Set Difference as Intersection with Relative Complement

Theorem
Let $$A, B \subseteq S$$.

Then the set difference between $$A$$ and $$B$$ can be expressed as the intersection with the relative complement with respect to $$S$$:


 * $$A \setminus B = A \cap \complement_S \left({B}\right)$$

Proof
$$ $$ $$