $1/2$ under action zero, while their action-one parameters are $1/10$ and $9/10$. Repeating action zero cannot distinguish them.","Let $P_{t,j}$ and $Q_{t,j}$ denote the true and proposed next-observation laws on the same actual history and action. The proposed law must be defined there. Since $0\\leq\\phi_{t,j}\\leq1$,","$$\n|m_{t,j}-p_{t,j}|\\leq\\|P_{t,j}-Q_{t,j}\\|_{\\mathrm{TV}}.\n$$","$$\n\\boxed{\n\\frac1t\\sum_{s=1}^t\\|P_{s,j}-Q_{s,j}\\|_{\\mathrm{TV}}\n\\geq\\frac{(|R_{t,j}|-b_{t,j})_+}{t}.\n}\n$$","**Proof.** The triangle inequality gives $|D_{t,j}|\\geq(|R_{t,j}|-b_{t,j})_+$. Also $|D_{t,j}|\\leq\\sum_s|m_{s,j}-p_{s,j}|\\leq\\sum_s\\|P_{s,j}-Q_{s,j}\\|_{\\mathrm{TV}}$. Combine these inequalities. $\\square$","The reference path has binary observations. Its true conditional probability of one is $4/5$ after a one and $1/5$ after a zero. A deliberately inadequate model predicts $1/2$ at every step. Its long-run aggregate frequency can be near one half even though its history-conditioned predictions are wrong.","For the supplied 4,096-step seeded path, the aggregate mean residual is approximately $0.00732$. The gated mean residual is approximately $0.14917$, while its time-uniform mean boundary is approximately $0.05114$. The lower certificate for average conditional total variation is approximately $0.09803$. The correctly predicted gated residual is approximately $-0.00303$.","Let $P$ be any causal process on a standard Borel observation space with finite actions. Let $Q$ be a proposed process. Both use the same measurable policy. Write $P_t(\\cdot\\mid h,a)$ and $Q_t(\\cdot\\mid h,a)$ for their next-observation kernels, and define","$$\nd_t(h,a)=\\|P_t(\\cdot\\mid h,a)-Q_t(\\cdot\\mid h,a)\\|_{\\mathrm{TV}}.\n$$","The proposed kernel must be specified on every history needed for the comparison, including histories impossible under $Q$ but possible under $P$. A declared fallback on a model-null history completes a causal kernel without changing $Q$'s own law. An implementation that rejects such a history must report that rejection rather than silently claim to have filtered it.","**Proposition 69. Observation-law error without a shared state space.** With the same initial history, the length-$T$ joint action-and-observation laws satisfy","$$\n\\boxed{\n\\|P_T-Q_T\\|_{\\mathrm{TV}}\n\\leq\\min\\left(1,\\sum_{t=0}^{T-1}\n\\mathbb E_{P,\\pi}[d_t(H_t,A_t)]\\right).\n}\n$$","An initial-history discrepancy adds its total variation. In particular, uniform bounds $d_t(h,a)\\leq\\epsilon_t$ give the upper bound $\\min(1,\\sum_{t<T}\\epsilon_t)$.","**Proof.** Introduce hybrid laws that follow $P$ for the first $t$ transitions and $Q$ for the rest, with the common policy throughout. Two successive hybrids have the same $P$ prefix distribution up to time $t$ and the same future $Q$ kernel after the next transition. Conditional on that prefix and its policy-chosen action, changing the next kernel costs $d_t(h,a)$ in total variation. A common continuation kernel cannot increase total variation, because integration of a test bounded between zero and one produces another such test. Average over the shared prefix, sum the adjacent-hybrid bounds, and cap by one. The same contraction argument handles a different initial law. $\\square$","on that prefix and its policy-chosen action, changing the next kernel costs $d_t(h,a)$ in total variation. A common continuation kernel cannot increase total variation, because integration of a test bounded between zero and one produces another such test. Average over the shared prefix, sum the adjacent-hybrid bounds, and cap by one. The same contraction argument handles a different initial law. $\\square$","**Proposition 70. Separate structural acceptance, approximation control and sequential rejection.** Consider a fitted polynomial model in a declared predictive coordinate system, and prospectively registered observation tests with total budget $\\delta$.","For any generating density on a common probability cone with a justified Lipschitz constant, a positive degree-$n$ approximation and certified coefficient budgets satisfy Propositions 64--65, hence a fixed-horizon upper law bound. If a generating process lies outside that representation, a separately justified bound on its conditional kernels instead invokes Proposition 69.","regularity and coefficient conditions, or alternatively the conditional-kernel discrepancy assumptions. The statistical conclusions hold on the simultaneous probability event of Proposition 66. None of these implications uses nonrejection as an upper bound or infers coverage from a finite determinant. Their conjunction therefore retains the separate hypotheses and conclusions exactly as stated. $\\square$","Let $\\mathcal L_d=\\{(m,v):m\\geq\\|v\\|_2\\}$, with mass row $\\lambda=(1,0,\\ldots,0)$. A mass preserving invertible matrix $T$ transports a proposed predictive domain to this cone. Its normalized base in the original coordinates is","$$\nK_T=\\{T^{-1}(1,v):\\|v\\|_2\\leq1\\}.\n$$","Write $F_a(u)=TB_a(u)T^{-1}$ for the transported operator density. Observations lie in $[-1,1]$ with reference measure $du/2$. All operator densities in this block are polynomial and all equalities involving integrals use that reference measure. For each action define","$$\n\\sigma_a(u)=\\min_{\\|v\\|_2\\leq1}\n\\left[(F_a(u)(1,v))_0-\\|(F_a(u)(1,v))_{\\rm sp}\\|_2\\right].\n$$","The subscript $\\rm sp$ denotes spatial coordinates, not a separate measured signal. This margin is expressed before normalization. It simultaneously controls mass orientation and distance from the curved boundary. It is coordinate dependent, so its uncertainty budget must use the same $T$.","**Proposition 71. Robust oriented margin.** Suppose $\\sigma_a(u)\\geq\\kappa_a(u)$ and an uncertain family has transported errors $\\Delta_a(u)$ satisfying, for every normalized input,","$$\n\\begin{aligned}\n|[\\Delta_a(u)(1,v)]_0|&\\leq e_{a,0}(u),\\\\\n\\|[\\Delta_a(u)(1,v)]_{\\rm sp}\\|_2&\\leq e_{a,1}(u).\n\\end{aligned}\n$$","If $\\kappa_a(u)\\geq e_{a,0}(u)+e_{a,1}(u)$, every uncertain branch preserves $\\mathcal L_d$. If the uncertain family has exact integrated mass and its initial state belongs to $K_T$, it defines a valid probability model at every finite horizon. With a strict inequality, the residual slack is at least $\\kappa_a-e_{a,0}-e_{a,1}$.","**Proof.** For $w=F_a(u)(1,v)$ and $d=\\Delta_a(u)(1,v)$,","$$\n(w_0+d_0)-\\|w_{\\rm sp}+d_{\\rm sp}\\|_2\n\\geq w_0-\\|w_{\\rm sp}\\|_2-|d_0|-\\|d_{\\rm sp}\\|_2.\n$$","nonnegative under the hypothesis. Homogeneity extends the assertion from normalized inputs to the entire cone. The only zero-mass cone element is zero. Integrated mass conservation and successive cone preservation give nonnegative normalized observation laws by the same induction as in Proposition 52. A positive-density branch may be normalized. A zero-density branch is not assigned a posterior. $\\square$","**Proposition 72. Boundary fragility under mass preserving perturbations.** Assume $d\\geq1$. Let $F$ preserve $\\mathcal L_d$. Suppose a normalized input $z=(1,v)$ has a boundary image $Fz=(m,w)$ with $m=\\|w\\|_2$. For every $\\epsilon>0$, there is a linear perturbation $\\Delta$ with $\\lambda\\Delta=0$ and","$$\n\\sup_{\\|v'\\|_2\\leq1}\\|\\Delta(1,v')\\|_{\\mathcal L_d}=\\epsilon\n$$","such that $(F+\\Delta)z\\notin\\mathcal L_d$. Here the base norm is $\\|(m,w)\\|_{\\mathcal L_d}=\\max(|m|,\\|w\\|_2)$.","**Proof.** If $w\\ne0$, put $e=w/\\|w\\|_2$. If $w=0$, choose any spatial unit vector $e$. Define $\\Delta=\\epsilon(0,e)\\lambda$. Its mass row vanishes and its value on every normalized input is $(0,\\epsilon e)$, of base norm $\\epsilon$. If $w\\ne0$, the new spatial norm is $m+\\epsilon$. If $w=0$, the original output mass is zero and the new spatial norm is $\\epsilon$. Either image lies outside the cone. $\\square$","oof.** If $w\\ne0$, put $e=w/\\|w\\|_2$. If $w=0$, choose any spatial unit vector $e$. Define $\\Delta=\\epsilon(0,e)\\lambda$. Its mass row vanishes and its value on every normalized input is $(0,\\epsilon e)$, of base norm $\\epsilon$. If $w\\ne0$, the new spatial norm is $m+\\epsilon$. If $w=0$, the original output mass is zero and the new spatial norm is $\\epsilon$. Either image lies outside the cone. $\\square$","The positive spatial dimension is essential. When $d=0$, the state is mass alone and $\\lambda$ is the identity on $\\mathbb R$. The condition $\\lambda\\Delta=0$ forces $\\Delta=0$, so a perturbation of prescribed norm $\\epsilon>0$ cannot exist. This explicit boundary was missing from the earlier wording. All previously worked examples have two spatial coordinates and are unaffected.","In the preceding exact continuous example, the direct operator at $u=0$ is $I/4$. Every normalized boundary state has a boundary image. There is therefore no positive unrestricted matrix-error radius on that domain, even though the original family is exactly valid. A robust theorem that silently treats its zero slack as positive would be false.","$$\nN_a(u)=g_a(u)e_0\\lambda,\\qquad\n g_a(u)\\geq g_{\\min}>0,\\qquad \\int g_a(u)\\,du/2=1.\n$$","This reference instrument reports a state-independent density and resets to the center. It is a mathematical model modification, not a new observation. Given an estimate $\\widehat F$, define","$$\n\\widetilde F_a(u)=(1-\\eta)\\widehat F_a(u)+\\eta N_a(u),\n\\qquad 0\\leq\\eta\\leq1.\n$$","**Proposition 73. Interior repair with an explicit error budget.** Suppose a reference family $F^*$ is valid on $\\mathcal L_d$, and the combined mass and spatial error of $\\widehat F-F^*$ is at most $d_a(u)$ on normalized inputs. Then the margin of $\\widetilde F_a(u)$ is at least","$$\n\\boxed{\\eta g_a(u)-(1-\\eta)d_a(u).}\n$$","An additional uncertainty family with combined output error at most $\\nu_a(u)$ is certified whenever","$$\n\\eta g_a(u)\\geq(1-\\eta)d_a(u)+\\nu_a(u).\n$$","Assume exact integrated mass for the estimate. If its integrated base-norm discrepancy from $F^*$ is at most $\\varepsilon$, the corrected discrepancy is at most","$$\n\\boxed{(1-\\eta)\\varepsilon+2\\eta.}\n$$","When the corrected model is valid on the same cone, with the same initial state, its length-$T$ law under any common policy differs from the reference law by at most","$$\n\\min\\left(1,\\frac T2[(1-\\eta)\\varepsilon+2\\eta]\\right).\n$$","**Proof.** The margin of $\\widehat F$ is at least $-d_a$ by Proposition 71's inequality applied to $F^*$. Adding the reference changes only output mass, by $\\eta g_a$, while scaling the remaining output by $1-\\eta$. This proves the first bound and the robust implication. Integrated mass is preserved by mixing.","$$\n\\widetilde F-F^*=(1-\\eta)(\\widehat F-F^*)+\\eta(N-F^*).\n$$","On a normalized input, both $N$ and $F^*$ distribute total cone mass one. Their integrated difference has base norm at most two. Integrate the triangle inequality to obtain the stated budget. The final implication is the continuous observation-law bound of Proposition 59. It requires validity of both comparison models on the same cone. $\\square$","For constant budgets, a sufficient repair weight is $(d+\\nu)/(g_{\\min}+d)$ when this number is at most one. This is not an optimality claim. The reference example instead fixes a repair weight and verifies the result independently. If an exact model already has a robust structured certificate, adding a reset may be unnecessary and would introduce avoidable bias.","The term $2\\eta$ is deliberately retained. A larger repair creates more room for uncertainty but can change future predictions more. The worst-case law bound can become uninformative at long horizons. Neither positivity nor mixing alone proves long-horizon predictive accuracy.","The exact identity $A^{\\mathsf T}HA=H$ need not survive estimation error. A useful proposal can still be obtained from a least-squares invariant form, but the residual must not be interpreted as a positive-cone guarantee.","Let $\\mathcal M_A$ map symmetric matrices into a Euclidean product space by","$$\n\\mathcal M_A(H)=(A^{\\mathsf T}HA-H,\\operatorname{tr}H).\n$$","Use the Frobenius norm on symmetric matrices and the induced product norm. Assume $\\mathcal M_{A^*}$ is injective and $\\mathcal M_{A^*}(H^*)=(0,1)=b$. This fixes the scale of a one-dimensional invariant-form space.","**Proposition 74. A posteriori proposal error.** Given $\\widehat H$, let $r=\\mathcal M_{\\widehat A}(\\widehat H)-b$. Suppose","$$\n\\|\\mathcal M_{\\widehat A}-\\mathcal M_{A^*}\\|\\leq e_M,\n\\qquad \\sigma_{\\min}(\\mathcal M_{\\widehat A})>e_M.\n$$","$$\n\\boxed{\n\\|\\widehat H-H^*\\|_F\n\\leq\n\\frac{\\|r\\|+e_M\\|\\widehat H\\|_F}\n{\\sigma_{\\min}(\\mathcal M_{\\widehat A})-e_M}.\n}\n$$","One may take $e_M$ from the bound","$$\n\\|\\mathcal M_{\\widehat A}-\\mathcal M_{A^*}\\|\\leq(\\|\\widehat A\\|_2+\\|A^*\\|_2)\\|\\widehat A-A^*\\|_2.\n$$","For fixed centers $c^*=(I-A^*)^{-1}b^*$ and $\\widehat c=(I-\\widehat A)^{-1}\\widehat b$, with $\\|\\widehat A-A^*\\|_2\\leq e_A$ and $\\|\\widehat b-b^*\\|_2\\leq e_b$, one has","$$\n\\|\\widehat c-c^*\\|_2\n\\leq\\frac{e_b+e_A\\|\\widehat c\\|_2}\n{\\sigma_{\\min}(I-\\widehat A)-e_A}\n$$","$$\n\\mathcal M_{A^*}(\\widehat H-H^*)\n=r-(\\mathcal M_{\\widehat A}-\\mathcal M_{A^*})\\widehat H.\n$$","The least singular value of $\\mathcal M_{A^*}$ is at least that of $\\mathcal M_{\\widehat A}$ minus $e_M$. Apply this lower bound and the triangle inequality. To bound the map perturbation, insert and subtract $\\widehat A^{\\mathsf T}HA^*$ and apply the Frobenius product inequality. The trace component is unchanged. Finally,","$$\n(I-A^*)(\\widehat c-c^*)=(\\widehat b-b^*)+(\\widehat A-A^*)\\widehat c.