If a set of sentences $\Delta$ is satisfiable and $\Delta \subseteq \Gamma$, then $\Gamma$ must also be satisfiable. Show that this does not hold.

Question

Let $\Delta$ and $\Gamma$ be sets of sentences. Definition of satisfiable: If a set of sentences is satisfiable, then it has model.

If a set of sentences $\Delta$ is satisfiable and $\Delta \subseteq \Gamma$, then $\Gamma$ must also be satisfiable. Show that this does not hold. Proof: There could be a model $\mathcal{M}$ s.t. $\mathcal{M} \models \Delta$ but $\mathcal{M} \not\models \Gamma$.

How's this for a proof?

For instance $\mathcal{M}$ could be s.t. for every sentence $\phi \in \Delta$, $\mathcal{M} \models \phi$. But because $\Delta \subseteq \Gamma$ it means there are sentences $\psi \in \Gamma$ and $\psi \not\in \Delta$. So now for every sentence $\psi \not\in \Delta$, but $\psi \in \Gamma$, $\mathcal{M} \not\models \psi$. Therefore, it's not the case that $\mathcal{M} \models \Gamma$, so $\mathcal{M}$ is a counterexample.

Answer 1

The counterexample asks for $\Delta \subseteq \Gamma$ such that $\Delta$ is satisfiable and $\Gamma$ is not. As an example you can take any satisfiable $\Delta$ (e.g. the empty theory) and let $\Gamma = \Delta \cup \{ \bot \}$, where $\bot$ is the symbol for falsity. If you do not like that symbol, just take anything that is tautologically false, e.g. $\exists x(x=x) \wedge \neg \exists x(x=x)$.

Answer 2

Assume $\Delta$ is satisfiable. Choose any sentence $\psi \in \Delta$, and let $\Gamma = \Delta \cup \{ \lnot \psi \}$.