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Tennenbaum's celebrated 1959 theorem (see here for a reference) is certainly one of the key theorems in mathematical logic. Not so much for its proof, but because it helps "isolating" $N$ as something very special: it is the only countable model of Peano arithmetic which is recursive.

But, does it?

Suppose I live inside any countable model $M$ of Peano: I think I am actually living in the standard natural numbers, and addition/multiplication are the standard ones (and so are all the recursive function, etc). So, it would seem that from the point of view of $M$, other PA-models are not recursive. To be a bit more precise, let me state this:

Internalized Tennenbaum Theorem (ITT): Let $M$ be a countable model of Peano. Then, for any other countable model $N$ not isomorphic to $M$, $N$ is not recursive in $M$, ie it is not $\Delta_1$-definable in M.

Question: Can ITT be proved in, say, ZFC? If not, what is the obstruction?

Post Scriptum.

Thanks to Emil Jerabek for his suggestion: rather than the original misleading name, Derived Tennenbaum, use Internal (or Internalized). The original name generated some confusion, see the comments of Francois Dorais, thus I decided to rename the question. The recursivity required is IN the model, not FROM the model.

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    $\begingroup$ The main obstruction would be that the statement is false. Maybe you meant to relativize "Peano" as well? If so, please explain how. (See the well-studied notion of PA-degrees.) $\endgroup$ Commented Aug 8, 2020 at 23:03
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    $\begingroup$ I’m pretty sure Tennenbaum’s theorem is provable in PA, which, translated to model-theoretic terms, means that no proper extension $M'$ satisfying PA of a model $M$ of PA is $\Sigma_1$-definable in $M$. (Note that such internally definable models are automatically end-extensions.) This should also hold for much weaker theories than PA. Perhaps there might be a version that applies to proper end-extensions that are merely externally relatively computable rather than $\Sigma_1$-definable, but I don’t think that any such thing stands a chance to hold when $M'$ is not an extension of $M$. $\endgroup$ Commented Aug 9, 2020 at 8:38
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    $\begingroup$ (In principle, I should have written “proper extension $M'$ of $M$ such that $M$ thinks $M'$ is a model of PA” rather than “proper extension $M'$ satisfying PA”, but since, as I mentioned, this should hold even for much weaker, finitely axiomatizable fragments of PA, the distinction does not matter.) $\endgroup$ Commented Aug 9, 2020 at 8:49
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    $\begingroup$ Rereading the question, it's starting to dawn on me that when you write "$M$-recursive", you perhaps do in fact mean " $\Sigma_1$-definable in $M$" rather than "computable with oracle $M$" (which is how François and me interpreted it). If so, you really, really shouldn't call it "relative" or "relativized" Tennenbaum theorem, but something like "internal" or "internalized" TT. "Relativized" unambiguously refers to computability with an oracle. $\endgroup$ Commented Aug 9, 2020 at 9:37
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    $\begingroup$ Ok, I will try to write it up as an answer, but it may take a while, as I’m out of office at the moment. $\endgroup$ Commented Aug 10, 2020 at 10:48

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To move this off the unanswered queue, let me summarize the situation as correctly explained by the comments above:

The standard proof of Tennenbaum's theorem goes through inside $\mathsf{PA}$: $\mathsf{PA}$ proves that there is no $\Delta_1$ description of a model of $I\Sigma_1$. (As usual, $\mathsf{PA}$ can be replaced with something vastly weaker here; at a glance, already $I\Sigma_1$ should be enough.)

One key point here is that $\mathsf{PA}$ can quantify over $\Delta_1$ descriptions since this only involves reference to a bounded truth predicate. (Something like "There is no definable structure such that [stuff]" would have to be expressed as a scheme, but that's not an issue here.)

On the other hand, $\mathsf{PA}$ cannot express "the structure defined by the formula tuple $\Phi$ satisfies $\mathsf{PA}$" (at least not as cleanly as one might hope - we'd need to talk about explicitly-Skolemized structures, and that's a whole annoying rabbit hole to go down). This is why I've used $I\Sigma_1$ as the "Tennenbaum target:" as a finitely axiomatizable theory, $\mathsf{PA}$ has no trouble talking about its satisfaction (or not) in a defined structure, again by virtue of the satisfactoriness of a bounded truth predicate.

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    $\begingroup$ let me process what you wrote (meanwhile you got already my vote, at the very least for trying to sum up the above). . The last comment of Emil was "Ok, I will try to write it up as an answer, but it may take a while,". $\endgroup$ Commented Mar 16, 2021 at 20:32

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