Let $\Phi(x,w) = $ the number integers $i$ where $i \le x$ and $\gcd(i,w)=1$

Let $a, n > 1, w$ be integers.

Does it follow that $\Phi(an+n,w) - \Phi(an,w) \ge \sum\limits_{d|w}\lfloor\frac{\mu(d)n}{d}\rfloor$

Here 's my reasoning:

By the Inclusion-Exclusion Principle:

$$\Phi(an+n,w) - \Phi(an,w) = \sum\limits_{d | w}\mu(d)\left\lfloor\frac{an+n}{d}\right\rfloor - \sum\limits_{d | w}\mu(d)\left\lfloor\frac{an}{d}\right\rfloor$$

$$= \sum\limits_{d | w}\mu(d)\left(\left\lfloor\frac{an+n}{d}\right\rfloor - \left\lfloor\frac{an}{d}\right\rfloor\right)$$

$$= \sum\limits_{d | w \text{ and }\mu(d)=1}\left(\left\lfloor\frac{an+n}{d}\right\rfloor - \left\lfloor\frac{an}{d}\right\rfloor\right) + \sum\limits_{d | w \text{ and }\mu(d)=-1}\left(\left\lfloor\frac{an}{d}\right\rfloor - \left\lfloor\frac{an+n}{d}\right\rfloor\right) \ge$$

$$\sum\limits_{d | w \text{ and }\mu(d)=1}\left(\left\lfloor\frac{n}{d}\right\rfloor\right) + \sum\limits_{d | w \text{ and }\mu(d)=-1}\left(\left\lfloor\frac{-n}{d}\right\rfloor\right) =\sum\limits_{d | w }\left(\left\lfloor\frac{\mu(d)n}{d}\right\rfloor\right) $$

Is this correct?

Edit 1: Thanks to Gerhard, I've fixed the mistake. I believe that it is now correct. I had overlooked that for $0 < a < 1$, $-\lfloor{a}\rfloor \ne \lfloor{-a}\rfloor$