Equivalence of Definitions of Symmetric Closure

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

Proof
First we note that from Union of Relation with Inverse is Symmetric Relation, $\RR \cup \RR^{-1}$ is a symmetric relation.

Definition $(1)$ implies Definition $(2)$
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$ by definition $1$.

Then by definition:
 * $\RR^\leftrightarrow = \RR \cup \RR^{-1}$

Let $\QQ$ be a symmetric relation such that:
 * $\RR \subseteq \QQ$

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

By definition of set union, either:
 * $\tuple {a, b} \in \RR$

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


 * Case $1$

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

Then by definition of subset:
 * $\tuple {a, b} \in \QQ$

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

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

Then by definition of subset:
 * $\tuple {b, a} \in \QQ$

But $\QQ$ is symmetric.

Hence:
 * $\tuple {a, b} \in \QQ$

So by Proof by Cases:
 * $\tuple {a, b} \in \RR \cup \RR^{-1} \implies \tuple {a, b} \in \QQ$

Hence $\RR \cup \RR^{-1} \subseteq \QQ$

It follows that $\RR \cup \RR^{-1}$ is the smallest symmetric relation the subset ordering.

Thus $\RR^\leftrightarrow$ is the symmetric closure of $\RR$ by definition $2$.

Definition $(2)$ implies Definition $(1)$
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$ by definition $2$.

Then by definition:
 * $\RR^\leftrightarrow$ is the smallest symmetric relation on $S$ which has $\RR$ as a subset.

Hence as $\RR \cup \RR^{-1}$ is a symmetric relation :


 * $\RR^\leftrightarrow \subseteq \RR \cup \RR^{-1}$

But also, for every symmetric relation $\QQ$ such that $\RR \subseteq \QQ$:
 * $\RR \cup \RR^{-1} \subseteq \QQ$

Hence:
 * $\RR \cup \RR^{-1} \subseteq \RR^\leftrightarrow$

Thus by set equality:
 * $\RR \cup \RR^{-1} = \RR^\leftrightarrow$

That is, $\RR^\leftrightarrow$ is the symmetric closure of $\RR$ by definition $1$.