🚀 go-pugleaf

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.