# Characterization of a theory whose model has elementary submodels as only its submodels

This is a problem (2.5.12) from Marker's Model Theory: An Introduction of showing that a model has only elementary submodels as its submodels if and only if for every formula is equivalent to some universal ($\Pi_1$) formula.

The "if" part is easy. I'm trying to show the "only if" part by induction. I try to show the induction step for a formula that starts with $\exists$ by finding the witness syntactically, but in vain. How can I do this?

• I always confuse these, but by the "only if" part you mean assuming that it has only elementary submodels? Oct 23, 2014 at 18:29
• Yes! (more keystrokes) Oct 23, 2014 at 18:30

Exercise 2.5.12 is actually a little different than what you've written. First, it's formula-by-formula, and second, it just says that truth of $$\phi$$ goes down to substructures, but not conversely.

For those who don't have a copy of Marker on hand, the text of the problem is:

Show that the following are equivalent:

i) There is a universal formula $$\psi(\overline{v})$$ such that $$T\models \forall\overline{v}\,(\varphi(\overline{v})\leftrightarrow \psi(\overline{v}))$$.

ii) If $$M$$ and $$N$$ are models of $$T$$ with $$M\subset N$$, $$\overline{a}\in M$$, and $$N\models \varphi(\overline{a})$$, then $$M\models \varphi(\overline{a})$$.

I'll give a Hint: Adapt the proof of Theorem 2.3.9: instead of taking $$\Gamma$$ to be all universal sentences which are consequences of $$T$$, take it to be all universal formulas which are consequences of $$\varphi(\overline{v})$$ mod $$T$$. Show that if $$\overline{a}\in M\models T$$ and $$M\models \Gamma(\overline{a})$$, then $$M\models \varphi(\overline{a})$$. Then by compactness, some finite subset of $$\Gamma(\overline{v})$$ implies $$\varphi(\overline{v})$$ mod $$T$$, and the conjunction of this finite subset will be a universal formula equivalent to $$\varphi$$.

But the statement you've asked about in the question ($$T$$ is model complete [if you don't know this term, see Def 3.1.13 in Marker] iff every formula is equivalent mod $$T$$ to a universal formula) follows easily from the exercise. In fact, this is Exercise 3.4.12 in Marker (except with "existential" instead of "universal"...but see below).

How? Well, if $$T$$ is model complete, then condition ii) above holds for all formulas, and hence every formula is equivalent to a universal formula.

Note that if every formula is equivalent to a universal formula, then for any formula $$\varphi$$, $$\lnot\varphi$$ is equivalent to a universal formula $$\psi$$, so $$\varphi \equiv \lnot(\lnot\varphi) \equiv \lnot\psi$$, and hence every formula is also equivalent to an existential formula.

Now for the converse, we have that every formula is equivalent to both a universal formula and an existential formula. Since truth of universal formulas goes down and truth of existential formulas goes up, this shows $$T$$ is model complete.