Not Preceding implies Exists Completely Irreducible Element in Algebraic Lattice

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.

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

Then
 * $\exists p \in S: p$ is completely irreducible $\land ~x \preceq p \land y \npreceq p$

Proof
By definition of algebraic:
 * $\forall z \in S: z^\ll$ is directed

and
 * $L$ satisfies axiom of approximation.

By Axiom of Approximation in Up-Complete Semilattice:
 * $\exists k \in S: k \ll y \land k \npreceq x$

By Algebraic iff Continuous and For Every Way Below Exists Compact Between:
 * $\exists z \in K\left({L}\right): k \preceq z \preceq y$

By definition of transitivity:
 * $z \npreceq x$

By definition of upper closure of element:
 * $x \notin z^\succeq$

By definition of difference:
 * $x \in S \setminus z^\succeq$

By definition of relative complement:
 * $x \in \complement_S\left({z^\succeq}\right)$

By definition of compact subset:
 * $z$ is a compact element.

By Upper Closure of Element is Way Below Open Filter iff Element is Compact:
 * $z^\succeq$ is a way below open filter on $L$.

By Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement:
 * $\exists p \in S: x \preceq p \land p = \max\complement_S\left({z^\succeq}\right)$

Thus by Maximal implies Completely Irreducible:
 * $p$ is completely irreducible.

Thus $x \preceq p$

Aiming for a contradiction suppose that
 * $y \preceq p$

By definition of transitivity:
 * $z \preceq p$

By definition of maximum element:
 * $p \in \complement_S\left({z^\succeq}\right)$

By definition of relative complement:
 * $p \notin z^\succeq$

Thus this contradicts $z \preceq p$ by definition of upper closure of element.