[Haskell-cafe] IO is not a monad
Lennart Augustsson
lennart at augustsson.net
Sat Feb 10 05:54:43 EST 2007
I'm not sure what you're asking. The (untyped) lambda calculus is
Turing complete.
How could seq improve that?
On Feb 8, 2007, at 11:18 , Yitzchak Gale wrote:
> Lennart Augustsson wrote:
>> I think seq is funny because it is not lambda definable.
>
> Does the set of computable functions on the natural
> numbers defined by the lambda calculus augmented
> with seq have higher Turing degree than the
> set of classical computable functions?
>
> -Yitz
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