Understanding the selection of $r$ as $r=q\wedge\neg p$ in a proof that $q$ is an atom My question comes from a part of a proof of the following Lemma:


If $p$ and $q$ are elements of a Boolean algebra such that $p\lt q$ and every $r\lt q$ satisfies $r\leq p$, then $p=0$ and $q$ is an atom.


In the ensuing discussion, an $r$ is selected, $r=q\wedge\neg p$, that is intended to be at most $q$ and unlikely to be at most $p$.
My questions pertain to this $r$.
I can see that as the greatest lower bound, it is at most $q$.
But what does it mean for $r$ as a $glb$ to be less than or equal to $\neg p$? 
And how is this "unlikely to be at most $p$"?
Lastly, in view $r\leq p$ in the hypothesis, (although the subsequent proof works nicely with $r$ as mentioned) why is this $r$ acceptable as compared to $r= q\wedge p$?
Thanks
 A: Since $p\wedge \neg p=0$, the only element that is less than or equal to both $p$ and $\neg p$ is $0$.  So if you want to find an element that is "unlikely" to be less than $p$, an obvious choice is an element which is less than or equal to $\neg p$.
The goal here is to extract as much information as possible from the hypothesis that "every $r<q$ satisfies $r\leq p$".  So, of all the possible choices of $r$, we want to choose one for which it is very difficult for it to be less than or equal to $p$.  That way, the fact that $r\leq p$ will tell us a lot of information.  If you chose $r=p\wedge q$ instead, you would get no information at all, since then $r\leq p$ would be true by definition of $r$.
Alternatively, this argument is very natural if you consider the special case that your Boolean algebra is the power set of some set.  In that case, we have a set $Q$ and a proper subset $P\subset Q$ such that every proper subset $R\subset Q$ is contained in $P$.  These hypotheses should sound fishy: what if you chose $R$ to consist of elements of $Q$ that are not in $P$?  Such an $R$ would be disjoint from $P$, so it can't be contained in $P$ unless it is empty.  So it is very natural here to consider the set $R=Q\setminus P$, which corresponds to $r=q\wedge \neg p$ in a general Boolean algebra.
