Equivalence of Definitions of Antisymmetric Relation

Definition 1 implies Definition 2
Let $\mathcal R$ be a relation which fulfils the condition:


 * $\left({x, y}\right) \in \mathcal R \land \left({y, x}\right) \in \mathcal R \implies x = y$

Let $\left({x, y}\right) \in \mathcal R$ such that $x \ne y$.

that $\left({y, x}\right) \in \mathcal R$.

Then $\left({x, y}\right) \in \mathcal R \land \left({y, x}\right) \in \mathcal R$.

By hypothesis, this implies that $x = y$.

From this contradiction it is concluded that $\left({y, x}\right) \notin \mathcal R$.

It follows that the condition:
 * $\left({x, y}\right) \in \mathcal R \land x \ne y \implies \left({y, x}\right) \notin \mathcal R$

holds for $\mathcal R$.

Definition 2 implies Definition 1
Let $\mathcal R$ be a relation which fulfils the condition:


 * $\left({x, y}\right) \in \mathcal R \land x \ne y \implies \left({y, x}\right) \notin \mathcal R$

Let $\left({x, y}\right) \in \mathcal R$ such that $\left({y, x}\right) \in \mathcal R$ also.

that $x \ne y$.

By hypothesis, this implies that $\left({y, x}\right) \notin \mathcal R$.

From this contradiction it is concluded that $\left({y, x}\right) \notin \mathcal R$.

It follows that the condition:
 * $\left({x, y}\right) \in \mathcal R \land \left({y, x}\right) \in \mathcal R \implies x = y$

holds for $\mathcal R$.