Question on Boolean algebras Let B be a Boolean algebra. Let a,b be elements of the universe of B such that a < 1 and b < 1, $a \bot b$(so neither $a \leq b$ nor $b \leq a$) and  $(\neg a \vee b) = 1$. Is it always the case that a=b?
 A: No. Consider the Boolean Algebra $2^2 = \{(x, y) \mid x, y \in 2\}$, with all operations occuring pointwise.
Then let $a = (0, 0) = 0$ and $b = (0, 1)$. Note that $a \neq 1$ and $b \neq 1$. Then $\neg a = (1, 1)$, so $\neg a \lor b = (1, 1) = 1$.
But clearly $a \neq b$.
Edit: the question was edited to add the requirement that $a$ and $b$ are incomparable. If $\neg a \lor b = 1$, then we must have $\neg a \lor (a \land b) = (\neg a \lor a) \land (\neg a \lor b) = 1 \land 1 = 1$. And we must have $\neg a \land (a \land b) = (\neg a \land a) \land b = 0 \land b = 0$. Therefore, we must have that $a \land b = \neg \neg a = a$, and thus $a \leq b$.
A: Boolean algebras are uniquely complemented lattices, thus if we have premises like $(\neg a \vee b) = 1$ and $(\neg a \land b) = 0$ then we must have $a=b$ due to its property of unique complement.
But here in your question you only have the first premise, thus it may not be unique at all, user @Mark Saving has already given you a counterexample.
