Proper and Prime iff Ultrafilter 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 a proper subset of $S$ and $F$ is a prime filter in $B$


 * $F$ is ultrafilter on $B$
 * $F$ is ultrafilter on $B$

Sufficient Condition
Let us assume
 * $F$ is a proper subset of $S$ and $F$ is a prime filter in $B$.

Thus
 * $F$ is a proper subset of $S$.

Let $G$ be a filter in $B$ such that
 * $F \subseteq G$ and $F \ne G$.

By definitions of subset and set equality:
 * $\exists x: x \in G \land x \notin F$

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

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

By Top in Filter:
 * $\top \in F$

By definition of prime filter:
 * $\neg x \in F$

By definition of subset:
 * $\neg x \in G$

By Filtered in Meet Semilattice:
 * $x \wedge \neg x \in G$

By definition of Boolean algebra:
 * $\bot \in G$

Thus by definition of subset:
 * $G \subseteq S$

By definition of set equality it remains to prove that
 * $S \subseteq G$

Let $y \in S$.

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

Thus by definition of upper set:
 * $y \in G$

Necessary Condition
Let $F$ be ultrafilter on $B$.

Thus by definition of ultrafilter:
 * $F$ is proper subset of $S$.

Let $x \in S$.

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

By Finite Infima Set and Upper Closure is Smallest Filter:
 * $F \cup \left\{ {x}\right\} \subseteq {\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$

By Set is Subset of Union:
 * $\left\{ {x}\right\} \subseteq F \cup \left\{ {x}\right\}$ and $F \subseteq F \cup \left\{ {x}\right\}$

By definition of singleton:
 * $x \in \left\{ {x}\right\}$

By definition of subset:
 * $x \in {\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$

By Finite Infima Set and Upper Closure is Filter:
 * ${\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$ is filter in $L$.

By Subset Relation is Transitive:
 * $F \subseteq {\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq$

By definition of ultrafilter:
 * ${\operatorname{fininfs}\left({F \cup \left\{ {x}\right\} }\right)}^\succeq = S$

By Finite Subset Bounds Element of Finite Infima Set and Upper Closure:
 * $\exists a \in S: a \in F \land \neg x \succeq a \wedge \inf \left\{ {x}\right\}$

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

By Infimum of Singleton:
 * $\inf \left\{ {x}\right\} = x$

By Meet Precedes Operands:
 * $a \wedge \neg x \preceq \neg x$

By definition of infimum:
 * $a \preceq \neg x$

By definition of upper set:
 * $\neg x \in F$

This contradicts $\neg x \notin F$

Thus by Proof by Contradiction:
 * $F$ is a prime filter by Filter is Prime iff For Every Element Element either Negation Belongs to Filter in Boolean Lattice.