Union as Symmetric Difference with Intersection

Theorem
Let $$A$$ and $$B$$ be sets.

Then:
 * $$A \cup B = \left({A * B}\right) * \left({A \cap B}\right)$$

where:
 * $$A \cup B$$ denotes set union;
 * $$A \cap B$$ denotes set intersection;
 * $$A * B$$ denotes set symmetric difference.

Proof
From the Symmetric Difference Alternative Definition, we have that:
 * $$\left({A * B}\right) * \left({A \cap B}\right) = \left({\left({A * B}\right) \cup \left({A \cap B}\right)}\right) - \left({\left({A * B}\right) \cap \left({A \cap B}\right)}\right)$$

Then we have, also from Symmetric Difference Alternative Definition:
 * $$\left({A * B}\right) \cap \left({A \cap B}\right) = \left({\left({A \cup B}\right) - \left({A \cup B}\right)}\right) \cap \left({A \cup B}\right)$$

But from Set Difference with Intersection, we have that $$\left({S - T}\right) \cap T = \varnothing$$.

Hence:
 * $$\left({\left({A \cup B}\right) - \left({A \cup B}\right)}\right) \cap \left({A \cup B}\right) = \varnothing$$

This leaves us with:

$$ $$ $$

Then we have:

$$ $$ $$ $$

Hence the result.