It is said that "Well-ordering Principle, Induction and Strong Induction" - each two can be proved from the other. However none can be proved independently.
Now, can't we say (without holding any axioms true) that a finite step process must end? (I have seen mathematics texts use this assumption in various proofs.) In that case:
Induction: We have: $P(0)$ holds and $P(k) => P(k+1)$. Now, let there exist an element $m∈N$ for which $P(m)$ is false. Now $P(0) \implies P(1)$ and now $P(1) \implies P(2)$ and now...(The process goes on.) Thus conducting $m-1$ steps, we ultimately achieve that $P(m)$ is true, since $P(1)$ is true in the first place. This is achievable since $m$ must be finite, and so should be $m-1$. This contradicts that $P(m)$ is false. Hence there must be no such $m$.
Well-Ordering: Let $S \subset \mathbb N, S\neq \varnothing$ . Now we choose an arbitrary element $k_0$ in it. Set $k = k_0$ Now:
Step 1: If $k$ is the least element in $S$, go to step 3. Else go to step 2.
Step 2: Since $k$ is not the least element, there must exist $x \in S : x < k$. Set $x = k$. Now go back to step 1.
Step 3: End
This is a finite step process(at most $k_0 - 1$ times looped back to step 1). Hence we will ultimately have a least element.
What's wrong here?