Data processing for finite quantum mutual information

First apply $\mathrm{\Phi }$ to $A$. Choose a Stinespring isometry $V:A\to A^{\mathrm{\prime }}\otimes E$ from [F3], and form ${\tilde{\rho }}_{A^{\mathrm{\prime }}EB}:=(V\otimes I_B){\rho }_{AB}(V^{\ast }\otimes I_B)$. This is a state: a spectral decomposition of ${\rho }_{AB}$ in [F4] shows that its nonzero eigenvalues remain the same under the isometric embedding. Its $A^{\mathrm{\prime }}E$ marginal is $V{\rho }_A{V}^{\ast }$ and has the same positive eigenvalues as ${\rho }_A$, while its $A^{\mathrm{\prime }}EB$ joint state has the same positive eigenvalues as ${\rho }_{AB}$. Thus [F1] gives $I(A^{\mathrm{\prime }}E:B{)}_{\tilde{\rho }}=I(A:B{)}_{\rho }$.

Reorder $A^{\mathrm{\prime }}EB$ to $E{A}^{\mathrm{\prime }}B$ by the unitary permutation in [F6]. Apply [F2] to the ordered tripartite state ${\tilde{\rho }}_{E{A}^{\mathrm{\prime }}B}$, with $A^{\mathrm{\prime }}=Y$, to obtain $S(E{A}^{\mathrm{\prime }})+S(A^{\mathrm{\prime }}B)\ge S(A^{\mathrm{\prime }})+S(E{A}^{\mathrm{\prime }}B)$. Rearranging by [F1] gives $I(E{A}^{\mathrm{\prime }}:B)\ge I(A^{\mathrm{\prime }}:B)$. The entropy identities and their logarithm product law are supplied by [F7]. The $A^{\mathrm{\prime }}B$ marginal is $(\mathrm{\Phi }\otimes {\mathrm{id}}_B)(\rho )$ by [F3] and the partial-trace convention; hence applying $\mathrm{\Phi }$ cannot increase mutual information.

Repeat step 1.1–2.1 with $\mathrm{\Psi }$ acting on the second register, now taking the first register to be $A^{\mathrm{\prime }}$. This gives $I(A^{\mathrm{\prime }}:B^{\mathrm{\prime }}{)}_{(\mathrm{\Phi }\otimes \mathrm{\Psi })(\rho )}\le I(A^{\mathrm{\prime }}:B{)}_{(\mathrm{\Phi }\otimes {\mathrm{id}}_B)(\rho )}$. Combine this with step 2.1 to obtain the claimed inequality. The entropy-law prerequisite remains [F7].