Every natural number is a transitive set and is not a member of itself

Base $P(0)$: $0=\mathrm{\varnothing }$ has no members, so it is vacuously transitive and $\mathrm{\varnothing }\mathrm{\notin }\mathrm{\varnothing }$.

Inductive hypothesis: $P(n)$ holds, i.e. $n$ is transitive and $n\mathrm{\notin }n$.

$\sigma (n)=n\cup \{n\}$ is transitive: take $x\in \sigma (n)$; if $x\in n$ then $x\subseteq n\subseteq \sigma (n)$ by the inductive hypothesis transitivity; if $x=n$ then $x=n\subseteq \sigma (n)$ since $n\subseteq n\cup \{n\}$; either way $x\subseteq \sigma (n)$.

$\sigma (n)\mathrm{\notin }\sigma (n)$: suppose $\sigma (n)\in \sigma (n)=n\cup \{n\}$, so $\sigma (n)\in n$ or $\sigma (n)=n$; if $\sigma (n)=n$ then $n\in n$ (as $n\in \{n\}\subseteq \sigma (n)=n$), contradicting the inductive hypothesis $n\mathrm{\notin }n$; if $\sigma (n)\in n$ then $\sigma (n)\subseteq n$ by transitivity of $n$, and since $n\in n\cup \{n\}=\sigma (n)\subseteq n$ we get $n\in n$, again contradicting $n\mathrm{\notin }n$; hence $\sigma (n)\mathrm{\notin }\sigma (n)$.

Hence $P(\sigma (n))$ holds; by induction [L1], $P(n)$ for all $n\in \mathbb{N}$: every natural number is transitive and satisfies $n\mathrm{\notin }n$.