# Preordering induces Ordering

## Theorem

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 relation defined on the quotient set $S / {\sim_\RR}$ by:

$\eqclass x {\sim_\RR} \preccurlyeq_\RR \eqclass y {\sim_\RR} \iff x \mathrel \RR y$

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

Then $\preccurlyeq_\RR$ is an ordering.

## Proof

First we note from Ordering Induced by Preordering is Well-Defined that $\preccurlyeq_\RR$ is a well-defined relation.

Checking in turn each of the criteria for an ordering:

### Reflexivity

Let $A \in S / {\sim_\RR}$.

By the definition of quotient set, $A$ is non-empty.

Thus there exists $a \in A$.

Since $\RR$ is a preordering, it is reflexive.

Hence:

$a \mathrel \RR a$

So by definition of $\preccurlyeq_\RR$:

$\eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass a {\sim_\RR}$

Thus $\preccurlyeq_\RR$ has been shown to be reflexive.

$\Box$

### Transitivity

Let $A, B, C \in S / {\sim_\RR}$ such that $A \preccurlyeq_\RR B$ and $B \preccurlyeq_\RR C$.

Then:

$\exists a \in A, b_1 \in B: a \mathrel \RR b_1$
$\exists b_2 \in B, c \in C: b_2 \mathrel \RR c$

By the definition of quotient set:

$b_1 \sim_\RR b_2$

By definition, $\sim_\RR$ is the equivalence on $S$ induced by $\RR$.

Hence:

$b_1 \mathrel \RR b_2$

We have:

$a \mathrel \RR b_1$
$b_1 \mathrel \RR b_2$
$b_2 \mathrel \RR c$
$\mathrel \RR$ is transitive

Hence from Transitive Chaining:

$a \precsim c$

Thus by the definition of $\mathrel \RR$:

$A \preccurlyeq_\RR C$

We have that $A$, $B$ and $C$ are arbitrary.

It follows that $\preccurlyeq_\RR$ is transitive.

$\Box$

### Antisymmetry

Let $A, B \in S / {\sim_\RR}$ such that $A \preccurlyeq_\RR B$ and $B \preccurlyeq_\RR A$.

Then:

$\eqclass a {\sim_\RR} \preccurlyeq_\RR \eqclass b {\sim_\RR}$
$\eqclass b {\sim_\RR} \preccurlyeq_\RR \eqclass a {\sim_\RR}$

That is:

By definition of $S / {\sim_\RR}$:

$a \mathrel \RR b$
$b \mathrel \RR a$

But by definition of $\sim_\RR$, that means:

$a \sim_\RR b$

Hence:

$a, b \in \eqclass a {\sim_\RR}$

and:

$a, b \in \eqclass b {\sim_\RR}$
$\eqclass a {\sim_\RR} = \eqclass b {\sim_\RR}$

So $\preccurlyeq_\RR$ has been shown to be antisymmetric.

$\Box$

$\preccurlyeq_\RR$ has been shown to be reflexive, transitive and antisymmetric.

Hence by definition it is an ordering.

$\blacksquare$