Empty Intersection iff Subset of Relative Complement

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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

$A \cap B = \varnothing$

if and only if

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

if and only if

$\forall x \in S:\lnot\left({x \in A \land x \in B}\right)$ 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

$\forall x \in S: x \in A \implies x \in \complement_S\left({B}\right)$ by definition of relative complement

if and only if

$A \subseteq \complement_S\left({B}\right)$ by Subset in Subsets.

$\blacksquare$


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