Let $X$ be a random variable and $\mathcal H$ be a $\sigma$-algebra.
The Doob martingale property says $E(X \mid E(X \mid \mathcal{H}))=E(X \mid \mathcal{H})$ (proved by using the tower property).
Now suppose that we have two $\sigma$-algebras $\mathcal H_1$, $\mathcal H_2$.
Can we say something about
$$E(X \mid E(X \mid \mathcal{H_1}), E(X \mid \mathcal{H_2}))$$
My intuition: the Doob martingale property says that the best approximation of $X$ knowing the approximation $E(X \mid \mathcal{H})$ is $E(X \mid \mathcal{H})$. That would mean that $E(X \mid E(X \mid \mathcal{H_1}), E(X \mid \mathcal{H_2})) = E(X \mid \mathcal{H_2})$ if $\mathcal H_1 \subseteq \mathcal H_2$ since $E(X \mid \mathcal{H_2})$ contains more information about $X$. What do you think?