Exercise 17.6 of Sacks' Saturated Model Theory I'd like to know whether my proof is correct. Exercise goes as follows.
17.6. Let $T$ be a model completion of some $\forall$-theory. Show there exists $T^* = T$ s.t. every member of $T^*$ is of the form $\forall x_1,\dots,x_n \exists y \phi(x_1,\dots,x_n, y)$, where $\phi$ has no quantifiers. 
PROOF: 
Suppose for a contradiction (i.e. $\bot$) that there is no such $T^* = T$ having the property, with $T$ a model completion of some universal or $\forall$-theory. Let $M$ be a model for $T$. Hence
$\Rightarrow$ There is a sentence of form $\forall x_1,\dots,x_n \exists y \phi(x_1,\dots,x_n,y) \in T^*$ s.t. $M\not\models \forall x_1,\dots,x_n\exists y \phi(x_1,\dots,x_n, y)$.
$\Rightarrow$ For some value of $\overline{x}$, there is no value of $y$ that validates the formula in $M$.
$\Rightarrow$ But $T$ is a model completion of some $T^\#$ s.t. $T^\#$ is $\forall$ -- by hypothesis.
$\Rightarrow$ So $T$ admits elimination of quantifiers by Theorem 13.2.
$\Rightarrow$ For every value of $\overline{x}$, there is a value of $y$ which validates the formula in $M$.
$\Rightarrow$ $\bot$.
END OF PROOF.

No, your proof needs some patching up. I see a few glaring issues:


*

*In the first step, you assume there is no theory $T^*$ of the specified form which is equivalent to $T$. Then you say "there is some sentence of the specified form in $T^*$ such that..." What is $T^*$? You've just assumed that no such $T^*$ exists.

*Maybe you meant to say: let $T^*$ be any theory of the specified form. Then we know that $T^*$ is not equivalent to $T$. But this doesn't imply that there is a sentence in $T^*$ not satisfied by some model of $M$. It could be that $T^*$ is strictly weaker than $T$ (take $T^*$ to be the empty theory, for example).

*Why does the fact that $T$ eliminates quantifiers imply that there is some witness $y$?
