# Set Difference Union Intersection/Proof 3

## Theorem

$S = \paren {S \setminus T} \cup \paren {S \cap T}$

## Proof

$S \setminus T \subseteq S$
$S \cap T \subseteq S$

Hence from Union is Smallest Superset:

$\left({S \setminus T}\right) \cup \left({S \cap T}\right) \subseteq S$

Let $s \in S$.

Either:

$s \in T$, in which case $s \in S \cap T$ by definition of set intersection

or

$s \notin T$, in which case $s \in S \setminus T$ by definition of set difference.

That is, $s \in \left({S \setminus T}\right) \cup \left({S \cap T}\right)$ by definition of set union, and so $S \subseteq \left({S \setminus T}\right) \cup \left({S \cap T}\right)$.

Hence the result by definition of set equality.

$\blacksquare$