The Integers Form an Ordered Ring Containing the Natural Numbers as Its Positive Elements

In the setting of The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion, let $\mathbb{Z}$, $[a,b]$, its operations and order, $0_{\mathbb{Z}}$, $1_{\mathbb{Z}}$ and $\iota$ be as in The Integers §integers, The Integers §operations, The Integers §constants and The Integers §embedding. Let $x,y,z\in\mathbb{Z}$ and $a,b\in\mathbb{N}$.

¶ $\mathbb{Z}$, with $+$, $\cdot$, $0_{\mathbb{Z}}$ and $1_{\mathbb{Z}}$, is a commutative ring, and $x+(-x)=0_{\mathbb{Z}}$.

¶ $\le$ is a total order on $\mathbb{Z}$, with $<$ as its strict relation, and $\mathbb{Z}$, with $+$, $\cdot$, $0_{\mathbb{Z}}$, $1_{\mathbb{Z}}$ and $\le$, is an ordered ring.

¶ $-(-x)=x$, $0_{\mathbb{Z}}\cdot x=0_{\mathbb{Z}}$ and $(-x)\cdot y=-(x\cdot y)$.

¶ $\iota$ is injective, $\iota(1)=1_{\mathbb{Z}}$, $\iota(a+b)=\iota(a)+\iota(b)$, $\iota(ab)=\iota(a)\,\iota(b)$, and $a<b$ if and only if $\iota(a)<\iota(b)$.

¶ $[a,b]=\iota(a)-\iota(b)$.

¶ Exactly one of the following holds: $x=\iota(n)$ for some $n\in\mathbb{N}$; $x=0_{\mathbb{Z}}$; $x=-\iota(n)$ for some $n\in\mathbb{N}$.

¶ $0_{\mathbb{Z}}<x$ if and only if $x=\iota(n)$ for some $n\in\mathbb{N}$.

¶ If $xy=0_{\mathbb{Z}}$, then $x=0_{\mathbb{Z}}$ or $y=0_{\mathbb{Z}}$.

¶ If $z\neq0_{\mathbb{Z}}$ and $xz=yz$, then $x=y$.

¶ If $x<y$ and $0_{\mathbb{Z}}<z$, then $xz<yz$.

¶ If $0_{\mathbb{Z}}<z$, then $x\le y$ if and only if $xz\le yz$.