Actually, you are missing the point. ;) The point of bisimulations is
that they are defined *coinductively*, so they let you work with
potentially infinite data structures. In your example, proving that
xs and ys are in the relation R really is that simple -- 1 = 1, and
then to complete the proof we are allowed to use the coinduction
hypothesis that xs and ys are in the relation R, since they are
guarded by a constructor (:).

Dan, does this help answer your original question? If not I can try
to give a more detailed answer in the morning.

I understand the coinduction principle for data structures like streams (e.g., Felipe's example) and finitely branching trees (from papers like "A calculus of binary trees"). In general, for lists and types constructed from arrow, product, and so on, it's easy to define conditions for a relation to be a bisimulation. For instance, I know that a relation R is a bisimulation over n-branching trees t1 and t2 (for some n) if their roots are equal and each of their subtrees are in R. My problem is, specifically, with the case of infinitely branching trees. In Haskell, these are modeled by the data type

T a = T a [T a]

and the possibility arises, of course, that the list [T a] is a stream. Clearly, we can't just say that a relation R is a bisimulation on trees t1 and t2 of type T a if their root values are equal and their lists of subtrees are equal. Because if the lists are infinite, we have to prove that they are bisimilar. And the coinduction principle for lists requires us to have established that the head of each list is equal. But this is what we're trying to prove!

So my questions is: what must be established to show that a relation is a bisimulation on an infinitely branching tree of type T a? This question arises because I'm having trouble "combining" the individual principles for trees and lists.