Complement of Reflexive Relation

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

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

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

Proof

 * If $$\mathcal{R} \subseteq S \times T$$ is reflexive then:

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

By the definition of complement of $$\mathcal{R}$$:

$$\left({x, y}\right) \in \mathcal{R} \implies \left({x, y}\right) \notin \mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)$$

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

Thus $$\mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)$$ is antireflexive

Similarly, by definition, $$\forall x \in S: \left({x, x}\right) \notin \mathcal{R} \implies \lnot \left({x, x}\right) \notin \mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)$$ and by Double Negation it follows that $$\left({x, x}\right) \in \mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)$$.

The converses of these results follow from the fact that $$\mathcal{C}_{S \times S} \left ({\mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)}\right) = \mathcal{R}$$ by Relative Complement of Relative Complement.