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A theory T has property M if the following holds: For A and B models of T, if A is a substructure of B then A is also an elementary substructure of B. I want to prove that if a theory admits Quantifier Elimination then it has property M.

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closed as off-topic by Emil Jeřábek, Tony Huynh, RP_, Chris Godsil, Michael Albanese Apr 7 '17 at 14:10

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  • $\begingroup$ The property is known as model completeness. This is not a research-level question, it belongs to math.stackexchange.com . $\endgroup$ – Emil Jeřábek Apr 7 '17 at 11:08
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The reason is that substructures agree on quantifier-free truth.

If both $A$ and $B$ model $T$ and $T$ admits quantifier-elimination, then for any formula $\varphi(x)$, there is a quantifier-free assertion $\psi(x)$ that $T$ proves is equivalent to $\varphi(x)$. So $A\models \psi(a)$ if and only if $B\models \psi(a)$, since $\psi$ is quantifier-free. Since $\psi$ is equivalent to $\varphi$ in models of $T$, it follows that $A\models\varphi(a)$ if and only if $B\models\varphi(a)$, and this is what it means for $A$ to be an elementary substructure.

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