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- Newsgroups: sci.logic
- Path: sparky!uunet!stanford.edu!CSD-NewsHost.Stanford.EDU!Sunburn.Stanford.EDU!pratt
- From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
- Subject: Re: A set cannot contain itself as a member
- Message-ID: <1992Oct8.033542.20684@CSD-NewsHost.Stanford.EDU>
- Keywords: recursion
- Sender: news@CSD-NewsHost.Stanford.EDU
- Organization: Computer Science Department, Stanford University.
- References: <1992Sep29.161016.19533@ghost.dsi.unimi.it> <1992Sep29.175554.12758@cs.rochester.edu> <Oct05.201328.96457@yuma.ACNS.ColoState.EDU>
- Date: Thu, 8 Oct 1992 03:35:42 GMT
- Lines: 25
-
- In article <Oct05.201328.96457@yuma.ACNS.ColoState.EDU> sa114984@longs.lance.colostate.edu (Steven Arnold) writes:
- >
- > I am not an expert in set theory or the predicate
- >calculus, but I was leafing through a book on the subject and the
- >following informal "proof" occured to me. Could any of you experts out
- >there tell me if this contains a major flaw -- and if so, what it
- >is? Thanks.
- >
- > The proposition to be proven is: A set can never contain
- >itself as a member.
-
- One more point about this "proof":
-
- If in your proof you replace "set" by "list", and write X = [g,X] in
- place of X = {g,X} to denote a list instead of a set, then you will
- have proved that such lists cannot exist. This contradicts what
- happens in Lisp. Lisp lists are necessarily finite, but may contain
- cycles. Peter Aczel's treatment of AFA does in careful detail for sets
- what Lisp long ago did for lists, albeit without the mandatory
- agonizing over whether this would blow the universe's fuse.
-
- --
- =================================================== In short, I am floundering
- Vaughan Pratt pratt@cs.Stanford.EDU 415-494-2545 around and don't take what
- =================================================== I say seriously.
-