Example: Input-independent marginals do not certify capacity

Let two qubits $M,F$ initially share

Encode $x_1,x_2$ by applying ${{\sigma }_x}^{x_1}{{\sigma }_z}^{x_2}$ to $M$. The four joint states are orthogonal Bell states, while both one-qubit marginals remain $I/2$.

A decoder with access to $M,F$ can identify both bits and answer either late query perfectly. The relevant cut contains two jointly accessible qubits, not one. Counting only input-dependent marginals would miss the resource.