# Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering

## Theorem

Let $S$ be a set.

Let $\RR$ be a transitive, antisymmetric relation on $S$.

Let $\RR^\ne$ denote the reflexive reduction of $\RR$.

Then $\RR^\ne$ is a strict ordering.

## Proof

To show that $\RR^\ne$ is a strict ordering, it is sufficient to show that $\RR^\ne$ is antireflexive and transitive.

### Antireflexive

Follows from Reflexive Reduction is Antireflexive.

$\Box$

### Transitive

Let $a, b, c \in S$.

Let $a \mathrel {\RR^\ne} b$ and $b \mathrel {\RR^\ne} c$.

By the definition of reflexive reduction:

$a \ne b$ and $a \mathrel \RR b$
$b \ne c$ and $b \mathrel \RR c$

Since $\mathrel \RR$ is transitive:

$a \mathrel \RR c$

Aiming for a contradiction, suppose $a = c$.

Since $a \mathrel \RR b$ it follows that $c \mathrel \RR b$.

Since $c \mathrel \RR b$, $b \mathrel \RR c$, and $\mathrel \RR$ is antisymmetric, $b = c$.

But this contradicts $b \ne c$.

The conclusion is that $a \ne c$.

Recall that $a \mathrel \RR c$.

By the definition of reflexive reduction:

$a \mathrel {\RR^\ne} c$

$\blacksquare$