# How do you determine the number of cases you need to check for floor/ceiling function proofs?

Especially when you have more than 2 variables and those terms are divided by an integer n within the ceiling/floor functions.

-

A partial answer to get an upper bound for individual floor/ceiling function.
There may be better solutions.

Due to the property of the floor function, you can always take out integers.
$\lfloor x+n\rfloor=\lfloor x\rfloor +n$, if $x$ is a real number and $n$ any integer.

This means for $\lfloor \dfrac{v_1+\dots+v_k}{n}\rfloor$, an upper bound is to check all $0\leq v_i < n$ combinations.
Any time $v_i=an+b$, you may take out $a$ and leave $b$ inside as a fraction.

Suppose you have $\lfloor\dfrac{v_1v_2\dots v_k}{n}\rfloor$.
Then you can write it as:
$\lfloor\dfrac{(an+b)v_2\dots v_k}{n}\rfloor$
$=\lfloor (av_2\dots v_k)+\dfrac{bv_2\dots v_k}{n}\rfloor$
$=(av_2\dots v_k)+\lfloor \dfrac{bv_2\dots v_k}{n}\rfloor$

and once again you may assume all the variables $0\leq v_i< n$.
This gives a maximum of $n^k$ combinations to be checked within each ceiling/floor function.

But in actual situation/working you can usually do better:
1) Some $v_i$ might not have the full range.
2) After setting some $v_i$, the remaining $v_i$ might not matter.
3) Due to interaction of different floor functions, some cases may be combined/simplified.

-