Symmetric Difference with Intersection forms Ring/Proof 2

Theorem

Let $S$ be a set.

Let:

$\symdif$ denote the symmetric difference operation

$\cap$ denote the set intersection operation

$\powerset S$ denote the power set of $S$.

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

This ring is not an integral domain.

Proof

From Power Set is Closed under Symmetric Difference and Power Set is Closed under Intersection, we have that both $\struct {\powerset S, \symdif}$ and $\struct {\powerset S, \cap}$ are closed.

Hence $\powerset S$ 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 $\struct {\powerset S, \cap}$.

We also note that set intersection is not cancellable, so $\struct {\powerset S, \symdif, \cap}$ is not an integral domain.

The result follows.

$\blacksquare$