Corollary: No-hiding and no independent duplicate recovery

If $\mathcal{N}$ is constant on the whole code, $B$ exactly recovers the code. If $\mathcal{E}$ is constant, $R$ exactly recovers it. For $d\ge 2$, the two outputs cannot both be constant, and cannot both independently recover the whole unknown code.

Proof. Interchange $R$ and $B$ in Theorem thm:exact. An exactly recoverable channel is not constant on two distinct code states. If one output is exactly recoverable, its complement is constant and is consequently not itself exactly recoverable.