On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$

$\mathbb{N}$ is a transitive set. Let $P(n)$ be "$n \subseteq \mathbb{N}$". $P(0)$ holds because $0 = \varnothing$ has no elements. If $n \subseteq \mathbb{N}$ then, since $n$ is itself an element of $\mathbb{N}$, the set $\sigma(n) = n \cup \{n\}$ is also a subset of $\mathbb{N}$, so $P(\sigma(n))$ holds. By induction $n \subseteq \mathbb{N}$ for every $n \in \mathbb{N}$, which is the transitivity of $\mathbb{N}$.

For every $n \in \mathbb{N}$ one has $n \in \sigma(n)$ and $n < \sigma(n)$. Indeed $n \in n \cup \{n\} = \sigma(n)$ directly; and taking $k = \sigma(0)$ gives $n + \sigma(0) = \sigma(n + 0) = \sigma(n)$, so $n \le \sigma(n)$, while $n \ne \sigma(n)$, whence $n < \sigma(n)$.

Mixed transitivity, in both directions. (i) If $m \le n$ and $n < p$ then $m < p$: transitivity of $\le$ gives $m \le p$; if $m = p$ then $p \le n$, and $n \le p$ holds because $n < p$, so antisymmetry gives $n = p$, contradicting $n \ne p$. Hence $m \ne p$ and $m < p$. (ii) If $m < n$ and $n \le p$ then $m < p$: transitivity of $\le$ again gives $m \le p$; if $m = p$ then $n \le p = m$, and $m \le n$ holds because $m < n$, so antisymmetry gives $m = n$, contradicting $m \ne n$. Hence $m \ne p$ and $m < p$.

No natural number satisfies $m < 0$. For every $m$ one has $0 + m = m$, so $0 \le m$; if also $m \le 0$ then antisymmetry gives $m = 0$, and $m < 0$ additionally demands $m \ne 0$.

For all $m, n \in \mathbb{N}$: $m < \sigma(n) \iff m \le n$. If $m \le n$ then, with $n < \sigma(n)$ from step 1.2, step 1.3(i) gives $m < \sigma(n)$. Conversely assume $m < \sigma(n)$ and suppose $m \le n$ fails; then $n < m$ by trichotomy, so $\sigma(n) \le m$ by discreteness, and step 1.3(i) applied to $\sigma(n) \le m$ and $m < \sigma(n)$ gives $\sigma(n) < \sigma(n)$, which irreflexivity forbids. Hence $m \le n$.

Membership implies order: for every $n \in \mathbb{N}$, every $m \in n$ satisfies $m < n$. Let $P(n)$ be that statement; $P(0)$ is vacuous since $0 = \varnothing$. Assume $P(n)$ and let $m \in \sigma(n) = n \cup \{n\}$. If $m \in n$ then $m < n$ by $P(n)$, and $n < \sigma(n)$ by step 1.2, so $m < \sigma(n)$ by step 1.3(i), whose hypothesis $m \le n$ follows from $m < n$. If $m = n$ then $m < \sigma(n)$ by step 1.2. So $P(\sigma(n))$ holds, and by induction $P(n)$ holds for every $n$; the elements $m$ involved are natural numbers by step 1.1, so the statement is about $\mathbb{N}$ throughout.

Order implies membership: for every $n \in \mathbb{N}$, every $m \in \mathbb{N}$ with $m < n$ satisfies $m \in n$. Let $Q(n)$ be that statement; $Q(0)$ holds vacuously by step 1.4. Assume $Q(n)$ and let $m < \sigma(n)$. By step 2.1, $m \le n$, that is $m < n$ or $m = n$. In the first case $m \in n$ by $Q(n)$; in the second $m \in \{n\}$. Either way $m \in n \cup \{n\} = \sigma(n)$, so $Q(\sigma(n))$ holds, and by induction $Q(n)$ holds for every $n$.

Steps 2.2 and 3.1 together give $m < n \iff m \in n$ for all $m, n \in \mathbb{N}$, which is claim 1; and since every element of $n$ is a natural number by step 1.1, this says exactly $n = \{\, m \in \mathbb{N} : m < n \,\}$.

If $m \le n$ then $m \subseteq n$: let $x \in m$; then $x \in \mathbb{N}$ by step 1.1 and $x < m$ by step 4.1, so $x < n$ by step 1.3(ii) applied to $x < m$ and $m \le n$, whence $x \in n$ by step 4.1.

If $m \subseteq n$ then $m \le n$: suppose $m \le n$ fails; then $n < m$ by trichotomy, so $n \in m$ by step 4.1, and $m \subseteq n$ would give $n \in n$, which is impossible. Hence $m \le n$.

For every $n \in \mathbb{N}$ one has $0 \le n$ by step 1.4; if in addition $n \ne 0$ then $0 < n$, hence $0 \in n$ by step 4.1.

The transitivity of $\mathbb{N}$ is step 1.1, claim 1 is step 4.1, claim 2 is steps 5.1 and 5.2 together, claim 3 is steps 1.2 and 2.1, and claim 4 is step 5.3; the description $n = \{\, m \in \mathbb{N} : m < n \,\}$ is part of step 4.1.