[Haskell-cafe] Re: Sending bottom to his room
Sorry, I think a and b are from the value domain. On Sat, 29 Dec 2007 13:14:09 +0200, Cristian Baboi [EMAIL PROTECTED] wrote: In The Implementation of Functional Programming Languages by S.P. Jones, section 2.5.3, page 32 it is written: Eval [[*]] a b = a x b Eval [[*]] _|_ b = _|_ Eval [[*]] a _|_ = _|_ but in section 2.5.2 it is said that _|_ is an element of the value domain. What business does it have on the left side of the '=' ? Can you help me send him to his room ? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Sending bottom to his room
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. -- (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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Sending bottom to his room
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 Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: Sending bottom to his room
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 ? None at all. A thing is nothing but itself and won't ever be anything else, identity is implied by existence. I used id = (\x - x) just as an arbitrary fixed point you can also recurse around to point back to lambda calculus, instead of eval. After all, id(eval) = eval, even if the first one is a compiler, the second one is an interpreter and the third one is your mind. you can also say (define (id x) (unquote (quote x))) or, in the esoteric domain, where -1 people can be in an elevator, (define (id x) (quote (unquote x))). If you dare, you can also write (define (id x) (car (cons x '())) It really doesn't matter, if you don't use map or fold or write or whatever you can just write x, and if you don't write an interpreter or something that needs to interpret on runtime, you can just write your code instead of eval. http://mitpress.mit.edu/sicp/full-text/sicp/book/node77.html In a moonlit night, turn your back to the screen and meditate about the funny annotated taichi pictured on top of the page. It pictures the unity in transcendence not the equivalence of opposites, by the way. You might also say that any expression of any axiomatic system revolves around the system in the void. Hell breaks loose here: http://en.wikipedia.org/wiki/Image:Lambda.svg Sorry, I got a significant part of my logic from a philosophy lexicon. -- (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. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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 [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 Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
