Set Difference as Intersection with Complement

Theorem

Set difference can be expressed as the intersection with the set complement:

$A \setminus B = A \cap \map \complement B$

Proof

This follows directly from Set Difference as Intersection with Relative Complement:

$A \setminus B = A \cap \relcomp S B$

Let $S = \Bbb U$.

Since $A, B \subseteq \Bbb U$ by definition of the universe, the result follows.

$\blacksquare$