# Condition for Relation to be Transitive and Antitransitive

## Theorem

Let $S$ be a set.

Let $\RR \subseteq S \times S$ be a relation in $S$.

Then:

$\RR$ is both transitive and antitransitive
$\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$

## Proof

### Necessary Condition

Suppose $\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$.

Then $\RR$ is both transitive and antitransitive vacuously.

$\Box$

### Sufficient Condition

Suppose $\RR$ is both transitive and antitransitive.

Aiming for a contradiction, suppose it is not the case that $\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$.

Then $\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z$.

By transitivity:

$x \mathrel {\RR} z$
$\neg \paren {x \mathrel {\RR} z}$

Hence we must have $\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$.
$\blacksquare$