$\omega$ is the least limit ordinal

For all $m, n \in \mathbb{N}$, $m \le n$ if and only if $m < \sigma(n)$: from $m \le n$ and $n < \sigma(n)$ ([L9]) transitivity gives $m < \sigma(n)$; conversely if $m < \sigma(n)$ and $m \le n$ failed, then $n < m$ by trichotomy, so $\sigma(n) \le m < \sigma(n)$ by [L3], which is impossible.

$\omega$ is a transitive set: the set $S = \{n \in \omega : n \subseteq \omega\}$ contains $0 = \varnothing$ and is closed under $\sigma$, since $n \subseteq \omega$ together with $n \in \omega$ gives $\sigma(n) = n \cup \{n\} \subseteq \omega$; so $S = \omega$ by [L2].

Claim (i): the set $T = \{n \in \omega : m \in n \iff m < n \text{ for every } m \in \omega\}$ contains $0$, because $m \in \varnothing$ is false and $m < 0$ is false, since $0 \le m$ always and $m < 0$ would give $m = 0$ by antisymmetry and then $m < m$; and $n \in T$ gives $\sigma(n) \in T$, because $m \in \sigma(n) \iff (m \in n \text{ or } m = n) \iff (m < n \text{ or } m = n) \iff m \le n \iff m < \sigma(n)$; hence $T = \omega$ by [L2].

Claim (ii) for natural numbers: fix $n \in \omega$; then $n$ is a transitive set by [L6], its elements are natural numbers by step 1.2, and on them membership is the strict order by step 2.1, so $\in$ is irreflexive, transitive and trichotomous on $n$ by [L4] and every nonempty subset of $n$ has an $\in$-least element by [L5]; hence $n$ is an ordinal.

Claim (ii) for $\omega$: $\omega$ is a transitive set by step 1.2 and membership is the strict order on it by step 2.1, so the same four properties hold by [L4] and [L5]; hence $\omega$ is an ordinal.

Claim (iii): $\omega \ne 0$ because $0 \in \omega$; and $\omega$ is not a successor ordinal, since $\omega = \beta^{+}$ would give $\beta \in \omega$ and hence $\sigma(\beta) = \beta^{+} = \omega \in \omega$ because $\omega$ is inductive, contradicting the fact that no ordinal is a member of itself; so $\omega$ is a limit ordinal.

Claim (iv), first half: the ordinals $\alpha$ with $\alpha \in \omega$ are exactly the natural numbers, each of which is $0$ or of the form $\sigma(m) = m^{+}$ with $m$ a natural number by [L7], hence $0$ or a successor ordinal; so no ordinal $\in$-below $\omega$ is a limit ordinal.

Claim (iv), second half, which is where "least" is more than $\in$-minimality: let $\gamma$ be any limit ordinal; $\omega$ is an ordinal by step 3.2, so by comparability of ordinals under inclusion [L10] either $\omega \subseteq \gamma$ or $\gamma \subseteq \omega$, and in the second case [L10] gives $\gamma \in \omega$ or $\gamma = \omega$; but $\gamma \in \omega$ would make $\gamma$ equal to $0$ or to a successor ordinal by step 5.1, contradicting the definition of a limit ordinal in [L8], so $\gamma = \omega$ and $\omega \subseteq \gamma$ again; hence $\omega \subseteq \gamma$, that is $\omega \le \gamma$ in the ordering of [L10], for every limit ordinal $\gamma$, and since $\omega$ is itself a limit ordinal by step 4.1 it is the least one.

Claims (i) to (iv) are established.