A property of L-structures related to model completeness Let $T$ be a an $L$-theory & suppose $ \mathcal{A}$ is a structure satisfying all existential consequences of $T$. I want to show that there is a model $\mathcal{M}⊨T$ embedding in an elementary extension $\mathcal{B}$ of $ \mathcal{A}$.
Here's what I've tried: Suppose for each $\mathcal{M}⊨T$, we have  $\text{Diag}(\mathcal{M}) \cup \text{CDiag}(\mathcal{A})⊨\bot $, then by compactness there are quantifier free $\varphi_1(\bar{x}),..,\varphi _m(\bar{x})$ s.t. $\mathcal{M}⊨\exists \bar x\land _i\varphi _i(\bar x)$ &  $\text{CDiag}(\mathcal{A})⊨\neg \exists \bar x\land _i\varphi _i(\bar x)$.
Then there must be $\mathcal{N}⊨T$ with $\mathcal{N}⊨\neg \exists \bar x\land _i\varphi _i(\bar x)$, but I can't see how to proceed to a contradiction.
I hope you can help with this question. If you also have some context for how this relates to model completeness, I would also appreciate that for background.
 A: If $T$ were complete, then the fact that $\mathcal{M}\models\exists\bar{x}\bigwedge_i\varphi_i(\bar{x})$ would mean also that $T\models\exists\bar{x}\bigwedge_i\varphi_i(\bar{x})$, and hence by hypothesis on $T$ that $\mathcal{A}$ also models this sentence, which would give the desired contradiction. So it suffices to find a completion $T'\supseteq T$ such that $\mathcal{A}$ satisfies all existential consequences of $T'$. We do this by Zorn's lemma; let $T'$ be a maximal $L$-theory with the property that $\mathcal{A}$ satisfies all of its existential consequences. If $T'$ is not complete, there is an $L$-sentence $\theta$ and quantifier-free formulas $\varphi_1(\bar{x})$ and $\varphi_2(\bar{x})$ such that neither $\varphi_i(\bar{x})$ is realized in $\mathcal{A}$ but $T'\cup\{\theta\}\models\exists\bar{x}\varphi_1(\bar{x})$ and $T'\cup\{\neg\theta\}\models\exists\bar{x}\varphi_2(\bar{x})$. But then $T'\models\exists\bar{x}\varphi_1(\bar{x})\vee\exists\bar{x}\varphi_2(\bar{x})$, hence $T'\models\exists\bar{x}\left[\varphi_1(\bar{x})\vee\varphi_2(\bar{x})\right]$. By hypothesis on $T'$, $\mathcal{A}$ now also models this sentence, contradicting the definition of $\varphi_1$ and $\varphi_2$.
