En02-A Finite-Resource Geometry for the Enter Calculus Cut Monotones, Certificate Costs, and Conditional Born Rigidity

We study destructive computational-basis readout under a finite live-qubit budget and a separately specified grammar of consumable geometric resources. The operational assumptions and their capacity restrictions are stated explicitly; results about the specified Born semantics are not imported into an unknown measurement strategy. Identified latent links and active edges give cut-indexed monotones, a bound on cross-cut activations, and decreasing potential reachability. A geometric diamond is proved for read–write independent, guard-stable updates, rather than inferred from quantum-name disjointness alone. For a precisely delimited decohered fine-graining certificate, reduced integer multiplicities give the exact register cost

This cost is not the unrestricted forcing threshold: the registered three-qubit permutation-symmetric witness already anchors $(1/3,2/3)$ at capacity three. Fine-graining and symmetry certificates yield finite rational nets and continuity-controlled approximation bounds wherever the requisite records fit. The principal new conditional result is exact Born readout on every sub-capacity input, at every capacity $N\ge 3$, when positive-branch residuals have the projective computational diagonal on those inputs. Its proof derives a cocycle on an explicit open domain and solves the resulting functional equation. Global diagonal faithfulness additionally removes the full-capacity boundary freedom in outcome laws. Unrestricted finite-capacity uniqueness remains open; finite partial slack models are not admissible counterexamples. A compactness theorem is proved under an additional uniform regularity assumption on selective outputs, without claiming an LP formulation or a computable convergence rate. A separate one-shot instrument ledger distinguishes coherent environment size from adaptive peak occupancy. A standalone computational appendix supplies exact arithmetic regressions and sampled diagnostics, not certificates of universally quantified statements.

The Enter calculus separates a specified concrete semantics from a characterization problem for an unknown measurement transition. In the concrete semantics, destructive readout has Born probabilities and projective residual rays by definition. In the characterization problem, those probabilities and residuals are unknown, while the qubit kinematics, coherent gates, allocation, typing, and observation interfaces remain fixed.

Manuscript I identifies the Born strategy within its declared structural class using composability, no-signaling, nullity, continuity, and a typed record equation I (labels ass:strategy-class, def:requirements, and thm:main). That theorem quantifies over unbounded finite resources. It is not, without a separate resource audit, a theorem about strategies defined only up to a fixed capacity.

This paper makes that audit explicit. It has three distinct subjects:

1. a combinatorial update grammar and its cut monotones;

2. finite-capacity operational constraints on unknown readout strategies;

3. register accounting and finite computational diagnostics.

No implication between these subjects is left implicit.

The following table records the relationship between the legacy “EC” locators used in the earlier draft and the stable labels of registered manuscript I. The legacy numbers are historical locators, not a second collection of independent premises.

The local proofs below state the assumptions they use and reproduce the finite arguments needed here. In particular, no unbounded characterization theorem is invoked to identify a finite-capacity strategy.

cost is not unrestricted forcing cost. The fine-graining construction for $(1/3,2/3)$ uses four qubits. Nevertheless, I's permutation-symmetric three-qubit witness anchors the same diagonal law using a different construction I (label prop:three). Consequently, four-qubit optimality within the fine-graining family does not establish a four-qubit unrestricted forcing threshold.

equality is not diagonal faithfulness. The registered record equation compares classical observations of two typed programs. Diagonal faithfulness is a condition on an unknown conditional residual state. They are related in the Born characterization, but they are not identical definitions. The conditional rigidity theorem below states diagonal faithfulness separately.

finite probe is not a quantified theorem. A finite collection of successful tests does not establish admissibility over every state, gate, continuation, and record instance. Conversely, one exact failed instance can reject a proposed strategy. The computational appendix respects this asymmetry.

The finite four-live-qubit class in X-Enter is an interface restriction on a specified family of first-order programs, including their input. It is not the fine-graining cost theorem, a rational-net resolution, or a field-to-qubit preparation theorem XEnter.

The cut quantities defined here are combinatorial resource counters. They are not a gravitational Noether current, a BV antibracket, a retarded field coefficient, or a physical causal cone. Likewise, the instrument ledger below does not instantiate the separately represented Models 1, 2, or 3 of manuscript IV. No gravitational or continuous-variable premise is imported.

For a finite set $L$ of live qubit names, let

An ordering of $L$ is used only to write coordinates. Global states are unit rays. For $S\subseteq L$ and $\kappa \in \{0,1{\}}^S$, define

On a positive Born branch the destructive projective residual is

The measured factors are removed, not retained.

For finite laws,

For unit rays we use

A capacity-$N$ strategy $\mathsf{S}=(\mathsf{P},\mathsf{U})$ supplies, for every $\left|L\right|\le N$ and $S\subseteq L$:

1. a normalized law ${{\mathsf{P}}^{\mathsf{S}}}_{\psi }$ on $\{0,1{\}}^S$;

2. a pure unit residual ray ${{\mathsf{U}}^{\mathsf{S}}}_{\psi }(\kappa )\in \mathbb{P}({\mathcal{H}}_{L\setminus S})$ whenever ${{\mathsf{P}}^{\mathsf{S}}}_{\psi }(\kappa )>0$.

