# 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:

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

Hence:

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

This leaves:

 $\displaystyle$  $\displaystyle \left({\left({A * B}\right) \cup \left({A \cap B}\right)}\right) \setminus \left({\left({A * B}\right) \cap \left({A \cap B}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\left({A * B}\right) \cup \left({A \cap B}\right)}\right) \setminus \varnothing$ $\displaystyle$ $=$ $\displaystyle \left({A * B}\right) \cup \left({A \cap B}\right)$ Set Difference with Empty Set is Self

Then:

 $\displaystyle$  $\displaystyle \left({A * B}\right) \cup \left({A \cap B}\right)$ $\displaystyle$ $=$ $\displaystyle \left({A \setminus B}\right) \cup \left({B \setminus A}\right) \cup \left({A \cap B}\right)$ Definition 1 of Symmetric Difference $\displaystyle$ $=$ $\displaystyle \left({\left({A \setminus B}\right) \cup \left({A \cap B}\right)}\right) \cup \left({\left({B \setminus A}\right) \cup \left({A \cap B}\right)}\right)$ Union is Idempotent, Union is Commutative and Union is Associative $\displaystyle$ $=$ $\displaystyle A \cup B$ Set Difference Union Intersection

Hence the result.

$\blacksquare$