Floor of x+m over n/Proof 1

Proof
First of all, subtract $\floor{x}$ from $x$ and add $\floor{x}$ to $m$.

This does not change either side of the claimed equality.

We now have
 * $(1): 0\le x<1$

and
 * $\floor{x}=0$.

Write
 * $(2): m=kn+r$ with $k\in\Z$ and $0\le r\le n-1$.

By $(1)$ and $(2)$:
 * $(3): 0\le r+x < n-1 + 1 = n$.

We have

as claimed.