Hilbert's style proof (FO logic) I am stuck with this question to check whether the following formulas are valid and if they are valid, then derive them using Hilbert's axiom schema and Modes Ponens for First Order Logic.
\begin{align}
\exists x(\alpha \rightarrow \beta) \rightarrow (\exists x \alpha \rightarrow \exists x \beta) , x \notin FV(\beta)  
\end{align}
\begin{align}
(\alpha \rightarrow \beta) \rightarrow (\exists x \alpha \rightarrow \exists x \beta) , x \notin FV(\beta)  
\end{align}
 The relevant axioms are:


*

*$\alpha \rightarrow (\beta \rightarrow \alpha)$

*$(\alpha \rightarrow (\beta \rightarrow \gamma)) \rightarrow ((\alpha \rightarrow \beta)\rightarrow(\alpha \rightarrow\gamma))$

*$\forall x \alpha \rightarrow \alpha[x/t]$, where $t$ is substitutable for $x$

*$\alpha \rightarrow \forall x \alpha, x \notin FV(\alpha)$ 

*$\forall x (\alpha \rightarrow \beta) \rightarrow (\forall x \alpha \rightarrow \forall x \beta)$



I am not really sure if any other axioms will be required or if all the above listed ones will be required. Any hints will be appreciated.

You can't prove it.
Consider the field of rationals. Take $\alpha(x)$ as $x=0$ and take $\beta$ as $\exists y(y\cdot 0=1)$. Then $\exists x (\alpha(x)\to\beta)$ holds (just take $x=1$) and $\exists x\alpha(x)$ also holds, but $\exists x\beta$ is false.
