Chad Scherrer wrote:
There must be a subtlety I'm missing, right?
What if the types are not instances of Eq?
Jason
Thanks, I figured it was something simple. Now I just to convince myself there's no way around that. Is there a proof around somewhere?
Yes, there is a proof that seq :: a -> b -> b with the semantics as described in the Haskell report cannot be defined in "Haskell minus seq". It goes as follows: If seq were so definable, then it would have to fulfill the free theorem derived from its type in "Haskell minus seq" (read: System F plus fix). But it doesn't. See Section 5 of http://wwwtcs.inf.tu-dresden.de/~voigt/seqFinal.pdf Ciao, Janis. -- Dr. Janis Voigtlaender http://wwwtcs.inf.tu-dresden.de/~voigt/ mailto:voigt@tcs.inf.tu-dresden.de