Relation is Antireflexive iff Disjoint from Diagonal Relation

Theorem
Let $\RR \subseteq S \times S$ be a relation on a set $S$.

Then:
 * $\RR$ is antireflexive


 * $\Delta_S \cap \RR = \O$

where $\Delta_S$ is the diagonal relation.

Necessary Condition
Let $\RR$ be an antireflexive relation.

Let $\tuple {x, y} \in \Delta_S \cap \RR$.

By definition:
 * $\tuple {x, y} \in \Delta_S \implies x = y$

Likewise, by definition:
 * $\tuple {x, y} \in \RR \implies x \ne y$.

Thus:
 * $\Delta_S \cap \RR = \set {\tuple {x, y}: x = y \land x \ne y}$

and so:
 * $\Delta_S \cap \RR = \O$

Sufficient Condition
Let $\Delta_S \cap \RR = \O$.

Then by definition:
 * $\forall \tuple {x, y} \in \RR: \tuple {x, y} \notin \Delta_S$

Thus:
 * $\not \exists \tuple {x, y} \in \RR: x = y$

Thus by definition, $\RR$ is antireflexive.