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?