⊢∃𝑥(𝜙⟹∀𝑥𝜙) using Hilbert-style proof [duplicate]

I am tasked to find a proof of $\vdash \exists x (\phi \implies \forall x \phi)$ by using the Hilbert axioms. I've been trying for about 2 hours. I tried using the results from the previous parts of the exercise below. I am also allowed to use substitution instances of tautologies (i.e $\vdash p \implies p$ implies $\vdash \phi \implies \phi$ where $p$ is a propsitional variable and $\phi$ is a first-order formula), that $\exists x$ is defined as $\neg \forall x \neg$, and the standard meta-theorems like proof by contradiction.

Basically I've tried to work backwards:

$$\vdash \exists x (\phi \implies \forall x \phi)$$ $$\vdash \forall x (\phi \implies \forall x \phi)$$ $$\vdash (\phi \implies \forall x \phi)$$

This last step seems impossible to prove, as it would have me show that if a property is true for some x then it is true for all $x$ (obviously not true if $\phi$ is taken to be "is even")

Similarly I tried invoking part (c)

$$\vdash \exists x (\phi \implies \forall x \phi)$$ $$\vdash \forall x (\phi \implies \forall x \phi)$$
$$\vdash (\exists x \phi \implies \forall x \phi) \implies \forall x(\phi \implies \forall x \phi)$$

Again, this would require to show that if a propery is true for some x then it is true for all $x$.

The only other way to introduce an $\exists$ as the "main connective" is by (a). But that seems to go nowhere as well.

Solution 1:

Hilbert system is well known to be very cumbersome to prove some straightforward result compared with natural deduction systems, especially for this non-trivial Drinker Paradox sentence and you don't even have the useful deduction meta-theorem. Since we know it'll be a little hard (as the last exercise in your book) so we may stand upon a known result from prenex normal form in predicate calculus that $\vdash (∀x\phi → \psi) \to ∃x(\phi → \psi)$, where $x$ does not occur free in $\psi$. So with this known theorem we can prove your result based solely on your system in only 7 steps as follows:

1. $\vdash (\forall x \phi → \forall x \phi) \to ∃x(\phi → \forall x \phi)~~~~~$ ...known theorem

2. $\vdash (\forall x \phi→((\psi→\forall x \phi) → \forall x \phi))→((\forall x \phi→(\psi→\forall x \phi))→(\forall x \phi→ \forall x \phi))~~~~~$ ...Ax 2

3. $\vdash \forall x \phi→((\psi→\forall x \phi)→\forall x \phi)~~~~~$ ...Ax 1

4. $\vdash (\forall x \phi→(\psi→\forall x \phi))→(\forall x \phi→\forall x \phi)~~~~~$ ...MP from 2, 3

5. $\vdash \forall x \phi→(\psi→\forall x \phi)~~~~~$ ...Ax 1

6. $\vdash \forall x \phi→\forall x \phi~~~~~$ ...MP from 4, 5

7. $\vdash \exists x (\phi \to \forall x \phi)~~~~~$ ...MP from 1, 6. Done!

Now sketch a proof of this known theorem in your specific formal system. As discussed in comments we only have one way to prove any result including $\exists$ quantifier in your system, so after plugin $\lnot \forall \lnot$ we 'd better use proof by negation technique to initally assume $\forall x \lnot (ϕ→ψ)$ where $x$ does not occur free in $ψ$, given the assumption $(∀xϕ→ψ)$. Now perhaps it's clear how to formally proceed with the given assumption and use your Axiom 4 (universal instantiation) to derive at $\bot$, thus our known theorem is proved...