# Do a von Neumann Algebra's subalgebras have unique “complements”?

Let $A \subset B \subset C$ be von Neumann algebras (and more specifically factors, if it helps) either all type II or all type III, acting on the same separable Hilbert space.

Call $A' \subset B$ the "unique complement" of $A$ in $B$, if $A'$ is the least sub-von-Neumann-algebra of $B$ such that the algebra generated by $A \cup A'$ is all of $B$.

Question 1: does $A$ always have a unique complement in $B$?

Question 2: when $A$ has unique complements $A'$ and $A''$, in $B$ and in $C$ respectively (with $B \subset C$), must we have $A' \subseteq A''$?

If anyone can point out examples/proofs I'd be very grateful!

Remark: with von Neumann algebras, using the notation $A'$ for anything other than the commutant is highly unusual.
Let $R$ be the hyperfinite II$_1$-factor and take $B=R\oplus R\oplus R=R\otimes\mathbb C^3$, and $A=R\otimes I$. So we think $$A=\{(a,a,a):\ a\in R\},\ \ B=\{(a,b,c):\ a,b,c\in R\}.$$ You can take a "complement" of $A$ to be $$A_1=\{(\lambda I,b,c):\ \lambda\in\mathbb C,\ b,c\in R\};$$ or $$A_2=\{(a,\lambda I, c):\ \lambda\in\mathbb C,\ a,c\in R\};$$ so no uniqueness.
Factors do not improve things. For instance you could think of $B=L(\mathbb F_2)=W^*(a,b)$ where $a,b$ are the usual generators of the group. Take $A=W^*(a)$. Now you can take the other canonical masa $A_1=W^*(b)$ as a complement, but also the radial masa $A_2=W^*(a+b)$.