Preceding and Way Below implies Way Below

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $u, x, y, z \in S$ such that
 * $u \preceq x \ll y \preceq z$

where $\ll$ denotes the way below relation.

Then
 * $u \ll z$

Proof
Let $D$ be a directed subset of $S$ such that
 * $D$ admits a supremum

and
 * $z \preceq \sup D$

By definition of transitivity:
 * $y \preceq \sup D$

By definition of way below relation:
 * $\exists d \in D: x \preceq d$

Thus by definition of transitivity:
 * $\exists d \in D: u \preceq d$

Thus by definition of way below relation:
 * $u \ll z$