# Empty Intersection iff Subset of Relative Complement

## Theorem

Let $S$ be a set.

Let $A, B$ be subset of $S$.

Then $A \cap B = \varnothing \iff A \subseteq \complement_S\left({B}\right)$

## Proof

$A \cap B = \varnothing$
$\forall x \in S: x \notin A \cap B$ by Empty Set as Subset
$\forall x \in S:\lnot\left({x \in A \land x \in B}\right)$ by definition of intersection
$\forall x \in S: x \notin A \lor x \notin B$ by De Morgan's Laws (Logic)/Disjunction of Negations
$\forall x \in S: x \in A \implies x \notin B$ by Rule of Material Implication
$\forall x \in S: x \in A \implies x \in \complement_S\left({B}\right)$ by definition of relative complement
$A \subseteq \complement_S\left({B}\right)$ by Subset in Subsets.

$\blacksquare$