# 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 $\RR$:

$x \sim_\RR y$ if and only if $x \mathrel \RR y$ 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:

 $\ds \eqclass a {\sim_\RR}$ $\preccurlyeq_\RR$ $\ds \eqclass b {\sim_\RR}$ $\ds \leadsto \ \$ $\ds a$ $\RR$ $\ds b$ Definition of $\preccurlyeq_\RR$ $\ds \leadsto \ \$ $\ds a'$ $\RR$ $\ds b$ $\RR$ is transitive: $a' \mathrel \RR a$ and $a \mathrel \RR b$ $\ds \leadsto \ \$ $\ds a'$ $\RR$ $\ds b'$ $\RR$ is transitive: $a' \mathrel \RR b$ and $b \mathrel \RR b'$ $\ds \leadsto \ \$ $\ds \eqclass {a'} {\sim_\RR}$ $\preccurlyeq_\RR$ $\ds \eqclass {b'} {\sim_\RR}$ Definition of $\preccurlyeq_\RR$

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.

$\blacksquare$