Relation is Antireflexive iff Disjoint from Diagonal Relation

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

Then:
 * $\mathcal R$ is antireflexive

iff
 * $\Delta_S \cap \mathcal R = \varnothing$

where $\Delta_S$ is the diagonal relation.

Necessary Condition
Let $\mathcal R$ be an antireflexive relation.

Let $\left({x, y}\right) \in \Delta_S \cap \mathcal R$.

By definition:
 * $\left({x, y}\right) \in \Delta_S \implies x = y$

Likewise, by definition:
 * $\left({x, y}\right) \in \mathcal R \implies x \ne y$.

Thus:
 * $\Delta_S \cap \mathcal R = \left\{{\left({x, y}\right): x = y \land x \ne y}\right\}$

and so:
 * $\Delta_S \cap \mathcal R = \varnothing$

Sufficient Condition
Let $\Delta_S \cap \mathcal R = \varnothing$.

Then by definition:
 * $\forall \left({x, y}\right) \in \mathcal R: \left({x, y}\right) \notin \Delta_S$

Thus:
 * $\not \exists \left({x, y}\right) \in \mathcal R: x = y$

Thus by definition, $\mathcal R$ is antireflexive.