Boolean Prime Ideal Theorem/Element Extension Lemma

Theorem

Let $\struct {B, \vee, \wedge, \le}$ be a Boolean lattice.

Let $F \subseteq B$ be a filter on $B$.

Let $a, x \in B$ such that:

$a \notin F$

Then either:

$a \vee x \notin F$

or:

$a \vee \neg x \notin F$

Aiming for a contradiction, suppose that both:

$a \vee x \in F$

and:

$a \vee \neg x \in F$

Then:

$\ds F$

$\ni$

$\ds \paren {a \vee x} \wedge \paren {a \vee \neg x}$

Filter is Closed under Meet

$\ds $

$=$

$\ds \paren {a \wedge \paren {a \vee \neg x} } \vee \paren {x \wedge \paren {a \vee \neg x} }$

Distributive Lattice Axiom $\paren {1'}$

$\ds $

$=$

$\ds a \vee \paren {\paren {x \wedge a} \vee \paren {x \wedge \neg x} }$

Meet Absorbs Join, Distributive Lattice Axiom $\paren 1$

$\ds $

$=$

$\ds a \vee \paren {\paren {a \wedge x} \vee \bot}$

Meet is Commutative, Definition of Complement (Lattice Theory)

$\ds $

$=$

$\ds a$

Definition 2 of Bottom of Lattice, Join Absorbs Meet

which contradicts the hypothesis.

$\blacksquare$