User blog comment:Alemagno12/Huge Ordinal Analysis I: Z 2/@comment-1605058-20180204192308/@comment-1605058-20180204201448

I assume here by < you mean the usual ordering. Take a recursive formula \(\varphi(n)\). Suppose \(\varphi(0)\) and \(\forall x:(\forall y<x:\varphi(x))\implies\varphi(y)\). Using a comprehension axiom, let \(S\) be the set of \(n\) such that \(\forall m\leq n:\varphi(m)\).

We assume \(\varphi(0)\), so \(\forall m\leq 0:\varphi(m)\), i.e. \(0\in S\). Suppose \(n\in S\), i.e. \(\forall m\leq n:\varphi(m)\), so \(\forall m<n+1:\varphi(m)\). By assumption, this implies \(\varphi(n+1)\), hence \(\forall m\leq n+1:\varphi(m)\), thus \(n+1\in S\).

Thus we have \(0\in S\) and \(n\in S\implies n+1\in S\), so the induction axiom guarantees \(\forall n:n\in S\), so \(\forall n\forall m\leq n:\varphi(m)\). In particular, \(\forall n:\varphi(n)\).

I have written out everything in more details in necessary, but I wanted it to be complete.