Let a tripartite system be given with pure state $\rho_{ABC}$. My adviser said that the entanglement of formation satisfies the identity
$${E_F}_{A(BC)}^2={E_F}_{AB}^2+{E_F}_{AC}^2$$
Where ${E_F}_{A(BC)}$ is the entanglement of formation considering the bipartite state with parts $A$ and $B+C$.
He further said that ${E_F}_{A(BC)}^2=S_A^2$ the Von-Neumman entropy, and that because of that if we know ${E_F}_{AB}$ we get ${E_F}_{AC}$ by means of
$${E_F}_{AC}^2=S_B^2-{E_F}_{AB}^2.$$
Now is this identity true? He couldn't point me the specific reference, and I've tried searching and found none.
That ${E_F}_{A(BC)}=S_A$ is obvious because the entanglement of formation reduces to the entanglement entropy for pure states. Since $\rho_{ABC}$ is pure, considered a bipartite state between $A$ and $B+C$ it is pure, and the entanglement of formation is the entropy.
My doubt is about the very first identity. If it is true, why is it the case?
