Could the Riemann hypothesis be provably unprovable? I don't know much about foundations and logic, so I ask forgiveness if my question is just plain stupid.
Assume we have a statement of the form:

There exist no $x\in X$ such that $P(x)$.

where $X$ is some kind of set (or class, or similar stuff) and $P$ is a set of properties. An example of such a statement could be the Riemann hypothesis:

There exist no $x\in\mathbb{C}$ such that $\Re(x)\neq\frac{1}{2}$ such that $x$ is not a negative even integer and $\zeta(x)=0$.

Can such a statement be provably unprovable?
Intuitively, I would say no, because to show that it is unprovable we should show that we cannot find $x\in X$ such that $P(x)$ (else finding such an $x$ would be a proof that the statement is false), but doing so would prove the statement to be true.
Is this correct, or am I missing something?

Edit: Please read the question correctly: it is not properly a question on the RH, but more a question on logic!
 A: You can check out the answers to this related MO question:
"Can the Riemann hypothesis be undecidable?"
A: If $p$ is a zero off the critical line, there are rational numbers $A<B, C<D$ such that the square with corners $A+iC, B+iC,B+iD, A+iD$ does not intersect the critical line and has a zero (namely $p$) inside it.  Thanks to the Argument Principle, this fact can be proven by computing a contour integral numerically 
with sufficient precision.  So if RH is false, it must be provably false. 
And so a proof of undecidability would be a proof that it is true.
A: Let us say that there exists some $x \in \Bbb{C}$ such that $x$ is neither a negative integer nor on the line of $\Re(x) = \frac{1}{2}$.  Let us further suppose that some inspired mathematician conjectured that $x$ could be expressed as the sum of some convergent but some god-awful complicated series.
Now "experiemental mathematicians" might well look at this $x$ and determine that 
$\zeta(x)$ is zero to a thousand decimal places; but of course that does not constitute a proof that $\zeta(x)=0$.  And it is certainly logically plausible that the theorem 
"$\zeta(x)$, with $x$ defined as $\sum_\infty (\mbox{this horribly complicated mess})$, is zero" could be true but unprovable. 
So it is plausible that the Riemann hypothesis could be undecidable within the usual mathematical axiomatic framework.  Contrast this with Fermat's last theorem (ignoring Wile's proof): there, any given purported counterexample can be tested algorithmically in a finite number of steps, so if you could prove it was undecidable, you would be proving it false.  (This argument would not say it can't be undecidable, just that it cannot be proven to be undecidable.)
A: No. The problem is in the definition of zeros. If we would not have 1,2,3,4,5,6... on the left side of the equation, which is the sum 

we could argue that we do not have enough information about zeros, hence some of them are not defined in some particular sense, hence Riemann cannot be decided. However, although we cannot calculate all of the zeros, there is nothing ambiguous about them. They are all perfectly defined objects on their own accord, so if the Riemann hypothesis is false there is the first zero off 1/2 mark.
Although I said this, hypothetically speaking it could be true that the set of zeros is not defined completely, by simply stating that they are zeros of the extended sum as above. This might mean that we would need to add something to the definition of natural numbers, some minuscule and yet not discovered condition that would change their nature in such a way that with one condition Riemann is true, yet with another false.
This last is not likely because we can already define the way of calculating any zero anywhere, although we do not have time and space to collect all of them.
So it is definitely true or false with Riemann.
A: The answer to the question is no: the Riemann hypothesis couldn't be provably unprovable. A quick way to show this is to start by observing that it cannot be false and provably unprovable, because if it is false there must be a counterexample that can be shown to be a counterexample, by which we could prove it false. Contradiction. So if we assume it's provably unprovable then it must be true, which is also a contradiction.
But it might be unprovably unprovable.
Note that it must be either true or false: either there is a counterexample or there isn't.
