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Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.