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.