= Solution
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 $\sum_kA_k^\dagger A_k=I$ and define the <isometry> in its <Stinespring dilation>
$$
V|\psi\rangle=\sum_kA_k|\psi\rangle\otimes|k\rangle_E.
$$
Then $V^\dagger V=I$ and $\Phi(\rho)=\operatorname{Tr}_E(V\rho V^\dagger)$. Acting on $B$ gives $\sigma_{AB'E}=(I_A\otimes V)\rho_{AB}(I_A\otimes V^\dagger)$. An <isometry> preserves the nonzero <eigenvalues> of a <density operator>. Hence $S(AB'E)=S(AB)$, $S(B'E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B'E)_\sigma=I(A:B)_\rho$.
Now discard $E$ and apply part 1. This proves <data processing for quantum mutual information>:
$$
\boxed{I(A:B')\le I(A:B'E)=I(A:B).}
$$
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 $\varepsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\varepsilon$ 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.
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