Antireflexive and Transitive Relation is Asymmetric

Theorem
Let $\mathcal R \subseteq S \times S$ be a relation which is not null.

Let $\mathcal R$ be antireflexive and transitive.

Then $\mathcal R$ is also asymmetric.

Proof
Let $\mathcal R \subseteq S \times S$ be antireflexive and transitive.

That is:

We have that $\mathcal R$ is not null.

Suppose $\mathcal R$ is not asymmetric.

So, by definition, $\exists \left({x, y}\right) \in \mathcal R: \left({y, x}\right) \in \mathcal R$.

Then from the transitivity of $\mathcal R$ that would mean $\left({x, x}\right) \in \mathcal R$.

But that would contradict the antireflexivity of $\mathcal R$.

Therefore $\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \notin \mathcal R$ and $\mathcal R$ has been shown to be asymmetric.

Also see

 * Null Relation is Antireflexive, Symmetric and Transitive for the case where $\mathcal R = \varnothing$.