If Every Element Pseudoprime is Prime then Way Below Relation is Multiplicative

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

Let every element $p \in S$: $p$ is pseudoprime $\implies p$ is prime.

Then $\ll$ is multiplicative

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

Proof
Let $a, x, y \in S$ such that
 * $a \ll x$ and $a \ll y$


 * $a \not\ll x \wedge y$

We will prove that
 * $\forall z \in S: z \in \paren {x \wedge y}^\ll \implies z \notin a^\succeq$

Let $z \in S$ such that
 * $z \in \paren {x \wedge y}^\ll$

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


 * $z \in a^\succeq$

By definition of upper closure of element:
 * $a \preceq z$

By Preceding and Way Below implies Way Below
 * $a \ll x \wedge y$

This contradicts $a \not\ll x \wedge y$

By definitions of empty set and intersection:
 * $\paren {x \wedge y}^\ll \cap a^\succeq = \O$

By Way Below Closure is Ideal in Bounded Below Join Semilattice:
 * $\paren {x \wedge y}^\ll$ is ideal in $L$.

By Upper Closure of Element is Filter:
 * $a^\succeq$ is filter on $L$.

By If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter:
 * there exists prime ideal $P$ in $L$:
 * $\paren {x \wedge y}^\ll \subseteq P$ and $P \cap a^\succeq = \O$

By definition of pseudoprime element:
 * $\sup P$ is pseudoprime.

By assumption:
 * $\sup P$ is prime.

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

By definition of upper closure of element:
 * $a \in a^\succeq$

By definition of axiom of approximation:
 * $\sup \paren {x \wedge y}^\ll = x \wedge y$

By definition of up-complete:
 * $\paren {x \wedge y}^\ll$ admits a supremum

and
 * $P$ admits a supremum.

By Supremum of Subset:
 * $x \wedge y \preceq \sup P$

By definition of prime element:
 * $x \preceq \sup P$ or $y \preceq \sup P$

By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
 * $a \in P$

By definition of intersection:
 * $a \in P \cap a^\succeq$

This contradicts $P \cap a^\succeq = \O$