Corollary: An ideal residual gap survives accurate implementation

Identical records can conceal residual outputs of trace distance $1/2$. If two implementations have errors ${\epsilon }_f,{\epsilon }_c$ on that witness, their mutual distance is at least

Proof. Take one coarse outcome, $T=\mathrm{id}$, and ${\mathcal{I}}_{\ast }=\mathrm{id}$. Fine computational projection produces dephasing. On ${\rho }_{+}=|+⟩⟨+|$, the residuals are ${\rho }_{+}$ and $I/2$. Their difference is $X/2$, with eigenvalues $1/2,-1/2$. Use $d_{\mathrm{tr}}={\Vert \cdot \Vert }_1/2$, then the triangle inequality.