Re: [Haskell-cafe] IO is not a monad

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