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

At most one of the three holds where equality is involved: $m < n$ forces $m \ne n$ and $m > n$ forces $n \ne m$, 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 $i, j \ne 0$, 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], $j + i = \sigma(a) + i = \sigma(a + i) \ne 0$ [L5], a contradiction.

For every $n$ one has $n < \sigma(n)$, since $n + \sigma(0) = \sigma(n + 0) = \sigma(n)$ by the recursion, with $\sigma(0) \ne 0$ [L5].

Base case $n = 0$: for arbitrary $m$, either $m = 0$, giving $m = n$, or $m \ne 0$ so $0 + m = m$ [L1] with $m \ne 0$ 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 $k \ne 0$, and the recursion gives $m + \sigma(k) = \sigma(m + k) = \sigma(n)$, so $\sigma(n) = m + \sigma(k)$ with $\sigma(k) \ne 0$ [L5], giving $m < \sigma(n)$; if $m = n$ then $m + \sigma(0) = \sigma(m + 0) = \sigma(m) = \sigma(n)$, so $\sigma(n) = m + \sigma(0)$ with $\sigma(0) \ne 0$, giving $m < \sigma(n)$; if $m > n$ then $m = n + i$ with $i \ne 0$, 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 $a \ne 0$ gives $\sigma(n) < m$; 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.