Set Difference and Intersection form Partition

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

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

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

Then $$S - T$$, $$T - S$$ and $$S \cap T$$ form a partition of $$S \cup T$$.

Proof
We have from Set Difference with Intersection that:
 * $$\left({S - T}\right) \cap T = \varnothing$$

and hence immediately from Intersection with Null:
 * $$\left({S - T}\right) \cap \left({S \cap T}\right) = \varnothing$$.

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

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

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

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

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

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

and the result follows.