Ordering Induced by Preordering is Well-Defined

Definition
Let $\struct {S, \RR}$ be a relational structure such that $\RR$ is a preordering.

Let $\sim_\RR$ denote the equivalence on $S$ induced by $\RR$:
 * $x \sim_\RR y$ $x \mathrel \RR y$ and $y \mathrel \RR x$

Let $\preccurlyeq_\RR$ be the ordering on the quotient set $S / {\sim_\RR}$ by $\RR$:
 * $\eqclass x {\sim_\RR} \preccurlyeq_\RR \eqclass y {\sim_\RR} \iff x \mathrel \RR y$

where $\eqclass x {\sim_\RR}$ denotes the equivalence class of $x$ under $\sim_\RR$.

Then $\preccurlyeq_\RR$ is a well-defined relation.

Proof
We need to demonstrate that when:
 * $a \sim_\RR a'$
 * $b \sim_\RR b'$

it follows that:
 * $\eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass b {\sim_\RR} \iff \eqclass {a'} {\sim_\RR} \preccurlyeq_\RR \eqclass {b'} {\sim_\RR}$

So, let:
 * $a \sim_\RR a'$
 * $b \sim_\RR b'$

for arbitrary $a, b, a', b' \in S$.

By definition of $\sim_\RR$, this means:


 * $a \mathrel \RR a'$ and $a' \mathrel \RR a$
 * $b \mathrel \RR b'$ and $b' \mathrel \RR b$

So:

That is:
 * $\eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass b {\sim_\RR} \implies \eqclass {a'} {\sim_\RR} \preccurlyeq_\RR \eqclass {b'} {\sim_\RR}$

The same argument is used to prove that:
 * $\eqclass {a'} {\sim_\RR} \preccurlyeq_\RR \eqclass {b'} {\sim_\RR} \implies \eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass b {\sim_\RR}$

Hence the result.