Considering the Well-Ordered Principle to prove the following:
$$ \text{Be } \mathbb{S} \text{ a set that belongs to the set of natural numbers, i.e. } \mathbb{S} \subset \mathbb{N}.\\ \text{ If } 0 \in \mathbb{S} \wedge (\forall k \in \mathbb{N} \{0,1,2,\cdots,k\} \subset \mathbb{S} \rightarrow k+1 \in S) \rightarrow \mathbb{S} = \mathbb{N} $$
I did a proof and I would like to know if it is in a good shape.
Proof reached:
Let's prove the stated proposition by contradiction. Therefore, we assume that $$ 0 \in \mathbb{S} \wedge (\forall k \in \mathbb{N} \{0,1,2,\cdots,k\} \subset \mathbb{S} \rightarrow k+1 \in S) $$ and $$ \mathbb{S} \neq \mathbb{N} $$ are true. Since $\mathbb{S} \neq \mathbb{N}$, there is a not empty set that follows from $$ \mathbb{N} - \mathbb{S}. $$ Be $\mathbb{W}$ the set resulting from the operation above. The set $\mathbb{W}$ if formed by the elements that are presentes in $\mathbb{N}$ and not in $\mathbb{S}$ and, by the Well-Ordered Principle, it has a minor element. Let $m$ be the minor element in question. We have that $$ 0 \in \mathbb{S} $$ and, therefore, $m > 0$. With $k$ as reference, we have that $$ \forall k \in \mathbb{N} \{0,1,2,\cdots,k\} \subset \mathbb{S} $$ and, therefore, $m > k$ or $m \geq k+1$. But, in this situation, $k+1 \in \mathbb{S}$ and $m > k +1$. With $k+1$ as reference, we have that $$ \forall k \in \mathbb{N} \{0,1,2,\cdots,k, k+1\} \subset \mathbb{S} $$ and, therefore, $m > k + 1$ or $m \geq k+2$. But, in this situation, $k+2 \in \mathbb{S}$ and $m > k + 2$. Well, from the exposed, one can see that m is not a minor value, since it is possible to always find a value that requires, for $\mathbb{S} \neq \mathbb{N}$ hold,to $m$ be bigger than it last possible minor value. This contradict the Well-Ordered principle and, therefore, proofs that $\mathbb{S} = \mathbb{N}$ for the situation stated.