# Is this entanglement of formation identity true?

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?

No. A counterexample is the tripartite GHZ state $$|000\rangle+|111\rangle$$.