Set Difference and Intersection form Partition/Corollary 1

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Corollary to Set Difference and Intersection form Partition

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

From Set Difference and Intersection form Partition:

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

From Set Difference Disjoint with Reverse:

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

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.

$\blacksquare$