Proving the Well-Ordering Principle for Natural Numbers I know the WOP is treated like axiom of a natural number, but I was curious if I can prove WOP defined for the set of natural numbers N by following:
Suppose A is a subset of N, which then obeys all axioms from Peano postulates, and let us assume that A does not have a least element.  Then, 1 in A must be a successor to another natural number, which is 0.  However, 0 is neither in A nor N.  (If 0 is in N, then we argue that 0 must be a successor to -1, which is not part of N).  Also, 1 in A being a successor contradicts with the Peano axiom (1 cannot be a successor).  Hence, our assumption that A having no least element is not true.  Therefore, A has a least element.  Without out loss o generality, every subset of the N has a least element.
I understand this is silly proof, but I wanted to make sure that I have reasoning for WOP in N before moving on.

Well ordering principle is a direct consequence of the Principle of Mathematical Induction and here we need the second version of Principle of Mathematical Induction stated below:
Principle of Mathematical Induction Second Version: If $A$ is a subset of $\mathbb{N}$ such that


*

*$1 \in A$

*If $1, 2, \ldots, n \in A$ then $(n + 1) \in A$
then $A = \mathbb{N}$.
Let $A$ be a non-empty subset of $\mathbb{N}$. If $1 \in A$ then obviously $A$ has a least member and hence let $1 \notin A$. Consider $B = \mathbb{N} - A$. Then we can see that $B$ is also a subset of $\mathbb{N}$ and $1 \in B$.
Let's assume that $A$ does not have a least member. Using this assumption I will prove that if $1, 2,\ldots, n \in B$ then $(n + 1) \in B$. If $1, 2,\ldots, n \in B$ then $1, 2, \ldots, n \notin A$ and hence any member in $A$ is greater than $n$. If $(n + 1) \in A$ then it would become the least member of $A$ and this is not allowed by our assumption. Hence $(n + 1) \in B$. It now follows by Principle of Mathematical Induction that $B = \mathbb{N}$ and hence $A = \mathbb{N} - B = \emptyset$ which is contrary to the assumption that $A$ is non-empty.
It follows from this contradiction that $A$ must have a least member. It is possible to prove the Principle of Mathematical Induction by assuming the truth of Well Ordering Principle. So in a sense both these principles are equivalent, but it appears that the principle of induction captures the essence of natural numbers in a more intuitive and obvious manner and perhaps that is the reason it was selected as one of the Peano's axioms.
