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

$\mathbb{N}$ is a transitive set. Let $P(n)$ be "$n\subseteq \mathbb{N}$". $P(0)$ holds because $0=\mathrm{\varnothing }$ 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 $n\mathrm{\ne }\sigma (n)$, 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 $n\mathrm{\ne }p$. Hence $m\mathrm{\ne }p$ 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 $m\mathrm{\ne }n$. Hence $m\mathrm{\ne }p$ 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 $m\mathrm{\ne }0$.

For all $m,n\in \mathbb{N}$: $m<\sigma (n)⟺m\le n$. 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 $0=\mathrm{\varnothing }$. 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 $m<n⟺m\in n$ 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 $n\mathrm{\ne }0$ 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.