Strong subadditivity for a pure GHZ tripartite state

By [F4], the vectors $g=\mathrm{\mid }\mathrm{G}\mathrm{H}\mathrm{Z}⟩$, $h=(\mathrm{\mid }000⟩-\mathrm{\mid }111⟩)\mathrm{/}\sqrt{2}$, and the six computational basis vectors other than $\mathrm{\mid }000⟩,\mathrm{\mid }111⟩$ form an orthonormal basis. Direct multiplication by ${\rho }_{ABC}=g{g}^{\ast }$ gives ${\rho }_{ABC}g=g$ and zero on the other seven basis vectors. Thus its spectrum is $(1,0,\dots ,0)$; it is non-negative by [F8] and has trace one, so it is a state by [F2]. Consequently $S({\rho }_{ABC})=-1\ln 1=0$ by [F1] and [F7].

Expanding ${\rho }_{ABC}$ gives the four terms $\frac{1}{2}\mathrm{\mid }000⟩⟨000\mathrm{\mid }$, $\frac{1}{2}\mathrm{\mid }000⟩⟨111\mathrm{\mid }$, $\frac{1}{2}\mathrm{\mid }111⟩⟨000\mathrm{\mid }$, and $\frac{1}{2}\mathrm{\mid }111⟩⟨111\mathrm{\mid }$. For any traced register, the first- or last-factor formula [F3], after the wire permutation [F9] when needed, kills both off-diagonal terms because $\mathrm{tr}(\mathrm{\mid }0⟩⟨1\mathrm{\mid })=\mathrm{tr}(\mathrm{\mid }1⟩⟨0\mathrm{\mid })=0$; the resulting two-register marginal is $\frac{1}{2}\mathrm{\mid }00⟩⟨00\mathrm{\mid }+\frac{1}{2}\mathrm{\mid }11⟩⟨11\mathrm{\mid }$. Tracing out two registers similarly gives $I\mathrm{/}2$ for each one-register marginal. These diagonal matrices have positive eigenvalues $1\mathrm{/}2,1\mathrm{/}2$, with two additional zero eigenvalues for each two-register marginal. Across each one-versus-two split, the displayed vector itself is a Schmidt expansion of the form [F5], with coefficients $1\mathrm{/}\sqrt{2},1\mathrm{/}\sqrt{2}$.

For each reduced state in step 1.2, [F1] gives entropy $-2(\frac{1}{2}\ln \frac{1}{2})=\ln 2$. Hence the left side of the claimed SSA instance is $2\ln 2$, and the right side is $\ln 2+0$. Since $2>1$ and $\ln 2>0$ by strict monotonicity in [F7], the inequality is strict and equals the stated comparison.