\n$$","The same singular-value argument proves the center bound. $\\square$","A rationalized invertible factor $L$ specifies $H=L^{\\mathsf T}L$ exactly. Positive definiteness is then an algebraic fact even when the factor was suggested by floating-point calculations. It still does not establish operator invariance.","For three predictive coordinates, the spatial unit circle admits two rational charts. For $s\\in\\{1,-1\\}$ and $t\\in[-1,1]$, put","$$\nd(t)=1+t^2,\\qquad\nr_s(t)=\\frac{s}{d(t)}(1-t^2,2t).\n$$","The charts cover the two semicircles, including their endpoints. For an operator $F_a(u)$ and required margin $\\kappa\\geq0$, form polynomial vectors and scalars","$$\n\\begin{aligned}\nw_{a,s}(u,t)&=F_a(u)(d(t),s(1-t^2),2st)^{\\mathsf T},\\\\\nf_{a,s}(u,t)&=w_{a,s,0}(u,t)-\\kappa d(t),\\\\\nh_{a,s}(u,t)&=f_{a,s}(u,t)^2-w_{a,s,1}(u,t)^2-w_{a,s,2}(u,t)^2.\n\\end{aligned}\n$$","**Proposition 75. Rational semicircle and Bernstein certificate.** If $f_{a,s}$ and $h_{a,s}$ are nonnegative on $[-1,1]^2$ for both charts and every action, then $\\sigma_a(u)\\geq\\kappa$ for every observation. For rational operators and geometry, nonnegative tensor-Bernstein coefficients on a complete rational subdivision of that square give a finite exact witness. If the actual margin is uniformly strictly greater than $\\kappa$, sufficiently fine dyadic subdivision succeeds.","l semicircle and Bernstein certificate.** If $f_{a,s}$ and $h_{a,s}$ are nonnegative on $[-1,1]^2$ for both charts and every action, then $\\sigma_a(u)\\geq\\kappa$ for every observation. For rational operators and geometry, nonnegative tensor-Bernstein coefficients on a complete rational subdivision of that square give a finite exact witness. If the actual margin is uniformly strictly greater than $\\kappa$, sufficiently fine dyadic subdivision succeeds.","**Proof.** The first polynomial inequality establishes output orientation after subtracting $\\kappa d$. The second then gives","$$\nw_{a,s,0}-\\|(w_{a,s,1},w_{a,s,2})\\|_2\\geq\\kappa d.\n$$","Divide by $d>0$ to obtain the required margin on the circle. For fixed $u$, the function consisting of affine output mass minus the norm of affine spatial output is concave in the input. Every point in the closed disk is a convex combination of circle points. Concavity therefore extends the lower bound to the disk.","On any rectangle mapped affinely to $[0,1]^2$, the tensor-Bernstein basis functions are nonnegative and sum to one. A polynomial with nonnegative coefficients in that basis is nonnegative throughout the rectangle. If the rectangles cover the square, their inequalities prove the global statement. The reference verifier checks coverage through a complete subdivision tree, not by accepting an arbitrary list of rectangles.","For termination under strict margin, both $f$ and $h$ are strictly positive on the compact square. Their minima are positive. For a fixed-degree polynomial on a small cell of side at most $\\delta$, its Taylor expansion at one corner has constant term equal to its value there and nonconstant coefficients bounded uniformly by $O(\\delta)$. The linear change from the fixed-degree power basis to the Bernstein basis is fixed and bounded. Thus every Bernstein coefficient differs from that constant value by $O(\\delta)$, uniformly over cells. At sufficiently small $\\delta$ all coefficients are positive. A finite dyadic subdivision attains that size. $\\square$","inima are positive. For a fixed-degree polynomial on a small cell of side at most $\\delta$, its Taylor expansion at one corner has constant term equal to its value there and nonconstant coefficients bounded uniformly by $O(\\delta)$. The linear change from the fixed-degree power basis to the Bernstein basis is fixed and bounded. Thus every Bernstein coefficient differs from that constant value by $O(\\delta)$, uniformly over cells. At sufficiently small $\\delta$ all coefficients are positive. A finite dyadic subdivision attains that size. $\\square$","all cell of side at most $\\delta$, its Taylor expansion at one corner has constant term equal to its value there and nonconstant coefficients bounded uniformly by $O(\\delta)$. The linear change from the fixed-degree power basis to the Bernstein basis is fixed and bounded. Thus every Bernstein coefficient differs from that constant value by $O(\\delta)$, uniformly over cells. At sufficiently small $\\delta$ all coefficients are positive. A finite dyadic subdivision attains that size. $\\square$","equal to its value there and nonconstant coefficients bounded uniformly by $O(\\delta)$. The linear change from the fixed-degree power basis to the Bernstein basis is fixed and bounded. Thus every Bernstein coefficient differs from that constant value by $O(\\delta)$, uniformly over cells. At sufficiently small $\\delta$ all coefficients are positive. A finite dyadic subdivision attains that size. $\\square$","$$\n\\Delta_a(u)=\\sum_{k=0}^{D}u^kE_{a,k},\\qquad\n|(E_{a,k})_{ij}|\\leq r_k.\n$$","$$\n\\boxed{e_0=3\\sum_{k=0}^{D}r_k,\\qquad e_1=6\\sum_{k=0}^{D}r_k.}\n$$","Only coefficient arrays obeying $\\lambda\\int\\Delta_a(u)\\,du/2=0$ are included in the normalized uncertainty family. A checked margin at least $e_0+e_1$ certifies every such array.","Alternatively, if finitely many complete polynomial operator families satisfy a common margin $\\kappa$ on the same domain, every convex combination of those families satisfies that margin. Initial states may also range over a certified convex hull.","For a fixed geometry, the constraints $\\sigma_a(u)\\geq\\kappa$, exact mass, and membership in a convex data-consistency set are convex constraints on the operator coefficients and initial state. Joint optimization of geometry and operators is not asserted to be convex.","**Proof.** Each output scalar is bounded by $3r_k$ on a normalized input. Bound the spatial Euclidean norm by the sum of the absolute values of its two coordinates, obtaining $6r_k$. Sum over coefficients using $|u|\\leq1$. Mass conservation is a separate linear constraint and follows for every family satisfying it.","For convex combinations, output mass is affine and the spatial norm is convex. Hence the margin of the combined output is at least the same combination of margins. A convex cone contains convex combinations of admissible initial states. For fixed geometry and input, the inequality affine mass minus a norm at least $\\kappa$ is convex in the coefficients. Intersect over all inputs and observations, then add the linear mass constraints and the convex data set. $\\square$","norm is convex. Hence the margin of the combined output is at least the same combination of margins. A convex cone contains convex combinations of admissible initial states. For fixed geometry and input, the inequality affine mass minus a norm at least $\\kappa$ is convex in the coefficients. Intersect over all inputs and observations, then add the linear mass constraints and the convex data set. $\\square$","**Proposition 77. Conditional search for strict rational curved certificates.** Consider finitely many rational polynomial operator families in three coordinates, each with exact mass, and finitely many rational normalized initial states. Suppose there exists a real mass preserving invertible $T_*$ such that every transformed initial state lies strictly inside the Lorentz cone and every transformed operator family has a common positive margin on the observation interval. Then a rational mass preserving invertible $T$, a positive rational margin, and finite rational Bernstein witnesses exist. A dovetailed enumeration of rational transforms, positive rational margins and subdivision depths eventually finds such a witness. Whenever it returns, the witness is sound for the convex hull of the supplied families.","al operator families in three coordinates, each with exact mass, and finitely many rational normalized initial states. Suppose there exists a real mass preserving invertible $T_*$ such that every transformed initial state lies strictly inside the Lorentz cone and every transformed operator family has a common positive margin on the observation interval. Then a rational mass preserving invertible $T$, a positive rational margin, and finite rational Bernstein witnesses exist. A dovetailed enumeration of rational transforms, positive rational margins and subdivision depths eventually finds such a witness. Whenever it returns, the witness is sound for the convex hull of the supplied families.","**Proof.** Inversion and matrix multiplication are continuous near $T_*$. Initial membership is strict, and the operator margin is continuous jointly in the transform, the observation, and the normalized input. The latter two range over compact sets and there are finitely many families. Thus the strict inequalities persist in a neighborhood of $T_*$. Rational mass preserving matrices are dense in that affine matrix space, and invertibility persists in a neighborhood. Choose a rational transform there and a rational positive margin smaller than the remaining minimum. Proposition 75 supplies finite subdivisions because there is still a strict gap above the chosen margin. Enumerating all finite rational descriptions and depths eventually reaches this finite witnes","**Proof.** Inversion and matrix multiplication are continuous near $T_*$. Initial membership is strict, and the operator margin is continuous jointly in the transform, the observation, and the normalized input. The latter two range over compact sets and there are finitely many families. Thus the strict inequalities persist in a neighborhood of $T_*$. Rational mass preserving matrices are dense in that affine matrix space, and invertibility persists in a neighborhood. Choose a rational transform there and a rational positive margin smaller than the remaining minimum. Proposition 75 supplies finite subdivisions because there is still a strict gap above the chosen margin. Enumerating all finite rational descriptions and depths eventually reaches this finite witness. Proposition 76 proves the convex-hull conclusion. $\\square$","and invertibility persists in a neighborhood. Choose a rational transform there and a rational positive margin smaller than the remaining minimum. Proposition 75 supplies finite subdivisions because there is still a strict gap above the chosen margin. Enumerating all finite rational descriptions and depths eventually reaches this finite witness. Proposition 76 proves the convex-hull conclusion. $\\square$","**Proposition 78. Data-selected geometry with honest coverage.** Suppose a random set $\\mathcal U$ of operators and initial states, in a declared fixed predictive coordinate system, contains the true model with probability at least $1-\\delta$. A data-dependent procedure may propose any transform $T$ and any domain. If its exact verifier proves domain membership and invariance for every member of the same set $\\mathcal U$, then on that coverage event the true model has the verified domain. No union bound over geometry proposals is needed for this deterministic implication.","78. Data-selected geometry with honest coverage.** Suppose a random set $\\mathcal U$ of operators and initial states, in a declared fixed predictive coordinate system, contains the true model with probability at least $1-\\delta$. A data-dependent procedure may propose any transform $T$ and any domain. If its exact verifier proves domain membership and invariance for every member of the same set $\\mathcal U$, then on that coverage event the true model has the verified domain. No union bound over geometry proposals is needed for this deterministic implication.","If original-coordinate perturbations satisfy $\\|\\Delta\\|_2\\leq e$, their combined mass and spatial output error after a known transform $T$ is at most","$$\n\\boxed{2\\|T\\|_2\\|T^{-1}\\|_2 e.}\n$$","The corresponding transformed uncertainty budget must be used even when $T$ was chosen from the data.","**Proof.** On the coverage event the true model is a member of the set quantified by the verifier. Its domain properties follow by direct inclusion, regardless of how $T$ was selected. For the error, a normalized Lorentz input has Euclidean norm at most $\\sqrt2$. Thus the Euclidean norm of $T\\Delta T^{-1}z$ is at most $\\sqrt2\\|T\\|_2\\|T^{-1}\\|_2e$. The sum of its mass magnitude and spatial norm is at most $\\sqrt2$ times its Euclidean norm. Multiply the two bounds. $\\square$","* On the coverage event the true model is a member of the set quantified by the verifier. Its domain properties follow by direct inclusion, regardless of how $T$ was selected. For the error, a normalized Lorentz input has Euclidean norm at most $\\sqrt2$. Thus the Euclidean norm of $T\\Delta T^{-1}z$ is at most $\\sqrt2\\|T\\|_2\\|T^{-1}\\|_2e$. The sum of its mass magnitude and spatial norm is at most $\\sqrt2$ times its Euclidean norm. Multiply the two bounds. $\\square$","t quantified by the verifier. Its domain properties follow by direct inclusion, regardless of how $T$ was selected. For the error, a normalized Lorentz input has Euclidean norm at most $\\sqrt2$. Thus the Euclidean norm of $T\\Delta T^{-1}z$ is at most $\\sqrt2\\|T\\|_2\\|T^{-1}\\|_2e$. The sum of its mass magnitude and spatial norm is at most $\\sqrt2$ times its Euclidean norm. Multiply the two bounds. $\\square$","The input to the new proposer contains only two perturbed polynomial operators and the initial vector. A separate fixture generator creates those inputs from the preceding continuous model by adding $\\pm\\delta E(u)$, with opposite signs for the two actions and","$$\n\\delta=\\frac1{2000},\\qquad\nE(u)=E_0+uE_1+u^2E_2,\n$$","$$\n\\begin{aligned}\nE_0&=\\begin{pmatrix}0&0&0\\\\-1&2&0\\\\0&1&-1\\end{pmatrix},&\nE_1&=\\begin{pmatrix}0&1&-1\\\\1&0&1\\\\1&-1&0\\end{pmatrix},&\nE_2&=\\begin{pmatrix}0&0&0\\\\0&-1&1\\\\1&0&1\\end{pmatrix}.\n\\end{aligned}\n$$","$$\nc=\\begin{pmatrix}1/2\\\\1/2\\end{pmatrix},\\qquad\nL=\\begin{pmatrix}123/37&-2\\\\0&109/41\\end{pmatrix},\n$$","$$\n\\boxed{\nH=L^{\\mathsf T}L=\n\\begin{pmatrix}15129/1369&-246/37\\\\-246/37&18605/1681\\end{pmatrix}.