The integers form a totally ordered ring

Suppose $(a,b)\sim (a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ and $(c,d)\sim (c^{\mathrm{\prime }},d^{\mathrm{\prime }})$, i.e. $a+b^{\mathrm{\prime }}=b+a^{\mathrm{\prime }}$ and $c+d^{\mathrm{\prime }}=d+c^{\mathrm{\prime }}$.

Suppose $a+d\le b+c$.

Totality: in $\mathbb{N}$ either $a+d\le b+c$ or $b+c\le a+d$, so $x\le y$ or $y\le x$.

Antisymmetry: $x\le y$ and $y\le x$ give $a+d\le b+c$ and $b+c\le a+d$, hence $a+d=b+c$, which is $x=y$ as classes.

Transitivity: from $a+d\le b+c$ and $c+f\le d+e$, adding gives $(a+d)+(c+f)\le (b+c)+(d+e)$; regrouping, $(a+f)+(c+d)\le (b+e)+(c+d)$; cancelling, $a+f\le b+e$, i.e. $x\le z$.

Compatibility with addition: $x+z\le y+z$ reads $(a+e)+(d+f)\le (b+f)+(c+e)$, which regroups to $(a+d)+(e+f)\le (b+c)+(e+f)$, equivalent to $a+d\le b+c$, i.e. $x\le y$.

Suppose $0<x$ and $0<y$, i.e. $b<a$ and $d<c$.

Adding $(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})$: $(a+d)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})\le (b+c)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})$.

Regrouping with the equations of step 1.1: $(b+c)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})=(b+a^{\mathrm{\prime }})+(c+d^{\mathrm{\prime }})=(a+b^{\mathrm{\prime }})+(d+c^{\mathrm{\prime }})=(a+d)+(b^{\mathrm{\prime }}+c^{\mathrm{\prime }})$.

Write $a=b+k$ and $c=d+m$ with nonzero naturals $k,m$.

So $(a+d)+(a^{\mathrm{\prime }}+d^{\mathrm{\prime }})\le (a+d)+(b^{\mathrm{\prime }}+c^{\mathrm{\prime }})$; cancelling $(a+d)$ gives $a^{\mathrm{\prime }}+d^{\mathrm{\prime }}\le b^{\mathrm{\prime }}+c^{\mathrm{\prime }}$. Applying the same argument with primed and unprimed representatives exchanged proves the converse implication, so the relation is independent of representatives.

Expanding both components of $xy=[(ac+bd,ad+bc)]$: $ac+bd=(b+k)(d+m)+bd=(bd+bm+kd+bd)+km$ and $ad+bc=(b+k)d+b(d+m)=bd+kd+bd+bm$, so $ac+bd=(ad+bc)+km$.

Since $k,m$ are nonzero, $0<k$ and $0<m$ by [L4]. Hence [L5] gives $0=0\cdot m<km$, and step 3.2 yields $ad+bc<ac+bd$, which is $0<xy$.

The order is well defined, total, compatible with addition, and positives are closed under multiplication; together with [L6], this makes $\mathbb{Z}$ a totally ordered commutative ring.