Set Difference Disjoint with Reverse

Theorem

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

Proof

We assume that $S, T \subseteq \mathbb U$ where $\mathbb U$ is the universe.

Then we can use the definition of Set Difference as Intersection with Complement.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle \left({S \backslash T}\right) \cap \left({T \backslash S}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({S \cap \complement \left({T}\right)}\right) \cap \left({T \cap \complement \left({S}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Difference as Intersection with Complement
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({S \cap \complement \left({S}\right)}\right) \cap \left({T \cap \complement \left({T}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Intersection is Associative and Intersection is Commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \varnothing \cap \varnothing\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Intersection with Complement
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \varnothing\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Empty Set Disjoint with Itself

$\blacksquare$