Equivalence of Definitions of Antisymmetric Relation

Theorem

The following definitions of the concept of Antisymmetric Relation are equivalent:

Definition 1

$\RR$ is antisymmetric if and only if:

$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$

that is:

$\set {\tuple {x, y}, \tuple {y, x} } \subseteq \RR \implies x = y$

Definition 2

$\RR$ is antisymmetric if and only if:

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

Proof

Definition 1 implies Definition 2

Let $\RR$ be a relation which fulfils the condition:

$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$

Let $\tuple {x, y} \in \RR$ such that $x \ne y$.

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

Then $\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR$.

This implies by hypothesis that $x = y$.

From this contradiction it is concluded that $\tuple {y, x} \notin \RR$.

It follows that the condition:

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

holds for $\RR$.

$\Box$

Definition 2 implies Definition 1

Let $\RR$ be a relation which fulfils the condition:

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

Let $\tuple {x, y} \in \RR$ such that $\tuple {y, x} \in \RR$ also.

Aiming for a contradiction, suppose that $x \ne y$.

This implies by hypothesis that $\tuple {y, x} \notin \RR$.

From this contradiction it is concluded that $\tuple {y, x} \notin \RR$.

It follows that the condition:

$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$

holds for $\RR$.

$\blacksquare$

Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $1 \ \text{(a)}$