id is well defined and there is only one of them. On Dec 29, 2007 3:13 PM, Cristian Baboi <[EMAIL PROTECTED]> wrote:
> 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 > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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