\n}\n$$","This is not the exact predecessor metric. The transformed family is not assumed conformal. The raw fitted family actually fails the proposed domain: for the direct action at $u=-1$ and Lorentz input $(1,1,0)$, its output mass is $106321/82000>0$, while the output quadratic form is","$$\n-\\frac{12009932651317769}{557059220496000000}<0.\n$$","The construction next mixes each operator with the center-reset reference using $\\eta=1/8$. The exact checker proves a margin of at least $1/16$ for the resulting family. It verifies eight polynomial inequalities across both actions and both semicircle charts. The witness contains 116 subdivision leaves and 1,780 exactly recomputed nonnegative Bernstein coefficients. These are arithmetic certificate sizes, not separate theorems.","The exported family admits mass and spatial perturbation budgets each equal to $1/128$, leaving certified slack $3/64$. In particular, it covers every degree-two transformed coefficient perturbation with each entry bounded by $1/4096$, subject to the exact integrated mass constraints. Proposition 76 verifies that the whole coefficient box lies inside the declared output budgets. No uncertainty vertices or future trajectories are sampled to establish that inclusion.","$$\n\\frac{11}{5000},\\qquad \\frac{137}{30000},\\qquad \\frac{37}{8000}.\n$$","Using $\\int_{-1}^1|u|^kdu/2=1/(k+1)$ gives the integrated upper budget $241/40000$. The same center-reset reference is valid there. Both models have the same initial state.","$$\n\\varepsilon_{\\rm repaired}\n\\leq\\frac78\\frac{241}{40000}+\\frac14\n=\\frac{81687}{320000}.\n$$","Consequently, for any common policy and any finite horizon $T$,","$$\n\\boxed{\n\\|P_T-\\widetilde P_T\\|_{\\rm TV}\n\\leq\\min\\left(1,\\frac{81687T}{640000}\\right).\n}\n$$","arison witness establishes the required common domain for the nominal repaired and reference operators. The absolute coefficient bounds dominate the integrated raw error by the triangle inequality. Proposition 73 then gives the displayed repair budget and law bound. All hypotheses are either exact identities in the supplied witnesses or direct norm bounds from the declared rational coefficients. $\\square$","Keep the fitted operators, proposed transform and initial state from Section 88 unchanged. In Lorentz coordinates use the center-reset operator $N=e_0\\lambda$, of constant density one under $du/2$. Write","$$\nF_\\eta=(1-\\eta)\\widehat F+\\eta N,\n\\qquad 0\\leq\\eta\\leq1.\n$$","Fix a required margin $0\\leq\\kappa\\leq1$. Feasibility means that every action, observation and normalized input has margin at least $\\kappa$. Integrated mass and initial membership remain separate requirements.","**Proposition 80. Convex repair bracketing.** The feasible repair weights form a closed interval $[\\eta_*,1]$. If a cone-positive functional $\\ell=(1,-v^{\\mathsf T})$, with $\\|v\\|_2\\leq1$, and a normalized input $z$ give $c=\\ell\\widehat F_a(u)z<\\kappa$, then every feasible weight satisfies","$$\n\\boxed{\\eta\\geq\\frac{\\kappa-c}{1-c}.}\n$$","A complete certificate at a weight $\\eta_U$ gives an upper bound on $\\eta_*$. A negative Bernstein coefficient without a negative evaluated point does not give a lower bound.","**Proof.** For a fixed input and observation, output mass minus spatial norm is a concave continuous function of the operator, hence of $\\eta$. Its superlevel set is closed and convex. Intersecting these sets preserves those properties. The reset has margin one, so the intersection contains one and is an interval ending there. If $F_\\eta$ has margin at least $\\kappa$, every such functional satisfies $\\ell F_\\eta z\\geq\\kappa$. Since $\\ell Nz=1$, this requires $(1-\\eta)c+\\eta\\geq\\kappa$. Here $1-c>0$, and solving gives the bound. A complete positive witness proves feasibility. A negative coefficient only says that a particular sufficient polynomial bound failed. $\\square$","oof.** For a fixed input and observation, output mass minus spatial norm is a concave continuous function of the operator, hence of $\\eta$. Its superlevel set is closed and convex. Intersecting these sets preserves those properties. The reset has margin one, so the intersection contains one and is an interval ending there. If $F_\\eta$ has margin at least $\\kappa$, every such functional satisfies $\\ell F_\\eta z\\geq\\kappa$. Since $\\ell Nz=1$, this requires $(1-\\eta)c+\\eta\\geq\\kappa$. Here $1-c>0$, and solving gives the bound. A complete positive witness proves feasibility. A negative coefficient only says that a particular sufficient polynomial bound failed. $\\square$","rvation, output mass minus spatial norm is a concave continuous function of the operator, hence of $\\eta$. Its superlevel set is closed and convex. Intersecting these sets preserves those properties. The reset has margin one, so the intersection contains one and is an interval ending there. If $F_\\eta$ has margin at least $\\kappa$, every such functional satisfies $\\ell F_\\eta z\\geq\\kappa$. Since $\\ell Nz=1$, this requires $(1-\\eta)c+\\eta\\geq\\kappa$. Here $1-c>0$, and solving gives the bound. A complete positive witness proves feasibility. A negative coefficient only says that a particular sufficient polynomial bound failed. $\\square$","spatial norm is a concave continuous function of the operator, hence of $\\eta$. Its superlevel set is closed and convex. Intersecting these sets preserves those properties. The reset has margin one, so the intersection contains one and is an interval ending there. If $F_\\eta$ has margin at least $\\kappa$, every such functional satisfies $\\ell F_\\eta z\\geq\\kappa$. Since $\\ell Nz=1$, this requires $(1-\\eta)c+\\eta\\geq\\kappa$. Here $1-c>0$, and solving gives the bound. A complete positive witness proves feasibility. A negative coefficient only says that a particular sufficient polynomial bound failed. $\\square$","ous function of the operator, hence of $\\eta$. Its superlevel set is closed and convex. Intersecting these sets preserves those properties. The reset has margin one, so the intersection contains one and is an interval ending there. If $F_\\eta$ has margin at least $\\kappa$, every such functional satisfies $\\ell F_\\eta z\\geq\\kappa$. Since $\\ell Nz=1$, this requires $(1-\\eta)c+\\eta\\geq\\kappa$. Here $1-c>0$, and solving gives the bound. A complete positive witness proves feasibility. A negative coefficient only says that a particular sufficient polynomial bound failed. $\\square$","intersection contains one and is an interval ending there. If $F_\\eta$ has margin at least $\\kappa$, every such functional satisfies $\\ell F_\\eta z\\geq\\kappa$. Since $\\ell Nz=1$, this requires $(1-\\eta)c+\\eta\\geq\\kappa$. Here $1-c>0$, and solving gives the bound. A complete positive witness proves feasibility. A negative coefficient only says that a particular sufficient polynomial bound failed. $\\square$","Use the previous transform with center $(1/2,1/2)$ and triangular factor","$$\nL=\\begin{pmatrix}123/37&-2\\\\0&109/41\\end{pmatrix}.\n$$","At the direct action, observation $u=-1$, normalized input $(1,1,0)$ and functional $\\ell=(1,7/25,-24/25)$, exact evaluation gives","$$\nc=-\\frac{17590411}{2332387500}.\n$$","For the required margin $\\kappa=1/64$, this proves","$$\n\\frac{864543451}{37599646576}\n\\leq\\eta_*\\leq\\frac1{32}.\n$$","The lower endpoint is approximately $0.02299$. The upper endpoint is not a sampled estimate. A complete exact Bernstein certificate proves that $\\eta=1/32$ works. The weight $1/64$ fails the requested $1/64$ margin at the displayed separating functional. This does not by itself prove that its nominal model is probability-invalid.","The new certificate uses 194 leaves and 3,058 recomputed coefficients. A separate comparison-domain certificate uses 185 leaves and 2,833 coefficients. Both retain the same degree-two coefficient box of radius $1/4096$ in their respective declared Lorentz coordinates, subject to exact integrated mass. The more general mass and spatial output budgets remain $1/128$ each. Their sum equals the certified margin, so that larger output-error family has a nonnegative lower margin, not a claimed strictly positive residual margin.","The reset weight has fallen from $1/8$ to $1/32$, while the deterministic uncertainty box is unchanged. No global optimization over geometry or repair operators is claimed. The current bracket concerns this fixed geometry, this reset path and this required margin.","The comparison domain is checked again before invoking the law theorem. Keeping the same raw discrepancy budget $241/40000$ from Section 89 gives","$$\n\\varepsilon_{\\mathrm{repair}}\n\\leq\\frac{31}{32}\\frac{241}{40000}+\\frac1{16}\n=\\frac{87471}{1280000}.\n$$","$$\n\\boxed{\n\\|P_T-\\widetilde P_T\\|_{\\mathrm{TV}}\n\\leq\\min\\left(1,\\frac{87471T}{2560000}\\right).\n}\n$$","Let $\\mathcal F_{t-1}$ contain all information available before observation $t$, including its already selected action. Let $\\phi_t\\in\\mathbb R^d$ be measurable with respect to that information. Assume","$$\nY_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t,\n\\qquad\n\\mathbb E[e^{h\\xi_t}\\mid\\mathcal F_{t-1}]\n\\leq e^{h^2\\sigma^2/2}\n$$","for every real $h$, with $\\sigma>0$, and a declared bound $\\|\\theta_*\\|_2\\leq S$. With ridge $\\gamma>0$, set","$$\nV_t=\\gamma I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T},\n\\qquad\n\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s.\n$$","**Proposition 81. Time-uniform regression confidence.** For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,","$$\n\\boxed{\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\n\\sigma\\sqrt{\n\\log\\frac{\\det V_t}{\\det(\\gamma I)}+2\\log\\frac1\\delta\n}\n+\\sqrt\\gamma S.\n}\n$$","**Proof.** Put $S_t=\\sum_{s\\leq t}\\phi_s\\xi_s$. For every fixed $q$, the conditional moment bound makes","$$\n\\exp\\left(q^{\\mathsf T}S_t/\\sigma\n-\\tfrac12q^{\\mathsf T}(V_t-\\gamma I)q\\right)\n$$","a nonnegative supermartingale. Integrate against the fixed Gaussian law $q\\sim N(0,(\\gamma I)^{-1})$. Conditional Tonelli and completing the square give the nonnegative supermartingale","$$\nM_t=\n\\left(\\frac{\\det(\\gamma I)}{\\det V_t}\\right)^{1/2}\n\\exp\\left(\\frac{\\|S_t\\|_{V_t^{-1}}^2}{2\\sigma^2}\\right),\n\\qquad M_0=1.\n$$","For completeness, stop at the first crossing of $1/\\delta$, then at any deterministic finite horizon. The supermartingale expectation is at most one, so the crossing probability up to that horizon is at most $\\delta$. Increasing the horizon proves the same bound for any finite crossing. Off this event, taking logarithms bounds $\\|S_t\\|_{V_t^{-1}}$ by the square-root term. Finally,","$$\n\\widehat\\theta_t-\\theta_*\n=V_t^{-1}(S_t-\\gamma\\theta_*).\n$$","The triangle inequality and $V_t\\succeq\\gamma I$ bound the ridge term by $\\sqrt\\gamma S$. The simultaneous event already covers every selected finite stopping time. $\\square$","This is the self-normalized mixture argument in [R23], not a new concentration principle. A bounded response in an interval of length $b-a$ has conditionally sub-Gaussian centered noise with $\\sigma=(b-a)/2$. One direct proof differentiates its tilted log moment-generating function twice. The second derivative is a variance bounded by $(b-a)^2/4$. Integrating from zero gives the stated moment bound.","The determinant term accounts for the observed design, but does not create excitation. If $\\phi_t$ never probes a direction, the set need not shrink in that direction. Regressors extracted after inspecting the same response are not predictable. An estimated hidden state cannot be substituted for $\\phi_t$ while silently retaining the theorem's true conditional-mean equation.","There is a direct connection to the unnormalized instrument. Assume that the normalized predictive state $Z_t$ is an exact, fixed, measurable function of the available observation history. For action $a$, suppose","$$\nP(U_t\\in du\\mid\\mathcal F_{t-1})\n=\\lambda B_a(u)Z_t\\,\\mu(du),\n\\qquad\nZ_{t+1}=\\frac{B_a(U_t)Z_t}{\\lambda B_a(U_t)Z_t}\n$$","on realized positive-density branches. Crucially, $Z_{t+1}$ must also be available from the observations using the specified state map. This is an accessibility assumption, not a consequence of successful regression.","Let $B_a(u)=\\sum_{k=0}^{D}\\psi_k(u)C_{a,k}$. Choose functions $g_j$ so that the known matrix $W_{jk}=\\int g_j(u)\\psi_k(u)\\,\\mu(du)$ is invertible.","Define $M_{a,j}=\\int g_j(u)B_a(u)\\,\\mu(du)$. Then","$$\n\\boxed{\n\\mathbb E[g_j(U_t)Z_{t+1}\\mid\\mathcal F_{t-1}]\n=M_{a,j}Z_t,\n\\qquad\nC_{a,k}=\\sum_j(W^{-1})_{kj}M_{a,j}.\n}\n$$","Under the rowwise noise and parameter-bound assumptions of Proposition 81, simultaneous confidence ellipsoids for the rows of $M_{a,j}$ give a confidence set for every $C_{a,k}$ by this known linear map. If row $i$ of $\\widehat M_{a,j}-M_{a,j}$ has $V_{a,t}$-norm at most $\\beta_{a,j,i}$, then for any direction $x$,","$$\n\\left|e_i^{\\mathsf T}(\\widehat C_{a,k}-C_{a,k})x\\right|\n\\leq\n\\sum_j |(W^{-1})_{kj}|\\,\n\\beta_{a,j,i}\\,\\|x\\|_{V_{a,t}^{-1}}.\n$$","**Proof.** In the conditional integral, multiply the normalized state by the observation density. The denominator cancels, giving $\\int g_j(u)B_a(u)Z_t\\,\\mu(du)$. The polynomial expansion gives $M_{a,j}=\\sum_kW_{jk}C_{a,k}$ and hence the inverse formula. Apply Proposition 81 to each row, using the predictable regressor $Z_t$ on visits to action $a$, or its action-gated version. A union bound over the predeclared finite rows and moments gives simultaneous coverage. Independence between these response coordinates is unnecessary. Weighted Cauchy--Schwarz and the triangle inequality give the displayed directional bound. $\\square$","nd hence the inverse formula. Apply Proposition 81 to each row, using the predictable regressor $Z_t$ on visits to action $a$, or its action-gated version. A union bound over the predeclared finite rows and moments gives simultaneous coverage. Independence between these response coordinates is unnecessary. Weighted Cauchy--Schwarz and the triangle inequality give the displayed directional bound. $\\square$","For $u\\in[-1,1]$, put","$$\nv(u)=(1,u,u^2)^{\\mathsf T},\n\\qquad L_2(u)=\\frac{3u^2-1}{2}.