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From: carlton@husc10.harvard.edu (david carlton)
Newsgroups: comp.lang.scheme
Subject: Re: eqness of procedures
Message-ID: <CARLTON.91May10135234@husc10.harvard.edu>
Date: 10 May 91 17:52:34 GMT
References: <9105080039.aa09228@mc.lcs.mit.edu> <1991May8.063427.25012@Think.COM>
	<ALAN.91May9100139@august.lcs.mit.edu>
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In-reply-to: alan@lcs.mit.edu's message of 9 May 91 14:01:39 GMT

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.

   My memory is that the -intent- of the Revised Report authors was that EQ?
   was to be just as undefined as EQV? in this case.

Doubtless.


david carlton
carlton@husc9.harvard.edu