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From: alan@lcs.mit.edu (Alan Bawden)
Newsgroups: comp.lang.scheme
Subject: Re: eqness of procedures
Message-ID: <ALAN.91May10232758@august.lcs.mit.edu>
Date: 11 May 91 04:27:58 GMT
References: <9105080039.aa09228@mc.lcs.mit.edu>
	<1991May8.063427.25012@Think.COM>
	<ALAN.91May9100139@august.lcs.mit.edu>
	<CARLTON.91May10135234@husc10.harvard.edu>
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In-Reply-To: carlton@husc10.harvard.edu's message of 10 May 91 17:52:34 GMT

In article <CARLTON.91May10135234@husc10.harvard.edu>
carlton@husc10.harvard.edu (david carlton) writes:

   In article <ALAN.91May9100139@august.lcs.mit.edu>
   alan@lcs.mit.edu (Alan Bawden) writes:

      This is true; if the EQV? case is undefined, it follows that the EQ?
      case -may- return #F.  But perhaps Olin is wondering if

	(eq? (lambda (x) x) (lambda (x) x))

      might be -required- to return #F (even though EQV? was allowed to be
      smarter and discover their equivalence).

   It would certainly be out of character, since they allow

   (eq? '(1 2) '(1 2))

   , say, to be true.

Well, but certainly

  (eq? (cons 1 2) (cons 1 2))

is required to be false.  I can imagine a programmer thinking it would be
convenient if every evaluation of a LAMBDA-expression resulted in a unique
(in the sense of EQ?) closure.