These data depend only on the current ray and measured names. They are equivariant under simultaneous relabeling of names, tensor factors, and outcome coordinates.

All finite-register unitaries and fresh $|0⟩$ allocation have their stipulated linear-algebraic meanings. Allocation is allowed only when the resulting live occupancy is at most $N$. No hidden memory, silent quantum deletion, or apparatus-dependent measurement parameter is included.

A finite experimental program uses allocation, gates, destructive readout, finite classical control, and return. Every quantum name is consumed or returned. Local programs exchange neither quantum names nor classical messages. Single programs use their prescribed evaluation order; bipartite experiments permit interleavings of their local orders.

For $S=\varnothing$, readout has its unique unit outcome and leaves the state unchanged.

This is the finite restriction of the relevant structural interfaces of manuscript I. It is not obtained from the operational requirements.

A strategy in Assumption ass:class is {operationally admissible at capacity $N$} if it satisfies the following requirements on every specified experiment whose entire execution remains within $N$ live qubits. Write $\mathsf{S}\in {\mathcal{A}}_N$.

{${\text{(C)}}_N$: composability.} For disjoint $S,T\subseteq L$, joint readout and both sequential orders have the same ordered joint law. Thus

on positive first branches, with value zero on a zero first branch; the corresponding equation holds in the reverse order.

{${\text{(NS)}}_N$: internal no-signaling.} For every bipartition, the terminal classical law of one local program is independent of the other local program and of the interleaving. Locally allocated qubits and returned quantum spectators are included. There is no exchange of names or messages.

{${\text{(N)}}_N$: nullity.}

{${\text{(K)}}_N$: continuity.} For each fixed $L,S$, the map

is continuous in total variation. No continuity of the unknown update is presumed.

{${\text{(R)}}_N$: typed records.} Let $L=S\bigsqcup B$, and let $u$ consume all of $B$, returning classical output and not using the first reported outcome. Compare:

1. read $S$, then run $u$ on the residual $B$, and report the ordered classical outputs;

2. allocate a fresh register $R$ of $\left|S\right|$ zero qubits, copy $S$ computationally to $R$ by bitwise CNOT, run the same $u$ on $B$, then read $R$, report the same ordered classical outputs, and return the original $S$.

Their classical observation laws must agree whenever both complete executions fit the budget.

The second program retains the original $S$ as quantum output. When $B\ne \varnothing$, a continuation that omits $B$ is not a legal record instance. This is the typed interface of I's :record, restated rather than replaced by silent discard I.

The strategy

belongs to ${\mathcal{A}}_N$ for every $N$.

Proof. Normalization follows from ${\sum }_{\kappa }({M^S}_{\kappa }{)}^{†}{M^S}_{\kappa }=I$. Disjoint destructive components commute after identifying surviving factors, proving composability. Nullity and continuity follow directly from squared norms.

For a finite local program, compose allocation isometries, unitaries, and destructive measurement bras along each history. Backward induction on the outcome tree gives a complete family of path maps: ${\sum }_h{K}_h^{†}{K}_h=I$. For bipartite histories the maps factor as $K_a\otimes L_b$. Summing over the remote histories gives

This proves the specified no-signaling requirement.

For records write $\psi ={\sum }_{\kappa }|\kappa {⟩}_S|{\xi }_{\kappa }{⟩}_B$. A continuation history $h$, with path map $K_h$, has joint weight ${\Vert K_h{\xi }_{\kappa }\Vert }^2$ in the direct program. The copied program has pre-readout vector

and record readout gives the same weight. Summing histories proves record equality. All compared paths obey the capacity restriction.

For $\mathsf{S}\in {\mathcal{A}}_N$:

1. composability implies

   $${{\mathsf{P}}^{\mathsf{S}}}_{\psi }={\mathrm{marg}}_S{{\mathsf{P}}^{\mathsf{S}\cup \mathsf{T}}}_{\psi };$$

2. applying a unitary outside $S$ does not change ${{\mathsf{P}}^{\mathsf{S}}}_{\psi }$;

3. adjoining fresh zero-state spectators outside $S$ does not change ${{\mathsf{P}}^{\mathsf{S}}}_{\psi }$, provided the enlarged configuration fits.

Proof. For (i), sum (eq:C) over $\lambda$ and use normalization. For (ii), compare the local readout of $S$ against a remote gate followed by return, and against remote return alone. For (iii), replace that remote gate experiment by allocation followed by return. These are capacity-expressible no-signaling comparisons.

Let two pure configurations have the same reduced state on a register $S$. If padding both environments to a common size fits capacity $N$, their $S$-laws agree.

