Question about the structure of the proof in how the well-ordering implies induction I am reading a text, and I have a question regarding the logical structure of the following proof.

$(1)$: $1\in X$ and $(2):$ If $k\in X$ for all $k<n$, then $n\in X$.


When we prove well-ordering $\implies$ mathematical induction by contradiction, what are we negating?

From my understanding, we have the logical structure of this proof as $p \implies q$, where $p$ denotes the well-ordering principle and $q$  denotes mathematical induction. Now, to derive a contradiction, we negate $p$ and show that $\neg p \implies (\neg c  \land c)$, whatever $c$ happens to be.
I am confused as to how this proof jumps to supposing that "$X$ is a subset of $\mathbb{N}$ satisfying both $(1)$ and $(2)$", doesn't $(1),(2)$ belong to $q$, which is a part of what we need to show as implied from $p$?
 A: Remember that the implication
$$
p\to q
$$
is logically equivalent to
$$
\neg(p\wedge\neg q).\tag{1}
$$
This is intuitive since the implication says that if the hypothesis $p$ is true then the conclusion $q$ should also be true, that is, it must NOT be the case that we have $p$ and NOT $q$.
Hence proving $p\to q$ by contradiction amounts to show $(1)$, that is, that $p\wedge \neg q$ is contradictory.
So, in the proof, they assume that $p$ is true, that is, that the well-ordering principle is true, AND that the principle of mathematical induction is NOT true. More precisely, if the latter principle is formulated like this:
$$
\text{"If $X$ is a subset of $\mathbb{N}$ satisfying $(1)$ and $(2)$, then $X=\mathbb{N}$"},\tag{2}
$$
then, once we take an arbitrary $X$ satisfying the hypothesis, we suppose that the consequence $X=\mathbb{N}$ is false, that is, that $X$ is a subset of $\mathbb{N}$ but that it is not equal to it (in other words, that it is a proper subset). Notice that $(2)$ is also of the form $r\to s$ and that it's negation is $\neg(\neg(r\wedge\neg s))$ which is logically equivalent to $r\wedge\neg s$.
A: If (2) alone holds for all $n \in N$ then $X=N$. PROOF: If not, there is a least $n \not \in X$.But then $\forall k<n (k \in X)$ which by (2) implies $n \in X$, a contradiction. Notice how the well-ordering is used : If at least one member of $ N$ has some property, there is a least member. Here the property is "not being in $X$". Note also that "$\forall k<n \text{ etc etc etc }$" does NOT imply that there exists anything less than $n$.
