Is $ \Phi \vdash \phi $, if and only if $ \Phi \cup \{\neg \phi \} $ being not satisfiable?

Question

In the first order logic, $ \Phi$ is a set of formulas, and $\phi$ is a formula. Is $ \Phi \vdash \phi $, if and only if $ \Phi \cup \{\neg \phi \} $ being not satisfiable?

Answer 1

The basis of this proof is the following natural deduction laws:

\begin{equation*} \frac{\Phi \cup \{ \phi \}\vdash \bot}{\Phi\vdash\neg\phi} \end{equation*}

and

\begin{equation*} \frac{\Phi\vdash\phi}{\Phi \cup \{\neg\phi\}\vdash \bot} \end{equation*}

$(\rightarrow)$ If $\Phi\vdash\phi$ then $\Phi\cup \{\neg \phi \}\vdash \phi$. But $\Phi\cup \{\neg \phi \}\vdash \neg\phi$ so it can't be satified since it proves a contradiction.

$(\leftarrow)$ If $\Phi\cup \{\neg \phi \}\vdash \phi$ can't be satisfied then it has no models and so $\Phi\cup \{\neg\phi\}\models\bot$. By using the semantic completeness theorem we get $\Phi\cup \{\neg \phi \}\vdash\bot$ which by using the rules of natural deduction results in $\Phi\vdash\phi$.

Observation: This is true even in the case in which $\Phi$ is incoherent (it can prove $\bot$).