# Peano axioms: S(1)=1?

I was reading through the Peano axioms here, and a question came up:

Can we define $$S(0)=1$$, and $$S(1)=1$$?

It seems to me (at least as it is stated) that it would satisfy all of the axioms listed. And I couldn't find any restrictions on the successor function, which would disallow this.

So would this be a valid Peano arithmetic or am I missing something here?

Thanks!

• If $S(1)=1=S(0)$, then $0=1$ by Axiom 7 in the link. – logarithm May 13 at 13:28
• This is still not a problem on its own, the contradiction arises with Axiom 8 which says that $0$ is not a successor, but $0=1$ implies $S(0)=1=0$. – logarithm May 13 at 13:37
• @logarithm: Yes, if we're not considering $=$ to be built into the logic (but the Wikipedia article is equivocal about that). – Henning Makholm May 13 at 13:41

1. For all natural numbers $$m$$ and $$n$$, $$m = n$$ if and only if $$S(m) = S(n)$$. That is, $$S$$ is an injection.
• $\{0\}$ with $S(0)=0$ satisfies Axiom 7. It is Axiom 8, the one that fails. – logarithm May 13 at 13:40