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Assuming that the Peano Axioms hold (without the axiom of induction), and assuming one of Robinson's Axioms, namely

Every natural number is either $0$ or the successor of a natural number.

It can be shown that you cannot use the above axiom to prove mathematical induction, since there's an inherent circularity, but I can't seem to pin down what will go wrong.

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    $\begingroup$ Can you write down your proof of the axiom of induction via Robinson's axiom, please? $\endgroup$ – Cave Johnson Jul 30 '16 at 14:53
  • $\begingroup$ I've edited my question. I know it cannot, but I can't seem to show why it can't be done. $\endgroup$ – Maxis Jaisi Jul 30 '16 at 14:57
  • $\begingroup$ What can you even do with such an axiomatic system? $\endgroup$ – mathreadler Jul 30 '16 at 14:57
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    $\begingroup$ Is that really Robinson's Axiom? It would seem to imply a universe with only two elements. $\endgroup$ – John Coleman Jul 30 '16 at 14:59
  • $\begingroup$ I've again edited my question; it should be "one" of Robinson's Axioms instead of "the" Robinson Axiom. $\endgroup$ – Maxis Jaisi Jul 30 '16 at 15:04
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Let $X$ be the set of polynomials $f(x)$ with integer coefficients which are either $0$ or have positive leading coefficient. This then satisfies your axioms. However, it does not satisfy induction. For instance, if it did, then it would have to satisfy $$\forall a\exists b (a=2b\vee a+1=2b),$$ since you can prove this statement by induction. But this statement is not true for $a=x$.

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