Example: Residual size is not a replacement estimate

For a qubit code, let

Both channels are computational dephasing, not replacement. For any replacement state ${\tau }_A$,

Indeed, the two accessible basis outputs have trace distance one; at least one has distance at least $1/2$ from ${\tau }_A$. Appending an arbitrarily large fixed residual spectator changes neither channel nor this bound. Conversely, an arbitrarily large fixed residual can accompany the identity accessible channel.