# Satisfiability in greater languages

Let $$\mathcal{L}, \mathcal{L}'$$ two languages in first order logic such that $$\mathcal{L}\subset \mathcal{L}'$$, $$\Gamma \subseteq Form_{\mathcal{L}}$$ and $$\varphi \in Form_{\mathcal{L}}$$. Prove that if $$\Gamma \vdash_{\mathcal{L}'} \varphi$$ then $$\Gamma \vdash_{\mathcal{L}}\varphi$$.

In my book, there is a proof when $$\mathcal{L}'$$ is an extension of $$\mathcal{L}$$ with only constant symbols, but this is the general case.

I am trying to proceed by contraposition. In case of $$\Gamma \nvdash_{\mathcal{L}}\varphi$$, then by the Completeness Theorem $$\Gamma \nvDash_{\mathcal{L}}\varphi$$ so $$\Gamma \cup \{\neg \varphi\}$$ is satisfiable.

There I stuck because I suspect that being satisfiable in $$\mathcal{L}$$ implies being satisfiable in $$\mathcal{L}'$$ but I don't know how to prove it or even it is true. Possible answers will be appreciated.

• Intuitively, the greater language adds new symbols but the formula are in the restricted one. But the underlying logic (axioms+rules) does not change. May 28 at 10:42
• More formally, by Soundness the antecedent implies $\Gamma \vDash \varphi$ and thus by Completeness the consequent follows. May 28 at 10:44