I saw a related answer here, but was unable to figure out what's going on.
If for all $x \in B$, either $a \leq x$ or $a \leq x^c$ then $a$ is an atom The converse holds as well. $B$ is a boolean algebra.
I was able to prove the converse. Here's how:
Consider $a \land x$, for $x \in B$ where $a$ is an atom. We know that $a \land x \leq a$ by definition of $\land$. Since $a$ is an atom, it is the minimal element in $B\backslash\{0\}$, so $a \land x \leq a$ means either $a \land x = 0$ or $a \land x = a$. From $a \land x = a$, we get (i) $a \leq x$ (since the g.l.b of $a$ and $x$ is $a)$ and from $a \land x = 0$ we get $a \leq x^c$. Is this correct?
To prove that: if for all $x \in B$, either $a \leq x$ or $a \leq x^c$ then $a$ is an atom, I'm trying to retrace my steps back in the above proof.
If (i) $a \leq x^c$, then $a \land x = 0$ & if (ii) $a \leq x$ then $a \land x = a$. I'm kinda stuck here, it seems.
Depending on your axioms for Boolean algebras, your proof of the converse may be complete or not.
Perhaps it is convenient to show that $a \wedge x = 0$ implies that $a \leq x'$: if $a \wedge x = 0$, then \begin{align} a &= a \wedge 1 = a \wedge 0' = a \wedge (a \wedge x)'\tag{by hypothesis}\\ &= a \wedge (a' \vee x') = (a \wedge a') \vee (a \wedge x')\tag{by distributivity}\\ &= 0 \vee (a \wedge x') = a \wedge x', \end{align} whence $a \leq x'$.
For the direct implication, suppose that $a \in B$ is such that for every $x \in B$ we have that $a \leq x$ or $a \leq x'$.
In order to prove that $a$ is an atom, take $b \in B$ such that $0 \leq b < a$.
It is enough to show that $b=0$.
As $b \in B$, it must be $a \leq b$ or $a \leq b'$ (because that's our hypothesis about $a$); but $b < a$ so $a \nleq b$, and hence $a \leq b'$. Now it follows that $$b = b \wedge a \leq b \wedge b' = 0,$$ and therefore $b=0$.
We conclude that $a$ is an atom.