Peano's axioms prove

Question

anonymous · Answer

Starting from Peano's axioms prove that if $$n \in \mathbb{Z}^+ and    n \ne1$$ then n is a successor, i.e. s(a)=n for some $$a \in \mathbb{Z}^+ $$

anonymous · Answer

[Let A=Im(s) U {1} and prove that A=Z+]

kinggeorge · Answer

I think the first could happen through induction. 

Base case: If $n=2$, then $s(1)=2$, so we're good.
Now we assume up to some $k<n$. 
Now we need to show it's true for $k+1$. But since $s(k)=k+1$ for $k\in\mathbb{Z}^+$, we're done.

I've got to go for a bit now. I'll come back to the second part later.

anonymous · Answer

which second part? whats in thee brackets is the hint

kinggeorge · Answer

If that's the hint, then we're completely done. For $n\in\mathbb{Z}_{\ge 2}$, we have that $n\in \text{im}(s)$, and since $\text{im}(s)\subseteq \mathbb{Z}^+$ by definition, we've proved that $A=\mathbb{Z}_{\ge2}$, so $A\cup \{1\}=\mathbb{Z}^+$.