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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.

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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 $y$?

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  • $\begingroup$ Thanks, Alex -- although it's been awhile since I've looked at Sacks (I put this problem away as "done"), I will think about your suggestions . . . I also appreciate your edits! $\endgroup$ – لويس العرب Feb 15 '14 at 23:14
  • $\begingroup$ Maybe I should try direct approach. Two ideas come to mind: one using chains and the other using a theorem of Robinson's proposition 9.3. The latter: Suppose T is a model completion of S, a universal theory. By 9.3, T is universal-existential . . . $\endgroup$ – لويس العرب Feb 16 '14 at 0:08

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