I want to prove that strong induction implies regular induction. Is my attempt correct:
Let $J\subseteq\mathbb{N}$ be a set such that $1\in J$ and $k\in J$ for all $k<n$ implies $n\in J$. Consider the set $S=\{n\in\mathbb{N}\mid n\in J\text{ and }(k<n\to k\in J)\}$. Clearly $1\in S$ so suppose $n\in S$. Then, let $i<n+1$. Clearly either $i=n$ or $i<n$. Either way, since $k<n$ implies $k\in J$, we see that $i\in J$ but then by our assumption at the beginning it follows that $n+1\in J$ which by regular induction implies $J=\mathbb{N}$.