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
If $$\mathcal R = \varnothing$$ then Null Relation is Antireflexive, Symmetric and Transitive applies instead.