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

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?

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.
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.