How to prove that $+$ is commutative on the natural numbers?

Question

Let $N$ be a non empty set. Let $s:N\to N$ a function satisfying:

1. there is only one element in $N-s(N)$ (denoted by $1$);

2. $s$ is injective;

3. for any subset $X\subset N$, if $1\in X$ and $(n\in N \Rightarrow s(n)\in N)$ then $X=N$.

We define a binary operation '$+$' on $N$ by $$m+n=s^n(m)$$ where $s^n$ is the iterated function. So $$m+1=s(m) \quad \text{and}\quad m+s(n)=s(m+n).$$

My problem is: how to prove (probably using the induction) that $$m+n=n+m.$$

Answer 1

Hint: Let $X_m:=\{n\in N\mid m+n=n+m\}$ for a fixed $m\in N$ and show that it satisfies 3, by induction on $m$.

For $X_1$, $1\in X_1$ by definition, and so on.

Answer 2

Hint: Show that the set $X=\{n\mid\forall m\leq n: m+n=n+m\}$ is inductive.

(Recall that $\leq$ is definable from $+$, $n\leq m\iff\exists k.n+k=m\lor n=m$)