Floor of x+m over n/Proof 1

Proof
Let $y=x-\floor x$ and $M=m+\floor x$.

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

and
 * $\floor y = 0$

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

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

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

We have:

Substituting $y$ and $M$, we obtain $\floor{\dfrac{x-\floor x+m+\floor x}n}=\floor{\dfrac{x+m}n}=\floor{\dfrac{m+\floor x}n}$ as claimed.