I am really puzzled right now.
To solve the issue, I need to prove this formular:
$$ \exists x P(x) \vdash P(x) $$
with the natural deduction rules for propositional and predicate logic.
I am sure this example should not be that difficult but, yep, now I am here.
For me, it should not be a counter example, so it should be solvable. I started to use exists elimination to get rid of the $\exists$, but than I end up with something like this:
\begin{align} \exists x P(x) \vdash P(x) \\ 1.&\ \exists x P(x) \ \text{(prem)} \\ 2.&\ P(x_0) \ \text{($x_0$ fresh/free)} \\ 3.&\ \dots (get \ x \ instead \ of \ x_0)\\ 4.&\ P(x) \end{align}
So I miss the little hint how I can transform the new variable to the existing one.
I hope someone can guide me.
Thank you.
As suggested by Henning Makholm's comment, $\exists x P(x) \vdash P(x)$ is not provable. If it were provable then you could take a derivation of $\exists x P(x) \vdash P(x)$ and, by applying the rule $\forall_I$ in your list for natural deduction, you would get a derivation of $\exists x P(x) \vdash \forall x P(x)$, which is not provable. Indeed, $\exists x P(x) \vdash \forall x P(x)$ means that if there exists something with the property $P$ then everything has the property $P$, which is clearly falsifiable: take $\mathbb{N}$ as domain, and let $P(x)$ be the property "$x$ is even", so that in this structure $\exists x P(x)$ is true but $\forall x P(x)$ is false.
Said differently, there is no way to fill the dots in your attempt of derivation, using natural deduction inference rules.