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.


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 $M$.

$\Rightarrow$ $\bot$.



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 $y$?

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.