Proof. Pad the smaller environment by zero spectators. Two purifications of the same reduced state with equal environment dimension are related by an environment unitary: choose matching Schmidt vectors on $S$, map one orthonormal environment family to the other, and extend that isometry to a unitary. Apply Lemma lem:remote.

Suppose $n=\left|L\right|$, $s=\left|S\right|$, and $n+s\le N$. Define the copied state

Then

Proof. Use the legal continuation $u=\mathsf{Enter}(B)$, retaining all its classical outputs. If $B$ is empty, use the unit continuation. The direct joint law has $S$-marginal ${{\mathsf{P}}^{\mathsf{S}}}_{\psi }$ by composability. In the deferred program, the marginal law of $R$ equals its law before the remote readout of $B$, by no-signaling. Record equality identifies these marginals. The density-operator formula follows by tracing the copied vector: the original computational labels in $S$ are orthogonal.

Lemma lem:record-marginal uses $s$ fresh record qubits and requires $n+s\le N$. I's budgeted stability theorem uses whole-register records, including further records of intermediate deferred states; its displayed conditions include $2n\le N$ and $2(n+s)\le N$ I (label thm:stability). The budgets price different comparisons. The present exact marginal argument does not remove any register from that approximate continuation theorem, nor does it supply an approximate update estimate.

Suppose

fits capacity $N$, where the strings $k$ are distinct and the computational strings ${\eta }_k$ are distinct. Then ${{\mathsf{P}}^{\mathsf{S}}}_{\psi }$ is uniform on $K$.

Proof. A diagonal unitary on $B$ removes the phases without changing the $S$-law. Write $p={{\mathsf{P}}^{\mathsf{S}}}_{\psi }$, $q={{\mathsf{P}}^{\mathsf{B}}}_{\psi }$, and $\nu ={{\mathsf{P}}^{\mathsf{S}\cup \mathsf{B}}}_{\psi }$. Nullity restricts $\nu$ to $(k,{\eta }_k)$; composability gives

For a permutation $\pi$ of $K$, apply the basis permutation on $S$ sending $k$ to $\pi (k)$. The resulting vector can also be obtained by a unitary on $B$ sending ${\eta }_k$ to ${\eta }_{{\pi }^{-1}(k)}$. Remote invariance therefore leaves the $S$-law equal to $p$ and the $B$-law equal to $q$. Nullity for the new pairing gives

Thus every permutation preserves $p$, and normalization makes it uniform. No extra qubits are allocated.

This is the finite version of I's :uniformity. It derives the required symmetry; it does not assume covariance of unknown probabilities under arbitrary computational-basis permutations I.

A geometry is

where:

1. $V$ is a fixed finite set of regions;

2. $\mathrm{cap}:V\to \mathbb{N}$ is a capacity function;

3. $L_{\mathrm{lat}}$ is a finite family of identified latent links $(\ell ,\{u,v\})$, with distinct endpoints in $V$;

4. $E_{\mathrm{act}}$ is a finite family of identified directed edges $(\ell ,u,v)$.

Identifiers are unique and no identifier occurs in both components. The directed graph underlying $E_{\mathrm{act}}$ is acyclic. Distinct identified edges may have the same endpoints.

Set

Identified parallel edges are retained because each activation consumes one identified token and creates one identified edge. If active edges were instead an unlabelled set of endpoint pairs, activating a second parallel token would not preserve the scalar count.

A configuration carries:

where $\lambda :L\to V$ labels live names and

Measurement removes names and their tensor factors. It does not remove the region containing them. Permanent capacity retirement is a separate update.

Mutable geometric cells comprise capacity cells and identified link cells. Their permissions are linearly owned. An endpoint region is not, by itself, a permission to modify every incident link.

A transition has a write footprint $W$, a read footprint $D$, and a guard. Its output on $W$ depends only on data in $D\cup W$, and it fixes all cells outside $W$.

Two transitions are independent if

and each guard remains true after the other transition.

An acyclicity check can read nonlocal graph data. Consequently, disjoint endpoint sets or disjoint quantum names do not automatically establish geometric independence.

Each latent link has one linear activation authority ${\mathsf{Send}}_{\text{ell@u}}$, with partner receive capability ${\mathsf{Recv}}_{\text{ell@v}}$. Only the authority can activate the link, from $u$ to $v$. The orientation and authority are preparation data. The receive capability is not a second activation authority.

Activation consumes ${\mathsf{Send}}_{\ell }$, replaces the latent link cell by the active edge with the same identifier, and is legal only if the resulting active graph is acyclic.

This avoids a two-party race in which two independently owned “halves” both attempt to consume the same global latent object. A different two-sided activation protocol would require its own synchronization and progress proof.

The only geometric updates are finite compositions of:

1. retire one capacity unit at an owned region, with $\mathrm{cap}(v)>o(v)$;

2. delete one owned identified active edge;

3. delete one owned identified latent link;

4. activate one owned latent link as in Definition def:authority.

No move creates a new identifier, changes endpoints, reverses an existing edge, or creates an edge without consuming its latent link.

