I have no idea how to begin constructing this proof, and I would appreciate any help!
I need to prove the following, and to make matters worse, without soundness or completness:
$$ \vdash (\forall x. \phi) \rightarrow (\exists x.\phi)$$
I can use the following axioms/theorems:
Ax1 $\langle \forall^\ast (\varphi \rightarrow (\psi \rightarrow \varphi)) \rangle$;
Ax2 $\langle \forall^∗ ((\varphi \rightarrow (\psi \rightarrow \eta)) \rightarrow ((\varphi \rightarrow \psi) \rightarrow (\varphi \rightarrow \eta))) \rangle$;
Ax3 $\langle \forall^∗ (((\neg \varphi) \rightarrow (\neg \psi)) \rightarrow (\psi \rightarrow \varphi)) \rangle$;
Ax4 $\langle \forall^\ast (\forall x.(\varphi \rightarrow \psi)) \rightarrow ((\forall x. \varphi) \rightarrow (\forall x. \psi)) \rangle$;
Ax5 $\langle \forall^\ast (\forall x. \varphi) \rightarrow \varphi^x_t \rangle$ for $t \in \rm{TS}$ a term;
Ax6 $\langle \forall^\ast (\varphi \rightarrow \forall x.\varphi) \rangle$ for $x \notin \operatorname{FV}(\varphi)$; and
MP $\langle \varphi, (\varphi \rightarrow \psi), \psi \rangle$.
and the lemma "Let $\Sigma \vdash \varphi$ and $x \notin \operatorname{FV}(\Sigma)$. Then $\Sigma \vdash \forall x. \varphi$."
Thank you.
EDIT [SN]: The original question did not mention the lemma in the final line. I added it based on another question that otherwise is an exact duplicate of this.