Set Difference and Intersection form Partition

Theorem
Let $S$ and $T$ be sets such that:
 * $S \setminus T \ne \varnothing$
 * $S \cap T \ne \varnothing$

where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.

Then $S \setminus T$ and $S \cap T$ form a partition of $S$.

Proof
From Set Difference Intersection with Second Set is Empty Set:
 * $\left({S \setminus T}\right) \cap T = \varnothing$

and hence immediately from Intersection with Empty Set:
 * $\left({S \setminus T}\right) \cap \left({S \cap T}\right) = \varnothing$

So $S \setminus T$ and $S \cap T$ are disjoint.

Next from Set Difference Union Intersection:
 * $S = \left({S \setminus T}\right) \cup \left({S \cap T}\right)$

Thus by definition $S \setminus T$ and $S \cap T$ form a partition of $S$.