Can a proof be constructed to show there is no distinguisher? Let's assume a simple algorithm like the Skein hash function.

Is it possible, given the algorithm, to construct a proof that it does not have a particular distinguisher, something like:

$P(xyz)$ is the probability that $xyz$ is truly random over some alphabet,

Given $\vert y \vert = l$, for some fixed length l, $z = f(x)$ (i.e., $z$ is dependent on $x$).

Not in general, of course, but for a particular such distinguisher.
There are a number of distinguishers that it it would be easy to prove are not present in a hash function.

For example, I can easily prove that Skein does not have the distinguisher "the 2nd bit in the output is equal to the first bit of the output with probability 1". The proof would be a simple example of a message whose digest does not have this property (which should be fairly easy to find).

For more interesting distinguishers, the problem becomes much harder (on a side note, is the distinguisher really interesting if I can prove it doesn't hold for a particular hash function?).

My gut feeling is that yes, one could come up with an "interesting" distinguisher and then show that it doesn't hold for the hash function, but I can not offer more than a gut feeling.