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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?

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Whenever $\mathfrak{M}^{1}$, $\mathfrak{M}^{2}$ are models of $T$ with a common $\mathcal{L}$-substructure $\mathfrak{A}$, then $\text{Th}(\mathfrak{M}_{A}^{1})=\text{Th}(\mathfrak{M}_{A}^{2})$.

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.

Now since all those models have a common $\mathcal{L}$-substructure, the equivalence of each theory of each model follows.

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?

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  • $\begingroup$ I think because $\mathfrak{M}$ embeds into every model of T? So the sentence would be satisfied for every model of T actually. SInce now $\mathfrak{M}$ is an elementary substructure oft those, that implies that T is complete, because in this case the theories of all models are equivalent? $\endgroup$ – craft Nov 20 '20 at 17:18
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    $\begingroup$ @craft Yes, that's right. $\endgroup$ – Alex Kruckman Nov 20 '20 at 18:11
  • $\begingroup$ Great, thanks so much $\endgroup$ – craft Nov 20 '20 at 18:23

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