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Let $$\mathfrak{A} = <A, \Sigma^f_A, \Sigma^r_A>, \mathfrak{B} = <B, \Sigma^f_B, \Sigma^r_B> $$ be a structures where $\Sigma^f_A, \Sigma^f_B $ set of function symbols, $\Sigma^r_A, \Sigma^r_B $ are set of relationships. Now, let $$\mathfrak{C} = \mathfrak{A} \cup \mathfrak{B} = <A \cup B, \Sigma^f_A \cup \Sigma^f_B, \Sigma^r_A \cup \Sigma^R_B> $$ Let $\Sigma^f_A \cup \Sigma^f_B = \emptyset $. Let $\mathfrak{A}, \mathfrak{B} \models \phi$ Prove that $\mathfrak{C} \models \phi$ or give a contrargument.

It seems that it is not true. But I cannot find an argument. Please help.

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  • $\begingroup$ $\Sigma^f_A$ should be a family of functions that interpret the function symbols, and likewise for $\Sigma^f_B$. Similarly, $\Sigma^r_A$ and $\Sigma^r_B$ are interpretations of the relation symbols. $\endgroup$ – Hans Hüttel Oct 25 '16 at 14:18
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Consider the non-transitivity property

$$ \exists x. \exists y. \exists z. R(x,y) \wedge R(y,z) \wedge \neg R(x,z)$$

It is easy to find two first-order structures $(A,\Sigma^f_A, \Sigma^r_A)$ and $(B,\Sigma^f_B,\Sigma^r_B)$ for which this property is true for the interpretation of $R$ (as described in $\Sigma^r_A$ and $\Sigma^r_B$, respectively) but such that the union of the structures fails to satisfy it.

Let $R_A = \{ (1,2), (2,3), (1,4) \}$ and $R_B = \{ (1,3), (3,4) \}$. Then $R_A$ and $R_B$ are both non-transitive, but $R_A \cup R_B$ is transitive.

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  • $\begingroup$ Thanks! That is what I want :) $\endgroup$ – user376326 Oct 25 '16 at 20:34
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$\mathfrak{C}$ isn't even well-defined without additional assumptions. Here's an example: Suppose that $A\cap B\neq\emptyset,$ and there is a relation symbol $R$ that is in both signatures. Let $a_0$ be some member of $A\cap B,$ and define the interpretations of $R$ in the two models so that $R(a_0)$ is true in $\mathfrak{A}$ but false in $\mathfrak{B}.$

Then what is the interpretation of $R$ in $\mathfrak{C}?$ Specifically, is $R(a_0)$ defined to be true or false in $\mathfrak{C}?$

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If you also assume that $A, B$ are disjoint, then $\mathfrak{C}$ is well-defined. The answer to your question now is no. HINT: there is a sentence $\varphi$ which is true in exactly those structures which have exactly 17 elements . . .

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  • $\begingroup$ I cannot see what do you mean. But, I found another example. And there is less elements than 17. I also cannot see why $A, B$ should be disjoint. $\endgroup$ – user376326 Oct 25 '16 at 20:33

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