$<$ on a preorder is a strict partial order

Question

Definition: Suppose $X$ is a preorder. Define $x < y$ as $x \le y$ and $y \not\le x$ for each $x, y \in X$.

Question: Show that this gives a strict partial order on $X$.

Answer 1

We must show that the relation $<$ is irreflexive and transitive.

Irreflexive: Suppose $x \in X$. Then $x \le x$, so $x \not< x$.

Transitive: Suppose $x, y, z \in X$ such that $x < y$ and $y < z$. Then $x \le y$, $y \not\le x$, $y \le z$, and $z \not\le y$. By the transitivity of $\le$, we immediately have $x \le z$. Now assume that $z \le x$. Then $z \le y$. This is a contradition.