Lennart Augustsson wrote:
I think seq is funny because it is not lambda definable.
I wrote:
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?
I guess I was not so clear here. What I mean is: Define a lazy lambda calculus, e.g., by restricting beta-reduction suitably. Does this still produce all general recursive functions? Now augment the lazy lambda calculus with seq. What functions does it produce now? I am pretty sure people have done this, but I am not up on the literature. My presumption is that lazy lambda calculus still produces all general recursive functions, and that adding seq makes no difference. In any case, what exactly is bothering you about seq not being a lambda? Regards, Yitz _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
