Exercise 2.5.10 Let \(T\) be an \(\cal{L}\)-theory and \(T_\forall\) be all of the universal sentences \(\phi\) such that \(T \vDash \phi\). Show that \(\cal{A}\) \(\vDash T_\forall\) if and only if there is \(\cal{M}\) \(\vDash T\) with \(\cal{A} \subseteq \cal{M}\).

Proof: (\(\Rightarrow\)) Consider the theory \(T' := T \cup \diag \cal{A}\). Any model of \(T'\) is an extension of \(\cal{A}\) which is also a model of \(T\). For a contradiction, assume that \(T'\) is unsatisfiable. By compactness, there exists a quantifier free \(\cal{L}\)-formula \(\phi(\bar{a})\) with \(\bar{a} \in A\) such that \(T \cup \{\phi(\bar{a})\} \vDash \bot\) implying \(T \vDash \forall \bar{x} \ \neg \phi(\bar{x})\). Therefore, \(\forall \bar{x} \ \neg \phi(\bar{x}) \in T_\forall\) and hence \(\cal{A} \vDash \forall \bar{x} \ \neg \phi(\bar{x})\). In particular, \(\cal{A} \vDash \neg \phi(\bar{a})\) which is a contradiction.