Lemma: Parity rigidity

If every proper reduced state of $\rho$ on $S$ is diagonal, then

For a positive family, a zero diagonal entry at even parity forces $c_S(\rho )\ge 0$, and one at odd parity forces $c_S(\rho )\le 0$. If both occur, $F_S(\rho )=\mathrm{diag}\rho$.

Proof. All proper-set coefficients vanish by anchoring. At a zero diagonal entry, positivity imposes the corresponding sign on the sole remaining correction.