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

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ 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 $L$: $p = \map \max {\relcomp S F}$
 * there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$

Sufficient Condition
Let $p$ be a prime element.

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

We will prove that:
 * $\forall x \in S: \paren {x \in \relcomp S {p^\preceq} \implies \exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x}$

Let $x \in S$ such that:
 * $x \in \relcomp S {p^\preceq}$

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:
 * $y \in \relcomp S {p^\preceq}$

Thus:
 * $\exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x$

By Prime Element iff Complement of Lower Closure is Filter:
 * $F := \relcomp S {p^\preceq}$ 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::
 * $\relcomp S F = p^\preceq$

By definitions of reflexivity and lower closure of element:
 * $p \in \relcomp S F$

Thus by definition of greatest element:
 * $p = \map \max {\relcomp S F}$

Necessary Condition
Let:
 * there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$

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.