Symmetric Difference with Intersection forms Ring/Proof 2

Theorem
Let $S$ be a set.

Then $\left({\mathcal P \left({S}\right), *, \cap}\right)$ is a commutative ring with unity, in which the unity is $S$.

This ring is not an integral domain.

Proof
From Power Set Closed under Symmetric Difference and Power Set Closed under Intersection, we have that both $\left({\mathcal P \left({S}\right), *}\right)$ and $\left({\mathcal P \left({S}\right), \cap}\right)$ are closed.

Hence $\mathcal P \left({S}\right)$ is a ring of sets, and hence a commutative ring.

From Intersection with Subset is Subset‎, we have $A \subseteq S \iff A \cap S = A$. Thus we see that $S$ is the unity.

Also during the proof of Power Set with Intersection is Monoid, it was established that $S$ is the identity of $\left({\mathcal P \left({S}\right), \cap}\right)$.

We also note that set intersection is not cancellable, so $\left({\mathcal P \left({S}\right), *, \cap}\right)$ is not an integral domain.

The result follows.