# Finitely valid sentences and other related things

I almost finish the section 2.6 in Enderton's A mathematical introduction to logic, but I still do not understand some thing.

(The first three question are closely related, so I hope that it does not a problem that I ask several questions in one topic.)

## (1) Finitely valid

At the beginning of the subsection Finite Models, he says

Some sentences have only infinite models, for example, the sentence saying that $$<$$ is an ordering with no largest element. The negation of such a sentence is finitely valid, that is, it is true in every finite structure.

• a sentence $$\sigma$$ is finitely valid iff $$\sigma$$ is true in every finite structure

If a sentence $$\sigma$$ has only infinite models (but not necessarily $$\sigma$$ is true in every infinite model, i.e., there can be some infinite structure such that the structure is not model of $$\sigma$$), then the negation of $$\sigma$$ is finitely valid. Am I right? Moreover, it is here some proof? (how can I be sure that there is no infinite models of $$\sigma$$)

Conversely, let $$\sigma$$ be a finitely valid sentence. Clearly the negation of $$\sigma$$ cannot be true in any finite structure (but necessarily $$\sigma$$ is true in every infinite structure). Am I right?

## (2) The class of all infinite structure is not $$EC$$ (second part of Corrolary 26B)

• a class of structures $$K$$ is in $$EC$$ iff there is some setence $$\sigma$$ such that $$\text{Mod }\sigma = K$$

Here is the proof given by Enderton:

If the class of all infinite structure is $$\text{Mod }\tau$$, then the class of all finite structures is $$\text{Mod }\neg\tau$$. But this class isn't even $$EC_\Delta$$, much less $$EC$$.

I accept that what he wrote, but I think that to finish the proof we must show that $$\text{Mod }\tau \in EC$$, but I do not see how it follows from that $$\text{Mod }\neg\tau$$ is not in $$EC$$. I think that something missing, but I am not able to finish the proof by myself.

## (3) Corollary 26E

Assume the language is finite, and let $$\Phi$$ be the set of setences true in every finite structure. Then its complement, $$\overline \Phi$$, is effectively enumerable.

Proof. For a sentence $$\sigma$$, $$\sigma \in \overline \Phi \Leftrightarrow (\neg\sigma) \text{ has a finite model.}$$ [...]

In other words, $$\Phi$$ is a set of finitely valid sentences. Thus its complement, $$\overline \Phi$$, is the set of sentences $$\sigma$$ such that $$\sigma$$ is not true in some finite structure. It does not imply that $$\sigma$$ has only infinite models ($$\sigma$$ can be true in some finite structure, but not in all finite structures). So, we cannot use the facts from (1). How can I get the equivalence?

2. For any sentence $$\sigma$$ whatsoever $$\operatorname{Mod}(\sigma)\in EC$$... this is the definition of $$EC.$$ So in particular $$\operatorname{Mod}(\lnot \tau)\in EC.$$ It doesn't follow from this that $$\operatorname{Mod}(\lnot \tau)\notin EC$$... that follows from the first part of the theorem where they show the class of finite structures is not in EC. This produces a contradiction that proves that a $$\tau$$ with the assumed properties can't exist.
3. This is a corollary to a theorem immediately preceding that shows that the set of all sentences that have a finite model is effectively enumerable (as you have said, $$\bar\Phi$$ is just the set of all sentences whose negations have a finite model). This in turn follows from a couple paragraphs up where they show that for any $$n$$, it is decidable if a sentence has a model of size $$n.$$ (All of this assuming a finite language.)