Way Below Closure is Lower Section

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

Let $x \in S$.

Then
 * $x^\ll$ is s lower set.

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

By definition of way below closure:
 * $y \ll x$

By definition of reflexivity:
 * $x \preceq x$

By Preceding and Way Below implies Way Below:
 * $z \ll x$

Thus by definition of way below closure:
 * $z \in x^\ll$

Thus by definition
 * $x^\ll$ is a lower set.