Set Difference Union Intersection/Proof 3

Proof
By Set Difference is Subset:
 * $S \setminus T \subseteq S$

By Intersection is Subset:
 * $S \cap T \subseteq S$

Hence from Union is Smallest Superset:
 * $\paren {S \setminus T} \cup \paren {S \cap T} \subseteq S$

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.