Inverse of Antireflexive Relation is Antireflexive
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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.
$\blacksquare$