Is Conditional Independence stable under unions regarding the condition?

Question

I stumbled across the question whether the following implication is true: Let $A,B,C,D$ be random variables. $$(A\perp B\mid C)\land (A\perp B \mid D)\Rightarrow (A\perp B\mid (C,D))$$ Intuitively, I'd say that this implication is true, but I failed to show it formally and I can not see how the statement can be derived by the usual properties of conditional independence (What does the decomposition, weak union and contraction rule mean for conditional probability and what are their proofs?), even though I suspect that Pearl uses this implication in a proof. My concern is that on the LHS, one conditions on the $\sigma$-field generated by $C$ or $D$, respectively, but on the RHS one conditions on $\sigma(\sigma(C)\cup\sigma(D))$, which might be bigger than the union of the $\sigma$-fields relevant for the LHS.

Answer 1

Consider three independent fair coin-tosses. Let $A$, $B$, $C$, $D$ be the indicator random variables for the events $$ \eqalign{ A: & \text{first toss heads}\cr B: & \text{second toss heads}\cr C: & \text{third toss heads}\cr D: & \text{an odd number of heads in the three tosses}\cr} $$ $A$ and $B$ are independent given $C$, and $A$ and $B$ are independent given $D$, but $A + B \equiv C + D \mod 2$ so $A$ and $B$ are not independent given $C$ and $D$.