[Haskell-cafe] Re: Sending bottom to his room

[Lennart Augustsson]

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>
> wrote:
>
> > "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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