Symmetric Difference with Union does not form Ring

Theorem
Let $S$ be a set.

Let:
 * $\symdif$ denote the symmetric difference operation
 * $\cup$ denote the set union operation
 * $\powerset S$ denote the power set of $S$.

Then $\struct {\powerset S, \symdif, \cup}$ does not form a ring.

Proof
For $\struct {S, \symdif, \cup}$ to be a ring, it is a necessary condition that $\cup$ be distributive over $*$.

Also, the identity element for set union and symmetric difference must be different.

However:
 * $(1): \quad$ the identity for union and symmetric difference is $\O$ for both operations
 * $(2): \quad$ set union is not distributive over symmetric difference:

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

The result follows.

Also see

 * Symmetric Difference with Intersection forms Ring