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Trace out $Q^{\prime }$ in the final state. With $q(x,y)=p(x)\mathrm{Tr}(E_y{\rho }_x)$, the remaining state is diagonal in the product classical basis:The Von Neumann entropy of a classical diagonal density operator equals the Shannon entropy of its diagonal probabilities, by its spectrum. Therefore $I(A^{\prime }:B^{\prime })=H(X)+H(Y)-H(X,Y)=I(X:Y)$, the mutual information between source symbol and measurement outcome.

For the input, put $\bar{\rho }={\sum }_{x}p(x){\rho }_x$. The marginal on $Q$ is $\bar{\rho }$, the marginal on $A$ has Von Neumann entropy $H(X)$, and the entropy of a classical-quantum state isTo justify the last formula, diagonalize each ${\rho }_x$ with eigenvalues ${\lambda }_{xj}$. The joint block-diagonal state has eigenvalues $p(x){\lambda }_{xj}$; expanding $-{\sum }_{x,j}p(x){\lambda }_{xj}{\log }_2[p(x){\lambda }_{xj}]$ yields exactly the displayed Von Neumann entropy. Thus $I(A:Q)=S(\bar{\rho })-{\sum }_{x}p(x)S({\rho }_x)$, the Holevo quantity of the ensemble. Part (ii) becomes the Holevo boundIt holds for every POVM, and hence also for the accessible information obtained by maximizing the mutual information over all measurements. The only extra Von Neumann entropy identities used were the classical-diagonal and classical–quantum block-spectrum formulas, both justified above; the information inequality itself followed from the requested strong-subadditivity argument.