Prove using real number axioms, that $n \ne m$ for distinct natural numbers $n, m \in N$.

Question

By distinct, 1+...+1 $\ne$ 1+...+1 if the number of ones on each side are not equal. We are asked to do a proof by induction. My attempt: Base case: 1<2 since 0<1 so 0+1<1+1 by order axiom. Assume $m<n$ for two distinct m,n. Then m+1 < n+1, therefore $n\ne m$ for two distinct numbers m and n.

Answer 1

Define $\sigma(n) = \nu^n(0)$ for all $n\geq 0$, where $\nu$ is the successor function mapping $n$ to $n+1$. Consider the set $A = \{\nu^n(0)\mid n\geq 0\}$. By the induction axiom, $A=N_0$. Define $T$ as the set of elements $\nu^m(0)$ such that $\nu^m(0)\ne\nu^n(0)$ for all $n\geq 0$ with $n>m$. By the induction axiom, $T=N_0$. This shows that all powers $\nu^n(0)$ are pairwise distinct.