Nonnegativity of finite-dimensional quantum conditional mutual information

Let $Q_{BC}:H_B\otimes H_C\to H_C\otimes H_B$ be the two-factor swap and let $P=I_A\otimes Q_{BC}:H_A\otimes H_B\otimes H_C\to H_A\otimes H_C\otimes H_B$. Put ${\tau }_{ACB}=P{\rho }_{ABC}{P}^{†}$. By [F3], $P$ is unitary; therefore an orthonormal eigenbasis of ${\rho }_{ABC}$ is carried to an orthonormal eigenbasis of ${\tau }_{ACB}$ with the same eigenvalues, and [F5] gives $S({\tau }_{ACB})=S({\rho }_{ABC})$. The $AC$ marginal of $\tau$ is exactly ${\rho }_{AC}$ by the definition in [F1].

In the product basis, tracing out A from $\tau$ gives ${\tau }_{CB}=Q_{BC}{\rho }_{BC}{Q}_{BC}^{†}$: its $(c,b),(c^{\mathrm{\prime }},b^{\mathrm{\prime }})$ coefficient is the sum over $a$ of the $\rho$ coefficients with indices $(a,b,c),(a,b^{\mathrm{\prime }},c^{\mathrm{\prime }})$, which is precisely the swapped coefficient. Thus [F3] and the eigenvalue argument in step 1.1 give $S({\tau }_{CB})=S({\rho }_{BC})$. Tracing out both A and B gives the same C marginal ${\tau }_C={\rho }_C$, since the finite sums over their basis indices are merely reordered.

Apply [F2] to the tripartite state $\tau$ with middle register C. It yields $S({\tau }_{AC})+S({\tau }_{CB})-S({\tau }_C)-S({\tau }_{ACB})\ge 0$. Substituting the equalities in steps 1.1 and 2.1 gives exactly the expression in [F1], so $I(A:B\mid C{)}_{\rho }\ge 0$.