The strict inequality in (m1) protects occupied slots. If measurement is followed by permanent retirement, name deletion is performed first and the resulting free unit is then retired.

Legal primitive moves preserve geometric well-formedness and the occupancy bound. Independent geometric transitions commute. For the specified Born quantum semantics, a pair of name-disjoint quantum redexes equipped with independent geometric transitions has the concrete probabilistic diamond.

Proof. Capacity retirement leaves capacity at least occupancy. Deletion preserves acyclicity. Activation preserves it by its guard and preserves identifier uniqueness by moving, rather than copying, the link identifier.

For independent transitions, neither writes data used by the other, and neither invalidates the other's guard. Each therefore produces the same update on its own write footprint in either order. Outside their union both act identically.

The Born quantum diamond follows from the commuting destructive components, disjoint gates, and fresh-name identification. Its joint branch weights are the same in both orders. Combine that equality with the geometric commutation just proved.

Manuscript I's :diamond concerns quantum-name disjointness in the specified concrete calculus. The geometric part of Proposition prop:diamond is new and needs the footprint and guard conditions above. It is not a theorem about arbitrary unknown updates merely because their names are disjoint.

Let $x,y\in \{0,1\}$ initially both be zero. One map writes $x:=1-y$ and leaves $y$ fixed; the other writes $y:=1-x$ and leaves $x$ fixed. Their write footprints are disjoint, but both read the other's cell. The two orders give $(1,0)$ and $(0,1)$.

Similarly, a collection of individually acyclic edge insertions may form a cycle when combined. A global acyclicity guard is therefore not automatically stable under disjoint writes. These are new counterexamples to insufficient geometric hypotheses, not inherited quantum counterexamples.

For a cut $X\mid Y$, with $Y=V\setminus X$, define

and

All counts include identifiers and hence multiplicity.

Deleting a crossing token $\{A,B\}$ and an internal token $\{B,B^{\prime }\}$, then creating edges $B\to A$ and $B^{\prime }\to A$, preserves the scalar count of links. It increases the number of crossing links from one to two. The grammar excludes the second edge because no consumed token had endpoints $\{B^{\prime },A\}$.

Under Definition def:moves:

1. $\mu$ decreases by one under (m1)–(m3) and is preserved by (m4);

2. for every cut, ${\mu }_{X\mid Y}$ and ${\sigma }_{X\mid Y}$ are non-increasing;

3. activation preserves ${\sigma }_{X\mid Y}$ for every cut: one crossing token becomes one crossing identified edge;

4. along a run, the number of cross-cut activations is at most the initial number of crossing latent links, and therefore at most ${\sigma }_{X\mid Y}(g_0)$.

Proof. Use $1_A$ for the indicator of a condition $A$. For a deleted or activated link with endpoints $u,v$, let $c=1_{\text{endpoints cross}}$. The changes are:

Every entry in the cut columns is non-positive. The activation row counts an identified edge, so it remains valid for parallel endpoint pairs. Finite composition proves (i)–(iii).

No move adds a latent link. Each crossing latent identifier can be activated at most once. This proves (iv).

The word “conservation” is appropriate for the latent-to-active conversion in (iii). The complete quantities are monotones, not constants along arbitrary runs.

including reflexive pairs and forgetting identifiers only when forming reachability.

For every legal run,

at the level of endpoint pairs. Each latent identifier activates at most once. If initially ${\sigma }_{X\mid Y}=0$, no crossing active edge or latent link is ever reachable.

Proof. Retiring capacity changes no reachability generator. Deleting structure removes generators. Activation replaces the two potential orientations of its latent link by one of those orientations. Thus its generator set only shrinks, and so does its reachability closure.

An activated identifier never returns to the latent component. The final assertion follows either from Theorem thm:E or directly from the absence of crossing generators.

Consider two local quantum programs with disjoint quantum ownership. Suppose their geometric actions leave the other party's private readable cells fixed, and neither program inspects shared link status, uses a cross-party communication primitive, or has its local enablement changed by the other's geometric actions. Then:

1. the specified Born semantics has program-independent unconditioned remote classical laws;

2. an operationally admissible unknown strategy has the same independence by ${\text{(NS)}}_N$;

3. the remote private geometric observation is unchanged.

Proof. Part (i) is the path-map marginalization in Lemma lem:born-inhabitation. Part (ii) is the operational requirement, applied to the stated local experiments. Part (iii) follows because no transition writes a remotely readable private cell or changes its local enablement.

An agent's view that includes the actual latent/active status of an incident link changes when that link activates. Literal equality of such views is false. Only an observation interface that cannot inspect or exploit that shared status before the authorized communication step supports an invisibility claim. The cut theorem itself does not prove physical no-signaling, nor does edge counting determine a channel's information capacity.

If $N\le N^{\prime }$ and $\mathsf{S}\in {\mathcal{A}}_{N^{\prime }}$, restricting its laws and residuals to inputs and experiments of peak occupancy at most $N$ gives a member of ${\mathcal{A}}_N$.

