Proving the antisymmetric property of the relation $R = \{(x, y) \in\Bbb R \mid x < y$ }

Question

The relation is $R = \{(x, y) \in\Bbb R^2 \mid x < y \}$

How can I prove antysimetric property for the relation?

Antisymmetric property states that IF $x R y$ and $y R x$, THEN $x = y$. The thing is, in this relation, the first part of the property is never true, so that's how I know that the relation IS antisymmetric. I just don't know how to prove it.

Answer 1

[Migrating a comment to serve as an answer.]

A reminder about logical implications of the form "If $P$, then $Q$" and denoted $P \rightarrow Q$:

Such an if-then statement is true if either both $P$ and $Q$ are true, or if $P$ is false.

You are in the latter case; as you write:

Antisymmetric property states that IF x R y and y R x, THEN x = y. The thing is, in this relation, the first part of the property is never true, so that's how I know that the relation IS antisymmetric.

Indeed, it cannot be the case that $xRy$ and $yRx$, for this would mean we had $x, y \in \mathbb{R}$ for which both $x<y$ and $y<x$; but, those inequalities yield $x < y < x$, which is a contradiction: $x \not < x$.

Since the antecedent (i.e., $P$) is false, the implication (i.e., $P \rightarrow Q$) is true. As one more piece of vocabulary: In such a scenario, one sometimes says that the statement is vacuously true, i.e., when an implication is true because its antecedent is false.