In the following $B$ denotes a Boolean algebra and $\bar{x}$ is the complement of $x$.
In my notes there is the following theorem:
If $U \subset B$ is an ultrafilter on $B$ then for every $x \in B$ we either have $x \in U$ or $\bar{x} \in U$.
The proof they give starts as follows:
"Assume neither $x$ nor $\bar{x}$ are in $U$. Then there exist $y_1 , \dots , y_n, z_1, \dots , z_m \in U$ such that $x \land y_1 \land \dots \land y_n = 0$ and $\bar{x} \land z_1 \land \dots \land z_m = 0$."
Question: I would like to know why they want more than just one $y$ and one $z$ because I think I can prove this claim as follows (thanks for pointing out where I go wrong):
Theorem: If $U \subset B$ is an ultrafilter on $B$ then for every $x \in B$ we either have $x \in U$ or $\bar{x} \in U$.
Proof: Assume neither $x$ nor $\bar{x}$ are in $U$. Then there is a $y \in U$ such that $x \land y = 0$ and a $z \in U$ such that $\bar{x} \land z = 0$. Using the fact that $x \land y = 0 \iff y \leq \bar{x}$ we then get $y \leq \bar{x}$ and $z \leq \bar{\bar{x}} = x$ and hence $y \land z \leq \bar{x} \land x = 0$ so $y \land z = 0$ which would be a contradiction to $U$ being an ultrafilter.
Claim used in proof: If neither $x$ nor $\bar{x}$ are in $U$. Then there is a $y \in U$ such that $x \land y = 0$ and a $z \in U$ such that $\bar{x} \land z = 0$.
Proof: Assume neither $x$ nor $\bar{x}$ are in $U$. Then we define a filter $\tilde{U}$ that properly contains $U$ as follows: $\tilde{U}:= U \cup \{x \} \cup \{ b \in B \mid b \geq x \} \cup \{ x \land b \mid b \in U \text{ or } b \geq x \}$.
(i) $\tilde{U}$ is a proper subset of $B$ since $\emptyset \notin \tilde{U}$.
(ii) $\tilde{U}$ is a proper superset of $U$ since $x \notin U$.
(iii) $\tilde{U}$ is upwards closed by construction
(iv) $\tilde{U}$ is closed with respect to $\land$ by construction
This would be a filter properly containing $U$ which would be a contradiction to $U$ being an ultrafilter.
Thanks for your help.