Corollary: Precise no-hiding statement

If $\mathcal{N}$ is constant on the entire code, the complete output $B$ exactly recovers the code. If $\mathcal{E}$ is constant, $R$ exactly recovers it. For $d\ge 2$, both complementary outputs cannot separately recover the entire unknown code, and both cannot be constant.

Proof. Interchange $R$ and $B$ in Theorem thm:exact. An exactly recoverable output has constant complement. A constant channel cannot recover two distinct logical states.