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This is the statement of Peano's axioms I will assume for this discussion:

  1. $1$ is a number.
  2. To every number $n$ there corresponds exactly one number $n^\prime.$
  3. $n^\prime=m^\prime\implies n=m.$
  4. $n^\prime\ne 1$
  5. Let $P\left[x\right]$ be a proposition (propositional form) containing the number variable $x$. If $P\left[1\right]$ holds and if $P\left[n^\prime\right]$ follows from $P\left[n\right]$ for every number n, then $P\left[x\right]$ holds for every number $x$.

Is Peano's axiom of induction (axiom #5) needed to show $n\ne n^\prime$ for $n\ne 1$? That is, nothing in any of the first four axioms individually or in combination seems to preclude the proposition $n\ne 1\land n= n^\prime.$

I thought I had shown this by the 5th axiom, but as I typed up this question I realized that my proof needs more work. I'm not asking for the proof. I'm merely asking for a confirmation that it requires the 5th axiom.

The answer seems clearly to be yes. But I've been wrong about such things before.

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For the new version of axiom 3, take $\Bbb N\cup\{*\}$ with $x'=x+1$ for $x\in\Bbb N$ and $*'=*$.

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