Inverse of Antireflexive Relation is Antireflexive

Theorem

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

If $\RR$ is antireflexive, then so is $\RR^{-1}$.

Proof

\(\ds x\)

\(\in\)

\(\ds S\)

\(\ds \leadsto \ \ \)

\(\ds \tuple {x, x}\)

\(\notin\)

\(\ds \RR\)

Definition of Antireflexive Relation

\(\ds \leadsto \ \ \)

\(\ds \tuple {x, x}\)

\(\notin\)

\(\ds \RR^{-1}\)

Definition of Inverse Relation

Hence the result by definition of antireflexive relation.

$\blacksquare$