Proof of Robinson's test I have been working with Tent and Ziegler's Model Theory.  
I am on the Quantifier elimination chapter, and there they mention Robinson's test. It says that, for an $L$-theory $T$ three statements are equivalent:


*

*$T$ is model complete

*Given two models $\mathfrak{A}_1$ and $\mathfrak{A}_2$ of $T$ such that $\mathfrak{A}_1\subset \mathfrak{A}_2$, and given any existential sentence $\varphi$, then $$ \mathfrak{A}_2\models\varphi \implies \mathfrak{A}_1\models \varphi.$$

*Each formula is, modulo $T$, equivalent to a universal formula. 


I understand the proofs of $3\implies 1$ and $1\implies 2$, but I can't do $2\implies 3$ (and their comments in the book didn't help). 
How can you proof that implication?
 A: Hint: Assume 2. Let $\phi$ be any sentence. Consider the set $\Sigma = \{\psi \colon \psi \text { is universal and } T \models \phi \to \psi\}$. Show that $T \cup \Sigma \cup \{\lnot \phi\}$ is not consistent. Deduce that for some finite subset $\Delta \subset \Sigma$ we have $T \models \Delta \to \phi$. Take as your universal formula the conjunction of formulas in $\Delta$.
A: First you need to use the hypothesis (2) and Corollary 3.1.5 (page 28) to prove the fact (call it Fact X) that existencial formulas are equivalent modulo $T$ to universal formulas. Then, as suggested in Model Theory by Chang and Keisler, page 187 (third edition), start the induction on the complexity of an arbitrary L-formula $\phi$ attempting to prove that $\phi\equiv^Tuniversal$. I believe you should only find trouble in the case $\phi=\exists x.\psi$, where $\psi\equiv ^Tuniversal$. In that case, notice that $\phi\equiv\neg\forall x.\neg \psi$, $\neg\psi\equiv ^Tuniversal$ by Fact X, $\phi\equiv ^T\neg universal\equiv existencial\equiv ^Tuniversal$ by Fact X once more.
