Sum of two structures in the first order logic. 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.
 A: $\mathfrak{C}$ 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 $\mathfrak{C}?$
A: 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 . . .
A: 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.
