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?


  • 2
    $\begingroup$ If $S(1)=1=S(0)$, then $0=1$ by Axiom 7 in the link. $\endgroup$ – logarithm May 13 at 13:28
  • $\begingroup$ 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 $S(0)=1=0$. $\endgroup$ – logarithm May 13 at 13:37
  • $\begingroup$ @logarithm: Yes, if we're not considering $=$ to be built into the logic (but the Wikipedia article is equivocal about that). $\endgroup$ – Henning Makholm May 13 at 13:41

Your structure fails to satisfy the axiom that the Wikipedia article gives number 7:

  1. For all natural numbers $m$ and $n$, $m = n$ if and only if $S(m) = S(n)$. That is, $S$ is an injection.
  • $\begingroup$ Ah! I see. Didn't see the "only if" part. Thanks! $\endgroup$ – Valdeminas May 13 at 13:34
  • $\begingroup$ @Valdeminas: It's actually the "if" direction that sinks your structure. $\endgroup$ – Henning Makholm May 13 at 13:39
  • $\begingroup$ $\{0\}$ with $S(0)=0$ satisfies Axiom 7. It is Axiom 8, the one that fails. $\endgroup$ – logarithm May 13 at 13:40

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