Prove Principle of Mathematical Induction and Principal of Well-ordering are equivalent.

Question

I need some help for verification please.

The principle of mathematical induction: Suppose $P(n)$ is a proposition depending on $n$ for each natural number $n$, such that $P(1)$ is true, and $P(n)$ implies $P(n+1)$ for all $n \in \mathbb{N}$, then $P(n)$ is true for all $n \in \mathbb{N}$.

The Well-ordering principle: If $A$ is a nonempty set of positive integers, then $A$ contains a smallest member.

I have proved they are equivalent, but I am not sure my proof are correct or not. I have put it on below, and can someone help me to verify? Thanks for your help.

This is my method:

Assume the principle of mathematical induction is true. Let $S$ be a nonempty set of natural number and let $P(n)$ be the proposition: If there exists $m \in S$ where $m \leq n$, then $S$ has a smallest element.

When $n=1$, $P(1)$ is true since $S$ only contains one element.

Assume $P(n)$ is true, then there exists $m \in S$ where $m \leq n+1$ since $P(n)$ implies $P(n+1)$. If $m \leq n$, then by induction hypothesis $S$ has the smallest element. If $m=n+1$, then either $m$ is the smallest element in $S$, or there exists $m' \in S$ such that $m' < m = n+1$. which is the case that $m'$ is the smallest element in $S$ by induction hypothesis. Then we have proved the well-ordering principle.

Conversely, assume the well-ordering principle is true. Suppose $P(n)$ is a proposition for each $n \in \mathbb{N}$ such that $P(1)$ is true and $P(n)$ implies $P(n+1)$ for each $n$.

Consider $F =\{m \in \mathbb{N} \,\vert\, P(m) \text{ is false}\}$. Suppose $F$ is a nonempty set, then by well-ordering principle there is a smallest element $m$. Namely, $m \geq 1$ and $P(m)$ is false. But since we assume $P(1)$ implies $P(2)$, $\ldots$, $P(m-1)$ implies $P(m)$ is true, hence it leads to a contradiction. Hence, $F = \emptyset$ and we have proved the principle of mathematical induction.

Answer 1

The second half of your proof isn't clear. Try this:

Consider $F=\{ m \in \Bbb N \mid \lnot P(m) \}$ and assume $F \neq \varnothing$. Then because $F$ is well ordered, it must have a least element $s$. We know that $P(1)$ is true, so $1 \notin F$, which means (since $s \in F$) that $s \neq 1$. Therefore, for some $r$ we have $s=r+1$.

But since $r \lt s$ and $s$ is the least element of $F$, we know that $r \notin F$, which means that $P(r)$ is true. But by hypothesis we also know that $P(r) \Rightarrow P(r+1)=P(s)$, so $P(s)$ is also true. But that, in turn, means that $s \notin F$, which is a contradiction. Thus, our assumption that $F \neq \varnothing$ must be false, so $P(m)$ is true for all $m \in \Bbb N$.