Proof. Every capacity-$N$ experiment is a capacity-$N^{\prime }$ experiment. Its state spaces, fixed gates, and observation interface are unchanged. Every required equation is therefore inherited.

For multiple regions one can make the same statement with pointwise occupancy and effect bounds, provided the experimental library is fixed independently of the current active graph.

Activation can enable a previously unavailable communication or gate interface. In such a graph-dependent language, pointwise capacity decrease does not imply that the new set of experiments is a subset of the old set. Run-invariant admissibility then needs a separate simulation or library-inclusion theorem. The present downward-closure claim is only the fixed-library claim of Lemma lem:restriction.

A total capacity sum is also not sufficient for local allocation. A register needed in one region must fit its local capacity, unless an explicitly permitted distributed placement implements the construction. All forcing results below concern a single laboratory of capacity $N$, with the coherent gates of Assumption ass:class.

Let a probability law on $s$ qubits have positive support $K\subseteq \{0,1{\}}^s$. Write

Zero weights are omitted. This is the primitive integer representation.

An $\mathsf{FG}$-certificate for this law uses disjoint registers $S,E^{\prime },E^{\prime \prime }$, with $\left|S\right|=s$, and a state

such that:

1. the labels $r(\alpha )$ are pairwise distinct;

2. the companions $(k(\alpha ),e(\alpha ))$ are pairwise distinct;

3. the diagonal marginal on $S$ is the target law.

Its certificate cost is the number of qubits simultaneously hosting $S,E^{\prime },E^{\prime \prime }$. Let $C_{\mathsf{FG}}(s,\vec{m})$ be the minimum within this specified family.

The family is intentionally narrower than all protocols capable of forcing a law. It is the register pattern of I's decohered fine-graining proof I (label thm:fine-anchor).

For primitive multiplicities,

Proof. For the upper bound, choose $a=\lceil {\log }_{2}M\rceil$ and $b=\lceil {\log }_2{\max }_k{m}_k\rceil$. Partition $M$ distinct $a$-bit labels into blocks of sizes $m_k$. Within block $k$, assign distinct $b$-bit labels. Use the resulting state with $M^{\prime }=M$.

For the lower bound, the number of terms in block $k$ is

Thus $M$ divides $m_k{M}^{\prime }$ for every $k$. Since the multiplicities have greatest common divisor one, an integer linear combination of them equals one; hence $M$ divides $M^{\prime }$, and $M^{\prime }\ge M$. Distinct refined labels require $2^{|E^{\prime }|}\ge M^{\prime }\ge M$. Within block $k$, distinct companions require

These two bounds are attained by the upper-bound construction.

Without reduction to primitive multiplicities the formula is not representation-invariant: $(2,2)/4=(1,1)/2$. The revised code reduces multiplicities before computing the cost.

If $C_{\mathsf{FG}}(s,\vec{m})\le N$, every $\mathsf{S}\in {\mathcal{A}}_N$ has the target law on any capacity-$N$ purification whose measured reduction is $\mathrm{diag}(\vec{m}/M)$.

Proof. On the certificate, Lemma lem:uniformity makes the $E^{\prime }$-law uniform on its $M$ labels. Nullity restricts joint $S,E^{\prime }$ outcomes to the prescribed pairs $(k(r),r)$. Composability therefore gives weight $1/M$ to each pair and marginal $m_k/M$ on $S$.

For another purification, pad the smaller environment to the larger of its size and the certificate size. Both fit $N$. Lemma lem:locality transfers the law.

No record register is included in (eq:fgcost). A separate coherent-to-record comparison on an $n$-qubit input adds $\left|S\right|$ live qubits. The two costs are not to be added when the corresponding witness configurations are used in separate experiments, but neither may be silently omitted from its own experiment.

For the same rational law, capacity $sM$ suffices to anchor its diagonal reduction, using composability, nullity, and the structural tensor-permutation equivariance, followed by purification locality.

Proof. Take $M$ blocks of $s$ qubits. Let $\mathcal{W}$ contain all sequences $(k_1,\dots ,k_M)$ with exactly $m_k$ occurrences of label $k$, and set

Block permutations are tensor-factor permutations. They preserve the vector and act transitively on its computational support. Equivariance and nullity make its full outcome law uniform on $\mathcal{W}$. Composability gives the first-block marginal

The first-block reduction is diagonal: completions corresponding to different first labels have different remaining multisets and hence disjoint support. Purification locality gives the stated transfer.

For every $\mathsf{S}\in {\mathcal{A}}_3$, measurement of the first qubit of any two-qubit purification of $\mathrm{diag}(1/3,2/3)$ has law $(1/3,2/3)$.

Proof. The state

is invariant under tensor-factor permutations, which act transitively on its three support strings. Its full law is uniform by equivariance and nullity. Its first-qubit marginal is $(1/3,2/3)$, and its reduced state is diagonal with those weights. Pad the two-qubit purification by one zero spectator and apply Lemma lem:locality.

