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

if and only if:

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

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:

$x \mathrel \RR z$

By antitransitivity:

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

This is a contradiction.

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

$\blacksquare$

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $7 \ \text{(a)}$