Under the Peano Axioms, I want to prove that if the Axiom of Induction is substituted with the well-ordering principle (every non-empty subset of $ \mathbb N$ has a minimum element), everything will be fine. That means I should prove that they are equivalent, as the proof can be found in various textbooks. However, I found a problem when proving that the well-ordering principle implies mathematical induction. The proof follows:
Let $P(0)$ be true and $P(n)\implies P(n^+)$. Suppose $A=\{x\in\mathbb N:P(x)\text{ is false.}\}$ is non-empty, then let $m$ be its least element. Since $P(0)$ is true, $m\neq 0$. Therefore, there must be some $n$ such that $m=n^+$. By definition of $m$, $P(n)$ is true, but this implies $P(n^+)=P(m)$ is true, a contradiction.
The statement in bold seems to be correct, but the Peano Axioms do not include it (every natural number is either 0 or a successor of a natural number). In fact, it's usually proven via mathematical induction, which we cannot use in the proof above.
Question: How can this flaw be fixed, or they (AI and WOP) are simply not equivalent?
Note: I found a paper claiming that they are not equivalent and most textbooks get it wrong. But I haven't read it. Link: https://link.springer.com/article/10.1007/s00283-019-09898-4