Boolean algebras- question

Question

Let B be a Boolean algebra. Let a,b be elements of the universe of B such that $\neg a$ < 1 and b < 1 and $(\neg a \vee b)= 1$. Is it always the case that $a \leq b$? If not could you provide a counterexample?

Answer 1

Yes, $\neg a \vee b = 1$ implies $a \leq b$ in any Boolean algebra. This does not even use the additional assumptions that $\neg a < 1$ and $b < 1$. To see this, we can follow Rushy's suggestion in the comments: $$ a = 1 \wedge a = (\neg a \vee b) \wedge a = (\neg a \wedge a) \vee (a \wedge b) = 0 \vee (a \wedge b) = a \wedge b. $$ So because $a = a \wedge b$ we conclude that indeed $a \leq b$.

There is actually a good intuitive explanation for this as well. If we think about propositional logic then the statement "$a$ implies $b$", usually written as $a \to b$, is logically equivalent to $\neg a \vee b$. We can view a Boolean algebra as a set of truth values. Obviously $\wedge$ and $\vee$ then represent the 'and' and 'or' operations respectively, while $1$ and $0$ give us 'true' and 'false'. The interpretation of $a \leq b$ is "$a$ implies $b$". So what we have just seen is that also in Boolean algebras we have that when $\neg a \vee b$ is true then "$a$ implies $b$" (i.e. $a \leq b$).

In fact, the converse is also true: if $a \leq b$ then $\neg a \vee b = 1$. This follows because $a \leq b$ implies $\neg b \leq \neg a$ and since $\neg b \vee b = 1$ we then have by monotonicity of $\vee$ that indeed $\neg a \vee b = 1$.