\n$$","$$\nK=\\{(1,x,y):x^2\\leq y\\leq1\\}.\n$$","$$\nC=\\left\\{(m,x,y):\n\\begin{pmatrix}m&x\\\\x&y\\end{pmatrix}\\succeq0,\n\\quad m-y\\geq0\\right\\},\n\\qquad\\lambda=(1,0,0).\n$$","Assume that the declared probability measure $\\mu$ on $[-1,1]$ has moments zero at orders one and three, $1/3$ at order two and $1/5$ at order four. Define a two-parameter degree-four operator density","$$\n\\boxed{\nB_{a,b}(u)=v(u)\\,(1,au,bL_2(u)).\n}\n$$","Here $a,b$ are parameters, not action labels. This example is autonomous. Its dependence on the observed signal is nonlinear because the state includes its square.","**Proposition 83. Structured curved-domain certificate.** The set $K$ is the convex hull of all $v(u)$. If $|a|+|b|\\leq1-\\rho$ for $0\\leq\\rho\\leq1$, then $B_{a,b}(u)$ preserves $C$ and has exact integrated mass. On a normalized state, its density is","$$\np_{a,b}(u\\mid x,y)=1+a x u+b yL_2(u)\\geq\\rho.\n$$","Every positive-density posterior is exactly $v(u)$. In particular, starting from $v(s)$ produces the observable Markov law","$$\np_{a,b}(u\\mid s)=1+a s u+b s^2L_2(u).\n$$","$$\n\\boxed{\n\\mathbb E[U_t\\mid U_{t-1}=s]=\\frac a3s,\n\\qquad\n\\mathbb E[L_2(U_t)\\mid U_{t-1}=s]=\\frac b5s^2.\n}\n$$","**Proof.** Every distribution on $[-1,1]$ has $x^2\\leq y\\leq1$. Conversely, fix $|x|<1$. A point mass at $x$ gives $(x,x^2)$, while the endpoint mixture with mean $x$ gives $(x,1)$. Their convex combinations give every intermediate $y$. The cases $x=\\pm1$ are immediate. Homogenizing yields the matrix and scalar inequalities, which also force the only mass-zero cone element to be zero.","On $K$, $|x|\\leq1$, $0\\leq y\\leq1$, $|u|\\leq1$ and $|L_2(u)|\\leq1$. Thus the density is at least $1-|a|-|b|$. Each operator output is that nonnegative scalar times $v(u)$, so lies in $C$. The first and second reference moments imply $\\lambda\\int B_{a,b}(u)\\,\\mu(du)=\\lambda$. Normalization cancels the scalar density, yielding $v(u)$. Symmetry through order three gives $\\int uL_2(u)d\\mu=0$. The second and fourth moments give $\\int L_2(u)^2d\\mu=1/5$. Integrating $u$ and $L_2$ against the density proves the last two formulas. $\\square$","\\leq1$, $|u|\\leq1$ and $|L_2(u)|\\leq1$. Thus the density is at least $1-|a|-|b|$. Each operator output is that nonnegative scalar times $v(u)$, so lies in $C$. The first and second reference moments imply $\\lambda\\int B_{a,b}(u)\\,\\mu(du)=\\lambda$. Normalization cancels the scalar density, yielding $v(u)$. Symmetry through order three gives $\\int uL_2(u)d\\mu=0$. The second and fourth moments give $\\int L_2(u)^2d\\mu=1/5$. Integrating $u$ and $L_2$ against the density proves the last two formulas. $\\square$","$. Thus the density is at least $1-|a|-|b|$. Each operator output is that nonnegative scalar times $v(u)$, so lies in $C$. The first and second reference moments imply $\\lambda\\int B_{a,b}(u)\\,\\mu(du)=\\lambda$. Normalization cancels the scalar density, yielding $v(u)$. Symmetry through order three gives $\\int uL_2(u)d\\mu=0$. The second and fourth moments give $\\int L_2(u)^2d\\mu=1/5$. Integrating $u$ and $L_2$ against the density proves the last two formulas. $\\square$","the density is at least $1-|a|-|b|$. Each operator output is that nonnegative scalar times $v(u)$, so lies in $C$. The first and second reference moments imply $\\lambda\\int B_{a,b}(u)\\,\\mu(du)=\\lambda$. Normalization cancels the scalar density, yielding $v(u)$. Symmetry through order three gives $\\int uL_2(u)d\\mu=0$. The second and fourth moments give $\\int L_2(u)^2d\\mu=1/5$. Integrating $u$ and $L_2$ against the density proves the last two formulas. $\\square$","t is that nonnegative scalar times $v(u)$, so lies in $C$. The first and second reference moments imply $\\lambda\\int B_{a,b}(u)\\,\\mu(du)=\\lambda$. Normalization cancels the scalar density, yielding $v(u)$. Symmetry through order three gives $\\int uL_2(u)d\\mu=0$. The second and fourth moments give $\\int L_2(u)^2d\\mu=1/5$. Integrating $u$ and $L_2$ against the density proves the last two formulas. $\\square$","The posterior ray lies on a curved boundary. Arbitrary perturbations of its output coordinates can destroy validity, as in Proposition 72. Parameter uncertainty inside the declared family behaves differently: it changes the likelihood but not the posterior ray. A certified rectangle in $(a,b)$ can therefore be robust even though a free matrix-entry ball has no positive validity radius.","This is an example of uncertainty that preserves the model's algebraic structure. It is not a claim that every real system obeys the two-parameter family. The feature dictionary $1,u,u^2$, its geometry, the reference measure and the parameter model are declared inputs to learning.","Both the continuous uniform measure $du/2$ and the following five-node measure satisfy the moment assumptions:","$$\n\\begin{array}{c|rrrrr}\nu&-1&-1/2&0&1/2&1\\\\\\hline\n\\mu\\{u\\}&7/90&16/45&2/15&16/45&7/90.\n\\end{array}\n$$","The reference experiment uses the **five-node measure**, not rounded observations from the continuous model. At each step, the conditional probability of a node is its reference weight times $p_{a,b}(u\\mid s)$. Exact rational probabilities are sampled by integer draws. The learner receives only the node sequence, the measure and the declared model. The true parameters are present only in the separate generator and verification code.","For a path of $n$ transitions, use two scalar regressions,","$$\n\\begin{aligned}\n\\phi_{a,t}&=U_{t-1}/3,&Y_{a,t}&=U_t,&\\sigma_a&=1,\\\\\n\\phi_{b,t}&=U_{t-1}^2/5,&Y_{b,t}&=L_2(U_t),&\\sigma_b&=3/4.\n\\end{aligned}\n$$","The noise scales follow from response ranges $[-1,1]$ and $[-1/2,1]$. The prior parameter bounds are $|a|,|b|\\leq1$. Each regression uses ridge one and failure probability $1/40$, giving a total failure budget $1/20$ for both parameters at every time.","**Proposition 84. Exact rational outer confidence intervals.** Write $V=1+\\sum\\phi_t^2$ and $\\widehat\\theta=\\sum\\phi_tY_t/V$. Let $k=\\lceil\\log_2 V\\rceil$ and $h=\\lceil\\log_2(1/\\delta)\\rceil$. A rational upper enclosure of","$$\n\\sigma\\sqrt{\\frac{(7/10)(k+2h)}{V}}+\\sqrt{\\frac1V}\n$$","is an upper bound on Proposition 81's scalar radius. Rounding the lower interval endpoint downward and the upper endpoint upward, then intersecting with $[-1,1]$, preserves its conditional coverage. Both intervals cover their true parameters simultaneously at every time with probability at least $19/20$.","**Proof.** The positive exponential series gives $e^{7/10}>1+7/10+(7/10)^2/2+(7/10)^3/6+(7/10)^4/24>2$. Hence $\\log 2<7/10$, so $\\log V+2\\log(1/\\delta)\\leq(7/10)(k+2h)$. Divide the scalar regression bound by $\\sqrt V$. For a nonnegative rational $x$, the smallest integer $j$ with $(j/10^{12})^2\\geq x$ supplies a rational upper square-root enclosure. Integer square root and an exact final comparison compute it. Outward rounding only enlarges the interval. Intersecting with a true prior constraint preserves coverage. Apply the two-regression union bound on their time-uniform events. $\\square$","egression bound by $\\sqrt V$. For a nonnegative rational $x$, the smallest integer $j$ with $(j/10^{12})^2\\geq x$ supplies a rational upper square-root enclosure. Integer square root and an exact final comparison compute it. Outward rounding only enlarges the interval. Intersecting with a true prior constraint preserves coverage. Apply the two-regression union bound on their time-uniform events. $\\square$","The seeded reference path uses $a=2/5$, $b=1/5$ and initial observation zero. It was first run to 65,536 transitions and then extended along the same seeded path to 262,144. Time-uniform coverage permits that extension without spending another time-selection budget. This is not a new independent coverage experiment.","$$\n\\boxed{\n\\frac{89}{250}\\leq a\\leq\\frac{231}{500},\n\\qquad\n\\frac{29}{250}\\leq b\\leq\\frac{29}{100}.\n}\n$$","$$\n\\rho=1-\\frac{231}{500}-\\frac{29}{100}=\\frac{31}{125}.\n$$","The rounded fitted center is $(409/1000,203/1000)$. It lies inside the rectangle. No reset or other coefficient correction is applied.","At earlier recorded checkpoints the sufficient density-floor test withheld a whole-box certificate. Its first successful recorded checkpoint was 32,768 transitions, with floor $1/100$. This is not a claim that 32,768 is the smallest possible sample size, or even the first passing individual step. Only the declared checkpoints were evaluated. A withheld whole-box certificate does not declare the nominal fit or true process invalid.","Consider two parameters in a certified family. For normalized $z=(1,x,y)$, their density difference is","$$\n\\Delta p(u)=\\Delta a\\,x u+\\Delta b\\,yL_2(u).\n$$","The state-update vector is the same $v(u)$ in both models after observing $u$. This does not make the models equivalent. Their likelihoods and hence their predicted observation laws can differ.","$$\nc_1=\\int|u|\\,d\\mu,\n\\qquad c_2=\\int|L_2(u)|\\,d\\mu.\n$$","If $|\\Delta a|\\leq d_a$ and $|\\Delta b|\\leq d_b$, the conditional observation laws on any common normalized state obey","$$\n\\boxed{\n\\|P(\\cdot\\mid z)-\\widehat P(\\cdot\\mid z)\\|_{\\mathrm{TV}}\n\\leq\\Delta:=\\frac{c_1d_a+c_2d_b}{2}.\n}\n$$","The same bound holds on every realized common observation history. With the same initial observation, complete future records of length $T$ have total variation at most $\\min(1,T\\Delta)$. For the five-node measure,","$$\nc_1=\\frac{23}{45},\\qquad c_2=\\frac{14}{45}.\n$$","For $du/2$, the respective constants are $1/2$ and $2/(3\\sqrt3)$.","**Proof.** The first claim follows by integrating $|\\Delta p|/2$, using $|x|\\leq1$ and $y\\leq1$. After the same observed value both models have exactly $v(u)$, so the conditional bound applies again without accumulated coordinate-state error. Telescoping conditional kernels gives the record bound, equivalently the common-history coupling argument of Proposition 69. Direct summation gives the discrete constants. For the continuous constant $c_2$, split the integral at $u=\\pm1/\\sqrt3$ and integrate the quadratic polynomial with its respective signs. $\\square$","ating $|\\Delta p|/2$, using $|x|\\leq1$ and $y\\leq1$. After the same observed value both models have exactly $v(u)$, so the conditional bound applies again without accumulated coordinate-state error. Telescoping conditional kernels gives the record bound, equivalently the common-history coupling argument of Proposition 69. Direct summation gives the discrete constants. For the continuous constant $c_2$, split the integral at $u=\\pm1/\\sqrt3$ and integrate the quadratic polynomial with its respective signs. $\\square$","|x|\\leq1$ and $y\\leq1$. After the same observed value both models have exactly $v(u)$, so the conditional bound applies again without accumulated coordinate-state error. Telescoping conditional kernels gives the record bound, equivalently the common-history coupling argument of Proposition 69. Direct summation gives the discrete constants. For the continuous constant $c_2$, split the integral at $u=\\pm1/\\sqrt3$ and integrate the quadratic polynomial with its respective signs. $\\square$","u)$, so the conditional bound applies again without accumulated coordinate-state error. Telescoping conditional kernels gives the record bound, equivalently the common-history coupling argument of Proposition 69. Direct summation gives the discrete constants. For the continuous constant $c_2$, split the integral at $u=\\pm1/\\sqrt3$ and integrate the quadratic polynomial with its respective signs. $\\square$","The use of $\\Delta$ here is an upper bound proved from a confidence set and a model class. It is not the lower discrepancy bound from a rejected residual test in Part VI. A test that does not reject cannot supply the displayed upper bound by itself.","$$\nd_a=\\frac{53}{1000},\\qquad d_b=\\frac{87}{1000},\n\\qquad\n\\boxed{\\Delta=\\frac{2437}{90000}.}\n$$","The previous record bound grows with $T$. A different question can have a uniform answer: how far apart are the distributions of the signal at a single future time? A lower density bound permits this distinction to be made exactly.","Let $P,Q$ be Markov kernels on a common state space. Suppose $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$ for every $s$, where $\\mu$ is one probability measure and $0<\\rho\\leq1$. Suppose also $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\mathrm{TV}}\\leq\\Delta$.","**Proposition 86. Uniform marginal and rolling-window stability.** For initial laws at distance $e_0$, the one-time marginal laws satisfy","$$\n\\boxed{\ne_t\\leq\\min\\left(1,\n(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n}\n$$","For $k$ consecutive observations starting at time $t$, their joint laws are at distance at most $\\min(1,e_t+(k-1)\\Delta)$. A uniform small marginal bound does not in general bound the total variation of arbitrarily long complete records away from one.","**Proof.