This question regards an alternative proof of the theorem of well-ordering of $\mathbb{N},$ using a modified principle of induction and contraposition, found on page 101 of Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra; Edited by H. Behnke, F. Bachmann, K. Fladt, W. Suess and H. Kunle.
The square brackets following quantifiers indicate the scope of the bound variables. The way I read my notation is, for example:
$$ \forall_{n<m}\left[n\in\mathcal{M}\right] $$
For all $n$ less than $m$, $n$ is an element of $\mathcal{M}$.
The question: Is the following proof correct?
The modified principle of induction is: If the number $m$ is contained in the number set $\mathcal{M}$ whenever $n\in\mathcal{M}$ for all numbers $n<m$, then $\mathcal{M}=\mathbb{N}.$
The theorem of well-ordering of the natural numbers is: Every nonempty set of natural numbers contains a smallest number.
This is what I am trying to accomplish:
As another method of proving the same theorem, we note that, if we replace $\mathcal{M}$ by the set $\mathbb{N}-\mathcal{M}$ of the numbers not in $\mathcal{M},$ our modified principle of induction can be transformed, by contraposition and other purely logical operations, into the desired theorem of well-ordering of the natural numbers.
One form of the contrapositive of implication is:
$$ \left(p\implies q\iff\lnot q\implies\lnot p\right). $$
The modified principle of induction is formalized as
$$ \forall_{m}\left[\forall_{n<m}\left[n\in\mathcal{M}\right]\Rightarrow m\in\mathcal{M}\right]\Rightarrow\mathcal{M}=\mathbb{N}. $$
Rename $\mathcal{M}\mapsto\overline{\mathcal{M}}$ and write the result as the contrapositive of the original expression, followed by some transformations
$$ \overline{\mathcal{M}}\neq\mathbb{N}\Rightarrow\lnot\left(\forall_{m}\left[\forall_{n<m}\left[n\in\mathcal{M}\right]\Rightarrow m\in\mathcal{M}\right]\right) $$
$$ \iff\overline{\mathcal{M}}\subset\mathbb{N}\Rightarrow\exists_{m}\left[\neg\left(\forall_{n<m}\left[n\in\overline{\mathcal{M}}\right]\Rightarrow m\in\overline{\mathcal{M}}\right)\right] $$
$$ \iff\overline{\mathcal{M}}\subset\mathbb{N}\Rightarrow\exists_{m}\left[\forall_{n<m}\left[n\in\overline{\mathcal{M}}\right]\land m\notin\overline{\mathcal{M}}\right]. $$
Now define $\overline{\mathcal{M}}\equiv\mathbb{N}-\mathcal{M},$ and replace $\overline{\mathcal{M}}\mapsto\mathcal{M}$ according to that meaning
$$ \emptyset\ne\mathcal{M}\subseteq\mathbb{N}\Rightarrow\exists_{m}\left[\forall_{n<m}\left[n\notin\mathcal{M}\right]\land m\in\mathcal{M}\right]. $$
This statement reads: That $\mathcal{M}$ is a non-empty subset of $\mathbb{N}$ implies the existence of $m$ such that for all $n<m$ we have $n\notin\mathcal{M}$ and $m\in\mathcal{M}.$ That is $m$ is the smallest element of $m.$
Apparently others affirm the validity of my proof, and I believe it to be formally correct, so my answer is: Yes the proof is correct.
For anybody who uses this proof, I urge him or her to explore its meaning by considering special cases such as $\mathcal{M}=\left\{n\backepsilon n>m\right\},$ etc.