Composite of Symmetric Relations is not necessarily Symmetric

Theorem
Let $A$ be a set.

Let $\RR$ and $\SS$ be symmetric relations on $A$.

Then their composite $\RR \circ \SS$ is also symmetric.

Proof
Recall the definition of composition of relations:

Hence in this particular context:
 * $\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$

Let $\tuple {x, z} \in \RR \circ \SS$ be arbitrary.

By definition of composition of relations:
 * $\exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR$

As $\RR$ and $\SS$ are symmetric relations:
 * $\exists y \in A: \tuple {y, x} \in \SS \land \tuple {z, y} \in \RR$

By definition of inverse relation:

As $\RR$ and $\SS$ are symmetric relations:
 * $\exists y \in A: \tuple {x, y} \in \SS^{-1} \land \tuple {y, z} \in \RR^{-1}$

where $\SS^{-1}$ and $\RR^{-1}$ are the inverses of $\SS$ and $\RR$ respectively.

Hence by definition of composition of relations:
 * $\tuple {x, y} \in \SS^{-1} \circ \RR^{-1}$

By Inverse of Composite Relation:
 * $\tuple {x, y} \in \paren {\RR \circ \SS}^{-1}$

Hence by definition of subset:
 * $\RR \circ \SS \subseteq \paren {\RR \circ \SS}^{-1}$

and so by definition, $\RR \circ \SS$ is symmetric.