Vtyjkyuk

$a,b,c,d,p,q$ are integers such that

$$c=b-aquad,quad 0leq dquad,quad 1leq aleq bleq d+1quad,quad 1leq pquad,quad 1leq q$$

How to argue that the cumbersome condition

$$leftlfloor fracp-1arightrfloor leq leftlfloor fracq+c-1brightrfloorlandundersethspace2.16cm
llapleftlfloor tfracp+0arightrfloor large textbf = leftlfloor tfracq+c+0brightrfloor land leftlfloor tfracp+1arightrfloor large textbf = leftlfloor tfracq+c+1brightrfloor
land largeldots land ,rlapleftlfloor tfracp+d-1arightrfloor large textbf = leftlfloor tfracq+c+d-1brightrfloorundersetunderbracetextequality conditionslandleftlfloor fracp+darightrfloor geq leftlfloor fracq+c+dbrightrfloor$$

is the same as the simpler condition

$$c leftlfloor fracp-1arightrfloor leq q-pleq c leftlfloor fracd+parightrfloor -cland (a=b lor dleq 2 a-(q bmod b))$$

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 $q$.