Way Above Closure is Upper

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

Let $x \in S$.

Then $x^\gg$ is upper

where $x^\gg$ denotes the way above closure of $x$.

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

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

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

Thus by definition of way above closure:
 * $z \in x^\gg$