Proposition: Record agreement is weaker than residual agreement

For finite instruments $\{{\mathcal{A}}_w\}$, $\{{\mathcal{B}}_w\}$, record equality on all inputs is equivalent to ${\mathcal{A}}_w^{\ast }(I)={\mathcal{B}}_w^{\ast }(I)$. Equality on all recorded residual observables requires the stronger identities ${\mathcal{A}}_w^{\ast }(B)={\mathcal{B}}_w^{\ast }(B)$ for every declared residual observable $B$.

Equal records can coexist with residual trace-distance difference $1/2$.

Proof. Pull each output effect back to the input and use separation by density matrices. For the example, append the same one-outcome record to identity and computational dephasing. On $|+⟩⟨+|$, their residuals are $|+⟩⟨+|$ and $I/2$. Their difference has eigenvalues $1/2,-1/2$.