[Haskell-cafe] IO is not a monad

[Lennart Augustsson]

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