# Equivalence of Definitions of Antisymmetric Relation

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## 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

$\mathcal R$ is antisymmetric if and only if:

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

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

By hypothesis, this implies 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 $x \ne y$.

By hypothesis, this implies 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 $\RR$.

$\blacksquare$