Corollary: No-hiding, no duplicate whole-code recovery, and rank

If $\mathcal{N}$ is replacement, $E$ exactly recovers the code. If $\mathcal{C}$ is replacement, $A$ exactly recovers it. For $d\ge 2$, both complementary outputs cannot independently recover the entire code. At exact recovery,

Proof. Interchange $A$ and $E$ in Theorem thm:exact. If $A$ recovers, its complement is constant and cannot itself recover two distinct states. The rank bound follows from the orthonormal family $\{u_{i\alpha }\}$.