The well-ordering principle

Suppose $S$ is nonempty but has no least element.

If $m < \sigma(n)$ then $m \le n$: by [L2] applied to $m$ and $\sigma(n)$ we get $\sigma(m) \le \sigma(n)$, say $\sigma(m) + i = \sigma(n)$; then $\sigma(m + i) = \sigma(m) + i = \sigma(n)$ [L5], so $m + i = n$ by injectivity [L4], hence $m \le n$.

First, $0 \in T$: if $0 \in S$ then $0 \le s$ for all $s \in S$, since $0 + s = s$ [L6], making $0$ a least element, contrary to the assumption; hence $0 \notin S$; moreover $m \le 0$ forces $m = 0$, because $m \ne 0$ would give $0 < m$ (as $0 + m = m$ [L6] with $m \ne 0$) while $m \le 0$ gives $m < 0$, and [L3] admits at most one of these; so every $m \le 0$ lies outside $S$.

Next, $T$ is closed under $\sigma$: assume $n \in T$; if $\sigma(n) \in S$ then no $s \in S$ satisfies $s < \sigma(n)$ (else $s \le n$ by step 1.2, so $s \notin S$ because $n \in T$, a contradiction), hence $\sigma(n) \le s$ for all $s \in S$ by [L3], making $\sigma(n)$ a least element, contrary to the assumption; therefore $\sigma(n) \notin S$, and any $m$ with $m \le \sigma(n)$ satisfies $m = \sigma(n)$ or $m \ne \sigma(n)$; in the second case $m \le \sigma(n)$ together with $m \ne \sigma(n)$ gives $m < \sigma(n)$ by the definition of the strict order, hence $m \le n$ by step 1.2 and $m \notin S$ because $n \in T$, while in the first case $m = \sigma(n) \notin S$; so every $m \le \sigma(n)$ lies outside $S$, giving $\sigma(n) \in T$.

By the induction principle [L1], from $0 \in T$ and closure under $\sigma$, $T = \mathbb{N}$; then for every $n$, taking $m = n \le n$ shows $n \notin S$, so $S = \varnothing$, contradicting nonemptiness; therefore $S$ has a least element.