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While I understand that the first-order theory of real-closed fields $(\langle \mathbb{R}, +, \cdot, < \rangle)$ is decidable via Tarski's theorem and quantifier elimination, I'm curious about the impact of adding uninterpreted functions.

I also know that by Nelson and Oppen, the quantifier-free fragment of the union of decidable theories is decidable. However, I am unsure if quantifier elimination is possible when the function's input is quantified.

Specifically, does the first-order theory over the reals extended with uninterpreted functions remain decidable?

Example:

To illustrate, consider the following statements involving uninterpreted functions $\alpha$ and $\beta$:

$$ \forall (t \in \mathbb{R}) \left( \alpha(t) = 1.0 \Rightarrow \exists (t_1 \in \mathbb{R}) \left( t_1 > t \land t_1 < (t + 10.0) \land \beta(t_1) = 2.0 \right) \right) $$

Would such statements be within a decidable theory when $\alpha$ and $\beta$ are uninterpreted functions?

I appreciate any insights or references to relevant literature on this topic!

bytemouse
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1 Answers1

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No, we do not keep decidability. (This uses a trick closely related to this old answer of mine.)

Letting $\alpha$ be a "fresh" function symbol, consider the sentence $\theta$ which says that $\alpha^{-1}(0)$ defines an integer part of the universe; that is, $\alpha^{-1}(0)$ is a discrete ordered ring and each $x$ is within $1$ (or even $1\over 2$) of some element of $\alpha^{-1}(0)$. Basically, $\theta$ "carves out" a part of the universe which looks like the integers. The theory of discrete ordered rings is known to be incomplete, essentially for the same reason as Robinson arithmetic (although we have to be careful since we don't have induction); in particular, letting $T$ be the theory in your post, we get that the set of $T$-theorems of the form $\theta\rightarrow\sigma$ for some $\sigma$ is not computable and so $T$ is not decidable.

(Note that all we really needed was a unary relation symbol.)

Noah Schweber
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