Let $\mathfrak{N} = \langle \mathbb{N}, +, \cdot\rangle$ be the usual natural numbers structure and $\text{Th}(\mathfrak{N})$ be the first-order theory of the natural numbers. Besides, let $\Sigma$ be a non-empty subset of $\text{Th}(\mathfrak{N})$ such that $\Sigma$ is closed under logical equivalence (that is, if $\varphi \in \Sigma$ and $\varphi \equiv \psi$, then $\psi \in \Sigma$). Is it possible to deduce from Godel's theorems that $\Sigma$ has to be undecidable? That is, there is no algorithm that given a first-order sentence $\varphi$ over the vocabulary $\{+, \cdot\}$, verifies whether $\varphi \in \Sigma$.
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What about Pressburger arithmetic? – Arturo Magidin Oct 30 '24 at 14:17
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@ArturoMagidin Since $\Sigma$ is closed under logical equivalence and the language contains multiplication, that doesn't work; see this old answer of mine. – Noah Schweber Oct 30 '24 at 16:49
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You are correct. You might wonder how we can reconcile this with the fact that Presburger arithmetic is a decidable complete subtheory of $\mathrm{Th}(\mathfrak{N})$. The issue is the change of language involved. While Presburger arithmetic is complete as an ${+}$-theory it is not even closed under equivalence as an ${+,\times}$-theory. These points are treated in more detail in this old answer of mine (while the wording is a bit different there the core ideas are exactly the same). – Noah Schweber Oct 30 '24 at 16:53
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On second thought this isn't quite a duplicate, since the argument in my linked answer needs to be tweaked a bit. – Noah Schweber Oct 30 '24 at 16:56
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This argument is very close to, but not quite the same as, this old answer of mine; on balance I think it's not a duplicate after all, but to avoid double-reputation gain for very-closely-related work I've made this answer CW.
Consider Robinson's arithmetic $Q$. The set of $Q$-theorems is undecidable (by Godel) but $Q$ is finitely axiomatizable (this was why Robinson introduced it). Let $\eta$ be the conjunction of some finite list of axioms for $Q$. Then for every $\{+,\times\}$-sentence $\tau$, the sentence $\eta\rightarrow\tau$ is a tautology iff $Q$ proves $\tau$.
How does this help us? Well, suppose $\Sigma$ is a nonempty deductively closed subset of $Th(\mathfrak{N})$ - or indeed any equivalence-closed consistent set of sentences at all! First let's suppose that $\Sigma$ contains a tautology, $\top$. Then if we could compute $\Sigma$ we could determine theoremhood for $Q$ since (by equivalence-closure) we have that $Q$ proves $\tau$ iff $\top\wedge(\eta\rightarrow\tau)\in\Sigma$.
This argument does need to be tweaked slightly in case $\Sigma$ doesn't contain a tautology (which is possible since $\Sigma$ is not required to be deductively closed in the OP). The issue is that, given $\sigma\in\Sigma$, we may not have the equivalence "$\sigma\wedge(\eta\rightarrow\tau)$ is equivalent to $\sigma$ iff $Q$ proves $\tau$." However, we do get this for equivalence for the theory axiomatized $Q+\sigma$ in place of $Q$. And it turns out this is enough, since - by a slightly stronger version of the incompleteness theorem - every consistent finitely axiomatizable extension of $Q$ (such as $Q+\sigma$) is also undecidable.
Noah Schweber
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I don't understand how this is supposed to work. First, why do you write $\top\land(\eta\to\tau)\in\Sigma$? By closure under equivalence, this is equivalent to just $(\eta\to\tau)\in\Sigma$ (irrespective if whether $\top$ is in $\Sigma$). But more importantly, even if I grant you that $\Sigma$ contains a tautology, $\top\land(\eta\to\tau)\in\Sigma$ does not imply that Q proves $\tau$. You seem to assume that $\Sigma$ only consists of one equivalence class, but that's not at all warranted by the question. – Emil Jeřábek Nov 26 '24 at 09:42
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1Oh, I see: it does not imply that Q proves $\tau$, but it implies that $\tau$ is true. So this shows undecidability by recursive inseparability of provability and refutability in Q. This argument works as long as $Q+\Sigma$ is consistent. (It is not enough for $\Sigma$ alone to be consistent: consider any consistent decidable theory in the language of arithmetic.) – Emil Jeřábek Nov 26 '24 at 11:40