$\mathcal{A} \vDash T_\forall$ iff there exists $\mathcal{M} \supseteq \mathcal{A}$ such that $\mathcal{M} \vDash T$

Question

Let $T$ be a theory in a first order language $\tau$. Let $T_\forall$ be set of all formulas $\varphi$ of the form $\forall x \psi$, where $\psi$ is quantifier-free, such that $T\vDash \varphi$. Then for all $\tau$-structures $\mathcal{A}$, $\mathcal{A} \vDash T_\forall$ iff there exists a $\tau$-structure $\mathcal{M}$ such that $\mathcal{A} \subseteq \mathcal{M}$ and $\mathcal{M} \vDash T$.

Maybe it is an easy problem, but I don't know how to deal with it. I can't seem to be able to construct a model $\mathcal{M}$ starting with $\mathcal{A}$.

Can anyone maybe give me some tips?

Answer 1

First, as written this is not true: you need $T_\forall$ to contain formulas with any number of universal quantifiers at the start (and no other quantifiers), not just a single universal quantifier. (Maybe you meant this implicitly, with the $x$ in $\forall x\psi$ actually standing for a tuple of variables.) Given the corrected statement, here is what you can do.

Typically the main tool to construct a model like this is compactness. In this case, to express that $\mathcal{M}$ contains $\mathcal{A}$ as a substructure you'll need to extend the language, and then you want to show that $T$ together with axioms saying that $\mathcal{A}$ is a substructure is finitely satisfiable.

A more detailed sketch of the proof is hidden below.

Add a constant symbol to $\tau$ for each element of $\mathcal{A}$. Let $S$ be the theory consisting of all quantifier-free sentences over this enlarged language that are true in $\mathcal{A}$. Note that a model of $S$ is exactly a $\tau$-structure that contains a substructure isomorphic to $\mathcal{A}$. So, we just have to show that $S\cup T$ has a model. But for any finite $S_0\subset S$, taking the conjunction of its elements and replacing its constants for elements of $\mathcal{A}$ with variables, we get a single quantifier-free formula $\psi$ over $\tau$ such that $\forall x_1\dots\forall x_n\neg\psi\not\in T_\forall$ (since $\mathcal{A}\models T_\forall$). Thus $T\not\models \forall x_1\dots\forall x_n\neg\psi$ and so $T\cup S_0$ is consistent.