Parity Addition is Commutative/Proof 2

Theorem
Let $R := \left({\left\{ {\text{even}, \text{odd} }\right\}, +, \times}\right)$ be the parity ring.

The operation $+$ is commutative:


 * $\forall a, b \in R: a + b = b + a$

Proof
Let $a, b \in R$.

That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.

By definition of odd:
 * $\text{odd} = 2 m + 1$

for some $m \in \Z$.

By definition of even:
 * $\text{even} = 2 n + 0$

for some $n \in \Z$.

Thus we can define the mapping $f: R \to \Z$ as:
 * $\forall x \in R: f \left({x}\right) := \begin{cases}

0 & : x \text { is even} \\ 1 & : x \text { is odd} \end{cases}$

Thus an element of $R$ can be expressed as an arbitrary integer of the form:
 * $x = 2 k + f \left({x}\right)$

where:
 * $k \in \Z$ is an integer
 * $f \left({x}\right)$ is either $0$ or $1$ according to whether $x$ is even or odd.

Let $+_2$ be used to denote the operation of $+$ in $R$, that is, the addition of two parities.

Then:

The result follows by the identification of $+_2$ to be used to denote the operation of $+$ in $R$:
 * $a + b = b + a$