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

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

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

$\sigma(n) = n \cup \{n\}$ 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 $x \subseteq \sigma(n)$.

$\sigma(n) \notin \sigma(n)$: 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 $n \notin n$; 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 $n \notin n$; hence $\sigma(n) \notin \sigma(n)$.

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