Proper Filter is Included in Ultrafilter in Boolean Lattice

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.

Let $F$ be a proper filter on $L$.

Then there exists ultrafilter $G$ on $L$: $F \subseteq G$

Proof
By Singleton of Bottom is Ideal:
 * $I := \set \bot$ is an ideal in $L$.

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

We will prove that
 * $F \cap I = \O$

Let $x \in I$.

By definition of singleton:
 * $x = \bot$

Thus by Bottom not in Proper Filter:
 * $x \notin F$

By If Ideal and Filter are Disjoint then There Exists Prime Filter Including Filter and Disjoint from Ideal:
 * there exists a prime filter $G$ on $L$: $F \subseteq G$ and $G \cap I = \O$

By definition of singleton:
 * $\bot \in I$

By definitions of empty set and intersection:
 * $\bot \notin G$

By Bottom not in Proper Filter:
 * $G$ is proper.

Thus by Proper and Prime iff Ultrafilter in Boolean Lattice:
 * $G$ is ultrafilter.

Thus $F \subseteq G$