Let $B$ be a finite boolean algebra.
Define for $a,b\in B$ $a\leq b$ if $ab=a$
If $x\in B$ and $a_1,\dots,a_k$ are the atoms of B (e.g. $a\neq 0$ and if $b\in B$ such that $0\leq b \leq a$ then $b=0$ or $b=a$) such that $\forall i$ $a_i\leq x$
Could anyone help me prove that $x=a_1+\dots +a_k$
So far I have
$(a_1+\dots +a_k)^2=a_1+\dots +a_k=a_1x+\dots +a_kx=(a_1+\dots +a_k)x$
Therefore $a_1+\dots +a_k\neq 0 \Rightarrow x=a_1+\dots +a_k$
Then I went to consider the case of $a_1+\dots +a_k= 0$ to show that $x=0$. Could anyone help with this second case? Or if i'm completely wrong help show me the correct way?
Addressing Michael Hardy's comment, note that the Boolean algebra is finite and hence atomic.
Let $a_1,\dots,a_k$ be the atoms below $x$. Note that since the BA is finite, there are only finitely many atoms below $x$. We have $a_1+\dots+a_k\leq x$. Now consider $b=x-(a_1+\dots+a_k)$. If $b=0$, then $x=a_1+\dots+a_k$. Otherwise there is an atom $a\leq b$. Now $a$ is disjoint and in particular different from $a_1,\dots,a_k$. But $a\leq b\leq x$, contradicting the fact that $a_1,\dots,a_k$ lists all atoms below $x$.
Does this help? Adressing your question more specifically, if $x=0$, then there is no atom below $x$ and the sum taken over an empty subset of a BA is usually defined to be $0$. This also follows automatically if you define the sum over a subset of a BA to be the least common upper bound of the set.