Given a non-empty set $X,$ let $\ast :\; P(X)\; \times\; P(X) \to P(X)$ be defined as $A \ast B = \; ( A-B)\; \cup \; (B-A),\; \forall A, B \in \; P(X).$. Show that the empty set $\emptyset$ is the identity for the operation $\ast$ and all the elemnets $A$ of $P(X)$ are invertible with $A^{-1} \;= A$.

Question

    $(Hint:\; (A- \emptyset )\; \cup \; (\emptyset - A)=A\; and ( A-A) \cup \; (A-A)= A \ast A = \emptyset)$