Mapping is Involution iff Bijective and Symmetric

Theorem
Let $S$ be a set.

Let $f: S \to S$ be a mapping on $S$.

Then $f$ is an involution iff $f$ is both a bijection and a symmetric relation.

Proof
By definition an involution on $S$ is a mapping such that:
 * $\forall x \in S: f \left({f \left({x}\right)}\right) = x$

Necessary Condition
Let $f$ be an involution.

By Involution is Permutation, $f$ is a permutation and therefore by definition a bijection.

Then:

Thus $f$, considered as a relation, is symmetric.

Thus it has been shown that if $f$ is an involution, it is both a bijection and a symmetric relation.

Sufficient Condition
Let $f$ be a mapping which is both a bijective mapping and a symmetric relation.

Then:

and so $f$ is shown to be an involution.