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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.

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

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