** For $\\rho<1$, write $P(s,\\cdot)=\\rho\\mu+(1-\\rho)R(s,\\cdot)$, where $R$ is a Markov kernel. Applying a Markov kernel cannot increase total variation, by integration of bounded test functions. Thus $\\|\\nu P-\\widetilde\\nu P\\|_{\\mathrm{TV}}\\leq(1-\\rho)\\|\\nu-\\widetilde\\nu\\|_{\\mathrm{TV}}$. Adding and subtracting $\\widetilde\\nu P$ yields $e_{t+1}\\leq(1-\\rho)e_t+\\Delta$. Iterate this inequality. The case $\\rho=1$ follows directly. Applying the finite-record telescoping argument from the initial time $t$ gives the rolling-window statement.","r $\\rho<1$, write $P(s,\\cdot)=\\rho\\mu+(1-\\rho)R(s,\\cdot)$, where $R$ is a Markov kernel. Applying a Markov kernel cannot increase total variation, by integration of bounded test functions. Thus $\\|\\nu P-\\widetilde\\nu P\\|_{\\mathrm{TV}}\\leq(1-\\rho)\\|\\nu-\\widetilde\\nu\\|_{\\mathrm{TV}}$. Adding and subtracting $\\widetilde\\nu P$ yields $e_{t+1}\\leq(1-\\rho)e_t+\\Delta$. Iterate this inequality. The case $\\rho=1$ follows directly. Applying the finite-record telescoping argument from the initial time $t$ gives the rolling-window statement.","a Markov kernel cannot increase total variation, by integration of bounded test functions. Thus $\\|\\nu P-\\widetilde\\nu P\\|_{\\mathrm{TV}}\\leq(1-\\rho)\\|\\nu-\\widetilde\\nu\\|_{\\mathrm{TV}}$. Adding and subtracting $\\widetilde\\nu P$ yields $e_{t+1}\\leq(1-\\rho)e_t+\\Delta$. Iterate this inequality. The case $\\rho=1$ follows directly. Applying the finite-record telescoping argument from the initial time $t$ gives the rolling-window statement.","h is a Markov kernel with a positive common minorization. At any time their marginal distance is the fixed parameter difference. For a full record, test whether its empirical mean lies above the midpoint between the parameters. Under each law, Chebyshev's inequality makes the error probability tend to zero as the record length grows. The total variation of the record laws therefore tends to one. $\\square$","With $e_0=0$, $\\rho=31/125$ and $\\Delta=2437/90000$, the reference confidence rectangle gives","$$\n\\boxed{\n\\sup_{t\\geq0}e_t\\leq\\frac{2437}{22320}\n\\approx0.10919.\n}\n$$","This is a nontrivial all-time bound on **one-time marginal observation distributions**. The complete-record bound is still $\\min(1,2437T/90000)$, which becomes trivial at integer horizon 37. The model does not exchange those two different claims. In live filtering after a shared observed value, its state is exactly known and the next-observation law has the separate uniform upper bound $2437/90000$.","**Proposition 87. Observable structured learning with a covered domain.** Suppose the scalar process obeys the conditional law of Proposition 83 with the declared reference measure and fixed parameters satisfying $|a|,|b|\\leq1$. Use the predictable regressions, fixed dictionary and confidence protocol of Proposition 84. At a data-selected finite time, form the full outward-rounded rectangle. If its density floor $\\rho$ is nonnegative, that rectangle has an exact all-horizon probability certificate on $C$. If $\\rho>0$, every chosen center inside it has the marginal and record bounds of Propositions 85--86 relative to every member.","Observable structured learning with a covered domain.** Suppose the scalar process obeys the conditional law of Proposition 83 with the declared reference measure and fixed parameters satisfying $|a|,|b|\\leq1$. Use the predictable regressions, fixed dictionary and confidence protocol of Proposition 84. At a data-selected finite time, form the full outward-rounded rectangle. If its density floor $\\rho$ is nonnegative, that rectangle has an exact all-horizon probability certificate on $C$. If $\\rho>0$, every chosen center inside it has the marginal and record bounds of Propositions 85--86 relative to every member.","the conditional law of Proposition 83 with the declared reference measure and fixed parameters satisfying $|a|,|b|\\leq1$. Use the predictable regressions, fixed dictionary and confidence protocol of Proposition 84. At a data-selected finite time, form the full outward-rounded rectangle. If its density floor $\\rho$ is nonnegative, that rectangle has an exact all-horizon probability certificate on $C$. If $\\rho>0$, every chosen center inside it has the marginal and record bounds of Propositions 85--86 relative to every member.","itional law of Proposition 83 with the declared reference measure and fixed parameters satisfying $|a|,|b|\\leq1$. Use the predictable regressions, fixed dictionary and confidence protocol of Proposition 84. At a data-selected finite time, form the full outward-rounded rectangle. If its density floor $\\rho$ is nonnegative, that rectangle has an exact all-horizon probability certificate on $C$. If $\\rho>0$, every chosen center inside it has the marginal and record bounds of Propositions 85--86 relative to every member.","With probability at least $19/20$ over the specified stochastic observation process, these implications cover its true parameter at every checked time, including a stopping time. Internal validity of a verified rectangle is deterministic. Statistical coverage remains conditional on the declared process, moment, parameter and observation assumptions.","ndpoints bound each parameter difference. Proposition 85 gives conditional and record error. If the floor is positive, all true conditional rows have the same lower reference component, so Proposition 86 applies to the scalar Markov process. Selection of the stopping time does not alter the already simultaneous event. None of these implications verifies the assumptions from a single sample path. $\\square$","midrule\nReconstruction & $SA=I_X$ & Can the accepted observation be recovered?","[5pt]\nTransport & $F_bT_{ba}=T_{ba}F_a$ & Do coordinate descriptions agree?\\\\[5pt]\nClosure & $\\gamma F_u=\\bar F_u\\gamma$ & Does the summary determine its own future?\\\\[5pt]\nProbability & $B_oC\\subseteq C$, $\\lambda\\sum_oB_o=\\lambda$ & Are every branch and every finite record valid?\\\\\\bottomrule\n\\end{tabular}\n\\vspace{10pt}\n\\textbf{Reading route.} Plates 1--4 build the representation and its predictive quotient. Plates 5--9 explain interactions, lost paths, memory and observable identification. Plates 10--16 develop behavior spaces, positivity, curved geometry and statistical guarantees. Plate 17 locates the formal proofs.\n\n\\textbf{Visual conventions.} Teal marks exact structure. Gold marks transported, refined or uncertain quantities. Red marks an obstruction or an omitted contribution. Every figure is drawn from a formula or an exact finite example. Curves of bounds are labeled as bounds.\n\n\\boundary{The atlas selects core mechanisms from the 87 registered propositions. It introduces no new numbered theorem, fitted model, benchmark result or empirical coverage claim. Its diagrams supp…","able coefficients form a subspace}{Exact storage includes a consistency condition.}{P1}\nFor linear analysis and synthesis maps $A:X\\to C$ and $S:C\\to X$, the left-inverse law $SA=I_X$ gives injectivity of $A$. The coefficient projector is $\\Pi=AS$, with\n\\[\n\\Pi^2=\\Pi,\\qquad \\im\\Pi=\\im A,\\qquad c\\text{ is realizable}\\iff \\Pi c=c.\n\\]\n\\diagram[0.90]{01_consistency}\n\\textbf{A redundant archive.} Take $X=\\R^2$, $C=\\R^3$ and\n\\[\nA(x_1,x_2)=(x_1,x_2,x_1+x_2),\\qquad S(c_1,c_2,c_3)=(c_1,c_2).\n\\]\nThen $\\Pi(c_1,c_2,c_3)=(c_1,c_2,c_1+c_2)$. The shaded plane is the set of realizable coefficients. The red segment shows synthesis followed by reanalysis of an inconsistent vector.\n\\claim{A left inverse determines an idempotent consistency map. Here it is oblique: $\\Pi$ is not self-adjoint. Orthogonality requires an additional hypothesis.}\n\\boundary{For bounded linear maps on normed spaces, $\\|S\\|$ controls amplification of coefficient errors. Exact invertibility by itself gives no conditioning guarantee.}\n\\nav{\\texttt{Realization.lean}; norm estimates in \\texttt{Normed.lean}.}","icients form a subspace}{Exact storage includes a consistency condition.}{P1}\nFor linear analysis and synthesis maps $A:X\\to C$ and $S:C\\to X$, the left-inverse law $SA=I_X$ gives injectivity of $A$. The coefficient projector is $\\Pi=AS$, with\n\\[\n\\Pi^2=\\Pi,\\qquad \\im\\Pi=\\im A,\\qquad c\\text{ is realizable}\\iff \\Pi c=c.\n\\]\n\\diagram[0.90]{01_consistency}\n\\textbf{A redundant archive.} Take $X=\\R^2$, $C=\\R^3$ and\n\\[\nA(x_1,x_2)=(x_1,x_2,x_1+x_2),\\qquad S(c_1,c_2,c_3)=(c_1,c_2).\n\\]\nThen $\\Pi(c_1,c_2,c_3)=(c_1,c_2,c_1+c_2)$. The shaded plane is the set of realizable coefficients. The red segment shows synthesis followed by reanalysis of an inconsistent vector.\n\\claim{A left inverse determines an idempotent consistency map. Here it is oblique: $\\Pi$ is not self-adjoint. Orthogonality requires an additional hypothesis.}\n\\boundary{For bounded linear maps on normed spaces, $\\|S\\|$ controls amplification of coefficient errors. Exact invertibility by itself gives no conditioning guarantee.}\n\\nav{\\texttt{Realization.lean}; norm estimates in \\texttt{Normed.lean}.}","subspace}{Exact storage includes a consistency condition.}{P1}\nFor linear analysis and synthesis maps $A:X\\to C$ and $S:C\\to X$, the left-inverse law $SA=I_X$ gives injectivity of $A$. The coefficient projector is $\\Pi=AS$, with\n\\[\n\\Pi^2=\\Pi,\\qquad \\im\\Pi=\\im A,\\qquad c\\text{ is realizable}\\iff \\Pi c=c.\n\\]\n\\diagram[0.90]{01_consistency}\n\\textbf{A redundant archive.} Take $X=\\R^2$, $C=\\R^3$ and\n\\[\nA(x_1,x_2)=(x_1,x_2,x_1+x_2),\\qquad S(c_1,c_2,c_3)=(c_1,c_2).\n\\]\nThen $\\Pi(c_1,c_2,c_3)=(c_1,c_2,c_1+c_2)$. The shaded plane is the set of realizable coefficients. The red segment shows synthesis followed by reanalysis of an inconsistent vector.\n\\claim{A left inverse determines an idempotent consistency map. Here it is oblique: $\\Pi$ is not self-adjoint. Orthogonality requires an additional hypothesis.}\n\\boundary{For bounded linear maps on normed spaces, $\\|S\\|$ controls amplification of coefficient errors. Exact invertibility by itself gives no conditioning guarantee.}\n\\nav{\\texttt{Realization.lean}; norm estimates in \\texttt{Normed.lean}.}","ar analysis and synthesis maps $A:X\\to C$ and $S:C\\to X$, the left-inverse law $SA=I_X$ gives injectivity of $A$. The coefficient projector is $\\Pi=AS$, with\n\\[\n\\Pi^2=\\Pi,\\qquad \\im\\Pi=\\im A,\\qquad c\\text{ is realizable}\\iff \\Pi c=c.\n\\]\n\\diagram[0.90]{01_consistency}\n\\textbf{A redundant archive.} Take $X=\\R^2$, $C=\\R^3$ and\n\\[\nA(x_1,x_2)=(x_1,x_2,x_1+x_2),\\qquad S(c_1,c_2,c_3)=(c_1,c_2).\n\\]\nThen $\\Pi(c_1,c_2,c_3)=(c_1,c_2,c_1+c_2)$. The shaded plane is the set of realizable coefficients. The red segment shows synthesis followed by reanalysis of an inconsistent vector.\n\\claim{A left inverse determines an idempotent consistency map. Here it is oblique: $\\Pi$ is not self-adjoint. Orthogonality requires an additional hypothesis.}\n\\boundary{For bounded linear maps on normed spaces, $\\|S\\|$ controls amplification of coefficient errors. Exact invertibility by itself gives no conditioning guarantee.}\n\\nav{\\texttt{Realization.lean}; norm estimates in \\texttt{Normed.lean}.}","stency}\n\\textbf{A redundant archive.} Take $X=\\R^2$, $C=\\R^3$ and\n\\[\nA(x_1,x_2)=(x_1,x_2,x_1+x_2),\\qquad S(c_1,c_2,c_3)=(c_1,c_2).\n\\]\nThen $\\Pi(c_1,c_2,c_3)=(c_1,c_2,c_1+c_2)$. The shaded plane is the set of realizable coefficients. The red segment shows synthesis followed by reanalysis of an inconsistent vector.\n\\claim{A left inverse determines an idempotent consistency map. Here it is oblique: $\\Pi$ is not self-adjoint. Orthogonality requires an additional hypothesis.}\n\\boundary{For bounded linear maps on normed spaces, $\\|S\\|$ controls amplification of coefficient errors. Exact invertibility by itself gives no conditioning guarantee.}\n\\nav{\\texttt{Realization.lean}; norm estimates in \\texttt{Normed.lean}.}",".\n\\]\nThen $\\Pi(c_1,c_2,c_3)=(c_1,c_2,c_1+c_2)$. The shaded plane is the set of realizable coefficients. The red segment shows synthesis followed by reanalysis of an inconsistent vector.\n\\claim{A left inverse determines an idempotent consistency map. Here it is oblique: $\\Pi$ is not self-adjoint. Orthogonality requires an additional hypothesis.}\n\\boundary{For bounded linear maps on normed spaces, $\\|S\\|$ controls amplification of coefficient errors. Exact invertibility by itself gives no conditioning guarantee.}\n\\nav{\\texttt{Realization.lean}; norm estimates in \\texttt{Normed.lean}.}","\\plate{2}{One operation, several coordinates}{A commuting square expresses coordinate independence.}{P2}\nLet $\\mathcal C_a=\\im A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S","\\plate{2}{One operation, several coordinates}{A commuting square expresses coordinate independence.}{P2}\nLet $\\mathcal C_a=\\im A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax","\\plate{2}{One operation, several coordinates}{A commuting square expresses coordinate independence.}{P2}\nLet $\\mathcal C_a=\\im A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy estab","\\plate{2}{One operation, several coordinates}{A commuting square expresses coordinate independence.}{P2}\nLet $\\mathcal C_a=\\im A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy establishes agreement.","\\plate{2}{One operation, several coordinates}{A commuting square expresses coordinate independence.}{P2}\nLet $\\mathcal C_a=\\im A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy establishes agreement. It does not establish correctness of the agreed predictor. Deterministic recodings of the same observation supply the same observed information.}\n\\nav{\\texttt{Realization.lean}. T","everal coordinates}{A commuting square expresses coordinate independence.}{P2}\nLet $\\mathcal C_a=\\im A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy establishes agreement. It does not establish correctness of the agreed predictor. Deterministic recodings of the same observation supply the same observed information.