Trichotomy of the order on $\mathbb{N}$

At most one of the three holds where equality is involved: $m<n$ forces $m\mathrm{\ne }n$ and $m>n$ forces $n\mathrm{\ne }m$, so $m=n$ is incompatible with either strict relation.

The two strict relations are incompatible: if $m<n$ and $m>n$ then $m+j=n$ and $n+i=m$ with $i,j\mathrm{\ne }0$, so $m+(j+i)=(m+j)+i=n+i=m=m+0$; by commutativity [L7], $(j+i)+m=0+m$, so $j+i=0$ by cancellation [L6]; yet $j=\sigma (a)$ [L4] gives, by the left successor law [L2], $j+i=\sigma (a)+i=\sigma (a+i)\mathrm{\ne }0$ [L5], a contradiction.

For any $x$ and nonzero $t$, one has $x+t\mathrm{\ne }x$: otherwise commutativity [L7] and $0+x=x$ [L1] give $t+x=0+x$, so cancellation [L6] gives $t=0$. Thus $x<x+t$ by the order definition. In particular $n+\sigma (0)=\sigma (n)$ by the recursion, and $\sigma (0)\mathrm{\ne }0$ [L5], so $n<\sigma (n)$.

Base case $n=0$: for arbitrary $m$, either $m=0$, giving $m=n$, or $m\mathrm{\ne }0$ so $0+m=m$ [L1] with $m\mathrm{\ne }0$ gives $0<m$, that is $n<m$; so at least one of $m<n$, $m=n$, $m>n$ holds.

Inductive hypothesis: fix $n$ and assume that for every $m$ at least one of $m<n$, $m=n$, $m>n$ holds.

Successor step: for arbitrary $m$, apply the hypothesis; if $m<n$ then $n=m+k$ with $k\mathrm{\ne }0$, and the recursion gives $m+\sigma (k)=\sigma (m+k)=\sigma (n)$, so $\sigma (n)=m+\sigma (k)$ with $\sigma (k)\mathrm{\ne }0$ [L5], giving $m<\sigma (n)$ by step 1.3; if $m=n$ then $m+\sigma (0)=\sigma (m+0)=\sigma (m)=\sigma (n)$, so $\sigma (n)=m+\sigma (0)$ with $\sigma (0)\mathrm{\ne }0$, giving $m<\sigma (n)$ by step 1.3; if $m>n$ then $m=n+i$ with $i\mathrm{\ne }0$, so $i=\sigma (a)$ [L4], and the recursion gives $n+\sigma (a)=\sigma (n+a)$ while the left successor law [L2] gives $\sigma (n)+a=\sigma (n+a)$, so $m=n+\sigma (a)=\sigma (n+a)=\sigma (n)+a$, whence $a=0$ gives $m=\sigma (n)$ and $a\mathrm{\ne }0$ gives $\sigma (n)<m$ by step 1.3; in every case at least one of $m<\sigma (n)$, $m=\sigma (n)$, $m>\sigma (n)$ holds.

By the induction principle with base 1.4, hypothesis 1.5, and step 2.1, comparability holds for all $m,n$, and with the incompatibilities 1.1 and 1.2 exactly one of $m<n$, $m=n$, $m>n$ holds, so the order is total.