Record agreement is weaker than residual agreement

For a single coarse outcome, compare the identity channel on a qubit with computational dephasing

Their coarse record laws agree on every input. On $\rho =|+⟩⟨+|$, their residuals are $|+⟩⟨+|$ and $I/2$, at trace distance $1/2$. Thus record agreement alone does not establish the complete instrument equality required in this paper. This is the observation-boundary distinction emphasized in En03-A and the supplied VIII-A records CheungEn03,CheungObservation.