Correct residual leakage in the flagged model

Consider

on occupied code plus vacancy. Replacement by ${\tau }_q=q{I}_C/d\oplus (1-q)$ gives distance $q{a}_d$.

For the lower bound, pinching any candidate replacement state into occupied and vacant blocks cannot increase its distance from this channel. Covariance under logical unitary conjugations, convexity, and averaging then reduce the optimization to states

On the Bell input, the occupied block difference is $q{\Phi }_d-s{I}_{QC}/d^2$, and the vacant block has trace norm $|s-q|$. Thus twice the tested trace distance is

For $s\le q$, this is $2q-2s/d^2\ge 2q{a}_d$. For $q\le s\le q{d}^2$, it is $2s{a}_d\ge 2q{a}_d$. For $s\ge q{d}^2$, it is $2(s-q)\ge 2q{a}_d$. Therefore

The reversed weight $(1-q)a_d$ is the residual recovery error, not its leakage. No contradictory equation is retained as a theorem statement in this manuscript.