Proposition: Ignored local operations do not change untouched outputs

Let $\Lambda :B\to B^{\prime }$ be any normal CPTP map. For every joint state,

In particular, unobserved residual readout-and-drop leaves ${\epsilon }_{\mathrm{rec}}(\mathcal{N})$ exactly unchanged.

Proof. The Heisenberg adjoint ${\Lambda }^{\ast }$ is unital. For every bounded observable $A$ on $QR$,

Thus all expectations on $QR$ are unchanged. The recovery task is determined by that unchanged channel.