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

Replies: 3   Last Post: Feb 17, 2013 9:03 PM

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Charlie-Boo

Posts: 1,585
Registered: 2/27/06
Re: I Bet $25 to your $1 (PayPal) That You Can¹t Pr
ove Naive Set Theory Inconsistent

Posted: Feb 17, 2013 9:03 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Feb 17, 7:16 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Feb 18, 10:07 am, Charlie-Boo <shymath...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>

> > On Feb 17, 5:59 pm, George Greene <gree...@email.unc.edu> wrote:
>
> > > > > That is NOT a real question.  Real questions about math have exactly
> > > > > zero to do with money.

>
> > > > But all questions about money have lots to do with math:
>
> > > SO WHAT?  THAT IS NOT the issue under discussion!
> > > I *SAID* that the questions about math had nothing to do with money!
> > > I DID NOT SAY anything about questions about MONEY!

>
> > > debt,
>
> > > > interest rates - because, after all, money is the formalization of
> > > > work.

> > > > > More to the point, this is no more a "question" than "how much is
> > > > > 2+2?".

> > > > False assumption.
>
> > > Liar.  I HAVEN'T MADE any assumptions here.
>
> > > > You have no proof.
>
> > > Liar.
> > > I CITED a proof. YOU CAN GOOGLE tons of proofs.  If you were NOT
> > > STUPID then
> > > YOU could state a proof.

>
> > > > The Frege-Russell argument is flawed.
>
> > > So  what??
> > > The argument that no binary relation has an element in its domain that
> > > bears the relation to
> > > all and only those things that don't bear it to themselves IS NOT
> > > flawed!
> > > It just requires YOU -- YOU -- to answer the question, "OK, if we DO
> > > have such an element,
> > > does it bear the relation TO ITSELF, OR NOT?"

>
> > > >  If you actually produced an attempted proof here, you would  see.
>
> > > Jeez; get over yourself.
> > > I JUST PRESENTED a proof, YET YOU do not see.
> > > Some forms of stupid apparently cannot be fixed.

>
> > If you present a valid proof here (before your cohorts) then you win
> > the $25.  Otherwise you're going to have to cough up a buck.

>
> > C-B
>
> You merely deny a machine parsed proof of exactly what you asked for.
>

> >   Axiom. If P(x) is a predicate with one and only one free variable
> > x, then {x | P(x)} is a set.

>
> [CB]
> Sure.
>
> 1
> LET
> P(x) <-> not(X e X)
>
> 2
> {x|P(X)} is a set
>
> 3
> X e rusl  <->  P(X)
>
> 4
> X e rusl <-> not(X e X)
>
> 5
> not(X e X) -> X e rusl
> AND
> X e rusl -> not(X e X)
>
> 6
> *** RUSSELLS SET ***
>  if( not(e(X,X)) , e(X,rusl) ).
>  if( e(X,rusl) , not(e(X,X)) ).
>
> 7
> ?- t( not(e(rusl,rusl)) , z(1) ).
> YES
> ?- t( e(rusl,rusl) , z(z(1)) ).
> YES
>
> 8
> CONTRADICTION!
>
> So either PAYPAL $25
> OR state which Step does not follow from the previous Steps!


Step 7 is no good because t and z are not defined.

Where's my $1?

C-B

> Herc
> --
> NEVER TAKE A BET WITH THE ADJUDICATOR!





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