Set Difference is Disjoint with Reverse

Theorem

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

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.

\(\ds \)

\(\)

\(\ds \paren {S \setminus T} \cap \paren {T \setminus S}\)

\(\ds \)

\(=\)

\(\ds \paren {S \cap \map \complement T} \cap \paren {T \cap \map \complement S}\)

Set Difference as Intersection with Complement

\(\ds \)

\(=\)

\(\ds \paren {S \cap \map \complement S} \cap \paren {T \cap \map \complement T}\)

Intersection is Associative and Intersection is Commutative

\(\ds \)

\(=\)

\(\ds \O \cap \O\)

Intersection with Complement

\(\ds \)

\(=\)

\(\ds \O\)

Empty Set Disjoint with Itself

$\blacksquare$