Asymmetric Relation is Antireflexive

Theorem

Let $S$ be a set.

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

Let $\RR$ be asymmetric.

Then $\RR$ is also antireflexive.

Let $\RR$ be asymmetric.

Then, by definition:

$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$

Aiming for a contradiction, suppose $\tuple {x, x} \in \RR$.

Then:

$\ds \tuple {x, x} \in \RR$

$\implies$

$\ds \tuple {x, x} \notin \RR$

Definition of Asymmetric Relation

$\ds \leadsto \ \ $

$\ds \tuple {x, x} \notin \RR$

$\ds $

Thus $\RR$ is antireflexive.

$\blacksquare$

1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: $163$
1979: John E. Hopcroft and Jeffrey D. Ullman: Introduction to Automata Theory, Languages, and Computation ... (previous) ... (next): Chapter $1$: Preliminaries: $1.5$ Relations: Properties of Relations