A partial order is an antisymmetric preorder, i.e. a binary relation $\leq$ on $X$ such that:
(Reflexivity) For all $x \in X$,
\[x \leq x.\](Transitivity) For all $x$, $y$, $z \in X$,
\[x \leq y \,\land\, y \leq z \enspace\rimp\enspace x \leq z.\](Antisymmetry) For all $x$, $y \in X$,
\[x \leq y \,\land\, y \leq x \enspace\rimp\enspace x = y.\]
A binary relation $<$ on $X$ is a strict partial order if:
(Irreflexivity) For all $x \in X$,
\[x \nless x.\](Transitivity) For all $x$, $y$, $z \in X$,
\[x < y \,\land\, y < z \enspace\rimp\enspace x < z.\](Asymmetry) For all $x$, $y \in X$,
\[x < y \enspace\rimp\enspace y \nless x.\]
A binary relation is a strict partial order as soon as it is irreflexive and transitive.
(This is why we do not talk about “strict preorder”.)
Proof. Irreflexivity and transitivity imply asymmetry: Let $<$ be an irreflexive and transitive relation on $X$. If $x < y$ and $y < x$, then $x < x$ by transitivity, contrary to irreflexivity.
$\leq$ is a partial order if and only if $<$ is a strict partial order.