Let $(\mathbb{B},\wedge,\vee,-,0,1,\leq)$ be a Boolean algebra. I wish to prove the following:
Claim: $a$ is an atom if and only if for each $x\in\mathbb{B}$ either $a\leq -x$, or $a\leq x$ (exclusively)
I managed to prove the $\Rightarrow$ part: Assume that both $a\leq -x$ and $a\leq x$ hold, then $a\leq x\wedge(-x)=0$, so $a\leq 0$, but then $a=0$, so $a$ is not an atom.
I am stuck on proving the second part, can someone help?
Assume that $a$ is not an atom. As you observed, $a=0$ is no good, so the indirect assumption means that there exists some $b$ with $0<b<a$. Clearly, $a\leq b$ is false. If $a\leq -b$, then $b=b\wedge a\leq b\wedge -b=0$, a contradiction with $x=b$.