If $X \perp Y ~|~ W, Z$ and $X \perp W ~|~ Z$, then $X \perp Y,W ~|~ Z$. proof: first assumption gives $p(x~|~y,w,z)=p(x~|~w,z)$. Second assumption gives $p(x ~|~ w,z)=p(x ~|~z)$. So chaining the two gives a sufficient condition/the definition of $X \perp Y,W ~|~Z$.