This is I's :three, included with its finite proof. In contrast,

Both statements are true; they optimize different certificate families.

For $a\in (0,1)$, put

Let $N_{\mathrm{force}}(a)$ be the least $N\ge 2$ for which every $\mathsf{S}\in {\mathcal{A}}_N$ satisfies ${{\mathsf{P}}^{\{0\}}}_{{\chi }_a}(0)=a$, if such an $N$ exists.

Lemma lem:restriction makes this forcing property upward monotone. Born inhabitation ensures that the admissible class is not empty.

In particular, a capacity-three strategy deviating at ${\chi }_{1/3}$ cannot satisfy the retained structural and operational requirements. The exact lower threshold is not determined here.

A proof that uses three qubits does not, by itself, exclude forcing at two. The earlier claimed equivalence between a four-qubit unrestricted threshold and a capacity-three deviant at this witness is therefore not retained.

For $N\ge s$, define

including zero coordinates at unsupported outcomes. Also set

The second term uses primitive multiplicities and the symmetry certificate of Proposition prop:symmetry. Both nets are finite. They are sufficient anchoring sets, not a classification of all forceable laws.

For either net $\mathcal{R}$, define its covering radius

For binary laws the parameter $p(0)$ identifies the simplex with $[0,1]$, and total variation is ordinary absolute distance. Thus the covering radius is half the largest gap between successive net points.

The fine-graining covering radii at capacities $4,5,6$ are

Including symmetry certificates gives

Proof. The binary fine-graining nets at capacities $4$ and $5$ are, respectively,

At capacity $6$, the net is

These lists follow directly from (eq:fgcost); their largest gaps are $1/3,1/4,1/5$.

For capacities $3\le N\le 6$, symmetry includes every grid point $j/N$, so all gaps in the union are at most $1/N$. The cost formula shows that fine-graining introduces no positive point below $1/N$ in these four cases. The initial gap is therefore exactly $1/N$.

For a fixed admissible strategy at capacity $N$, there is a nondecreasing function ${\omega }_N$, tending to zero at zero, such that

for every same-size pair with at most $N$ qubits.

Proof. Each fixed ray space is compact, so continuity implies uniform continuity. Up to name relabeling there are finitely many register sizes and measured subsets at capacity $N$. Take the maximum of their uniform moduli.

Let $\mathsf{S}\in {\mathcal{A}}_N$, let $n=\left|L\right|$, and let $s=\left|S\right|\ge 1$.

1. Every diagonal measured reduction whose weights belong to ${\mathcal{R}}_{\mathrm{cert}}(N,s)$ is exactly Born.

2. If $n+s\le N$, then

   $$d_{\mathrm{TV}}({{\mathsf{P}}^{\mathsf{S}}}_{\psi },{\mathrm{Born}}_{\psi }^S)\le {\omega }_N\left(\sqrt{2{\delta }_{\mathrm{cert}}(N,s)}\right)+{\delta }_{\mathrm{cert}}(N,s).$$

   The same bound holds with the fine-graining net and its radius.

Proof. Part (i) is Propositions prop:fg-anchor and prop:symmetry, including purification locality.

For (ii), use Lemma lem:record-marginal. Write the copied vector with the record first as

where the $v_{\kappa }$ are orthonormal on positive weights. The environment has $n\ge s$ qubits, so extend this family orthonormally to every record label. Choose $q$ in the certified net with $d_{\mathrm{TV}}(p,q)\le {\delta }_{\mathrm{cert}}$, and define

Then

The $R$-law at ${\Psi }_q$ is $q$ by (i). Uniform continuity and the triangle inequality give

The record identity transfers this to direct $S$-readout.

A residual law close to Born on its own state need not be close to the Born conditional law of the original input: the unknown update may change that residual state. Consequently, (eq:meshbound) alone does not justify replacing sequential conditionals by Lüders conditionals at full capacity. No support-graph classification or state-uniform boundary $O(\epsilon )$ estimate is claimed here.

For fixed $s$, the certified radii tend to zero as $N$ grows. For example, taking denominator $2^h$ with $h=\lfloor (N-s)/2\rfloor$ gives a fine-graining grid whose rounding radius is at most

The argument of ${\omega }_N$ is the square root of this probability radius. There is no strategy-uniform rate without a common modulus assumption.

A strategy satisfies ${\mathrm{DF}}_{<N}$ if, for every input on $n\le N-1$ qubits and every positive-probability measurement branch,

Nullity ensures that the Lüders residual exists on such a branch. The diagonals are taken on the surviving full computational basis.

The condition ${\mathrm{DF}}_{\le N}$ imposes the same equality on inputs with $n\le N$. Neither condition identifies residual phases.

Diagonal faithfulness isolates the projective-diagonal property needed in the cocycle proof. A continuation record argument that has already identified Born future-test laws can imply this property by taking computational diagonals. The registered record equality itself, however, is an equality of classical observations of typed programs, not this residual equation. We therefore retain Definition def:df as an explicit conditional hypothesis.

