# 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

 $$\displaystyle A \setminus B$$ $$=$$ $$\displaystyle \left\{ {x : x \in A \land x \notin B}\right\}$$ $\quad$ Definition of Set Difference $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle \left\{ {x : \left(x \in A \land x \in X\right) \land x \notin B}\right\}$$ $\quad$ Definition of subset; Modus Ponens and Rule of Conjunction $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle \left\{ {x : x \in A \land \left(x \in X \land x \notin B\right)}\right\}$$ $\quad$ Rule of Association for Conjunction $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle \left\{ {x : x \in A \land x \in \complement_S \left({B}\right)}\right\}$$ $\quad$ Definition of Relative Complement $\quad$ $$\displaystyle$$ $$=$$ $$\displaystyle A \cap \complement_S \left({B}\right)$$ $\quad$ Definition of Set Intersection $\quad$

$\blacksquare$