I need to prove the statement but am not sure about my proof. Here it is:
First we recall some definitions and basic results that we will use to prove the statement. At theory is model complete if for all models $\mathfrak{M}^{1}$ and $\mathfrak{M}^{2}$ of $T$, if $\mathfrak{M}^{1} \subset \mathfrak{M}^{2}$ then $\mathfrak{M}^{1} \prec \mathfrak{M}^{2}$. A theory $T$ is complete if and only if all models of $T$ are elementarily equivalent. Whenever $\mathfrak{M}^{1}$, $\mathfrak{M}^{2}$ are models of $T$ with a common $\mathcal{L}$-substructure $\mathfrak{A}$, then Th($\mathfrak{M}_{A}^{1}$)=Th($\mathfrak{M}_{A}^{2}$). We can now say that $\mathfrak{M}$ is a common elementary substructure since it embeds into every other model of $T$. Now since all those models have a common $\mathcal{L}$-substructure, the equivalence of each theory of each model follows. Hence $T$ is complete.
Could anyone validate it or give me a hint on how to tackle it?
Best Answer
This is false in general! It's equivalent to saying that $T$ has quantifier elimination, and there are model complete theories which do not have quantifier elimination.
Instead of using the fact that $\mathfrak{M}^1$ and $\mathfrak{M}^2$ have a common $\mathcal{L}$-substructure, use the (much stronger) fact that they have a common elementary substructure.
Suppose a sentence $\varphi$ is true in $\mathfrak{M}^1$. Then it's true in $\mathfrak{M}$. Then it's true in $\mathfrak{M}^2$. Do you see why? And do you see why this implies $T$ is complete?