Equivalence of Definitions of Ordering on Integers

Proof
Let $x, y \in \Z$ such that $x \le y$.

Let $x$ and $y$ be defined as from the formal definition of integers:


 * $x = \eqclass {x_1, x_2} {}$ and $y = \eqclass {y_1, y_2} {}$ where $x_1, x_2, y_1, y_2 \in \N$.

$(1)$ implies $(2)$
Let $\le$ be an ordering on integers by definition $1$.

Then by definition:
 * $y - x$ is non-negative

That is:
 * $\map \PP {y - x}$

where $\PP$ is the positivity property.

Thus:
 * $\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is non-negative

By definition of integer subtraction:


 * $\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is non-negative

and by the formal definition of integers:


 * $\eqclass {y_1 + x_2, y_2 + x_1} {}$ is non-negative

We have that $y_1 + x_2$ and $y_2 + x_1$ are natural numbers.

Thus by definition of natural number ordering:


 * $y_1 + x_2 \ge y_2 + x_1$

Thus $\le$ is an ordering on integers by definition $2$.

$(2)$ implies $(1)$
Let $\le$ be an ordering on integers by definition $2$.

Then by definition:
 * $x_1 + y_2 \le x_2 + y_1$

That is:
 * $x_2 + y_1 \ge x_1 + y_2$

Hence by the formal definition of integers:


 * $\eqclass {y_1 + x_2, y_2 + x_1} {}$ is non-negative

By definition of integer addition:


 * $\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is non-negative

By definition of integer subtraction:


 * $\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is non-negative

That is:
 * $\map \PP {y - x}$

where $\PP$ is the positivity property.

Thus $\le$ is an ordering on integers by definition $1$.