}\n\\nav{\\texttt{Realization.lean}. The formal construction uses a nonemp…","A_a$ and $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy establishes agreement. It does not establish correctness of the agreed predictor. Deterministic recodings of the same observation supply the same observed information.}\n\\nav{\\texttt{Realization.lean}. The formal construction uses a nonempty chart family.}","nd $T_{ba}=A_bS_a$. Coordinate realizations are $F_a=A_aFS_a$ on $\\mathcal C_a$. Both paths through the square represent the same underlying operation.\n\\diagram{02_transport}\n\\[\nT_{cb}T_{ba}=T_{ca},\\qquad T_{aa}=I_{\\mathcal C_a},\\qquad\nF_bT_{ba}=T_{ba}F_a.\n\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy establishes agreement. It does not establish correctness of the agreed predictor. Deterministic recodings of the same observation supply the same observed information.}\n\\nav{\\texttt{Realization.lean}. The formal construction uses a nonempty chart family.}","\\]\nConversely, a compatible family $G_a$ on the realizable coefficient spaces determines one operation. With a supplied reference chart $r$,\n\\[\nF=S_rG_rA_r,\\qquad G_a=A_aFS_a.\n\\]\n\\claim{Compatibility is a descent condition: local coordinate descriptions belong to one object-space operation precisely when they agree under transport.}\nA fitted family's discrepancy can be measured in object space:\n\\[\nS_b\\bigl(G_bA_bx-T_{ba}G_aA_ax\\bigr)\n=S_bG_bA_bx-S_aG_aA_ax.\n\\]\n\\boundary{Zero chart discrepancy establishes agreement. It does not establish correctness of the agreed predictor. Deterministic recodings of the same observation supply the same observed information.}\n\\nav{\\texttt{Realization.lean}. The formal construction uses a nonempty chart family.}","\\plate{3}{The obstruction lives in a fiber}{Equal present summaries must have equal next summaries.}{P6--P7}\nA reduced map $\\bar F:\\gamma(X)\\to\\gamma(X)$ exists exactly when\n\\[\n\\gamma x=\\gamma x'\\quad\\Longrightarrow\\quad\\gamma Fx=\\gamma Fx'.\n\\]\nFor linear $L$ and linear $\\gamma$, this is $L(\\ker\\gamma)\\subseteq\\ker\\gamma$.\n\\diagram{03_fibers}\n\\textbf{The running rational example.} Throughout the deterministic plates, use\n\\[\nL=\\begin{pmatrix}1/2&1","ives in a fiber}{Equal present summaries must have equal next summaries.}{P6--P7}\nA reduced map $\\bar F:\\gamma(X)\\to\\gamma(X)$ exists exactly when\n\\[\n\\gamma x=\\gamma x'\\quad\\Longrightarrow\\quad\\gamma Fx=\\gamma Fx'.\n\\]\nFor linear $L$ and linear $\\gamma$, this is $L(\\ker\\gamma)\\subseteq\\ker\\gamma$.\n\\diagram{03_fibers}\n\\textbf{The running rational example.} Throughout the deterministic plates, use\n\\[\nL=\\begin{pmatrix}1/2&1\\\\1/4&1/3\\end{pmatrix},\\qquad\n\\gamma=(1,0),\\qquad x_0=(1,0)^{\\mathsf T}.\n\\]\nThe two states $(0,0)$ and $(0,1)$ have summary zero. Their next summaries are $0$ and $1$. Thus no scalar autonomous update reproduces every state's first-coordinate dynamics.\n\\claim{Predictive sufficiency is a property of entire summary fibers. A lossless archive of each full state does not make its chosen scalar summary sufficient.}\n\\boundary{The diagram is a counterexample for the declared operator. In a stochastic model, the corresponding condition concerns equality of next-summary distributions.}\n\\nav{\\texttt{PredictiveClosure.lean}; \\texttt{LinearClosure.lean}.}","1/4&1/3\\end{pmatrix},\\qquad\n\\gamma=(1,0),\\qquad x_0=(1,0)^{\\mathsf T}.\n\\]\nThe two states $(0,0)$ and $(0,1)$ have summary zero. Their next summaries are $0$ and $1$. Thus no scalar autonomous update reproduces every state's first-coordinate dynamics.\n\\claim{Predictive sufficiency is a property of entire summary fibers. A lossless archive of each full state does not make its chosen scalar summary sufficient.}\n\\boundary{The diagram is a counterexample for the declared operator. In a stochastic model, the corresponding condition concerns equality of next-summary distributions.}\n\\nav{\\texttt{PredictiveClosure.lean}; \\texttt{LinearClosure.lean}.}","\\plate{4}{Refinement closes an observable space}{Add the directions generated by known dynamics.}{P8, P14}\nFor a known finite family $L_u$ on a $d$-dimensional space, start from the rows of a required observation $C_0$ and form\n\\[\nW_{n+1}=W_n+\\sum_uW_nL_u,\\qquad W_0=\\operatorname{row}(C_0).\n\\]\n\\diagram{04_refinement}\nEvery strict inclusion raises dimension. After at most $d-\\dim W_0$ strict enlargements, the limit $W_*$ is the least common invariant row space containing $W_0$.","t closes an observable space}{Add the directions generated by known dynamics.}{P8, P14}\nFor a known finite family $L_u$ on a $d$-dimensional space, start from the rows of a required observation $C_0$ and form\n\\[\nW_{n+1}=W_n+\\sum_uW_nL_u,\\qquad W_0=\\operatorname{row}(C_0).\n\\]\n\\diagram{04_refinement}\nEvery strict inclusion raises dimension. After at most $d-\\dim W_0$ strict enlargements, the limit $W_*$ is the least common invariant row space containing $W_0$.","d by known dynamics.}{P8, P14}\nFor a known finite family $L_u$ on a $d$-dimensional space, start from the rows of a required observation $C_0$ and form\n\\[\nW_{n+1}=W_n+\\sum_uW_nL_u,\\qquad W_0=\\operatorname{row}(C_0).\n\\]\n\\diagram{04_refinement}\nEvery strict inclusion raises dimension. After at most $d-\\dim W_0$ strict enlargements, the limit $W_*$ is the least common invariant row space containing $W_0$.","For the running example, the rows of\n\\[\nO=\\begin{pmatrix}1&0\\\\1/2&1\\end{pmatrix},\\qquad\nD=OLO^{-1}=\\begin{pmatrix}0&1\\\\1/12&5/6\\end{pmatrix}\n\\]\nspan $W_*=\\R^{2\\,*}$. The state $Ox_s=(y_s,y_{s+1})^{\\mathsf T}$ therefore closes.\n\\claim{The added coordinate is forced by the action of $L$ on the requested observable. In a larger system, unobserved directions unrelated to its future can remain outside $W_*$.}\n\\boundary{Computing $Ox_s$ uses a forward observation. It becomes an available delay state $(y_{t-1},y_t)$ only after a one-step warm-up. Known-operator completion does not establish sensor access to arbitrary new observables.}\n\\nav{\\texttt{InvariantCompletion.lean}; \\texttt{PredictiveCompletion.lean}; \\texttt{DelayClosure.lean}.}","1/12&5/6\\end{pmatrix}\n\\]\nspan $W_*=\\R^{2\\,*}$. The state $Ox_s=(y_s,y_{s+1})^{\\mathsf T}$ therefore closes.\n\\claim{The added coordinate is forced by the action of $L$ on the requested observable. In a larger system, unobserved directions unrelated to its future can remain outside $W_*$.}\n\\boundary{Computing $Ox_s$ uses a forward observation. It becomes an available delay state $(y_{t-1},y_t)$ only after a one-step warm-up. Known-operator completion does not establish sensor access to arbitrary new observables.}\n\\nav{\\texttt{InvariantCompletion.lean}; \\texttt{PredictiveCompletion.lean}; \\texttt{DelayClosure.lean}.}","\\plate{5}{Packets decompose operations too}{Resolution changes preserve identities when every block is retained.}{P3--P5, P20}\nFor complete input and output projector families,\n\\[\nx=\\sum_iP_ix,\\qquad L=\\sum_{j,i}Q_jLP_i.\n\\]\nRefining either family subdivides these transfer blocks without changing their total. Orthogonal packet families also give the corresponding squared-norm decomposition.\n\\diagram{05_packets}\nFor a polynomial operation, multilinearity produces every ordered packet interaction:\n\\[\nQ_jB_m(x,\\ldots,x)=\\sum_{i_1,\\ldots,i_m}\nQ_jB_m(P_{i_1}x,\\ldots,P_{i_m}x).\n\\]\nIn the scalar square, the mixed contribution is $2uv$. With $u=1$, it is linear in omitted amplitude $v$, whereas the omitted packet square is quadratic.\n\\claim{A nonlinear operation acts on interactions between packets. Reconstruction energy alone does not bound an omitted packet's influence on a future quantity.}\n\\boundary{Complete refi","retained.}{P3--P5, P20}\nFor complete input and output projector families,\n\\[\nx=\\sum_iP_ix,\\qquad L=\\sum_{j,i}Q_jLP_i.\n\\]\nRefining either family subdivides these transfer blocks without changing their total. Orthogonal packet families also give the corresponding squared-norm decomposition.\n\\diagram{05_packets}\nFor a polynomial operation, multilinearity produces every ordered packet interaction:\n\\[\nQ_jB_m(x,\\ldots,x)=\\sum_{i_1,\\ldots,i_m}\nQ_jB_m(P_{i_1}x,\\ldots,P_{i_m}x).\n\\]\nIn the scalar square, the mixed contribution is $2uv$. With $u=1$, it is linear in omitted amplitude $v$, whereas the omitted packet square is quadratic.\n\\claim{A nonlinear operation acts on interactions between packets. Reconstruction energy alone does not bound an omitted packet's influence on a future quantity.}\n\\boundary{Complete refinement and truncation are different operations. Truncation requires an explicit omitted-interaction estimate on its declared domain. Projectors from different atlases need not commute.}\n\\nav{\\texttt{PacketTree.lean}; \\texttt{TreeEnergy.lean}; \\texttt{Transfer.lean}; \\texttt{Polynom…","ng either family subdivides these transfer blocks without changing their total. Orthogonal packet families also give the corresponding squared-norm decomposition.\n\\diagram{05_packets}\nFor a polynomial operation, multilinearity produces every ordered packet interaction:\n\\[\nQ_jB_m(x,\\ldots,x)=\\sum_{i_1,\\ldots,i_m}\nQ_jB_m(P_{i_1}x,\\ldots,P_{i_m}x).\n\\]\nIn the scalar square, the mixed contribution is $2uv$. With $u=1$, it is linear in omitted amplitude $v$, whereas the omitted packet square is quadratic.\n\\claim{A nonlinear operation acts on interactions between packets. Reconstruction energy alone does not bound an omitted packet's influence on a future quantity.}\n\\boundary{Complete refinement and truncation are different operations. Truncation requires an explicit omitted-interaction estimate on its declared domain. Projectors from different atlases need not commute.}\n\\nav{\\texttt{PacketTree.lean}; \\texttt{TreeEnergy.lean}; \\texttt{Transfer.lean}; \\texttt{Polynomial.lean}.}","mily subdivides these transfer blocks without changing their total. Orthogonal packet families also give the corresponding squared-norm decomposition.\n\\diagram{05_packets}\nFor a polynomial operation, multilinearity produces every ordered packet interaction:\n\\[\nQ_jB_m(x,\\ldots,x)=\\sum_{i_1,\\ldots,i_m}\nQ_jB_m(P_{i_1}x,\\ldots,P_{i_m}x).\n\\]\nIn the scalar square, the mixed contribution is $2uv$. With $u=1$, it is linear in omitted amplitude $v$, whereas the omitted packet square is quadratic.\n\\claim{A nonlinear operation acts on interactions between packets. Reconstruction energy alone does not bound an omitted packet's influence on a future quantity.}\n\\boundary{Complete refinement and truncation are different operations. Truncation requires an explicit omitted-interaction estimate on its declared domain. Projectors from different atlases need not commute.}\n\\nav{\\texttt{PacketTree.lean}; \\texttt{TreeEnergy.lean}; \\texttt{Transfer.lean}; \\texttt{Polynomial.lean}.}","hout changing their total. Orthogonal packet families also give the corresponding squared-norm decomposition.\n\\diagram{05_packets}\nFor a polynomial operation, multilinearity produces every ordered packet interaction:\n\\[\nQ_jB_m(x,\\ldots,x)=\\sum_{i_1,\\ldots,i_m}\nQ_jB_m(P_{i_1}x,\\ldots,P_{i_m}x).\n\\]\nIn the scalar square, the mixed contribution is $2uv$. With $u=1$, it is linear in omitted amplitude $v$, whereas the omitted packet square is quadratic.\n\\claim{A nonlinear operation acts on interactions between packets. Reconstruction energy alone does not bound an omitted packet's influence on a future quantity.}\n\\boundary{Complete refinement and truncation are different operations. Truncation requires an explicit omitted-interaction estimate on its declared domain. Projectors from different atlases need not commute.}\n\\nav{\\texttt{PacketTree.lean}; \\texttt{TreeEnergy.lean}; \\texttt{Transfer.lean}; \\texttt{Polynomial.lean}.}","\\plate{6}{Compression can erase a return path}{A two-step effect can survive when both compressed steps vanish.}{P7}\nLet $P^2=P$ and $Q=I-P$. Inserting $I=P+Q$ gives the exact defect\n\\[\n\\boxed{PLMP-(PLP)(PMP)=PLQMP.}\n\\]\n\\diagram{06_return_path}\nTake $P=\\diag(1,0)$ and\n\\[\nL=M=\\begin{pmatrix}0&1","\\plate{6}{Compression can erase a return path}{A two-step effect can survive when both compressed steps vanish.}{P7}\nLet $P^2=P$ and $Q=I-P$. Inserting $I=P+Q$ gives the exact defect\n\\[\n\\boxed{PLMP-(PLP)(PMP)=PLQMP.}\n\\]\n\\diagram{06_return_path}\nTake $P=\\diag(1,0)$ and\n\\[\nL=M=\\begin{pmatrix}0&1\\\\1&0\\end{pmatrix}.\n\\]\nThen $PLP=PMP=0$, while $PLMP=P$. The missing term $PLQMP=P$ is the full return through the discarded coordinate.","1&0\\end{pmatrix}.\n\\]\nThen $PLP=PMP=0$, while $PLMP=P$. The missing term $PLQMP=P$ is the full return through the discarded coordinate.","The running example exhibits the same mechanism in $PL^2P$:\n\\[\nPL^2P-(PLP)^2\n=\\begin{pmatrix}1/4&0","The running example exhibits the same mechanism in $PL^2P$:\n\\[\nPL^2P-(PLP)^2\n=\\begin{pmatrix}1/4&0\\\\0&0\\end{pmatrix}.\n\\]\nStarting from $(1,0)$, separate scalar compression predicts $y_2=1/4$. The exact value is $1/2$.\n\\claim{The obstruction is an omitted route through the operator composition. Its size depends jointly on outgoing and returning transfer blocks.}\n\\boundary{The algebra requires idempotence. An orthogonal split additionally gives the familiar Hilbert-space energy interpretation.}\n\\nav{\\texttt{LinearClosure.lean}; norm bounds in \\texttt{Normed.lean}.}","0&0\\end{pmatrix}.\n\\]\nStarting from $(1,0)$, separate scalar compression predicts $y_2=1/4$. The exact value is $1/2$.\n\\claim{The obstruction is an omitted route through the operator composition. Its size depends jointly on outgoing and returning transfer blocks.}\n\\boundary{The algebra requires idempotence. An orthogonal split additionally gives the familiar Hilbert-space energy interpretation.}\n\\nav{\\texttt{LinearClosure.lean}; norm bounds in \\texttt{Normed.lean}.