Theorem: Equal quantum marginals do not give equal local input

In the exceptional-block model there are runs with identical initial states and identical channels on all original-metric inner blocks for $(Q,t_0+1)$, but different readout laws at that event. In both runs the non-selective quantum marginal at $Q$ is $I/2$.

Proof. Prepare a Bell pair with halves at $P,Q$, using a declared prehistory before the exceptional update is activated. In the exceptional block, measure the $P$-half destructively and deliver its record $b$ to $Q$. Use the $Z$ basis in the first run and the $X$ basis in the second. These are admissible choices of complete exceptional-block instruments.

The resulting record–qubit states at $Q$ are

Their quantum marginals are both $I/2$, but ${\omega }_Z\ne {\omega }_X$.

Read $b$, measure the qubit in the $Z$ basis with outcome $a$, and output $a\oplus b$. The law is ${\delta }_0$ on ${\omega }_Z$ and uniform on $\{0,1\}$ on ${\omega }_X$. The exceptional block is not original-metric inner because $d(P,Q)>r$; all inner channels and all initial data agree.