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In 1949 Julia Robinson showed the undecidability of the first order theory of the field of rationals by demonstrating that the set of natural numbers $\Bbb{N}$ is first order definable in $(\Bbb{Q}, +, \cdot$).

It is not hard to see that Robinson's result can be reformulated in the following symmetric form.

Theorem A. The structures ($\Bbb{N}, +, \cdot$) and $(\Bbb{Q}, +, \cdot$) are bi-interpretable.

The following generalization of Theorem A is considered folkore (I am not aware of a published reference).

Theorem B. If $(M, +, \cdot)$ is a model of $PA$ (Peano arithmetic), then the field of rationals $\Bbb{Q}^M$ of $(M, +, \cdot)$ is bi-interpretable with $(M, +, \cdot )$.

Let $EFA$ denote the exponential function arithmetic fragment of $PA$, a fragment also known as $I\Delta_{0}+exp$.

Based on a posteriori evidence classical theorems of Number Theory do not require the full power of $PA$ since they can be already verified in $EFA$ (indeed Harvey Friedman has conjectured that even FLT can be verified in $EFA$, with a proof that would be very different from Wiles').

This suggests that in Theorem B one should be able to weaken $PA$ to $EFA$, hence my question:

Question. Is there a published reference for the strengthening of Theorem B, where $PA$ is weakened to $EFA$?

P.S. The following paper provides an excellent expository account of Robinson's theorem (and related results).

D. Flath and S. Wagon, How to Pick Out the Integers in the Rationals: An Application of Logic to Number Theory, American Mathematical Monthly, Nov. 1991.

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Ali: Suppose $M\models PA$. To interpret $M$ in $\mathbb{Q}^M$ does Julia Robinson's formula with the operations of $\mathbb{Q}^M$ do the job, or is there something fancier going on? Is your question about how much of the argument in Robinson's paper goes through in models of $I\Delta_0+exp$? –  SJR May 22 '11 at 5:51
    
SJR: The answer to your first question is positive, i.e., nothing fancier is going on. Regarding the second one: the arguments of Robinson's paper should go through in $EFA$ (based on $EFA$'s track record in handling "elementary" number theory); my question is whether anyone has actually shown - in a publishd source - that this is indeed the case. –  Ali Enayat May 22 '11 at 16:37
    
@AliEnayat: Have you found an answer to your question yet? I would be very interested to know. –  Samuel Reid Mar 11 '12 at 0:14
    
@Samuel Ried: No I have not (sorry for the tardy reply, I have been "away" for a long while). –  Ali Enayat Apr 11 '12 at 1:51
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