}","\\plate{7}{Memory is an exact elimination}{The discarded state leaves a kernel and an initial transient.}{P10, P19}\nWrite the split dynamics as $y_{t+1}=Ay_t+Bw_t$ and $w_{t+1}=Cy_t+Dw_t$. Eliminating $w_t$ gives\n\\[\ny_{t+1}=Ay_t+\\sum_{j=0}^{t-1}BD^jC\\,y_{t-1-j}+BD^tw_0.\n\\]\n\\diagram{07_memory}\nFor the running example, $A=1/2$, $B=1$, $C=1/4$, $D=1/3$ and $w_0=0$. Thus $K_j=\\tfrac14\\,3^{-j}$. If $|y_t|\\leq1$, retaining the last $H$ history terms leaves a same-history tail at most $\\Delta_H=\\tfrac38\\,3^{-H}$.","limination}{The discarded state leaves a kernel and an initial transient.}{P10, P19}\nWrite the split dynamics as $y_{t+1}=Ay_t+Bw_t$ and $w_{t+1}=Cy_t+Dw_t$. Eliminating $w_t$ gives\n\\[\ny_{t+1}=Ay_t+\\sum_{j=0}^{t-1}BD^jC\\,y_{t-1-j}+BD^tw_0.\n\\]\n\\diagram{07_memory}\nFor the running example, $A=1/2$, $B=1$, $C=1/4$, $D=1/3$ and $w_0=0$. Thus $K_j=\\tfrac14\\,3^{-j}$. If $|y_t|\\leq1$, retaining the last $H$ history terms leaves a same-history tail at most $\\Delta_H=\\tfrac38\\,3^{-H}$.","n initial transient.}{P10, P19}\nWrite the split dynamics as $y_{t+1}=Ay_t+Bw_t$ and $w_{t+1}=Cy_t+Dw_t$. Eliminating $w_t$ gives\n\\[\ny_{t+1}=Ay_t+\\sum_{j=0}^{t-1}BD^jC\\,y_{t-1-j}+BD^tw_0.\n\\]\n\\diagram{07_memory}\nFor the running example, $A=1/2$, $B=1$, $C=1/4$, $D=1/3$ and $w_0=0$. Thus $K_j=\\tfrac14\\,3^{-j}$. If $|y_t|\\leq1$, retaining the last $H$ history terms leaves a same-history tail at most $\\Delta_H=\\tfrac38\\,3^{-H}$.","Here $0\\leq y_t\\leq1$ and the truncated recursion has gain\n\\[\na_H=\\frac12+\\sum_{j=0}^{H-1}K_j\n=\\frac78-\\frac38\\,3^{-H}<1.\n\\]\nWith the same initial state, recursive propagation therefore gives\n\\[\n\\sup_t|y_t-\\widehat y_t|\\leq\\frac{\\Delta_H}{1-a_H}\n=\\frac{3}{3^H+3}.\n\\]\n\\boundary{P10's tail estimate uses the same resolved history. P19 supplies the separate stability argument for a predictor using its own history. For other initial states, retain the term $BD^tw_0$.}\n\\nav{\\texttt{Memory.lean}; \\texttt{MemoryBounds.lean}; \\texttt{MemoryStability.lean}.}","in\n\\[\na_H=\\frac12+\\sum_{j=0}^{H-1}K_j\n=\\frac78-\\frac38\\,3^{-H}<1.\n\\]\nWith the same initial state, recursive propagation therefore gives\n\\[\n\\sup_t|y_t-\\widehat y_t|\\leq\\frac{\\Delta_H}{1-a_H}\n=\\frac{3}{3^H+3}.\n\\]\n\\boundary{P10's tail estimate uses the same resolved history. P19 supplies the separate stability argument for a predictor using its own history. For other initial states, retain the term $BD^tw_0$.}\n\\nav{\\texttt{Memory.lean}; \\texttt{MemoryBounds.lean}; \\texttt{MemoryStability.lean}.}","\\plate{8}{A local defect needs a propagation law}{Small one-step residuals can accumulate or contract.}{P18}\nAllow a representation $\\gamma_t$ to change with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{","\\plate{8}{A local defect needs a propagation law}{Small one-step residuals can accumulate or contract.}{P18}\nAllow a representation $\\gamma_t$ to change with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analyt","\\plate{8}{A local defect needs a propagation law}{Small one-step residuals can accumulate or contract.}{P18}\nAllow a representation $\\gamma_t$ to change with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not","\\plate{8}{A local defect needs a propagation law}{Small one-step residuals can accumulate or contract.}{P18}\nAllow a representation $\\gamma_t$ to change with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured predi","\\plate{8}{A local defect needs a propagation law}{Small one-step residuals can accumulate or contract.}{P18}\nAllow a representation $\\gamma_t$ to change with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured prediction errors.}\n\\boundary{Both the true represented state and pre","\\plate{8}{A local defect needs a propagation law}{Small one-step residuals can accumulate or contract.}{P18}\nAllow a representation $\\gamma_t$ to change with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured prediction errors.}\n\\boundary{Both the true represented state and predicted state must remain in the Lipschitz domain. Stability …","with time. Define the typed defect\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured prediction errors.}\n\\boundary{Both the true represented state and predicted state must remain in the Lipschitz domain. Stability of each action's eigenvalues alone does not certify stability under switching.}\n\\nav{\\texttt{DefectPropagation.lean}.}","t\n\\[\nr_t(x)=\\gamma_{t+1}F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured prediction errors.}\n\\boundary{Both the true represented state and predicted state must remain in the Lipschitz domain. Stability of each action's eigenvalues alone does not certify stability under switching.}\n\\nav{\\texttt{DefectPropagation.lean}.}","F_t(x)-\\widehat f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured prediction errors.}\n\\boundary{Both the true represented state and predicted state must remain in the Lipschitz domain. Stability of each action's eigenvalues alone does not certify stability under switching.}\n\\nav{\\texttt{DefectPropagation.lean}.}","f_t\\gamma_t(x).\n\\]\nIf $\\widehat f_t$ is $\\lambda_t$-Lipschitz on the relevant domain and $\\|r_t(x_t)\\|\\leq\\delta_t$, the prediction error satisfies\n\\[\ne_T\\leq\\left(\\prod_{j=0}^{T-1}\\lambda_j\\right)e_0\n+\\sum_{k=0}^{T-1}\\left(\\prod_{j=k+1}^{T-1}\\lambda_j\\right)\\delta_k.\n\\]\n\\diagram[0.94]{08_defect}\nFor a fixed representation $R$, a reference reduced operator $B$ and fitted operator $\\widehat B$,\n\\[\nRL-\\widehat BR=(RL-BR)+(B-\\widehat B)R.\n\\]\nThe first term is representational nonclosure. The second is estimation error. They can enter the same propagation inequality after being bounded in the state norm.\n\\claim{The suffix products determine whether earlier defects are damped or amplified. The chart shows analytical envelopes, not measured prediction errors.}\n\\boundary{Both the true represented state and predicted state must remain in the Lipschitz domain. Stability of each action's eigenvalues alone does not certify stability under switching.}\n\\nav{\\texttt{DefectPropagation.lean}.}","\\plate{9}{What four observations identify}{Data span, closure and coverage are separate checks.}{P14--P17}\nThe running example begins $1,\\tfrac12,\\tfrac12,\\tfrac{11}{24},\\tfrac{61}{144}$. Its first four outputs give\n\\[\nZ_-=\\begin{pmatrix}1&1/2","\\plate{9}{What four observations identify}{Data span, closure and coverage are separate checks.}{P14--P17}\nThe running example begins $1,\\tfrac12,\\tfrac12,\\tfrac{11}{24},\\tfrac{61}{144}$. Its first four outputs give\n\\[\nZ_-=\\begin{pmatrix}1&1/2\\\\1/2&1/2\\end{pmatrix},\\quad\nZ_+=\\begin{pmatrix}1/2&1/2\\\\1/2&11/24\\end{pmatrix},\\quad\nZ_+Z_-^{-1}=\\begin{pmatrix}0&1\\\\1/12&5/6\\end{pmatrix}.\n\\]\n\\diagram{09_identification}\nThe determinant $\\det Z_-=1/4$ establishes coverage of the \\emph{declared} two-dimensional delay space. The resulting recurrence is\n\\[\ny_{t+1}=\\frac56y_t+\\frac1{12}y_{t-1},\\qquad t\\geq1.\n\\]\nFor general paired states, a linear fit exists exactly when $\\ker Z_-\\subseteq\\ker Z_+$. It is unique on $W=\\im Z_-$, and maps $W$ to itself exactly when $\\im Z_+\\subseteq W$.\n\\claim{These data identify the transition on their input span. The known generating system separately certifies that this span is the required full predictive state space.}\n\\boundary{Finite sampled rank does not exclude a larger unexcited state space. Finite continuation agreement does not establish un","1/12&5/6\\end{pmatrix}.\n\\]\n\\diagram{09_identification}\nThe determinant $\\det Z_-=1/4$ establishes coverage of the \\emph{declared} two-dimensional delay space. The resulting recurrence is\n\\[\ny_{t+1}=\\frac56y_t+\\frac1{12}y_{t-1},\\qquad t\\geq1.\n\\]\nFor general paired states, a linear fit exists exactly when $\\ker Z_-\\subseteq\\ker Z_+$. It is unique on $W=\\im Z_-$, and maps $W$ to itself exactly when $\\im Z_+\\subseteq W$.\n\\claim{These data identify the transition on their input span. The known generating system separately certifies that this span is the required full predictive state space.}\n\\boundary{Finite sampled rank does not exclude a larger unexcited state space. Finite continuation agreement does not establish unrestricted discovery. Perturbations additionally require a c","1/12&5/6\\end{pmatrix}.\n\\]\n\\diagram{09_identification}\nThe determinant $\\det Z_-=1/4$ establishes coverage of the \\emph{declared} two-dimensional delay space. The resulting recurrence is\n\\[\ny_{t+1}=\\frac56y_t+\\frac1{12}y_{t-1},\\qquad t\\geq1.\n\\]\nFor general paired states, a linear fit exists exactly when $\\ker Z_-\\subseteq\\ker Z_+$. It is unique on $W=\\im Z_-$, and maps $W$ to itself exactly when $\\im Z_+\\subseteq W$.\n\\claim{These data identify the transition on their input span. The known generating system separately certifies that this span is the required full predictive state space.}\n\\boundary{Finite sampled rank does not exclude a larger unexcited state space. Finite continuation agreement does not establish unrestricted discovery. Perturbations additionally require a conditioning bound.}\n\\nav{\\texttt{DelayClosure.lean}; \\texttt{DataSpan.lean}; see the exact module map at the end.}","\\plate{10}{Behavior can supply the coordinates}{Future tests define a predictive state without naming hidden variables.}{P27--P28}\nFor a controlled process, arrange population probabilities into a behavior table: columns index past histories and rows index future tests. Its full column span is the linear predictive behavior space.\n\\[\n\\begin{array}{c|ccc}\n & h_1 & h_2 & \\cdots\\\\\\hline\n\\tau_1 & p(h_1\\tau_1)&p(h_2\\tau_1)&\\cdots\\\\\n\\tau_2 & p(h_1\\tau_2)&p(h_2\\tau_2)&\\cdots\\\\\n\\vdots &\\vdots&\\vdots&\n\\end{array}\n\\]\nAssume the full behavior space has rank at most $r$. Choose an invertible $r$-by-$r$ population core $H$. Let $G_s$ be the corresponding table with a declared action-observation symbol $s$ inserted. With a compatible row/column convention,\n\\[\n\\boxed{B_s=G_sH^{-1}.}\n\\]\nThe empty-test row supplies the mass functional and a covered history supplies the initial predictive state. The factorization then transports every future test through the observable coordinates.\n\\claim{The dimension is determined by independent predictive behaviors. A global behavior-rank bound and an invertible co…","\\vdots &\\vdots&\\vdots&\n\\end{array}\n\\]\nAssume the full behavior space has rank at most $r$. Choose an invertible $r$-by-$r$ population core $H$. Let $G_s$ be the corresponding table with a declared action-observation symbol $s$ inserted. With a compatible row/column convention,\n\\[\n\\boxed{B_s=G_sH^{-1}.}\n\\]\nThe empty-test row supplies the mass functional and a covered history supplies the initial predictive state. The factorization then transports every future test through the observable coordinates.\n\\claim{The dimension is determined by independent predictive behaviors. A global behavior-rank bound and an invertible core establish that the chosen coordinates cover the full behavior space.}","\\plate{11}{Probability meaning is geometric}{An operator family needs a domain, a mass map and a covered initial state.}{P29, P39--P40}\nLet $C\\subseteq\\R^d$ be a convex cone and $\\lambda$ a linear mass map strictly positive on $C\\setminus\\{0\\}$. Set $K=\\{z\\in C:\\lambda z=1\\}$. Linear observation operators must satisfy\n\\[\nB_oC\\subseteq C,\\qquad \\lambda\\sum_o B_o=\\lambda,\\qquad z_0\\in K.\n\\]\n\\diagram{10_domain}\nA finite record $o_1\\cdots o_T$ has probability\n\\[\n\\lambda B_{o_T}\\cdots B_{o_1}z_0\\geq0.\n\\]\nSumming over every branch repeatedly applies the mass identity and gives total probability one. A positive-probability posterior is $B_oz/(\\lambda B_oz)$.","g is geometric}{An operator family needs a domain, a mass map and a covered initial state.}{P29, P39--P40}\nLet $C\\subseteq\\R^d$ be a convex cone and $\\lambda$ a linear mass map strictly positive on $C\\setminus\\{0\\}$. Set $K=\\{z\\in C:\\lambda z=1\\}$. Linear observation operators must satisfy\n\\[\nB_oC\\subseteq C,\\qquad \\lambda\\sum_o B_o=\\lambda,\\qquad z_0\\in K.\n\\]\n\\diagram{10_domain}\nA finite record $o_1\\cdots o_T$ has probability\n\\[\n\\lambda B_{o_T}\\cdots B_{o_1}z_0\\geq0.\n\\]\nSumming over every branch repeatedly applies the mass identity and gives total probability one. A positive-probability posterior is $B_oz/(\\lambda B_oz)$.","y needs a domain, a mass map and a covered initial state.}{P29, P39--P40}\nLet $C\\subseteq\\R^d$ be a convex cone and $\\lambda$ a linear mass map strictly positive on $C\\setminus\\{0\\}$. Set $K=\\{z\\in C:\\lambda z=1\\}$. Linear observation operators must satisfy\n\\[\nB_oC\\subseteq C,\\qquad \\lambda\\sum_o B_o=\\lambda,\\qquad z_0\\in K.\n\\]\n\\diagram{10_domain}\nA finite record $o_1\\cdots o_T$ has probability\n\\[\n\\lambda B_{o_T}\\cdots B_{o_1}z_0\\geq0.\n\\]\nSumming over every branch repeatedly applies the mass identity and gives total probability one. A positive-probability posterior is $B_oz/(\\lambda B_oz)$.","\\{0\\}$. Set $K=\\{z\\in C:\\lambda z=1\\}$. Linear observation operators must satisfy\n\\[\nB_oC\\subseteq C,\\qquad \\lambda\\sum_o B_o=\\lambda,\\qquad z_0\\in K.\n\\]\n\\diagram{10_domain}\nA finite record $o_1\\cdots o_T$ has probability\n\\[\n\\lambda B_{o_T}\\cdots B_{o_1}z_0\\geq0.\n\\]\nSumming over every branch repeatedly applies the mass identity and gives total probability one. A positive-probability posterior is $B_oz/(\\lambda B_oz)$.","\\textbf{A finite-generator witness.} Let $V$ have full row rank and normalized generator columns. Nonnegative lifts $A_o$ with $B_oV=VA_o$ and $\\mathbf1^{\\mathsf T}\\sum_oA_o=\\mathbf1^{\\mathsf T}$ certify the generated cone. The initial state must be a normalized nonnegative mixture of those columns.