On Sat, 29 Dec 2007 16:01:51 +0200, Achim Schneider <[EMAIL PROTECTED]> wrote:
"Cristian Baboi" <[EMAIL PROTECTED]> wrote:
It appears as if lambda calculus is defined by lambda calculus.
Yes. id (lambda calculus) = lambda calculus. You might try to point
back to yourself when being asked who you are to see the advantage of
this technique.
The next question is if id is well defined.
There is such a function ?
How many of them ?
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
