If
then
. Thus, if
_j\mid m\right\}" alt="A_{j,n}=\left\{m\leqslant n
_j\mid m\right\}" /> we see that
where the square cup just means that
forms a partition of
. Thus,
. But, by the inclusion exclusion principle
where
. Thus, noticing that any subcollection of
is a family of coprime numbers we see that
_s\mid m\text{ for all }s\in S\right\}=\left\{m\leqslant n:\prod_{s\in S}p_s\mid m\right\}" alt="\displaystyle \bigcap_{s\in S}A_{s,n}=\left\{m\leqslant n
_s\mid m\text{ for all }s\in S\right\}=\left\{m\leqslant n:\prod_{s\in S}p_s\mid m\right\}" />
From where simple reasoning leads us to
Thus,
From where it quickly follows that
At least we know now that