Set Difference Union Intersection/Proof 3

Theorem

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

$S \setminus T \subseteq S$

$S \cap T \subseteq S$

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

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, by definition of set union:

$s \in \paren {S \setminus T} \cup \paren {S \cap T}$

and so by definition of subset:

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

Hence the result by definition of set equality.

$\blacksquare$