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I want to show that $\{0\}, <, ×,+$, and $S$ are definable in $\langle \mathbb{N}; E \rangle$. Here are the steps I tried to take:

  1. Since $(x^{a})^{b} = x^{ab}$, I define $×$ by: $$\forall x((xEa)Eb = xEc) $$

  2. Now given $\times$, since $x^{a}x^{b} = x^{a+b}$, I define $+$ by: $$\forall x (xEa \times xEb =xEc)$$

  3. Given $+$, I define $\{0\}$ by: $$a + a = a$$

  4. Given $0$, since $a+x^{0}=a+1 = S(a),$ I define $S$ by: $$\forall x( a+ xE0 = b)$$

  5. Given $S$, I define $<$ by: $$\exists x ( a + Sx = b)$$

Assuming any of this is close to being correct to begin with, is the way I defined $S$ problematic since one of the substitution instances of $x$ would be $0$? If it is a problem, what would be a better way to define $S$? Should I take an intermediate step to define $\{1\}$ by $\forall x (a × x = a)$ and then define $S$ by $a+1 = b$?

Any feedback or corrections are greatly appreciated!

Nika
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    It looks like by $E$ you mean the function $xEy = x^y$, is that correct? – Chris Eagle Dec 04 '19 at 21:25
  • @ChrisEagle Yes! Sorry if I wasn't clear enough about that. – Nika Dec 04 '19 at 21:53
  • What is $0E0$? Also, you have to prove that the formulas lead to well defined functions (i.e. there exists exactly one such $c$ or $b$). – Berci Dec 04 '19 at 22:12
  • @Berci This is essentially what my commentary at the bottom of the question is about; $0E0 = 0^{0}$ which often is defined to be $1$, but I know in some contexts that is seen as problematic and so $0E0$ is left undefined. If it is problematic to assume that $0E0 = 1$, my question is whether it is ok take an intermediate step to define ${1}$ and use that to define $S$, or if there is a better way to define $S$. Could you say a little more about what you mean by 'you have to prove that the formulas lead to a well defined function'? How would you show that, say, $<$ is well defined here? – Nika Dec 04 '19 at 22:41
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    @Berci If you prove the formula expresses, say, addition, you have proved it is a well-defined function. – spaceisdarkgreen Dec 04 '19 at 23:33
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    Just assume $0E0=1$. It is problematic to leave it undefined here and this gives the right behavior. – spaceisdarkgreen Dec 04 '19 at 23:36
  • @spaceisdarkgreen Well, yes, indeed. I was thinking some underlying axioms of $E$ instead. – Berci Dec 05 '19 at 08:48

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