A general Sieve method

Question

I am studying "Sieve Methods" by Halberstam and Richert and I'm kinda stuck. In chapter 2 "The Combinatorial Sieve", in the first chapter about a general Sieve method we want to prove a relation which acts as sort of an improvement of the Eratosthenes-Legendre Sieve, mainly relation (1.8) of the chapter: $$ S(A;B,z)=\sum_{d\mid{P(z)}}\mu(d)\chi(d)|{A_d}|-\sum_{d\mid{P(z)}}\sum_{\substack{p\mid{P(z)}\\ p<q(d)}}\mu(d)({\chi(d)-\chi(pd))}S(A_{pd};B,z)$$ where $\mu(d)$ is the well-known Mobius function, $\chi(d)$ any arithmetic function with $\chi(1)=1$, $$\sigma(n)=\sum_{d\mid{n}}\mu(d)\chi(d)$$ and $S(A;B,z)$ is the standard sieve setting: $A$ is a set of natural numbers, $A_d$ is all the elements in $A$ that are $\equiv0(\text{mod }d)$ and $B$ the set of primes such that we exclude all natural numbers in $A$ that have prime factors in $B$ that are less than $z$. It's easy to see that if we assume relation (1.5) $$S(A;B,z)=\sum_{d\mid{P(z)}}\mu(d)\chi(d)|{A_d}|-\sum_{\substack{d>1\\ d\mid{P(z)}}}\sigma(d)S(A_d;B^{(d)},z)$$ where $B^{(d)}=\{p\in{B}:p\nmid{d}\}$, then we need only prove that $$\sum_{\substack{d>1\\ d\mid{P(z)}}}\sigma(d)S(A_d;B^{(d)},z)=\sum_{d\mid{P(z)}}\sum_{\substack{p\mid{P(z)}\\ p<q(d)}}\mu(d)({\chi(d)-\chi(pd))}S(A_{pd};B,z)$$ The authors provide a proof but it doesn't really make sense for me. Specifically they use the facts that, for a prime $p$ dividing $d$: $$\sigma(d)=\sum_{\tau\mid{d/p}}\mu(\tau)({\chi(\tau)-\chi(p\tau)})$$ and for $2\leq{z_1}\leq{z}$, if $P(z_1,z)=P(z)/P(z_1)$ then $$S(A;B,z_1)=\sum_{\tau\mid{P(z_1,z)}}S(A_{\tau};B^{(\tau)},z)$$ From these two they prove relation (1.8) in a way that makes absolutely no sense to me. I have proven everything mentioned here on my own without issue except for relation (1.8). Can anybody explain how the facts I mentioned are meant to give us the result we want? Any help is appreciated, I've been stuck on this for three days now.