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.