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

Also from the definition of Symmetric Difference:


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

From Set Difference Intersection Second Set is Empty Set:
 * $\left({S \setminus T}\right) \cap T = \varnothing$

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

This leaves:

Then:

Hence the result.