Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of $x$,$x^*$ belongs to $U$. I have been stuck with this problem for a while now. I have a proof that letting $U$ be an ultrafilter, exactly one of $x,x^*$ belongs to $U$ for all $x$ in $B$, I did this by showing that both belong to $U$ implies that $U=B$ which cannot be the case, and if neither long to $U$ then a contradiction can be derived by de-Morgan's laws. However, I'm stuck with the reverse implication, I need to prove that:
"If $U$ is a subset of a Boolean algebra $B$ such that $\forall x \in B$, exactly one of $x \in U$ and $x^* \in U$ is true, U is an ultrafilter.".
Is this even the case? Any help would be greatly appreciated, thanks in advance.
 A: A correct statement would be

If $U\subseteq B$ is a filter and for all $x\in B$ exactly one of $x\in U$ or $x^*\in U$ is true, then $U$ is an ultrafilter.

The converse holds as well, and this is sometimes taken as the definition of an ultrafilter. I presume the one you've been given is that an ultrafilter is a maximal (proper) filter.
To see that the statement holds, note first that the fact that both $x$ and $x^*$ aren't in $U$ implies $U$ is a proper filter. Assume that $F$ is a filter extending $U$ and let $x\in F\setminus U.$ Then $x^*\in U,$ so $x^*\in F$ and since both $x$ and $x^*$ are in $F,$ $F$ is improper.
Without the provision that $U$ is a filter, the statement becomes very badly false. Just randomly choose elements of a Boolean algebra, randomly label with zero or one (indicators as to whether the element is inside or outside $U$), and assign the complement the opposite label. Odds are you hit a counterexample. (To guarantee you do so, just make sure you start out labeling the $0$ element with a one and the $1$ element with a zero).
