Set Difference and Intersection form Partition

Theorem
Let $S$ and $T$ be sets such that: where $S \setminus T$ denotes set difference and $S \cap T$ denotes set intersection.
 * $S \setminus T \ne \varnothing$
 * $S \cap T \ne \varnothing$

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

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

Then $S \setminus T$, $T \setminus S$ and $S \cap T$ form a partition of $S \cup T$, the union of $S$ and $T$.

Proof
We have from Set Difference Intersection Second Set is Empty Set that:
 * $\left({S \setminus T}\right) \cap T = \varnothing$

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

So we have that $S \setminus T$ and $S \cap T$ are disjoint.

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

That is all we need to assert that $S \setminus T$ and $S \cap T$ form a partition of $S$.

Proof of Corollary
From above, we have that:
 * $S \setminus T$ and $S \cap T$ form a partition of $S$
 * $T \setminus S$ and $S \cap T$ form a partition of $T$

Finally we note from Set Difference Disjoint with Reverse that $\left({S \setminus T}\right) \cap \left({T \setminus S}\right) = \varnothing$.

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

and the result follows.