Consider a probability space $(\Omega, \mathbb{F}, P)$. Let $\mathbb{F}_1$, $\mathbb{F}_2$ and $\mathbb{F}_3$ be sub-$\sigma$-algebras of $\mathbb{F}$. Assume that $\mathbb{F}_1$ and $\mathbb{F}_2$ are conditionally independent given $\mathbb{F}_3$, i.e. $$ P(F_1 \cap F_2 \ | \ \mathbb{F}_3) = P(F_1 \ | \ \mathbb{F}_3) P(F_2 \ | \ \mathbb{F}_3), \quad \forall F_1 \in \mathbb{F}_1 , F_2 \in \mathbb{F}_2 $$
where $P(F \ | \ \mathbb{F}_3) = E(1_F \ | \ \mathbb{F}_3)$.
Does then
$$ P(F_{13} \cap F_{23} \ | \ \mathbb{F}_3) = P(F_{13} \ | \ \mathbb{F}_3) P(F_{23} \ | \ \mathbb{F}_3), \quad \forall F_{13} \in \sigma(\mathbb{F}_1 , \mathbb{F}_3), F_{23} \in \sigma(\mathbb{F}_2 , \mathbb{F}_3) $$
where $\sigma(\mathbb{F}_1, \mathbb{F}_3)$ is the smallest $\sigma$-algebra containing both $\mathbb{F}_1$ and $\mathbb{F}_3$?
I'm fairly sure this is correct and intuitively it makes sense that since we condition on $\mathbb{F}_3$, it doesn't really matter if the set is measurable wrt. $\mathbb{F}_3$. A proof eludes me however, any ideas?
It seems to be true. Here are some steps towards a proof.
We first treat the case where $F_{13}\in\mathbb F_1$. Define for a fixed $G_1\in\mathbb F_1$ the collection of sets $$\mathcal A:= \{A\mid \Pr\left(G_1\cap A\mid\mathbb F_3 \right)=\Pr\left(G_1 \mid\mathbb F_3 \right)\Pr\left( A\mid\mathbb F_3 \right) \mbox{ a.s.}\}.$$ We can check that $\mathcal A$ is a $\lambda$-system which contains the $\pi$-systems which consists of the sets of the form $G_2\cap G_3$, $G_i\in\mathbb F_i$.
Define now for a fixed $F_{23}\in\mathbb F_{23}$ the collection of sets $$\mathcal B:= \{B\mid \Pr\left(B\cap F_{23}\mid\mathbb F_3 \right)=\Pr\left(B \mid\mathbb F_3 \right)\Pr\left( F_{23}\mid\mathbb F_3 \right) \mbox{ a.s.}\}. $$ Then $\mathcal A$ is a $\lambda$-system which contains (by 1.) the $\pi$-systems which consists of the sets of the form $G_1\cap G_3$, $G_i\in\mathbb F_i$.