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Topic: I Bet \$25 to your \$1 (PayPal) That You Can¹t Pr
ove Naive Set Theory Inconsistent

Replies: 4   Last Post: Mar 8, 2013 4:44 PM

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 Graham Cooper Posts: 4,495 Registered: 5/20/10
Re: I Bet \$25 to your \$1 (PayPal) That You Can¹t Pr
ove Naive Set Theory Inconsistent

Posted: Feb 27, 2013 9:04 PM
 Plain Text Reply

On Feb 28, 9:10 am, Charlie-Boo <shymath...@gmail.com> wrote:
> On Feb 27, 5:24 pm, Rupert <rupertmccal...@yahoo.com> wrote:

>  > For every formula with exactly one free variable phi(x), NST proves
> {x:phi(x)} exists. It doesn't mean anything to ask whether NST proves
> the existence of a set not defined by a formula, there is no way to
> express that in the language of NST.
>
> No way to express exactly what and how do you know?
>
> The question is whether you can prove it yourself and that is the
> subject of the wager.  If you cannot, then you don't know if phi(x)
> exists or not due to possible inconsistency in your definitions, just
> as there is inconsistency in defining a set to be expressed by x~ex.
>
> C-B
>

Possible inconsistency in your definitions??

OK Charlie Boo wins!

No known system has that capability.

Of course, NO PROOF of ANYTHING exists in Charlie's framed world.

Charlie, would you accept the AXIOMS OF PROVABLE_SET_THEORY?

ALL(X) ALL(p(X))
E(S) S= {x|p(x)}
IFF
provable( ALL(X) ALL(p(X))
E(S) S= {x|p(x)} )

ALL(thm)
( not(thm) IFF not(provable(thm) )

---------------------------

i.e. a Set Exists only if that set not existing is not true

A(X) ALL(P)

E(S) [XeS <-> P(X)]
<->
~(~E(S) [XeS <-> P(X)] )

Since: ~E(RS) [XeRS <-> X~eX]
The RHS of <-> is FALSE
so the LHS : EXIST(RS) is also false

Herc
--
www.BLoCKPROLOG.com

Date Subject Author
2/27/13 Graham Cooper
2/28/13 Graham Cooper
3/8/13 Charlie-Boo

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