Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.

Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, $\mathcal{B}'$ are also ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$.

*Examples*:

(1) Take $\mathcal{B} = \mathcal{A}'$ then $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$ by definition of a factor.

(2) Take $(\mathcal{A}' \subset \mathcal{B})$ an irreducible subfactor, then $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$ by definition of irreducibility.

Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the *notation* $\langle S \rangle := (S \cup S^* \cup \mathbb{C}) ''$.

**Question**: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?

Else, what are counterexamples?

*Remark*: It's true for the examples (1) and (2).