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

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

But from Set Difference Intersection Second Set is Null, we have that $\left({S \setminus T}\right) \cap T = \varnothing$.

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

This leaves us with:

Then we have:

Hence the result.