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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




Re: I Bet $25 to your $1 (PayPal) That You Can¹t Pr ove Naive Set Theory Inconsistent
Posted:
Feb 17, 2013 7:16 PM


On Feb 18, 10:07 am, CharlieBoo <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 FregeRussell 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. > > CB
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 {xP(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!



