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

[Christian Sternagel]

> 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