$\le$ 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 $j \ne 0$ then $j = \sigma(i)$ and $k + j = k + \sigma(i) = \sigma(k + i) \ne 0$ [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 $\mathbb{N}$.