Let $\phi :[0,1]\to [0,1]$ be continuous, with

Suppose

for all

Then $\phi (x)=x$ on $[0,1]$.

Proof. We first prove positivity. Choose $a=x$ and $m=2xu$. This is allowed when

and gives

If $\phi (x_0)=0$ for $x_0\in (0,1)$, then $\phi$ vanishes on

If it vanishes on $(0,\lambda )$ with $\lambda \le 1/2$, then for $x\in (\lambda /2,\lambda )$ we have $x<1/2$, hence $U(x)=1$; (eq:doubling) makes it vanish on $(0,2x)$. Taking the union gives vanishing on $(0,2\lambda )$. Finite repetition produces an interval containing $1/2$, contradicting the anchor. Thus $\phi >0$ on $(0,1)$.

Put $H=\log \phi$. For $x,y\in (0,1)$, set $p=xy$. The admissible intervals for the factorizations $(x,y)$ and $(\sqrt{p},\sqrt{p})$ are

Their intersection is nonempty. Choosing $m$ in that intersection and applying (eq:cocycle) twice gives

Hence, with $f(t)=2H(\sqrt{t})$,

Write $G(s)=H(e^{-s})$ and $F(t)=f(e^{-t})$. Then

For every $\tau >0$,

The common increment $\Delta (\tau )$ is continuous and additive on positive arguments. Additivity gives $\Delta (q)=q\Delta (1)$ for positive rational $q$; continuity gives the same formula for every positive real argument. Thus

Continuity at one and $\phi (1)=1$ give $d=0$. The half anchor gives $2^{-c}=1/2$, so $c=1$. The endpoint anchors complete the result.

Let $N\ge 3$ and $\mathsf{S}\in {\mathcal{A}}_N$. If $\mathsf{S}$ satisfies ${\mathrm{DF}}_{<N}$, then on every input with $n\le N-1$ and every measured register $S$,

Proof. 1: one marginal function. Define

For any single-qubit readout on an $n\le N-1$ input, let $a=⟨0|{\rho }_S|0⟩$. One record qubit fits. By Lemma lem:record-marginal, its law equals the direct law, and its reduction is $\mathrm{diag}(a,1-a)$. Pad ${\chi }_a$ with $n-1$ zero spectators. The padded reference and copied input are equal-size purifications of that same record reduction, both on $n+1\le N$ qubits. Purification locality and spectator invariance therefore give

Continuity of $a\mapsto {\chi }_a$ and ${\text{(K)}}_N$ make $\phi$ continuous.

Only the reference and cocycle witnesses below have a fixed three-qubit maximum. The general marginal comparison uses $n+1\le N$, not a universal three-qubit bound.

2: anchors. Nullity at ${\chi }_0,{\chi }_1$ gives $\phi (0)=0$, $\phi (1)=1$. Uniform correlations at $(|00⟩+|11⟩)/\sqrt{2}$ give $\phi (1/2)=1/2$.

3: the exact witness domain. Fix $a,u\in (0,1)$ and $m\in (au,au+1-a)$. Define

All four squared amplitudes are strictly positive and sum to one. The computational diagonals of its one-qubit reductions are

The reductions themselves generally have nonzero off-diagonal entries; no diagonality claim is needed.

The projective residual after $q_0=0$ has zero-outcome weight $u$; after $q_1=0$ it has zero-outcome weight $au/m$. Both quantities lie in $(0,1)$. The input occupies two qubits and hence is sub-capacity at every $N\ge 3$. Its one-qubit residuals are sub-capacity as well.

If the first probability is positive, ${\mathrm{DF}}_{<N}$ makes the strategy's residual diagonal equal to the corresponding projective diagonal, so Step 1 evaluates the second law by $\phi$. If the first probability is zero, the joint probability is zero and the same product expression remains valid. Composability in the two orders therefore gives

on precisely the stated open domain.

4: solve the scalar equation. Lemma lem:functional gives $\phi =\mathrm{id}$. Thus every single-qubit readout on a sub-capacity input is Born.

5: arbitrary registers. Induct on $s=\left|S\right|$. For $s\ge 2$, write $S=\{i\}\bigsqcup S^{\prime }$. Composability gives

A positive first branch has its projective residual diagonal by ${\mathrm{DF}}_{<N}$. The residual occupies $n-1\le N-2$ qubits. The induction hypothesis makes its $S^{\prime }$-law Born, and that Born law depends only on the residual computational diagonal. The Born chain rule then gives the joint Born law. Zero first branches contribute zero to both sides. The empty register has its unique outcome.

If an admissible capacity-$N$ strategy, $N\ge 3$, deviates from Born on a sub-capacity input, it violates ${\mathrm{DF}}_{<N}$. Equivalently, under ${\mathrm{DF}}_{<N}$, any outcome-law deviation must be at full capacity.

