Question: 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?

Answer: 
Use the Chinese remainder theorem, $A = A/(\prod \mathfrak m\_i)\simeq A/(\mathfrak m\_1^{n\_1})\times\cdots\times A/(\mathfrak m\_i^{n\_i})$. It's enough to show $A/\mathfrak m\_j^{n\_j}$ is Artinian. (if you assume $\mathfrak m\_i$ are distinct, then $A$ is just a finite product of fields, hence Artinian, without assuming it's already Noetherian.) This follows from $\mathfrak m\_j + \mathfrak m\_j^{n\_j}$ is the only prime ideal, as it consists of only nilpotent elements, so it has Krull dimension $0$.

Note that this is false if $A$ is not assumed to be Noetherian in the first place. Let $A=F[x\_1, x\_2, \cdots]/\mathfrak m^2$ where $\mathfrak m=(x\_1, x\_2, \cdots)$. Then $\overline{\mathfrak m}^2=0$, but $A$ is not Artinian: $(\bar x\_1, \bar x\_2, \bar x\_3, \cdots)\supsetneq (\bar x\_2, \bar x\_3, \cdots)\supsetneq(\bar x\_3, \cdots)\supsetneq \cdots$ form an infinite chain of decreasing ideals of $A$. In particular, $\prod\_{i=1}^{j-1}\mathfrak{m}\_j/\prod\_{i=1}^{j}\mathfrak{m}\_j$ is not finite-dimensional vector spaces over $A/\mathfrak{m}\_j=F$.
We shall not pay attention on refining the chain to be composition chain. We only need to use the factors in the chain. We can prove as follows:

Suppose $A$ is Noetherian. Then it is clear that $A/\mathfrak{m}\_1...\mathfrak{m}\_n$ is Noetherian as for $A$-module. Consider the following exact sequences: $$0\rightarrow \mathfrak{m}\_1...\mathfrak{m}\_{n-1}/\mathfrak{m}\_1...\mathfrak{m}\_n \rightarrow A/\mathfrak{m}\_1...\mathfrak{m}\_n \rightarrow A/\mathfrak{m}\_1...\mathfrak{m}\_{n-1}\rightarrow 0$$
We can deduce that both $\mathfrak{m}\_1...\mathfrak{m}\_{n-1}/\mathfrak{m}\_1...\mathfrak{m}\_n$ and $A/\mathfrak{m}\_1...\mathfrak{m}\_{n-1}$ are Noetherian as for $A$-module. Repeating using this methods, one can find that for any $1\le i\le n-1$, $\mathfrak{m}\_1...\mathfrak{m}\_i/\mathfrak{m}\_1...\mathfrak{m}\_{i+1}$ is Noetherian as for $A$-module, hence it is also Noetherian as for $A/\mathfrak{m}\_{i+1}$-vector space (If $IM=0$, then $A$-module $M$ is equivalent to $A/I$-module $M$.). Hence $\mathfrak{m}\_1...\mathfrak{m}\_i/\mathfrak{m}\_1...\mathfrak{m}\_{i+1}$ is Artinian as for $A/\mathfrak{m}\_{i+1}$-vecor space, so does as for $A$-module. Inductively back to deduce that $A/\mathfrak{m}\_1...\mathfrak{m}\_i$ is Artinian as for $A$-module for all $i$, hence $A$ is Artinian.