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

id is well defined and there is only one of them. On Dec 29, 2007 3:13 PM, Cristian Baboi wrote:

On Sat, 29 Dec 2007 16:01:51 +0200, Achim Schneider wrote:

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"Cristian Baboi" wrote:

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It appears as if lambda calculus is defined by lambda calculus.

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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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