Set of Meet Irreducible Elements Excluded Top is Order Generating

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a continuous complete lattice.

Let $X = \mathit{IRR}\left({L}\right) \setminus \left\{ {\top}\right\}$

where $\mathit{IRR}\left({L}\right)$ denotes the set of all meet irreducible element of $S$,
 * $\top$ denotes the top of $L$.

Then $X$ is order generating.

Proof
We will prove that
 * $\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p }\right)$

Let $x, y \in S$ such that
 * $y \npreceq x$

By Not Preceding implies There Exists Meet Irreducible Element Not Preceding
 * $\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq p$

By definition of greatest element:
 * $p \ne \top$ and $p \in \mathit{IRR}\left({L}\right)$

By definitions of difference and singleton:
 * $p \in X$

Thus
 * $\exists p \in X: x \preceq p \land y \npreceq p$

Hence by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding:
 * $X$ is order generating.