Boolean Prime Ideal Theorem/Extension Lemma

Theorem
Let $(B, \vee, \wedge, \neg, \bot, \top)$ 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
Suppose for the sake of contradiction that neither 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 $\top = (q \vee z) \wedge (q \vee \neg z) = (q \wedge q) \vee (q \wedge \neg z) \vee (q \wedge z) \vee (z \wedge \neg z) = q \wedge (q \vee \neg z \vee z \vee \top) = q$.

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