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Let $\mathcal{N}=(\mathbb{N},+,\cdot,0,1)$. I want to show that $\{\#F\mid\mathcal{N}\models F \text{ and $F$ is an $\exists$-formula} \}$ is a recursive set. Here $\#F$ is the Gödel number of $F$ and I think that $\exists$-formula just means a formula on the form $\exists v_0 \dots\exists v_n F$, without any occurrence of universal quantifiers. The problem is from an old mock-exam from a second-level logic course given at my university.

This was my original attempt, which, as Carl Mummert pointed out, is clearly wrong:

My idea is to show that $\{\#F\mid\mathcal{N}\models F \}$ is recursive and then just multiply its characteristic funtion with something like $f(x)=\left[1\,\dot{-}\,sg(\vert \beta_3^3(x)-7\vert )\right]$, i.e. something that takes the Gödel number of $x$ and returns $0$ if the last digit is not $7$ and $1$ if it is (i.e. if it is an $\exists$-formula). That would give me the characteristic funtion for the original set. But then my problem is that I don't know how to show that $\{\#F\mid\mathcal{N}\models F \}$ is recursive.

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    $\begingroup$ The set of true formulas of arithmetic - $\{ \#\phi : \mathbb{N} \vDash \phi\}$ - is very far from being recursive. So that method will not work. Instead you would need to look at the special structure of a $\exists$ formula and use that to show that there is a way specific to those formula to tell if they are true. $\endgroup$ – Carl Mummert Jan 15 at 16:58
  • $\begingroup$ You could improve the post by adding a little more background. What book is the problem from? What is your specific definition of a $\exists$ formula? $\endgroup$ – Carl Mummert Jan 15 at 16:59
  • $\begingroup$ Yes of course, now that you phrase it that way... The problem is from an old mock-exam from a logic course given at my university. Now, unfortunately this is the first time I see the term "$\exists$-formula" but I am assuming they just mean that it is on the form $\exists v_0 G$. But thanks, I'll add that to my question. $\endgroup$ – KurtKnödel Jan 15 at 17:07
  • $\begingroup$ \mid produces the correct spacing. $\endgroup$ – Asaf Karagila Jan 15 at 18:09
  • $\begingroup$ Are you sure the text you got the problem from intended you to show the set is recursively decidable, rather than recursively enumerable (which is semidecidable) ? $\endgroup$ – DanielV Jan 15 at 21:10
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It's not recursive.

Whether this is easy or very hard to show depends on exactly what "$\exists$-sentence" means. Specifically, the issue is whether "bounded quantifiers" (like $\forall x<5$, $\exists y<z+w$) are permitted. Formulas of the form $\exists x_1,...,x_n G$ where $G$ uses only bounded quantifiers are called "$\Sigma_1$" (or "$\Sigma^0_1$"); to me, "$\exists$-formula" sounds like a stronger condition, namely we require that the matrix $G$ be genuinely quantifier-free (and this is your interpretation in the comments above).

But in each case, the resulting set is non-recursive.


Showing that the $\Sigma_1$ theory of $\mathcal{N}$ is non-recursive isn't too hard. There is a $\Sigma_1$ formula $Halt(x)$ such that for each $n\in\mathbb{N}$ we have $\mathcal{N}\models Halt(\underline{n})$ iff $\varphi_n(n)\downarrow$ - writing "$\varphi_e(x)\downarrow$" for "the $e$th Turing machine halts on input $x$" and "$\underline{k}$" for the numeral corresponding to $k$ - but this means that the $\Sigma_1$-theory of $\mathcal{N}$ is at least as complicated as (in fact, it's equivalent to) the halting problem, which is non-recursive.

Gauging the complexity of the genuinely $\exists$-theory of $\mathcal{N}$ is harder, but ultimately still the same. This is the MRDP theorem: that the set of Diophantine equations with solutions (which is about as $\exists$ as it gets) is non-recursive, and in fact has complexity again equal to that of the halting problem.

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  • $\begingroup$ Thank you very much for this! This explains why it's been so hard to prove it... Actually, the original problem was to show that $\{F\,\vert\,\mathcal{N}\models F \text{ and $F$ is an $\exists$-formula} \}\cup PA$ is an incomplete theory. Do you reckon this is still possible (using the first incompleteness theorem), despite the theory not being recursive? $\endgroup$ – KurtKnödel Jan 15 at 17:56
  • $\begingroup$ @KurtKnödel Ah, that's a very different problem. It is indeed correct, and I have two hints. $(1)$ For a "nuke," use Craig's trick and the strong version of GIT ("no consistent recursively axiomatizable extensionof PA is complete"). $(2)$ For the optimal solution, show that PA already proves every true $\exists$-sentence, so you don't even need strong GIT. $\endgroup$ – Noah Schweber Jan 15 at 17:58
  • $\begingroup$ Your second suggestion was actually my idea all along, but then somewhere someone mentioned recursive sets and I just lost it at that point. Thank you for clarifying things! $\endgroup$ – KurtKnödel Jan 15 at 18:06

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