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I can see how a theory can be complete but undecidable.

But I have trouble with the decidable but incomplete.

I saw some examples, but is there a property or how?

Edit: I saw the question marked as duplicate, but I am hoping to learn what property makes theories both decidable and incomplete rather than examples. The answers there focus more on complete and undecidable

Bram28
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csp2018
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    I just reopened this question. I agree with the OP that the marked duplicate question mostly focused on complete but undecidable theories, rather than incomplete but decidable. And, it wasn't really explained how it is that the one example given there is decidable but incomplete. Indeed, the intuition of the OP is: how is it possible for something to still be decidable, when it is already Incomplete? Incompleteness seems to say: there are some things that are true but not provable .... so how can you still decide whether or not it is true? I think the OP is looking for an explanation of that – Bram28 Nov 06 '19 at 14:50
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    Here's the link to the related question: https://math.stackexchange.com/questions/3017207/example-of-incomplete-but-decidable-theory-and-of-complete-and-undecidable-the. – Noah Schweber Nov 06 '19 at 18:58
  • @Bram28 Yes, that sums up my confusion. Thanks! – csp2018 Nov 06 '19 at 20:19

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For simplicity, by "theory" I mean "consistent, but not necessarily deductively closed, set of sentences."

There is an easy recipe for whipping up decidable incomplete theories in a given language:

Suppose $\Sigma$ is a finite language (= finitely many function, constant, and relation symbols). Let $T_\Sigma$ be the $\Sigma$-theory consisting of the single sentence $$\forall x,y,z(x=y\vee y=z\vee x=z).$$ The models of $T$ are exactly the $\Sigma$-structures with at most two elements.

  • Decidability: Since $\Sigma$ is finite, there are (up to isomorphism) only finitely many models of $T$ (this is a fancy way of saying that up to isomorphism there are only finitely many structures of bounded finite size in a given finite language). Fix representatives $\mathcal{M}_i$ ($1\le i\le n$). Each $\mathcal{M}_i$ is finite, and we can computably check whether a sentence is true in a finite structure (this is a good exercise), so to tell whether $T\models\varphi$ we just ask "Is $\varphi$ true in each $\mathcal{M}_i$?"

    • That is: $T$ is decidable since it has finitely many models, each of which has a decidable theory.

  • Incomplete: $T$ does not decide whether the sentence $$\forall x,y(x=y)$$is true, since (this is a good exercise) $T$ has at least one model of size exactly $1$ and at least one model of size exactly $2$.

Basically, $T$ is strong enough to limit us to an "easily-checkable" set of possibilities, but still leaves some variety amongst those possibilities. Knowing all the options doesn't mean there aren't any options.

The following generalization of the above is a good exercise:

Suppose $T_1,...,T_n$ are decidable theories in a given language. Then the theory $$S=\{\varphi_1\vee...\vee\varphi_n: \varphi_1\in T_1,...,\varphi_n\in T_n\}$$ is decidable. (HINT: show the models of $S$ are exactly the structures satisfying some $T_i$.)

This lets us build incomplete decidable theories from complete decidable theories: simply take as our $T_i$s a bunch of decidable, complete, but mutually inconsistent theories.

It is worth noting that there are simple-looking theories which are not in fact decidable. For example, consider the empty theory $E$ in the language of arithmetic. There is a finitely axiomatizable undecidable theory $Q$ in this language; letting $\varphi$ be the conjunction of the axioms of $Q$, we have (by the deduction theorem) $$Q\models \theta\quad\iff\quad E\models\varphi\rightarrow\theta.$$ Basically, the set of consequences of the empty theory in a given language is at least as complicated as the set of consequences of any finitely axiomatizable theory in that language. (See also here for some notes on the behavior of finitely axiomatizable theories which are somewhat relevant to this question.)

Noah Schweber
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  • Thanks! I will have to do more research to fully grasp. Would "this sentence is false" make a theory both undcidable & incomplete, while "this sentence is unprovable" make a theory incomplete(syntatically) but not necessarily undecidable(semantic)? – csp2018 Nov 07 '19 at 14:27
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    @csp2018 No, that's not related. Neither of those sentences is generally expressible in first-order logic. The first never is, and the second only is in certain situations - but even when it is expressible, it need not be undecidable (consider PA + "PA is inconsistent," which proves that it proves everything while not actually proving everything :P) – Noah Schweber Nov 07 '19 at 16:15