\n\\claim{All-horizon validity follows by induction from a finite exact witness. Statistical proximity to a generating process is a further obligation.}\n\\boundary{A zero-probability branch does not define a normalized posterior by division. A finite positive lift is sufficient, but some valid low-dimensional processes require curved domains.}\n\\nav{\\texttt{CanonicalCone.lean}; \\texttt{PositiveLift.lean}; \\texttt{PositiveLiftTopology.lean}.}","\\plate{12}{A curved domain can have a finite witness}{A matrix embedding changes the certificate language.}{P46, P50--P51}\nThe Lorentz cone and its normalized disk admit the representation\n\\[\nC=\\{(m,x,y):m\\geq\\sqrt{x^2+y^2}\\},\\qquad\nJ(m,x,y)=\\frac12\\begin{pmatrix}m+x&y\\\\y&m-x\\end{pmatrix}.\n\\]\n\\diagram[0.96]{11_psd}\nSince the eigenvalues of $J$ are $(m\\pm\\sqrt{x^2+y^2})/2$, membership in $C$ is equivalent to $J\\succeq0$. Also $\\operatorname{tr}J=m$.","y&m-x\\end{pmatrix}.\n\\]\n\\diagram[0.96]{11_psd}\nSince the eigenvalues of $J$ are $(m\\pm\\sqrt{x^2+y^2})/2$, membership in $C$ is equivalent to $J\\succeq0$. Also $\\operatorname{tr}J=m$.","For the exact spatial rotation $R=\\bigl(\\begin{smallmatrix}3/5&-4/5\\\\4/5&3/5\\end{smallmatrix}\\bigr)$ and $T=\\diag(1,R)$,\n\\[\nJ(Tz)=\\frac15WJ(z)W^{\\mathsf T},\\qquad\nW=\\begin{pmatrix}2&-1","4/5&3/5\\end{smallmatrix}\\bigr)$ and $T=\\diag(1,R)$,\n\\[\nJ(Tz)=\\frac15WJ(z)W^{\\mathsf T},\\qquad\nW=\\begin{pmatrix}2&-1","4/5&3/5\\end{smallmatrix}\\bigr)$ and $T=\\diag(1,R)$,\n\\[\nJ(Tz)=\\frac15WJ(z)W^{\\mathsf T},\\qquad\nW=\\begin{pmatrix}2&-1\\\\1&2\\end{pmatrix}.\n\\]\n\\claim{A rational congruence identity certifies the unchanged rotation on the curved cone. In the full P46 model, the canonical domain has no finite normalized positive-generator certificate.}\n\\boundary{The drawn orbit is finite. Density and the finite-generator obstruction are proved in the volume, not established by its plotted points. The full instrument also needs its observation, mass and initial-state identities. No quantum implementation is asserted.}\n\\nav{\\texttt{RotationP46.lean}; \\texttt{MatrixEmbedding.lean}; \\texttt{MatrixRotation.lean}.}","\\plate{13}{Integration retains a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\bounda","\\plate{13}{Integration retains a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices i","\\plate{13}{Integration retains a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal a","\\plate{13}{Integration retains a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The fa","\\plate{13}{Integration retains a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav","tegration retains a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and …","a covariance term}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","m}{Averaging an operator before quadratic action changes the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","es the instrument.}{P56--P59}\nLet $X\\succeq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","eq0$, $K(u)=K_0+uK_1$ and $\\bar K=\\mathbb E[K(U)]$. For a scalar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","alar $U$ of finite second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","second moment,\n\\[\n\\mathbb E[K(U)XK(U)^{\\mathsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","hsf T}]-\\bar KX\\bar K^{\\mathsf T}\n=\\operatorname{Var}(U)K_1XK_1^{\\mathsf T}\\succeq0.\n\\]\n\\diagram{12_covariance}\nFor the uniform probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","m probability measure $\\mu(du)=du/2$ on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","on $[-1,1]$, take\n\\[\nK(u)=\\tfrac12\\diag(1+3u,1-3u),\\qquad X=I/2.\n\\]\nThen $\\bar K=I/2$. The averaged congruence is $I/2$, with trace one. Congruence by the averaged matrix is $I/8$, with trace $1/4$. The omitted covariance correction is $3I/8$.\n\\claim{The missing term is positive semidefinite and carries three quarters of the mass in this one-bin example. With $n$ equal bins, the missing mass is $3/(4n^2)$.}\n\\boundary{The matrices in this example are diagonal and commute. The failure concerns the order of averaging and forming a quadratic action. A density value is interpreted relative to its declared measure.}\n\\nav{See P57--P58 in the formal coverage map for exact integral and bin identities.}","\\plate{14}{Uncertainty on a curved domain}{Parameter uncertainty and arbitrary entrywise noise have different geometry.}{P72, P83}\nFor $u\\in[-1,1]$, set $v(u)=(1,u,u^2)^{\\mathsf T}$. Its convex hull is\n\\[\nK=\\{(1,x,y):x^2\\leq y\\leq1\\},\\qquad\nC=\\left\\{(m,x,y):\\begin{pmatrix}m&x","\\plate{14}{Uncertainty on a curved domain}{Parameter uncertainty and arbitrary entrywise noise have different geometry.}{P72, P83}\nFor $u\\in[-1,1]$, set $v(u)=(1,u,u^2)^{\\mathsf T}$. Its convex hull is\n\\[\nK=\\{(1,x,y):x^2\\leq y\\leq1\\},\\qquad\nC=\\left\\{(m,x,y):\\begin{pmatrix}m&x\\\\x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the","x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, no","x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMom","x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMomentGeometry.lean}; \\t","x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMomentGeometry.lean}; \\texttt{CurvedMomentInstru","x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMomentGeometry.lean}; \\texttt{CurvedMomentInstrument.lean}; exact P83 map","x&y\\end{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMomentGeometry.lean}; \\texttt{CurvedMomentInstrument.lean}; exact P83 map in \\texttt{COVERAGE.json}.}","{pmatrix}\\succeq0,\\ m-y\\geq0\\right\\}.\n\\]\n\\diagram{13_moments}\nChoose a reference probability measure on $[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMomentGeometry.lean}; \\texttt{CurvedMomentInstrument.lean}; exact P83 map in \\texttt{COVERAGE.json}.}","[-1,1]$ with moments $\\mathbb E U=\\mathbb E U^3=0$, $\\mathbb E U^2=1/3$ and $\\mathbb E U^4=1/5$. With $L_2(u)=(3u^2-1)/2$, define\n\\[\nB_{a,b}(u)=v(u)(1,au,bL_2(u)).\n\\]\nIts normalized-state density is $p_{a,b}(u\\mid x,y)=1+axu+byL_2(u)$. If $|a|+|b|\\leq1-\\rho$ for $\\rho\\in[0,1]$, it is at least $\\rho$, and integrated mass is exact.\n\\claim{Every positive-density posterior is the same boundary point $v(u)$ throughout this parameter family. Uncertainty changes its probability without moving its posterior ray.}\n\\boundary{This protection depends on the supplied algebraic family. Free perturbations of output coordinates can leave the cone. The three-dimensional wireframe shows the homogeneous geometry, not sampled evidence.}\n\\nav{\\texttt{CurvedMomentGeometry.lean}; \\texttt{CurvedMomentInstrument.lean}; exact P83 map in \\texttt{COVERAGE.json}.}","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} par","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous c","\\plate{15}{A confidence set can meet an exact certificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence sta…","ificate}{The statistical event and the internal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or pr…","nternal validity check do different work.}{P66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed f…","66, P68, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed facts.}\n\\boundary{Predictable residual reject…","8, P81--P84, P87}\nAssume $Y_t=\\theta_*^{\\mathsf T}\\phi_t+\\xi_t$ with predictable $\\phi_t$, conditional sub-Gaussian noise of scale $\\sigma>0$ and $\\|\\theta_*\\|_2\\leq S$. For ridge $\\eta>0$, define $V_t=\\eta I+\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed facts.}\n\\boundary{Predictable residual rejection ca…","\\sum_{s\\leq t}\\phi_s\\phi_s^{\\mathsf T}$ and $\\widehat\\theta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed facts.}\n\\boundary{Predictable residual rejection can give a lower discrepancy certificate. Nonrejection alone supplies no upper accuracy guarantee. The drawn rectangle is schematic, not an estimated confidence set.}\n\\nav{\\texttt{RegressionConfidence.lean}; \\te…","eta_t=V_t^{-1}\\sum_{s\\leq t}\\phi_sY_s$. For $0<\\delta<1$, with probability at least $1-\\delta$, simultaneously for every finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed facts.}\n\\boundary{Predictable residual rejection can give a lower discrepancy certificate. Nonrejection alone supplies no upper accuracy guarantee. The drawn rectangle is schematic, not an estimated confidence set.}\n\\nav{\\texttt{RegressionConfidence.lean}; \\texttt{PredictableResidualP66.lean};\\newline\\texttt{Struct…","very finite $t$,\n\\[\n\\|\\widehat\\theta_t-\\theta_*\\|_{V_t}\n\\leq\\sigma\\sqrt{\\log\\frac{\\det V_t}{\\det(\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed facts.}\n\\boundary{Predictable residual rejection can give a lower discrepancy certificate. Nonrejection alone supplies no upper accuracy guarantee. The drawn rectangle is schematic, not an estimated confidence set.}\n\\nav{\\texttt{RegressionConfidence.lean}; \\texttt{PredictableResidualP66.lean};\\newline\\texttt{StructuredLearningCapstone.lean}.}","\\eta I)}+2\\log(1/\\delta)}+\\sqrt\\eta S.\n\\]\n\\diagram{14_uncertainty}\nIn the structured moment family, a nonempty returned rectangle $\\Theta$ is internally valid when its sufficient floor\n\\[\n\\rho_\\Theta=1-\\sup_{(a,b)\\in\\Theta}(|a|+|b|)\n\\]\nis nonnegative. Statistical coverage places the true parameter in $\\Theta$ on a simultaneous event. The exact domain argument applies to \\emph{every} parameter in $\\Theta$.\n\\claim{A data-selected time is allowed by the simultaneous confidence statement. It does not convert the model, noise or predictability assumptions into observed facts.}\n\\boundary{Predictable residual rejection can give a lower discrepancy certificate. Nonrejection alone supplies no upper accuracy guarantee. The drawn rectangle is schematic, not an estimated confidence set.}\n\\nav{\\texttt{RegressionConfidence.lean}; \\texttt{PredictableResidualP66.lean};\\newline\\texttt{StructuredLearningCapstone.lean}.}","\\plate{16}{Choose the future quantity before its bound}{A long record and one distant observation are different objects.}{P85--P87}\nStart two processes from the same history or initial law and use the same action policy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\","\\plate{16}{Choose the future quantity before its bound}{A long record and one distant observation are different objects.}{P85--P87}\nStart two processes from the same history or initial law and use the same action policy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constan","cord and one distant observation are different objects.}{P85--P87}\nStart two processes from the same history or initial law and use the same action policy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that r…","ne distant observation are different objects.}{P85--P87}\nStart two processes from the same history or initial law and use the same action policy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember th…","rent objects.}{P85--P87}\nStart two processes from the same history or initial law and use the same action policy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember the entire history require their o…","-P87}\nStart two processes from the same history or initial law and use the same action policy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember the entire history require their own comparison argum…","licy. If their conditional next-observation laws differ by at most $\\Delta$ at every common history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember the entire history require their own comparison argument.}\n\\nav{\\texttt{CurvedMomentRecordTV.lean}; \\texttt{MarkovMarginals.lean}; \\newline\\te…","history, telescoping gives the record bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember the entire history require their own comparison argument.}\n\\nav{\\texttt{CurvedMomentRecordTV.lean}; \\texttt{MarkovMarginals.lean}; \\newline\\texttt{MarkovWindows.lean}; \\texttt{BernoulliRecordSeparation.lean}.}","d bound $\\min(1,T\\Delta)$ in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember the entire history require their own comparison argument.}\n\\nav{\\texttt{CurvedMomentRecordTV.lean}; \\texttt{MarkovMarginals.lean}; \\newline\\texttt{MarkovWindows.lean}; \\texttt{BernoulliRecordSeparation.lean}.}","in total variation.\n\\diagram{15_horizons}\nFor Markov kernels with a common probability measure $\\mu$ and $P(s,\\cdot)\\geq\\rho\\mu(\\cdot)$, $0<\\rho\\leq1$, and $\\sup_s\\|P(s,\\cdot)-Q(s,\\cdot)\\|_{\\TV}\\leq\\Delta$, their one-time law distances satisfy\n\\[\ne_t\\leq\\min\\!\\left(1,(1-\\rho)^te_0+\n\\frac{\\Delta[1-(1-\\rho)^t]}{\\rho}\\right).\n\\]\nA length-$k$ rolling window starting at time $t$ has distance at most $\\min(1,e_t+(k-1)\\Delta)$.\n\\claim{Contraction can keep a single marginal close indefinitely while complete-record distinguishability tends to one. Two distinct independent Bernoulli laws already exhibit that separation.}\n\\boundary{The plotted curves are analytical illustrations with declared constants. The marginal result needs actual Markov and minorization hypotheses. Arbitrary policies that remember the entire history require their own comparison argument.}\n\\nav{\\texttt{CurvedMomentRecordTV.lean}; \\texttt{MarkovMarginals.lean}; \\newline\\texttt{MarkovWindows.lean}; \\texttt{BernoulliRecordSeparation.lean}.}","bottomrule\n\\end{tabular}}\n\\par\\vspace{12pt}\n\\textbf{A small notation key.} $A,S$ denote analysis and synthesis; $\\mathcal C=\\im A$ denotes realizable coefficients; $\\gamma$ denotes a predictive summary; $P,Q$ denote retained and discarded projections; $W_*$ denotes the completed observable row space; $C$ denotes a probability cone in the stochastic plates; $\\lambda$ denotes its mass map. Symbols local to a plate are introduced there."]}