Complement of Reflexive Relation

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

Then $\mathcal R$ is reflexive its complement $\complement_{S \times S} \left ({\mathcal R}\right) \subseteq S \times S$ is antireflexive.

Likewise, $\mathcal R$ is antireflexive its complement $\complement_{S \times S} \left ({\mathcal R}\right) \subseteq S \times S$ is reflexive.

Proof
Let $\mathcal R \subseteq S \times T$ be reflexive.

Then:
 * $\forall x \in S: \left({x, x}\right) \in \mathcal R$

By the definition of complement of $\mathcal R$:
 * $\left({x, y}\right) \in \mathcal R \implies \left({x, y}\right) \notin \complement_{S \times S} \left ({\mathcal R}\right)$

The same applies to $\left({x, x}\right)$, and thus:: $\forall x \in S: \left({x, x}\right) \notin \complement_{S \times S} \left ({\mathcal R}\right)$

Thus $\complement_{S \times S} \left ({\mathcal R}\right)$ is antireflexive.

Similarly, by definition:
 * $\forall x \in S: \left({x, x}\right) \notin \mathcal R \implies \neg \left({x, x}\right) \notin \complement_{S \times S} \left ({\mathcal R}\right)$

By Double Negation it follows that:
 * $\left({x, x}\right) \in \complement_{S \times S} \left ({\mathcal R}\right)$

The converses of these results follow from the fact that:
 * $\complement_{S \times S} \left ({\complement_{S \times S} \left ({\mathcal R}\right)}\right) = \mathcal R$

by Relative Complement of Relative Complement.