$a,b,c,d,p,q$ are integers such that
$$c=b-a\quad,\quad 0\leq d\quad,\quad 1\leq a\leq b\leq d+1\quad,\quad 1\leq p\quad,\quad 1\leq q$$
How to argue that the cumbersome condition
$$\left\lfloor \frac{p-1}{a}\right\rfloor \leq \left\lfloor \frac{q+c-1}{b}\right\rfloor\land\underset{{\hspace{2.16cm} \llap{\left\lfloor \tfrac{p+0}{a}\right\rfloor {\large \textbf{ = }}\left\lfloor \tfrac{q+c+0}{b}\right\rfloor\ \land\ \left\lfloor \tfrac{p+1}{a}\right\rfloor {\large \textbf{ = }}\left\lfloor \tfrac{q+c+1}{b}\right\rfloor} \ \ \land \ \ {\large\ldots}\ \ \land\ \,\rlap{\left\lfloor \tfrac{p+d-1}{a}\right\rfloor {\large \textbf{ = }}\left\lfloor \tfrac{q+c+d-1}{b}\right\rfloor}}}{\underset{}{\underbrace{\text{equality conditions}}}}\land\left\lfloor \frac{p+d}{a}\right\rfloor \geq \left\lfloor \frac{q+c+d}{b}\right\rfloor$$
is the same as the simpler condition
$$c \left\lfloor \frac{p-1}{a}\right\rfloor \leq q-p\leq c \left\lfloor \frac{d+p}{a}\right\rfloor -c\land (a=b \lor d\leq 2 a-(q \bmod b))$$
I wanted to do it by induction wrt. $d$, but I can't see how to do it even when $d=0$. Any ideas?
I think the $d=0$ case should be fine: In this case, the equality conditions are void, and we have $c=d=0$ as well as $a=b=1$. But then the left-hand side of the equivalence is $\lfloor p-1 \rfloor \le \lfloor q-1 \rfloor \land \lfloor p \rfloor \ge \lfloor q \rfloor$, where we may drop the floor brackets since $p$ and $q$ are integers. Hence the left-hand side just says that $p=q$. For the right-hand side one obtains $0 \le q-p \le 0 \land \big( a=b \lor 0 \le -(q \bmod 1) \big)$, whose left conjunct simplifies to $p=q$ while the right conjunct is trivial since we have $a=b=1$ and $q \bmod 1 = 0$ for any positive integer $q$.