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For a channel on $B$, use a Stinespring dilation $B\to B^{\prime }E$. Isometry invariance gives $I(A:B)=I(A:B^{\prime }E)$, so the loss on discarding $E$ is $I(A:B)-I(A:B^{\prime })=I(A:E\mid B^{\prime })$. This quantum conditional mutual information is nonnegative by Strong subadditivity of Von Neumann entropy. The identity holds for arbitrary mixed inputs and identifies the correlations lost to the discarded environment.

Let the quantum channel send $B$ to $B^{\prime }$. From its Kraus representation, choose a Stinespring representation $V:B\to B^{\prime }E$ and formThe channel output is ${\sigma }_{A{B}^{\prime }}={\mathrm{Tr}}_E{\sigma }_{A{B}^{\prime }E}$. Isometric invariance of Von Neumann entropy givesTherefore the quantum mutual information satisfies $I(A:B{)}_{\rho }=I(A:B^{\prime }E{)}_{\sigma }$. Expanding its loss after discarding the environment givesThe final expression is the quantum conditional mutual information $I(A:E\mid B^{\prime })$. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. HenceThis 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.

Take a Stinespring dilation $V:B⟶B^{\prime }E$ of the local quantum channel and define ${\tau }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. Tracing out $E$ gives the prescribed output, with $A^{\prime }=A$. An isometry preserves the nonzero eigenvalues of a density operator; therefore $S(B^{\prime }E{)}_{\tau }=S(B{)}_{\rho }$, $S(A{B}^{\prime }E{)}_{\tau }=S(AB{)}_{\rho }$, and $S(A{)}_{\tau }=S(A{)}_{\rho }$.

Consequently $I(A:B^{\prime }E{)}_{\tau }=I(A:B{)}_{\rho }$. The loss of quantum mutual information isThe last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. HenceThis mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.

Use two results: the isometric Stinespring dilation of a quantum channel, and Strong subadditivity of Von Neumann entropy. Let $V:B\to B^{\prime }E$ dilate the given operation, and defineAn linear isometry of Hilbert spaces preserves the nonzero eigenvalues, so $S(B^{\prime }E{)}_{\omega }=S(B{)}_{\rho }$ and $S(A{B}^{\prime }E{)}_{\omega }=S(AB{)}_{\rho }$. Subtracting the two coherent information expressions givesThe last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus the data-processing inequality for coherent information isThe lost coherent information is precisely the quantum conditional mutual information between the reference $A$ and discarded environment $E$, conditional on the retained output $B^{\prime }$.