Theorem: I: sub-capacity diagonal-faithful rigidity

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.