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Let's say you have a structure $A$ of some language and a set of all sentences that are true in such structure $Th(A)$ of the same language.

Then $Th(A)$ is supposedly a complete theory. I fully believe that $A$ is a model of $Th(A)$ but have hard time understanding why should it be the only one.

I mean, why can't we have a structure $B$ in which all $Th(A)$ sentences are true but where one additional sentence is true. Then it would still be a model of $Th(A)$ (as $Th(A)$ is subset of $Th(B)$) which means $Th(A)$ wouldn't be a complete theory (it would have two non-equivalent models).

BrianO
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Petrroll
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1 Answers1

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Let $\varphi$ be any sentence of the language $L$, and let $A$ be an $L$-structure. Then one of $\varphi$ or $\lnot\varphi$ is true in $A$. Hence $\text{Th}(A)$ is complete.

We look in particular at your structure $B$. Let $\psi$ be the "additional" sentence of $L$ which is true in $B$. If $\psi$ was not true in $A$, then $\lnot\psi$ was true in $A$, so is an axiom of $\text{Th}(A)$. Since $B$ is a model of $\text{Th}(A)$, it follows that $\psi$ cannot be true in $B$.

It is perfectly possible by extending $L$ to a new language $L'$, to produce a new sentence $\psi$ and an $L'$-structure $B$ in which $\psi$ is true. But this can only be done with a sentence that is not a sentence of $L$.

André Nicolas
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  • Could you elaborate a bit more on the "Hence" part and how exactly does it counter my argument in 3rd paragraph? I mean, I believe that it works the way you wrote it but I'd very much prefer to understand why and all implications of it. – Petrroll Dec 25 '15 at 19:13
  • Ok, understand it now. Thanks :)! – Petrroll Dec 25 '15 at 19:26
  • @Petrroll: You are welcome. I hope the added material helps. In principle the first paragraph is enough, since a theory is by definition complete iff for any $\varphi$ of the language, one of $\varphi$ or $\lnot\varphi$ is a theorem. – André Nicolas Dec 25 '15 at 19:30
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    Just as a footnote, it is possible to have $Th(A) = Th(B)$ for two nonisomorphic models $A, B$ — for example, if both are models of a complete theory, such as real closed fields. First order logic can't characterize the cardinality of models (by the Lowenheim-Skolem theorem), so $A$ an $B$ might even have different sizes. – BrianO Dec 25 '15 at 20:31
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    This statement has always confused me: "Then one of $\varphi$ or $\lnot\varphi$ is true in $A$." Is there some implicit platonism going on here preventing $\varphi$ from being independent of $A$? (Are we abandoning formal proof?) – Andrew Szymczak Sep 08 '23 at 02:16
  • @AndréNicolas why can you say that either $\varphi$ or $\neg \varphi$ is true in $A$? I think that this is the point of the question. – Mayk Alves de Andrade Sep 12 '23 at 19:56