<s>[INST] Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$$
for all $n,m \in \mathbb{Z}$, $m>0$.
This is what I came up with. Is my proof correct?
Proof:

[see edit below]
Case 1: $m=1$ $$\left\lfloor \frac{n}{1} \right\rfloor = \lfloor n \rfloor = n$$ $$\left\lceil \frac{n-1+1}{1} \right\rceil = \lceil n \rceil = n$$
Case 2: $m>1$

If $\frac{n}{m}$ is an integer, then
$$\left\lceil \frac{n-m+1}{m} \right\rceil = \left\lceil \frac{n}{m} -1 + \frac{1}{m}\right\rceil = \frac{n}{m} -1 + \left\lceil \frac{1}{m} \right\rceil = \frac{n}{m} -1 + 1 = \frac{n}{m} = \left\lfloor \frac{n}{m} \right\rfloor$$
If $\frac{n}{m}$ is NOT an integer, then
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} \right\rceil - 1 = \left\lceil \frac{n}{m} + \frac{1}{m} -1 \right\rceil = \left\lceil \frac{n-m+1}{m} \right\rceil$$
$\blacksquare$
If it's correct but you know a simpler/better way to prove it, please include that in your answer. Thank you.

EDIT: As pointed out by user peterwhy, "Case 1: $m=1$" is simply a special case of "$\frac{n}{m}$ is an integer" and therefore is not needed; hence I have grayed it out, and we don't need to separate the $m=1$ and $m>1$ cases anymore.
 [/INST] Another succinct proof is
$$\left\lfloor\frac nm\right\rfloor = \left\lfloor\frac{n + \frac12}m\right\rfloor = \left\lceil\frac{n - m + \frac12}m\right\rceil = \left\lceil\frac{n - m + 1}m\right\rceil,$$
since $\frac{n + \frac12}m$ and $\frac{n - m + \frac12}m$ are always non-integers with a difference of $1$.
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