Complement of Symmetric Relation

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

Then $$\mathcal{R}$$ is symmetric iff its complement $$\mathcal{C}_{S \times S} \left ({\mathcal{R}}\right) \subseteq S \times S$$ 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 $$\mathcal{C}_{S \times S} \left ({\mathcal{R}}\right) \subseteq S \times S$$ is not symmetric.

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

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 $$\left({x, y}\right) \in \mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)$$, so it follows that $$\mathcal{C}_{S \times S} \left ({\mathcal{R}}\right)$$ is symmetric after all.

The converse follows similarly.