Set Difference Union Intersection/Proof 3

Theorem

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

Proof
We have that:
 * From Set Difference Subset: $S \setminus T \subseteq S$
 * From Intersection Subset: $S \cap T \subseteq S$

Hence from Union Smallest it follows that $\left({S \setminus T}\right) \cup \left({S \cap T}\right) \subseteq S$.

Now let $s \in S$.

Either: or
 * $s \in T$, in which case $s \in S \cap T$ by definition of set intersection
 * $s \notin T$, 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 Equality of Sets.