Set Difference and Intersection form Partition

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

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:
 * $\paren {S \setminus T} \cap T = \O$

and hence immediately from Intersection with Empty Set:
 * $\paren {S \setminus T} \cap \paren {S \cap T} = \O$

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

Next from Set Difference Union Intersection:
 * $S = \paren {S \setminus T} \cup \paren {S \cap T}$

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