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:
 * $(1): \quad 0 \le x < 1$

and
 * $\floor x = 0$

Write:
 * $(2): \quad m = k n + r$

with $k \in \Z$ and $0 \le r \le n - 1$.

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

We have:

as claimed.