Symmetric Difference with Intersection forms Ring/Proof 1

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 Symmetric Difference on Power Set forms Abelian Group, $\left({\mathcal P \left({S}\right), *}\right)$ is an abelian group, where $\varnothing$ is the identity and each element is self-inverse.

From Power Set with Intersection is Monoid, $\left({\mathcal P \left({S}\right), \cap}\right)$ is a commutative monoid whose identity is $S$.

Also Intersection Distributes over Symmetric Difference.

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

From Intersection with Empty Set:
 * $\forall A \in \mathcal P \left({S}\right): A \cap \varnothing = \varnothing = \varnothing \cap A$

Thus $\varnothing$ is indeed the zero.

However, from Set Intersection Not Cancellable, it follows that $\left({\mathcal P \left({S}\right), *, \cap}\right)$ is not an integral domain.