Exercise 2.5.12 Let \(\phi(v)\) be an \(\cal{L}\)-formula. Show that the followings are equivalent.

i. There is a universal formula \(\psi(v)\) such that \(T \vDash \forall v ( \phi(v) \leftrightarrow \psi(v) )\).

ii. If \(\cal{M}\) and \(\cal{N}\) are models of \(T\) with \(\cal{M} \subseteq \cal{N}\), \(a \in M\), and \(\cal{N} \vDash \phi(a)\), then \(\cal{M} \vDash \phi(a)\).

Proof: (i \(\Rightarrow\) ii) Let \(\psi(v)\) be a universal formula such that \(T \vDash \forall v ( \phi(v) \leftrightarrow \psi(v) )\). Let \(\cal{M}\) and \(\cal{N}\) be models of \(T\) with \(\cal{M} \subseteq \cal{N}\), \(a \in M\), and \(\cal{N} \vDash \phi(a)\). We have:

(ii \(\Rightarrow\) i) Consider the following collection:

By compactness and by the fact that \(\Gamma\) is closed under conjunctions, \(T \cup \Gamma \vDash \phi(v)\) implies that \(T \cup {\psi(v)} \vDash \phi(v)\), and hence, \(T \vDash \forall v (\phi(v) \leftrightarrow \psi(v))\) for some \(\psi(v) \in \Gamma\). To that end, let \(\cal{M} \vDash T \cup \Gamma\). We will show that \(\cal{M} \vDash \phi(v)\). Consider the theory \(T' := T \cup \diag \cal{M} \cup \{\phi(v)\}\). For a contradiction, assume \(T' \vDash \bot\). By compactness, there exists \(a \in M\), \(\chi(a, v) \in \diag \cal{M}\) such that \(T \cup \{\phi(v), \chi(a, v)\} \vDash \bot\). This means \(T \cup \{\phi(v)\} \vDash \forall x \, \neg \chi(x, v)\) showing \(\forall x \, \neg \chi(x, v) \in \Gamma\). Thus, \(\cal{M} \vDash \forall x \, \neg \chi(x, v)\) but \(\cal{M} \vDash \chi(a, v)\), a contradiction. Therefore, \(T'\) is satisfiable. Let \(\cal{N} \vDash T'\). Then, by ii, \(\cal{M} \vDash \phi(v)\).