Prime Element iff There Exists Way Below Open Filter which Complement has Maximum

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

Let $p \in S$ such that
 * $p \ne \top$

where $\top$ denotes the top of $L$.

Then
 * $p$ is a prime element


 * there exists a way below open filter $F$ in $L$: $p = \max \left({ \complement_S\left({F}\right) }\right)$

Sufficient Condition
Let $p$ be a prime element.

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

We will prove that
 * $\forall x \in S: \left({ x \in \complement_S\left({p^\preceq}\right) \implies \exists y \in S: y \in \complement_S\left({p^\preceq}\right) \land y \ll x }\right)$

Let $x \in S$ such that
 * $x \in \complement_S\left({p^\preceq}\right)$

By definition of relative complement:
 * $x \notin p^\preceq$

By definition of lower closure of element:
 * $x \npreceq p$

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

By definition of lower closure of element:
 * $y \notin p^\preceq$

By definition of relative complement:
 * $y \in \complement_S\left({p^\preceq}\right)$

Thus
 * $\exists y \in S: y \in \complement_S\left({p^\preceq}\right) \land y \ll x$

By Prime Element iff Complement of Lower Closure is Filter:
 * $F := \complement_S\left({p^\preceq}\right)$ is way below open filter in $L$.

By definitions of antisymmetry and lower closure of element:
 * $\lnot \exists y \in S: y \in F \land y \prec p$

By Relative Complement of Relative Complement::
 * $\complement_S\left({F}\right) = p^\preceq$

By definitions of reflexivity and lower closure of element:
 * $p \in \complement_S\left({F}\right)$

Thus by definition of greatest element:
 * $p = \max \left({\complement_S\left({F}\right)}\right)$

Necessary Condition
Let
 * there exists a way below open filter $F$ in $L$: $p = \max \left({ \complement_S\left({F}\right) }\right)$

By Maximal Element of Complement of Filter is Meet Irreducible:
 * $p$ is meet irreducible.

Thus by Prime Element iff Meet Irreducible in Distributive Lattice:
 * $p$ is a prime element.