Symmetric Difference with Intersection forms Ring

Theorem
Let $S$ be a set.

Then $\struct {\powerset S, *, \cap}$ is a commutative ring with unity, in which the unity is $S$.

This ring is not an integral domain.

Comment
The same does not apply to symmetric difference and union unless $S = \O$.

For a start, the identity for union and symmetric difference is $\O$ for both, and the only way the identity of both operations in a ring can be the same is if the ring is null.

Unless $S \ne \O$, then this can not be the case.

Also note that union is not distributive over symmetric difference.

From Symmetric Difference of Unions:
 * $\paren {R \cup T} * \paren {S \cup T} = \paren {R * S} \setminus T$