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