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I know that for the language of ordered rings $\{0,1,+,-,\cdot,<\}$, the theory of $\mathbb{R}$ admits quantifier elimination and is decidable. I also know that for the language of rings (without order) $\{0,1,+,-,\cdot\}$, the theory of $\mathbb{R}$ does not admit quantifier elimination. However, I can't seem to find anywhere whether it is decidable or not. Can anyone tell me?

mathstudent
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1 Answers1

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Below, unless stated otherwise all languages are finite.

Throwing things out of the language can never make things more complicated. A sentence $\varphi$ in the language $(0,1,+,-,\cdot)$ is also a sentence in the language $(0,1,+,-,\cdot,<)$, and to tell whether $\varphi$ is true about $\mathbb{R}$ as a $(0,1,+,-,\cdot)$-structure we just ask whether $\varphi$ is true about $\mathbb{R}$ as a $(0,1,+,-,\cdot,<)$-structure.

More snappily: the restriction of a decidable theory to a sublanguage is also decidable, and in particular any reduct of a decidable structure is also decidable.

Hagen von Eitzen's comment is really addressing the opposite direction: if $\mathcal{A}$ is decidable and $\mathcal{A}'$ is an expansion of $\mathcal{A}$, under what conditions is $\mathcal{A}'$ also decidable? (A structure is decidable if its theory is decidable.) The point of Hagen's comment is that if $\mathcal{A}'$ is an expansion-by-definitions of $\mathcal{A}$ - that is, every new relation/function/constant is definable in the original structure - then decidability is inherited upwards. So we have (focusing for simplicity on theories of the form $Th(\mathcal{M})$, that is, complete theories):

  • Reducts of decidable structures are decidable.

  • Expansions-by-definitions of decidable structures are decidable.

Let me end by pointing out a couple things about QE by way of contrast:

  • First, it's important to remember that QE doesn't actually imply decidability! Any structure $\mathcal{M}$ has an expansion-by-definitions - admittedly in an infinite language - which has QE: just add new relations corresponding to every old formula. The result structure $\hat{\mathcal{M}}$, called the Morleyization of $\mathcal{M}$, is decidable iff $\mathcal{M}$ was. When used as a tool to prove decidability, QE has to be paired with a proof that the quantifier-free fragment of the theory in question is decidable (EDIT: as well as a proof that the QE itself is decidable, that is, that we have a decidable way of finding a quantifier-free $\psi$ equivalent to a given $\varphi$). By itself, QE is really a model-theoretic, rather than computability-theoretic, criterion. (See also this old MSE question (EDIT: and also this one).)

  • We can also babble a bit about the qualitative differences between QE and decidability - what does each property "care about," formulaically speaking? QE can be gained by passing to an expansion-by-definitions (or dually, lost by passing to a reduct), and this is because QE "pays attention to syntax" in a way that decidability doesn't: passing to an expansion-by-definitions of a structure, we take old formulas and make them equivalent to new ones of possibly nicer syntactic form. In this sense decidability is less finicky than QE, in that decidability doesn't care about what a formula "looks like." But that's only a first approximation: decidability cares very much about having a fixed ordering on formulas, since we need to pass from formulas to natural numbers for computability-theoretic questions to even make sense. So the right thing to say I think is that there's a "local vs. global" distinction going on here. QE cares about what individual formulas "look like," while decidability doesn't care at all what individual formulas "look like" but cares very much about how we organize the set of all formulas (which in turn QE couldn't give a hooey about).

Noah Schweber
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