I want to prove that well-orderedness on the integers implies induction. The proof is the classical "assume a contradiction" and see what happens.
So begin with an intended contradiction:
\begin{align*} &\quad A\subset\mathbb{Z} \;\wedge\; 0\in A \;\wedge\; (n\in A \implies (n+1)\in A) \;\;\implies\;\; \neg \mathbb{N}\subset A \\ &\quad\text{...unpacking definition of $\neg\mathbb{N}\subset A$...} \\ &= A\subset\mathbb{Z} \;\wedge\; 0\in A \;\wedge\; (n\in A \implies (n+1)\in A) \;\;\implies\;\; \exists n\in \mathbb{N} \;s.t. \neg n\in A \\ &\quad\text{...definition of bunches...} \\ &= A\subset\mathbb{Z} \;\wedge\; 0\in A \;\wedge\; (n\in A \implies (n+1)\in A) \;\;\implies\;\; \exists B\subset \mathbb{N} \;s.t. |B| > 0 \;\wedge\; &&B\cap A = \emptyset \end{align*}
Working within the context of what has been proven so far, we let $c$ be the minimal element of the non-empty set $B$:
\begin{align*} &\quad (c = \min B) \\ &\quad\text{...$0\in A \wedge B\cap A = \emptyset$...}\\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; c > 0 \\ &\quad\text{...algebra...}\\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; c - 1 > -1 \\ &\quad\text{...properties of $>$ and $\geq$...}\\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; c - 1 \geq 0 \\ &\quad\text{...definition of $\geq$...} \\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; (c - 1 = 0 \vee c > 0) \\ &\quad \text{...$0\in A$...} \\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; [(c - 1 = 0 \implies c - 1\in A) \vee c > 0] \\ &\quad\text{...$n-1\in A \implies n\in A$...} \\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; [(c - 1 = 0 \implies c - 1\in A \implies c\in A) \vee c > 0] \\ &\quad\text{...transitivity of $\implies$...}\\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \;\wedge\; [(c - 1 = 0 \implies c\in A) \vee c > 0]\\ &\quad\text{...distributivity of $\wedge$ over $\vee$...}\\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \wedge (c - 1 = 0 \implies c\in A) \vee \neg (c\in A) \wedge c > 0 \\ &\quad\text{...contrapositive law...}\\ &= c = \min B \;\;\implies\;\; \neg (c\in A) \wedge (\neg(c\in A) \implies \neg(c - 1 = 0)) \vee \neg (c\in A) \wedge c > 0 \\ &\quad\text{...law of discharge...} \\ &= c = \min B \;\;\implies\;\; \neg(c\in A) \wedge \neg(c - 1 = 0) \vee \neg (c\in A) \wedge c > 0 \\ &\quad\text{...De Morgan's Law...} \\ &= c = \min B \;\;\implies\;\; \neg(c \in A \vee c - 1 = 0) \;\vee\; \neg (c \in A) \wedge c > 0 \\ &\quad\text{...De Morgan's Law...} \\ &= c = \min B \;\;\implies\;\; \neg(c \in A \vee c - 1 = 0) \;\vee\; \neg ((c \in A) \vee c \leq 0) \\ &\quad\text{...De Morgan's Law...} \\ &= c = \min B \;\;\implies\;\; \neg[(c \in A \vee c - 1 = 0) \wedge (c \in A \vee c \leq 0)] \\ &\quad\text{...factoring out $c:A$...} \\ &= c = \min B \;\;\implies\;\; \neg[c \in A \vee (c - 1 = 0 \wedge c \leq 0)] \\ &\quad\text{...algebra...} \\ &= c = \min B \;\;\implies\;\; \neg[c \in A \vee (c = 1 \wedge c \leq 0)] \\ &\quad\text{...integer properties...} \\ &= c = \min B \;\;\implies\;\; \neg[c \in A \vee \bot] \\ &\quad\text{...definition of $\vee$...} \\ &= c = \min B \;\;\implies\;\; \neg(c \in A) \end{align*}
I am supposed to get a simpler contradiction, which is that $c = \min B \wedge c \in A$. Formally, I do not get this. What is my error?