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:[EMAIL PROTECTED]
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