Relation is Antireflexive iff Disjoint from Diagonal Relation

Theorem
A relation $\mathcal R \subseteq S \times S$ is antireflexive iff it is disjoint from the diagonal relation: $\Delta_S \cap \mathcal R = \varnothing$.

Proof

 * Suppose $\mathcal R$ is 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 $\Delta_S \cap \mathcal R = \varnothing$.


 * Now suppose $\Delta_S \cap \mathcal R = \varnothing$.

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.