$\le$ is a linear order on $\mathbb{N}$

Reflexive: $m+0=m$, so $m\le m$.

Transitive: if $m\le n$ and $n\le p$, say $m+k=n$ and $n+j=p$, then $m+(k+j)=(m+k)+j=n+j=p$, so $m\le p$.

A sum is zero only if both summands are: if $k+j=0$ with $j\mathrm{\ne }0$ then $j=\sigma (i)$ and $k+j=k+\sigma (i)=\sigma (k+i)\mathrm{\ne }0$ [L6], a contradiction, so $k=j=0$.

Total: by trichotomy one of $m<n$, $m=n$, $m>n$ holds, and each of these gives $m\le n$ or $n\le m$.

Antisymmetric: if $m\le n$ and $n\le m$, say $m+k=n$ and $n+j=m$, then $m+(k+j)=(m+k)+j=n+j=m=m+0$; commuting both sides [L5] gives $(k+j)+m=0+m$, and cancelling the common right summand $m$ [L2] gives $k+j=0$, hence $k=j=0$ by step 1.3 and $m=n$.

Reflexivity, antisymmetry, transitivity, and totality all hold, so $\le$ is a linear order on $\mathbb{N}$.