Filter is Prime iff For Every Element Element either Negation Belongs to Filter in Boolean Lattice

Theorem
Let $B = \left({S, \vee, \wedge, \neg, \preceq}\right)$ be a Boolean lattice.

Let $F$ be a filter in $B$.

Then
 * $F$ is prime


 * $\forall x \in S: x \in F \lor \left({\neg x}\right) \in F$

Sufficient Condition
Let $F$ be prime.

Let $x \in S$.

By definition of Boolean lattice:
 * $x \vee \neg x = \top$

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

By definition of non-empty set:
 * $\exists y: y \in F$

By definition of greatest element:
 * $y \preceq \top$

By definition of upper set:
 * $\top \in F$

Thus by definition of prime filter:
 * $x \in F$ or $\neg x \in F$

Necessary Condition
Assume that
 * $\forall x \in S: x \in F \lor \left({\neg x}\right) \in F$

Let $a, b \in S$ such that
 * $a \vee b \in F$

Aiming for a contradiction suppose that
 * $a \notin F$ and $b \notin F$

By assumption:
 * $\neg a \in F$ and $\neg b \in F$

By Filtered in Meet Semilattice:
 * $\left({\neg a}\right) \wedge \left({\neg b}\right) \in F$

By De Morgan's Laws (Boolean Algebras):
 * $\neg\left({a \vee b}\right) \in F$

By Filtered in Meet Semilattice:
 * $\left({a \vee b}\right) \wedge \neg\left({a \vee b}\right) \in F$

By definition of Boolean lattice:
 * $\bot \in F$

where $\bot$ denotes the bottom of $B$.

By definition of smallest element:
 * $\bot \preceq a$

By definition of upper set:
 * $a \in F$

This contradicts $a \notin F$

Thus by Proof by Contradiction
 * the result holds.