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

"Cristian Baboi" 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. -- (c) this sig last receiving data processing entity. Inspect headers for past copyright information. All rights reserved. Unauthorised copying, hiring, renting, public performance and/or broadcasting of this signature prohibited.