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I'm aware that a semantically complete theory can be undecidable. (I believe it's because only logically valid sentences need be provable for a theory to be semantically complete.)

But is it possible for a syntactically/negation complete theory to be undecidable? And do "syntactically complete" and "negation complete" mean the same thing?

I've heard that:

Any consistent, negation-complete, effectively axiomatized formal theory T is decidable.

So for a negation complete theory to be undecidable, it'd have to be either inconsistent or not effectively axiomatized. But I have no idea how a theory could be negation-complete without those two other criteria in the first place.

csp
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    Yes, "theory" is just a set of sentences, it makes no assumption about computability. I think this is an important part of Gödel's incompleteness theorems, that is often not directly mentioned, you can have two out of three: Completeness, decidability and soundness, iirc. – fweth Jun 13 '23 at 15:38
  • Thanks @fweth ... When you say a theory is just a set of sentences, do you mean that since theories can lack an effectively decidable set of axioms, they can be undecidable? But then how could those theories also be negation complete? – csp Jun 13 '23 at 15:46
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    Sorry, I'm not an expert here, I just think that we can take any model and then form the set of sentences satisfied by the model, so we can e.g. prove in ZFC that there exists the set of sentences satisfied by e.g. the natural numbers, and we can prove that this set is negation complete. But I hope somebody who knows this stuff better than me will chime in! – fweth Jun 13 '23 at 15:51
  • @fweth You're exactly right. See my answer. – Noah Schweber Jun 13 '23 at 16:37

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This answer is basically just an expansion of fweth's comments above; I've marked it CW to avoid rep gain, and if they provide an answer of their own I'll delete this one.

Any structure $\mathfrak{S}$ gives rise to a consistent-and-negation-complete theory $$Th(\mathfrak{S}):=\{\varphi: \mathfrak{S}\models\varphi\}.$$ Some particularly simple choices of $\mathfrak{S}$ yield low-complexity theories (e.g. the fields of real or complex numbers), but in general there's no bound on how nasty $Th(\mathfrak{S})$ can be. In particular, Godel's incompleteness theorem implies that $Th(\mathbb{N};+,\times)$ is not c.e. since it's a complete consistent theory extending Robinson arithmetic.

  • In case you're concerned about how we can even talk about theories which are not c.e., note that the above paragraph can be formalized and proved inside $\mathsf{ZFC}$. (See this old answer of mine for an outline of how one might do this.) So even if you're a strict formalist - which I sometimes am, FWIW - the above paragraph is contentful.

Separately, note that by the completeness theorem every negation complete theory is $Th(\mathfrak{S})$ for some structure $\mathfrak{S}$. So the focus on the above construction is really not losing any generality here.

Noah Schweber
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  • Hey, happy to hear that my comment was correct, I don't know what CW means but feel free to provide you answer the "normal" way so it can get accepted :) – fweth Jun 13 '23 at 16:40
  • Thanks! I'm still confused how any structure gives rise to a consistent-and-negation-complete theory, so I'll read up on that link to your previous answer. – csp Jun 13 '23 at 16:51
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    @csp For that issue I don't recommend my previous answer (which is really only worth looking at after you understand the "Platonistic" approach). Instead, just think in terms of the definition of $\models$ itself, specifically the negation clause. We can't have $\mathfrak{S}\models\varphi\wedge\neg\varphi$ since $$\mathfrak{S}\models\neg\varphi\implies\mathfrak{S}\not\models\varphi,$$ and we can't have $\mathfrak{S}\not\models\varphi\vee\neg\varphi$ since $$\mathfrak{S}\not\models\varphi\implies\mathfrak{S}\models\neg\varphi.$$ – Noah Schweber Jun 13 '23 at 17:17
  • @fweth CW ("community wiki") just means that it doesn't generate any Magical Internet Points for me. It can still be upvoted (or downvoted for that matter) and accepted. – Noah Schweber Jun 13 '23 at 17:18
  • @NoahSchweber thanks, but hmm... I think I'm confused about something. By your example, is it "proving" via assigning truth values OUTSIDE of the system? I thought the whole point of syntactic or negation incompleteness is that you can't prove or disapprove every statement within the system. So this kind of feels like changing the definition to call this negation complete yet undecidable? – csp Jun 14 '23 at 12:25