Hello in one class my instructor once gave us a sketch proof that involves the following equality regarding conditional independence: $$ P \perp B \mid C \quad \land \quad A \perp B \mid (C \cup P) \quad \implies \quad (A \cup P) \perp B \mid C $$ where $\perp$ indicates independence and $C$ and $P$ are disjoint. I cannot remember all the details but I believe $A$ and $P$ are not disjoint.
Is this equality true? I tried to prove it by showing $p(a, p, b \mid c) = p(a, p \mid c) p(b \mid c)$ but could not proceed further since I do not know how to incorporate $A \perp B \mid (C \cup P)$.
Thank you for the help.