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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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Graham Cooper

Posts: 4,253
Registered: 5/20/10
Re: I Bet $25 to your $1 (PayPal) That You Can¹t Pr
ove Naive Set Theory Inconsistent

Posted: Feb 17, 2013 7:16 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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!


Herc
--
NEVER TAKE A BET WITH THE ADJUDICATOR!




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