[Haskell-cafe] Re: Sending bottom to his room
lennart at augustsson.net
Sat Dec 29 09:43:56 EST 2007
id is well defined and there is only one of them.
On Dec 29, 2007 3:13 PM, Cristian Baboi <cristian.baboi at gmail.com> wrote:
> On Sat, 29 Dec 2007 16:01:51 +0200, Achim Schneider <barsoap at web.de>
> > "Cristian Baboi" <cristian.baboi at gmail.com> 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 ?
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