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I am working on the following exercise from Boolos' Computability and Logic:

Problem. Suppose an axiomatizable theory $T$ has only infinite models. Suppose $T$ is not complete, [yet has] two isomorphism types of denumerable models. Show that $T$ is decidable.

We are told to use the following two results:

Prop. 1 If an axiomatizable theory $T$ is complete, then $T$ is decidable.

Prop. 2 If $\Gamma$ is a denumerably categorical set of sentences having no finite models, then $\Gamma$ is complete.

Here is how I would proceed. Take a sentence $A$ which for which neither $A$ nor $\neg A$ is a theorem of $T$. Then take $\Gamma = T\cup\{A\}$. If we can show that $\Gamma$ is a denumerably categorical set, then by Props. 1 and 2 we are finished.

Any hints, though, on how to show that $T\cup\{A\}$ is a denumerably categorical set? In other words, why is it the case that adding $A$ to $T$ causes the isomorphism type to drop from two to one?

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    $\begingroup$ Here's one example. Consider the language $\{R\}$ and the axioms stating "if $R$ is anti-symmetric, then $R$ is a dense linear order without endpoints" and "if $R$ is not anti-symmetric, then $R$ is the trivial equivalence relation". $\endgroup$ – Asaf Karagila Apr 22 at 17:54
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If neither $A$ nor $\neg A$ is a theorem of $T$, then both $T\cup\{\neg A\}$ and $T\cup\{A\}$ must be satisfiable - by the Lowenheim-Skolem theorem, that means that there are denumerable models.

Now when we add (say) $A$ to $T$, we "lose" a(n isomorphism type of a) denumerable model - namely, the model of $T\cup\{\neg A\}$. Since $T$ only had two denumerable models to begin with, how many does that leave?

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  • $\begingroup$ But how do we know that adding a sentence $A$ to a theory $T$ does not increase (rather than decrease) the number of unique denumerable models? $\endgroup$ – Doubt Apr 24 at 2:49
  • $\begingroup$ @Doubt You can't gain models by adding sentences. Adding sentences makes it harder, not easier, to satisfy the whole theory: if $\Theta\subseteq\Gamma$ then the class of models of $\Gamma$ is a subclass of the class of models of $\Theta$. Contrapositively, you can't lose models by removing sentences: if $\mathcal{M}\models\Gamma$ and $\Theta\subseteq\Gamma$ then $\mathcal{M}\models\Theta$. $\endgroup$ – Noah Schweber Apr 24 at 2:52
  • $\begingroup$ Thanks Noah. But suppose $M_1$ and $M_2$ are non-isomorphic models for $T$. Isn't it possible that $M_1$ and $M_2$ are both models of $T\cup\{A\}$, but neither is a model of $T\cup\{\neg A\}$? $\endgroup$ – Doubt Apr 24 at 16:21
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    $\begingroup$ @Doubt No, that's not possible if $T$ has only two countable models and $T\cup\{\neg A\}$ is consistent: since $T\cup\{\neg A\}$ is consistent, it has a countable model, which can't be isomorphic to either $M_1$ or $M_2$ in the situation you describe (you can't be isomorphic to something satisfying $A$ while satisfying $\neg A$ yourself). But then we get too many isomorphism types of countable models. $\endgroup$ – Noah Schweber Apr 24 at 16:23

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