Changing order of summation with Möbius function

Question

Let $\mu(d)$ be the Möbius function, and $\mu_r(d)$ be the modified Möbius function which satisfies $\mu_r(d)=0$ if $d$ has strictly more than $r$ distinct prime factors. Let $\psi_r(n)=\sum_{d\mid n}\mu_r(d)$. Finally, we let $P_z$ be the product of all primes less than or equal to $z$. Then, what I am humbly requesting help with proving, is $$ \sum_{d\mid P_z}\frac{1}{d}\sum_{\delta\mid d}\mu(d/\delta)\psi_r(\delta) = \sum_{\delta\mid P_z}\frac{\psi_r(\delta)}{\delta}\sum_{d\mid P_z/\delta} \frac{\mu(d)}{d}. $$ Thank you for the help!

Answer 1

Note $\newcommand\1{\mathbf 1}\newcommand\id{\text{id}}$that $\psi_r=\1*\mu_r$, so $\mu*\psi_r=\mu_r$ by Möbius inversion. Therefore the LHS equals $$\begin{align}\sum_{d\mid P_z}\frac1d\cdot\mu_r(d) &=\frac1{P_z}\sum_{d\mid P_z}\mu_r(d)\frac{P_z}d\\ &=\frac1{P_z}(\mu_r*\id)(P_z).\end{align}$$

The RHS is

$$\begin{align}\sum_{\delta\mid P_z}\frac{\psi_r(\delta)}{P_z}\sum_{d\mid P_z/\delta}\mu(d)\frac{\frac{P_z}\delta}d &=\sum_{\delta\mid P_z}\frac{\psi_r(\delta)}{P_z}(\mu*\id)(P_z/\delta)\\ &=\frac1{P_z}(\psi_r*(\mu*\id))(P_z)\\ &=\frac1{P_z}((\psi_r*\mu)*\id)(P_z)\\ &=\frac1{P_z}(\mu_r*\id)(P_z).\end{align}$$

This is exactly the same as the LHS.