Strict Ordering is Asymmetric

Theorem

Let $S$ be a set.

Let $\mathcal R$ be a strict ordering on $S$.

Then $\mathcal R$ is asymmetric, that is:

$\forall a, b \in S: a \mathcal R b \implies \neg b \mathcal R a$

Proof

By definition of a strict ordering:

$(1): \quad \mathcal R$ is irreflexive, that is, $\forall a \in S: \neg a \mathcal R a$

$(2): \quad \mathcal R$ is transitive, that is, $\forall a, b, c \in S: a \mathcal R b \land b \mathcal R c \implies a \mathcal R c$

Suppose $\exists a, b \in S: a \mathcal R b \land b \mathcal R a$.

As $\mathcal R$ is transitive, it follows that $a \mathcal R a$.

That would contradict the fact that $\mathcal R$ is irreflexive.

So there can be no such $a$ and $b$, and so by definition $\mathcal R$ is asymmetric.

$\blacksquare$