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Let L be an arbitrary language and $T_1$ and $T_2$ be non-empty L-theories. Suppose that the L-theory $T_1 \cup T_2$ is inconsistent. Proof that there is an L-sentence $\phi$ such that $T_1$ $\vDash$ $\phi$ and $T_2$ $\vDash$ $\neg$ $\phi$.

Is the following proof correct?

Because of the compactness theorem, there exist two subtheories $\Delta_1$ $\subset$ $T_1$ and $\Delta_2$ $\subset$ $T_2$ such that $\Delta_1$ $\cup$ $\Delta_2$ $\models$ $\bot$ $\equiv$ $\neg$ $\phi$ $\wedge$ $\phi$. So we found our sentence?

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The idea is right. By Compactness, there exist finite subsets $\Delta_1,\Delta_2$ of $T_1$, $T_2$ such that $\Delta_1\cup \Delta_2$ is inconsistent.

Let $\varphi_1$ be the conjunction of the sentences of $\Delta_1$, and let $\varphi_2$ the conjunction of the sentences of $\Delta_2$. Continue.

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No, but by compactness we can choose $\Delta _1 \subseteq T_1$ and $\Delta _2 \subseteq T_2$ both finite such that $\Delta _1 \cup \Delta_2 \models \bot.$ Let $\phi_1$ be the conjunction of all sentences in $\Delta_1$ and $\phi_2$ the conjunction of all sentences in $\Delta_2$, then $\phi_1 \wedge \phi_2 \models \bot \Longleftrightarrow \phi_1\models \neg \phi_2.$ Let $\phi=\phi_1.$

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