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

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.