Complement of Symmetric Relation

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

Then $\mathcal R$ is symmetric iff its complement $\complement_{S \times S} \left ({\mathcal R}\right) \subseteq S \times S$ is also symmetric.

Proof
Let $\mathcal R \subseteq S \times S$ be symmetric.

Then from Symmetry of Relations is Symmetric:
 * $\left({x, y}\right) \in \mathcal R \iff \left({y, x}\right) \in \mathcal R$

Suppose $\complement_{S \times S} \left ({\mathcal R}\right) \subseteq S \times S$ is not symmetric.

Then:
 * $\exists \left({x, y}\right) \in \complement_{S \times S} \left ({\mathcal R}\right): \left({y, x}\right) \notin \complement_{S \times S} \left ({\mathcal R}\right)$

But then by definition of complement of $\mathcal R$, $\left({y, x}\right) \in \mathcal R$ and so $\left({x, y}\right) \in \mathcal R$.

This contradicts the fact that $\left({x, y}\right) \in \complement_{S \times S} \left ({\mathcal R}\right)$, so it follows that $\complement_{S \times S} \left ({\mathcal R}\right)$ is symmetric after all.

The converse follows similarly.