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.
closed as offtopic by Emil Jeřábek, Tony Huynh, RP_, Chris Godsil, Michael Albanese Apr 7 '17 at 14:10
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Emil Jeřábek, Tony Huynh, RP_, Chris Godsil, Michael Albanese

$\begingroup$ The property is known as model completeness. This is not a researchlevel question, it belongs to math.stackexchange.com . $\endgroup$ – Emil Jeřábek Apr 7 '17 at 11:08
The reason is that substructures agree on quantifierfree truth.
If both $A$ and $B$ model $T$ and $T$ admits quantifierelimination, then for any formula $\varphi(x)$, there is a quantifierfree 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 quantifierfree. 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.