This is a statement about the strategy's update conditions, not a claim that the observed deviating branch itself must be the branch on which diagonal faithfulness fails.

Let $N\ge 3$, $\mathsf{S}\in {\mathcal{A}}_N$, and assume ${\mathrm{DF}}_{\le N}$. Then every outcome law on every input with $n\le N$ is Born.

Proof. Theorem thm:I handles $n<N$. Take $n=N$ and first suppose every computational amplitude of $\psi$ is nonzero. Partition its names into nonempty $A,B$. Every first-readout residual has fewer than $N$ qubits and therefore has Born laws by Theorem thm:I. Global diagonal faithfulness identifies its computational diagonal with the original input's projective conditional diagonal. Consequently,

The formula remains true when a strategy first weight is zero. All Born weights in the display are positive, so

for every pair. These ratios have one common value. Normalization makes it one, and hence the full law is Born. Composability gives every measured marginal.

Full-support rays are dense: perturb a vector in a direction with all coordinates nonzero, avoid the finitely many perturbation values that cancel a coordinate, and normalize. Continuity of the strategy laws and Born laws extends the conclusion to every ray.

The distinction between ${\mathrm{DF}}_{<N}$ and ${\mathrm{DF}}_{\le N}$ matters. The weaker condition supports Corollary cor:K's boundary alternative. Under the stronger, globally diagonal-faithful condition, Theorem thm:boundary removes that alternative for outcome laws. No exceptional GHZ support classification is needed.

Under the hypotheses of Theorem thm:boundary, a readout of $S$ on an $n$-qubit input has its destructive projective residual on every positive branch whenever $n+\left|S\right|\le N$. If no qubits survive, the residual ray is unique without this budget condition.

Proof. Write $\psi ={\sum }_{\kappa }|\kappa {⟩}_S|{\xi }_{\kappa }{⟩}_B$. For a fixed unitary $V$ on all of $B$, compare the direct continuation $\mathsf{Enter}(V(B))$ with its typed deferral. The copied input fits by hypothesis. All outcome laws involved are Born by Theorem thm:boundary. Record equality gives, for $b_{\kappa }={\Vert {\xi }_{\kappa }\Vert }^2>0$,

Choose $V$ sending ${\xi }_{\kappa }/\sqrt{b_{\kappa }}$ to a fixed basis vector. The equality makes the strategy residual have unit overlap with that ray. Thus the rays agree. The chosen continuation is fixed within each experiment and does not depend on the first reported outcome.

For a fixed $N\ge 3$, do the requirements defining ${\mathcal{A}}_N$, without diagonal faithfulness, force every capacity-$N$ outcome law and every positive residual ray to be Born and destructive projective, respectively?

A witness-specific version asks whether an admissible strategy can deviate at a specified input not already anchored by a certificate. For capacity three, one possible diagnostic is

No claim is made that this diagnostic captures every possible deviation.

The diagnostic at ${\chi }_{1/3}$ is identically zero on ${\mathcal{A}}_3$ by Corollary cor:third. It is therefore unsuitable as an open capacity-three witness.

I's static appendix assigns marginal probabilities without the full operational record equation and without a finite-capacity protocol model. It is not a capacity-three admissible strategy. The operational question here additionally constrains conditional residuals, sequential experiments, null branches, and records. Neither a static deformation nor a finite partial assignment settles Open Problem op:G.

The following example replaces an unsupported claim of a globally admissible compensated strategy by an explicitly partial construction.

For $0<\eta \le 1/2$, set

It has anchors at $0,1/3,1/2,2/3,1$ and obeys ${\phi }_{\eta }(1-x)=1-{\phi }_{\eta }(x)$. The derivative of the added polynomial has absolute value at most one on $[0,1]$: the product rule bounds its terms by $4/9$, $2/9$, and $1/3$. Thus ${\phi }_{\eta }^{\prime }\ge 1-\eta >0$, and it maps $[0,1]$ increasingly onto itself.

At $\eta =1/2$,

However, the cocycle instance

has defect

The associated Born table is

One can nevertheless satisfy both sequential orders on this one strictly supported input by setting the proposed joint table to

After the first-qubit readout, assign a one-qubit residual with computational zero-weight $m$; after the second-qubit readout, assign one with zero-weight $a$. If one-qubit readout uses ${\phi }_{\eta }$, both sequential products equal $J_{ij}$.

The first assigned residual has weight $m=1/4$, whereas the projective residual after first outcome zero has weight $u=1/2$. The compensation is therefore diagonally unfaithful.

This is an exact partial assignment at the displayed state and continuations. It is not a strategy on all rays. It does not prove continuity across support boundaries, all record equations, or no-signaling for arbitrary local programs.

A record-law equality can be assigned on a finite copied family as well, but extending all these assignments simultaneously is precisely an additional feasibility obligation. A passing selected record test is not that extension.

Continuity of outcome laws alone does not compactify unknown update maps. A correct compactness statement must control selective outputs.

Represent a strategy branch by