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!

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.
