= Solution
Let the <quantum channel> send $B$ to $B'$. From its <Kraus representation>, choose a <Stinespring representation> $V:B\to B'E$ and form
$$
\sigma_{AB'E}=(I_A\otimes V)\rho_{AB}(I_A\otimes V^\dagger).
$$
The channel output is $\sigma_{AB'}=\operatorname{Tr}_E\sigma_{AB'E}$. Isometric invariance of <Von Neumann entropy> gives
$$
S(B'E)_\sigma=S(B)_\rho,\quad S(AB'E)_\sigma=S(AB)_\rho,\quad S(A)_\sigma=S(A)_\rho.
$$
Therefore the <quantum mutual information> satisfies $I(A:B)_\rho=I(A:B'E)_\sigma$. Expanding its loss after discarding the environment gives
$$
\begin{aligned}
I(A:B)_\rho-I(A:B')_\sigma
&=I(A:B'E)_\sigma-I(A:B')_\sigma\\
&=S(AB')_\sigma+S(B'E)_\sigma-S(B')_\sigma-S(AB'E)_\sigma.
\end{aligned}
$$
The final expression is the <quantum conditional mutual information> $I(A:E\mid B')$. The <Strong subadditivity of Von Neumann entropy> states exactly that it is nonnegative. Hence
$$
\boxed{I(A:B')\leq I(A:B).}
$$
This proves <data processing for quantum mutual information> for any mixed input and any local <quantum channel>, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.
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