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A quantum channel on the receiving system cannot increase coherent information: $I(A⟩B{)}_{\rho }\ge I(A⟩B^{\prime }{)}_{\sigma }$. Dilate the channel to a linear isometry of Hilbert spaces $B\to B^{\prime }E$. The difference is $I(A:E|B^{\prime })$ of the dilated state, nonnegative by Strong subadditivity of Von Neumann entropy.

In the state-picture convention, a Stinespring representation of a completely positive map $\mathcal{N}:\mathcal{L}({\mathcal{H}}_Q)\to \mathcal{L}({\mathcal{H}}_{Q^{\prime }})$ consists of an auxiliary Hilbert space ${\mathcal{H}}_E$ and a linear map $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q^{\prime }}\otimes {\mathcal{H}}_E$ such thatThe partial trace discards the environment. For example, from a Kraus representation $\mathcal{N}(X)={\sum }_j{K}_{j}X{K}_j^{†}$, take $V={\sum }_j{K}_j\otimes |j{⟩}_E$. The adjoint, or observable-picture, form is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$.

A general completely positive map does not require $V$ to be an linear isometry of Hilbert spaces. If $\mathcal{N}$ is trace preserving, then $V^{†}V={\sum }_j{K}_j^{†}{K}_j=I$, so $V$ is an linear isometry of Hilbert spaces. This is the Stinespring dilation of a quantum channel, which will be used in the data-processing proof.

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 $B^{\prime }$.

In finite dimensions, a completely positive map has the form $\mathcal{N}(X)={\mathrm{Tr}}_E(VX{V}^{†})$ for $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q\prime }\otimes {\mathcal{H}}_E$. From a Kraus representation take $V={\sum }_j{K}_j\otimes |j⟩$. The observable-picture adjoint is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$. A general completely positive map need not have isometric $V$; it is a linear isometry of Hilbert spaces precisely when the map is trace preserving.