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:
\(\ds \top\) | \(=\) | \(\ds \paren {q \vee z} \wedge \paren {q \vee \neg z}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {q \wedge q} \vee \paren {q \wedge \neg z} \vee \paren {q \wedge z} \vee \paren {z \wedge \neg z}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds q \wedge \paren {q \vee \neg z \vee z \vee \top}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds q\) |
This contradicts the fact that $J$ has the finite join property.
$\blacksquare$