True or false: For random variables $x,y,z,w$,
$$x\perp \!\!\! \perp y|z \text{ and } x\perp \!\!\! \perp y|w\implies x\perp \!\!\! \perp y|(z,w)$$
According to this link, this statement is false. Their reasoning is as follows. Consider a direct graph model $X\rightarrow Z\leftarrow U\rightarrow W\leftarrow Y$. Then, if exactly one of $Z$ or $W$ is observed, $X$ and $Y$ are d-separated, but if both are observed, they are no longer d-separated, and thus the conditional independence $x\perp \!\!\! \perp y|(z,w)$ cannot be guaranteed.
I suspect that this statement is indeed false, based on playing around with probability density functions. However, I'm still not entirely convinced with the directed graph model approach above. For example, if we take an undirected graph approach, where $x\perp \!\!\! \perp y|z$ means all paths from node $X$ to node $Y$ must path through node $Z$ and $x\perp \!\!\! \perp y|w$ means all paths from $X$ to $Y$ must path through $W$, doesn't this imply all paths from $X$ and $Y$ must pass through a node $N\in\{W,Z\}$, implying $x\perp \!\!\! \perp y|(z,w)$? What is wrong with this reasoning?
Also, if someone could give a concrete counterexample (e.g. explicitly construct $x,y,z,w$ for which the statement is false), that would be very insightful! Any comments are appreciated! Thanks!
To use the undirected approach, you need to moralise the DAG first. This means that along with making the directed edges undirected, you also need to introduce edges between any two nodes that share a common child. Please refer to the early chapters of Koller and Friedman for a primer on these things. In your case, this means that you need to introduce edges $\{X,U\}$ and $\{Y,U\}$ which together give a path not going through $\{Z,W\}$.
As for an example, the structure specified in the proof gives a natural one: take $(X,Y,U)$ to be jointly independent uniform bits, and define $Z = X \oplus U, W = Y \oplus U,$ where $\oplus$ denotes the XOR operation. I'll leave it to you to show that $X \perp Y | Z$ (and given $W,$ by symmetry). But given $(W,Z)$ we can determine $X \oplus Y = W\oplus Z$ meaning $X$ and $Y$ are dependent.