Prove Craig's interpolation theorem by Robinson's consistency theorem

Question

Let $ \varphi $ be a $ \tau_1 $-sentence, $ \psi $ be a $ \tau_2 $-sentence. $ \tau=\tau_1\cap\tau_2 $. Suppose that $ \varphi\models\psi $. We need to show there is $ \tau $-sentence $ \theta $ such that $ \varphi\models\theta $ and $ \theta\models\psi $

Suppose $ \Gamma=\lbrace\sigma|\varphi\models\sigma,\sigma\ is\ a\ \tau_1-sentence\rbrace $, We need to show $ \Gamma\models\psi $. For the contradiction, Suppose that $ \Gamma\nvDash\psi $. Let $ \cal M $ be a model such that $ \cal M\models$ $ \Gamma $$\cup\lbrace\neg\psi\rbrace $ . Let T=Th$_\tau (\cal M) $=$ \lbrace\gamma|\cal M\models\gamma, $$\gamma$ is a $ \tau $- sentence$\rbrace $. Then T is a complete theory. So T$ \cup\lbrace\neg\psi\rbrace $ can be satisfied. If we can show that T$ \cup\lbrace\varphi\rbrace $ can be satisfied, then by Robinson Interpolation theorem, T$ \cup\lbrace\varphi\rbrace\cup \lbrace\neg\psi\rbrace $ can be satisfied. But it's contradict with $ \varphi\models\psi $. Then finished.

But how to show that T$ \cup\lbrace\varphi\rbrace $ can be satisfied?

Answer 1

Your set $\Gamma$ should be a set of $\tau$-sentences. I assume this was a typo?

Suppose $T\cup\{\varphi\}$ is inconsistent. By compactness, there is a sentence $\sigma\in T$ such that $\sigma\models \lnot\varphi$. Equivalently, $\varphi\models\lnot\sigma$. But then $\lnot\sigma\in\Gamma\subseteq T$, contradicting consistency of $T$.

Answer 2

Let $ \varphi $ be a $ \tau_1 $-sentence, $ \psi $ be a $ \tau_2 $-sentence, and $ \tau=\tau_1\cap\tau_2 $. Suppose that $ \varphi\models\psi $. We need to show there is a $ \tau $-sentence $ \theta $ such that $ \varphi\models\theta $ and $ \theta\models\psi $

Define $ \Gamma=\lbrace\sigma|\sigma\mbox{ is\ a\ }\tau\mbox{-sentence and }\varphi\models\sigma \rbrace $, We need to show $ \Gamma\models\psi $. Suppose by contrary that $ \Gamma\nvDash\psi $. Let $ \cal M $ be a model such that $ \cal M\models$ $ \Gamma $ $\cup\lbrace\neg\psi\rbrace $ . Let $$T=\mbox{Th}_\tau (\cal M) = \lbrace\gamma|\gamma\mbox{ is a } \tau \mbox{- sentence and }\cal M\models\gamma\rbrace.$$ Then $T$ is a complete $\tau$-theory. Let $T_1=T\cup\{\varphi\}$, $T_2=T \cup\lbrace\neg\psi\rbrace $. Clearly ${\cal M}$ is a model of $T_2$. Next we show that $T_1$ also has a model. Suppose otherwise, by compactness, there are $\sigma_1, \cdots, \sigma_n\in T$ such that $\{\sigma_1,\cdots,\sigma_n,\varphi\}$ has no model, that is， $\varphi\models \neg (\sigma_1\wedge\cdots\wedge\sigma_n)$, hence $\neg (\sigma_1\wedge\cdots\wedge\sigma_n)\in \Gamma$. Then ${\cal M}\models \neg (\sigma_1\wedge\cdots\wedge\sigma_n)$, contradicts the fact that ${\cal M}\models \sigma_1\wedge\cdots\wedge\sigma_n$. Therefore, $T_1$ has a model.

By Robinson consistency theorem, $T_1\cup T_2$ has a model, hence $\varphi\wedge\neg\psi$ can be satisfied. But this contradicts the fact that $ \varphi\models\psi $. Thus $\Gamma\models \psi$. By compactness, there are $\sigma_1,\cdots,\sigma_k\in\Gamma$ such that $\sigma_1\wedge \cdots\wedge\sigma_k\models \psi$. Clearly, $\sigma_1\wedge\cdots\wedge\sigma_k$ is a required interpolation sentence.