Empty Intersection iff Subset of Relative Complement

Theorem

Let $S$ be a set.

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

Then $A \cap B = \O \iff A \subseteq \relcomp S B$

Proof

$A \cap B = \O$

if and only if

$\forall x \in S: x \notin A \cap B$ by Empty Set as Subset

if and only if

$\forall x \in S: \neg \paren {x \in A \land x \in B}$ by definition of intersection

if and only if

$\forall x \in S: x \notin A \lor x \notin B$ by De Morgan's Laws (Logic)/Disjunction of Negations

if and only if

$\forall x \in S: x \in A \implies x \notin B$ by Rule of Material Implication

if and only if

$\forall x \in S: x \in A \implies x \in \relcomp S B$ by definition of relative complement

if and only if

$A \subseteq \relcomp S B$ by Subset in Subsets.

$\blacksquare$

Sources

• Mizar article SUBSET_1:23