Boolean Prime Ideal Theorem/Extension Lemma

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Theorem

Let $\struct {B, \vee, \wedge, \neg, \bot, \top}$ be a Boolean algebra.

Let $J \subseteq B$ have the finite join property.

Let $z \in B$.


Then either $J \vee z$ or $J \vee \neg z$ also has the finite join property.


Proof

Aiming for a contradiction, suppose that neither $J \vee z$ nor $J \vee \neg z$ has the finite join property.

Then there are $x_1, \dots, x_n, y_1, \dots, y_m \in J$ such that $x_1 \vee \dots \vee x_n \vee z = y_1 \vee \dots \vee y_m \vee \neg z = \top$.

Let $q = x_1 \vee \dots \vee x_n \vee y_1 \vee \dots \vee y_m$.

Then $q \vee z = q \vee \neg z = \top$.

Thus:

\(\displaystyle \top\) \(=\) \(\displaystyle \paren {q \vee z} \wedge \paren {q \vee \neg z}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {q \wedge q} \vee \paren {q \wedge \neg z} \vee \paren {q \wedge z} \vee \paren {z \wedge \neg z}\)
\(\displaystyle \) \(=\) \(\displaystyle q \wedge \paren {q \vee \neg z \vee z \vee \top}\)
\(\displaystyle \) \(=\) \(\displaystyle q\)

This contradicts the fact that $J$ has the finite join property.

$\blacksquare$