Corollary: Destination qualification

The residual alone recovers the whole code exactly if and only if the accessible channel is a replacement channel. A merely nonconstant complement obstructs exact recovery from $R$; it does not establish recovery from $B$. For $d\ge 2$, both factors cannot independently recover the whole code.

Proof. Interchange the output factors in Theorem thm:exact. A replacement channel cannot recover two distinct inputs.