Way Below Relation is Multiplicative implies Pseudoprime Element is Prime

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice such that
 * $\ll$ is multiplicative relation

where $\ll$ denotes the way below relation of $L$.

Let $p \in S$.

Then $p$ is a pseudoprime element is a prime element.

Proof
Let $p$ be a pseudoprime element.

Aiming for a contradiction suppose that
 * $p$ is not a prime element.

By definition of prime element:
 * $\exists x, y \in S: x \wedge y \preceq p$ and $x \npreceq p$ and $y \npreceq p$

By definition of continuous:
 * $\forall z \in S: z^\ll$ is directed.

and
 * $L$ satisfies axiom of approximation.

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

and
 * $\exists v \in S: v \ll y \land v \npreceq p$

By Way Below Relation is Auxiliary Relation:
 * $\ll$ is auxiliary relation.

By Multiplicative Auxiliary Relation iff Congruent:
 * $u \wedge v \ll x \wedge y$

By definition of transitivity:
 * $u \wedge v \ll p$

By Characterization of Pseudoprime Element when Way Below Relation is Multiplicative:
 * $u \preceq p$ or $v \preceq p$

This contradicts $u \npreceq p$ and $v \npreceq p$