Modulo Addition is Well-Defined/Proof 1

Proof
We need to show that if:


 * $\left[\!\left[{x'}\right]\!\right]_m = \left[\!\left[{x}\right]\!\right]_m$
 * $\left[\!\left[{y'}\right]\!\right]_m = \left[\!\left[{y}\right]\!\right]_m$

then:
 * $\left[\!\left[{x' + y'}\right]\!\right]_m = \left[\!\left[{x + y}\right]\!\right]_m$

Since:
 * $\left[\!\left[{x'}\right]\!\right]_m = \left[\!\left[{x}\right]\!\right]_m$

and:
 * $\left[\!\left[{y'}\right]\!\right]_m = \left[\!\left[{y}\right]\!\right]_m$

it follows from the definition of set of integers modulo $m$ that:
 * $x \equiv x' \pmod m$

and:
 * $y \equiv y' \pmod m$

By definition, we have:


 * $x \equiv x' \pmod m \implies \exists k_1 \in \Z: x = x' + k_1 m$
 * $y \equiv y' \pmod m \implies \exists k_2 \in \Z: y = y' + k_2 m$

which gives us:
 * $x + y = x' + k_1 m + y' + k_2 m = x' + y' + \left({k_1 + k_2}\right) m$

As $k_1 + k_2$ is an integer, it follows that, by definition:
 * $x + y \equiv \left({x' + y'}\right) \pmod m$

Therefore, by the definition of integers modulo $m$:
 * $\left[\!\left[{x' + y'}\right]\!\right]_m = \left[\!\left[{x + y}\right]\!\right]_m$