Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mnForE;

It is given that *: P(X) × P(X) → P(X) is defined as

A * B = (A − B) ∪ (B − A) &mnForE; A, B ∈ P(X).

Let A ∈ P(X). Then, we have:

$A^{*} \Phi=(A-\Phi) \cup(\Phi-A)=A \cup \Phi=A$

$\Phi^{*} A=(\Phi-A) \cup(A-\Phi)=\Phi \cup A=A$

$\therefore A^{*} \Phi=A=\Phi^{\star} A . \& m n F o r E ; A \in \mathrm{P}(X)$

Thus, $\Phi$ is the identity element for the given operation $^{\star}$.

Now, an element $A \in \mathrm{P}(X)$ will be invertible if there exists $B \in \mathrm{P}(X)$ such that

$A^{*} B=\Phi=B^{*} A .($ As $\Phi$ is the identity element $)$

Now, we observed that $A * A=(A-A) \cup(A-A)=\phi \cup \phi=\phi \forall A \in \mathrm{P}(X)$.

Hence, all the elements A of P(X) are invertible with A−1 = A.