# Trichotomy is Antireflexive

## Theorem

Let $\RR$ be a trichotomy.

Then $\RR$ is an antireflexive relation.

## Proof

Let $\RR$ be a trichotomy on a set $S$.

Let $x \in S$.

By definition of a trichotomy, for all $a, b \in S$, either:

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

As $x = x$ it follows directly that $x \not < x$.

Hence the result by definition of antireflexive relation.

$\blacksquare$