Re: [Haskell-cafe] Is (==) commutative?

It's a classically valid inference, so you're "safe" in that respect, and it is true that evaluating x == y requires traversing x and y, so that if x or y "are" bottom, (x == y) and (y == x) will both be bottom. Well, (x == y) could result in bottom, even if neither x nor y are bottom, e.g., [undefined] == [undefined]. -cheers chris