Union of Relation with Inverse is Symmetric Relation

Theorem

Let $\RR$ be a relation on a set $S$.

Let $\RR^{-1}$ denote the inverse of $\RR$.

Then $\RR \cup \RR^{-1}$, the union of $\RR$ with $\RR^{-1}$, is symmetric.

Proof

Let $\tuple {a, b} \in \RR \cup \RR^{-1}$.

By definition of union, either:

$\tuple {a, b} \in \RR$

or:

$\tuple {a, b} \in \RR^{-1}$

Case 1

If $\tuple {a, b} \in \RR$, then by definition of inverse relation:

$\tuple {b, a} \in \RR^{-1}$

But from Set is Subset of Union:

$\tuple {b, a} \in \RR \cup \RR^{-1}$

$\Box$

Case 2

If $\tuple {a, b} \in \RR^{-1}$, then by definition of inverse relation:

$\tuple {b, a} \in \paren {\RR^{-1} }^{-1}$

From Inverse of Inverse Relation:

$\tuple {b, a} \in \RR$

But from Set is Subset of Union:

$\tuple {b, a} \in \RR \cup \RR^{-1}$

$\Box$

Hence by Proof by Cases:

$\tuple {a, b} \in \RR \cup \RR^{-1} \implies \tuple {b, a} \in \RR \cup \RR^{-1}$

and $\RR \cup \RR^{-1}$ is symmetric by definition.

$\blacksquare$