# Complement of Symmetric Relation

## Theorem

Let $\RR \subseteq S \times S$ be a relation.

Then $\RR$ is symmetric if and only if its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is also symmetric.

## Proof

Let $\RR \subseteq S \times S$ be symmetric.

Then from Symmetry of Relations is Symmetric:

$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$

Aiming for a contradiction, suppose $\relcomp {S \times S} \RR \subseteq S \times S$ is not symmetric.

Then:

$\exists \tuple {x, y} \in \relcomp {S \times S} \RR: \tuple {y, x} \notin \relcomp {S \times S} \RR$

But then by definition of complement of $\RR$:

$\tuple {y, x} \in \RR$

As $\RR$ is symmetric it follows that:

$\tuple {x, y} \in \RR$

This contradicts the premise that $\tuple {x, y} \in \relcomp {S \times S} \RR$.

Hence by Proof by Contradiction it follows that $\relcomp {S \times S} \RR$ is symmetric.

$\blacksquare$

The converse follows similarly.