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Let $\Sigma_1$ and $\Sigma_2$ be sets of $L$-sentences such that no symbol of $L$ occurs in both $\Sigma_1$ and $\Sigma_2$. Suppose $\Sigma_1$ and $\Sigma_2$ have infinite models. Then $\Sigma_1 \cup \Sigma_2$ has a model.

We have some facts about back and forth systems and Vaught's test, and Löwenheim-Skolem's theorem, and by that last one I can say that $\Sigma_1, \Sigma_2$ have models of cardinality equal to the cardinality of $L$, but I don't know where to go from there. For instance, we don't necessarily want to show that $\Sigma_1 \cup \Sigma_2$ is complete, so things like Vaught's test seem irrelevant.

Does this just follow from a slick argument about compactness?

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(1) Are you treating equality as a primitive concept? (2) Is Gödel's completeness theorem available? –  Henning Makholm Oct 24 '11 at 1:11
    
So, for instance, what if you take a unary operator $U$ and let the sentence $\forall x Ux$ this can certainly be true in a model of $\Sigma_1$, but in the union it is not true in general, since no model of $\Sigma_2$ has any restrictions based on $U$. Would "closing it under Skolem functions" change that? I don't quite know what that means. –  JeremyKun Oct 24 '11 at 1:12
    
@HenningMakholm, we do have the completeness/compactness theorems available. And equality is not considered part of the language. This is all within predicate logic, and I think this is what you mean by equality being primitive. –  JeremyKun Oct 24 '11 at 1:13
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I asked because it is possible to formulate predicate calculus such that equality is not a logical concept, but one that a theory has to define itself with appropriate proper axioms if it needs it. (For example, Mendelson develops set theory with $a=b$ being an abbreviation for $\forall x.x\in a\leftrightarrow x\in b$). –  Henning Makholm Oct 24 '11 at 1:18
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My point is that while the choice in your course is by all means a common and respectable one, it probably cannot be said to be the standard one. –  Henning Makholm Oct 24 '11 at 1:22

2 Answers 2

up vote 5 down vote accepted

Because the theories have infinite models, you can find a $\kappa$ large enough that both theories have models of size $\kappa$. What happens if you take two such models, one from each theory, and simply identify the domains with each other?

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Let $M_1$ and $M_2$ be a models of $\Sigma_1$ and $\Sigma_2$. By Skolem-Löwenheim we can assume without loss of generality that $M_1$ and $M_2$ have the same cardinality. And then, again without loss of generality, we can assume that they have the same set of individuals and just come with different interpretations of their respective parts of the language ...

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