Strong subadditivity of finite-dimensional quantum entropy

View ${\rho }_{ABC}$ first across the grouped bipartition $A\mathrm{\mid }BC$: [F9] makes ${\rho }_A={\mathrm{Tr}}_{BC}{\rho }_{ABC}$ and ${\rho }_{BC}={\mathrm{Tr}}_A{\rho }_{ABC}$ states. Applying [F9] across $AB\mathrm{\mid }C$ also makes ${\rho }_{AB}={\mathrm{Tr}}_C{\rho }_{ABC}$ a state. Applying it once more to ${\rho }_{AB}$ across $A\mathrm{\mid }B$ makes ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$ a state. This is also ${\mathrm{Tr}}_C{\rho }_{BC}$: in the ordered product basis [F6]–[F7], both have coefficient ${\sum }_{a,c}({\rho }_{ABC}{)}_{(a,b,c),(a,b^{\mathrm{\prime }},c)}$, since the finite sums may be reordered. The operator ${\rho }_A\otimes {\rho }_{BC}$ is also a state: spectral decompositions of the two positive factors give nonnegative product eigenvalues, and their sum is $\mathrm{tr}({\rho }_A)\mathrm{tr}({\rho }_{BC})=1$. Applying [F2] to the bipartite states ${\rho }_{A\mathrm{\mid }BC}$ and ${\rho }_{AB}$ gives support inclusion for both relative entropies $D({\rho }_{ABC}\mathrm{\parallel }{\rho }_A\otimes {\rho }_{BC})$ and $D({\rho }_{AB}\mathrm{\parallel }{\rho }_A\otimes {\rho }_B)$; [F1] then makes both finite.

Set $X={\rho }_{ABC}$ and $Y={\rho }_A\otimes {\rho }_{BC}$. By [F6] and the ordered product basis [F7], ${\mathrm{Tr}}_{C}X={\rho }_{AB}$ and ${\mathrm{Tr}}_{C}Y={\rho }_A\otimes {\mathrm{Tr}}_C{\rho }_{BC}={\rho }_A\otimes {\rho }_B$: the latter equality follows entrywise by summing the C-diagonal block of ${\rho }_{BC}$ while the A-factor matrix coefficient remains fixed. Apply [F4] to this pair to obtain $D({\rho }_{AB}\mathrm{\parallel }{\rho }_A\otimes {\rho }_B)\le D({\rho }_{ABC}\mathrm{\parallel }{\rho }_A\otimes {\rho }_{BC})\mathrm{.}$

Both sides in step 2.1 are finite by step 1.1, so [F3] identifies them as $S({\rho }_A)+S({\rho }_B)-S({\rho }_{AB})$ and $S({\rho }_A)+S({\rho }_{BC})-S({\rho }_{ABC})$, respectively. Cancel the common finite term $S({\rho }_A)$ and rearrange the inequality to obtain $S({\rho }_{AB})+S({\rho }_{BC})\ge S({\rho }_B)+S({\rho }_{ABC})$.