🚀 go-pugleaf

In article <Oct05.201328.96457@yuma.ACNS.ColoState.EDU>, sa114984@longs.lance.colostate.edu (Steven Arnold) writes:
> [...some of proof deleted...]
>
>		X = {g,X}		(1)
>
>(Of course, g may be the empty set.)
>	Eq. 1 can be expanded as follows:
>
>		X = {g, X'}
>		        /\
>		       {g,X''}
>		          /\
>			 {g,X'''}
>
>...and so on forever.  (The ' symbol is merely meant to
>distinguish an X of one iteration from that of another.)  X'',
>for example, is equivalent to {g,X'''}; X''' is equivalent to
>{g,X''''}, etc etc.
>	The number of X's below X'' is infinite; nonetheless, it
>is certainly true that X' contains X'' as well as all the X's
>contained by X''; therefore X' contains one more X than X''!  It [...]

(I'm ignoring the fact that g should not be a single element, for we could
quite easily replace it by g1,...,gn with out affecting the rest.)

I think one mistake you've made here (and unfortunately it only takes one
to render a proof invalid :-) is saying:
	X' contains X'' as well as all the X's contained by X''
If we expand X (partially) using formal notation rather than your tree
diagram we get:
	X = {g, {g, {g, {g, X}}}}
	                -X'''-
	            -----X''---
	        --------X'------
	    --------X------------

So it is true that X' contains X'' but it does not contain X'''. What is
true is that X'' contains X'''. So by saying X(m) contains any X(n) (where
(X(p) is X with p dashes) is not true unless n=m+1). So you can't then go
on to say that X' contains more than one copy of X.

--
Nik ``nik@scs.leeds.ac.uk'' Silver