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!
You can check out the answers to this related MO question:

"Can the Riemann hypothesis be undecidable?"
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.