Union as Symmetric Difference with Intersection

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

Then:
 * $A \cup B = \paren {A \symdif B} \symdif \paren {A \cap B}$

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

Proof
From the definition of symmetric difference:
 * $\paren {A \symdif B} \symdif \paren {A \cap B} = \paren {\paren {A \symdif B} \cup \paren {A \cap B} } \setminus \paren {\paren {A \symdif B} \cap \paren {A \cap B} }$

Also from the definition of symmetric difference:


 * $\paren {A \symdif B} \cap \paren {A \cap B} = \paren {\paren {A \cup B} \setminus \paren {A \cap B} } \cap \paren {A \cup B}$

From Set Difference Intersection with Second Set is Empty Set:
 * $\paren {S \setminus T} \cap T = \O$

Hence:
 * $\paren {\paren {A \cup B} \setminus \paren {A \cup B} } \cap \paren {A \cup B} = \O$

This leaves:

Then:

Hence the result.