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Topic: I Bet \$25 to your \$1 (PayPal) That You Can’t P
rove Naive Set Theory Inconsistent

Replies: 7   Last Post: Feb 18, 2013 1:48 PM

 Messages: [ Previous | Next ]
 Charlie-Boo Posts: 1,600 Registered: 2/27/06
Re: I Bet \$25 to your \$1 (PayPal) That You Can’t P
rove Naive Set Theory Inconsistent

Posted: Feb 17, 2013 5:17 PM

On Feb 17, 4:35 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Feb 16, 7:56 am, Charlie-Boo <shymath...@gmail.com> wrote:
>

> > Agreement:
>
> > I, the owner of email account shymathgu...@aol.com, do hereby agree to
> > wager \$25 against \$1 from anyone, payable through PayPal, that they
> > cannot prove Naïve Set Theory inconsistent, subject to the condition
> > that the person states here that they enter into this wager within 24
> > hours after this offer appears and they are the first to give their
> > proof as part of this wager.

>
> > C-B
>
> Here is a Small Depth Limited Formal Set Theory in PROLOG!
>
> *** t( THEOREM , LEVEL )  ***
>
> t(1,z(1)).
> not(0).
>
> *** PREDICATE CONSTRUCTION ***
>
> if(   and(X,Y)                ,    or(X,Y)            ).
> if(   and(not(X),Y)           ,    or(X,Y)            ).
> if(   and(X,not(Y))           ,    or(X,Y)            ).
> if(   and(not(X),not(Y))      ,    not(or(X,Y))       ).
>
> if(   and(not(X),not(Y))      ,    not(and(X,Y))      ).
> if(   and(not(X),Y)           ,    not(and(X,Y))      ).
> if(   and(X,not(Y))           ,    not(and(X,Y))      ).
>
> if(   and(X,Y)                ,    if(X,Y)            ).
> if(   and(not(X),not(Y))      ,    if(X,Y)            ).
> if(   and(not(X),Y)           ,    if(X,Y)            ).
> if(   and(X,not(Y))           ,    not(if(X,Y))       ).
>
> if(   and(X,Y)                ,    iff(X,Y)           ).
> if(   and(not(X),not(Y))      ,    iff(X,Y)           ).
> if(   and(not(X),Y)           ,    not(iff(X,Y))      ).
> if(   and(X,not(Y))           ,    not(iff(X,Y))      ).
>
> *** NEGATION ***
>
> if(   not(and(X,Y))           ,    or(not(X),not(Y))  ).
> if(   not(or(X,Y))            ,    and(not(X),not(Y)) ).
> if(   not(xor(X,Y))           ,    iff(X,Y)           ).
> if(   not(not(X))             ,    X                  ).
> if(   X                       ,    not(not(X))        ).
>
> *** TRANSITIVE RELATIONS ***
>
> if(   and(if(A,B),if(B,C))    ,    if(A,C)            ).
> if(   and(or(A,B),if(B,C))    ,    or(A,C)            ).
> if(   and(and(A,B),if(B,C))   ,    and(A,C)           ).
>
> *** ASSOCIATIVE RELATIONS ***
>
> if(   and(A,B)                ,    and(B,A)           ).
> if(   or(A,B)                 ,    or(B,A)            ).
>
> *** THEOREMHOOD ***
>
> t(if(X,Y),z(1)) :- if(X,Y).
> t(not(X),z(1)) :- not(X).
> t(X,z(Z)) :- t(X,Z).
>
> *** CARTESIAN JOIN ON THEOREM PAIRS ***
>
> t(  and(X,Y)           , z(Z))  :-  t(X,Z), t(Y,Z).
> t(  and(X,not(Y))      , z(Z))  :-  t(X,Z), not(Y).
> t(  and(not(X),Y)      , z(Z))  :-  not(X), t(Y,Z).
> t(  and(not(X),not(Y)) , z(Z))  :-  not(X), not(Y).
>
> *** SETHOOD ***
>
> t(e(A,B),z(1)) :- e(A,B).
>
> *** DEMO SET ***
>
> if( e(X,X) , e(X,selfish) ).
> e( ideas, abstract ).
> e( abstract, abstract ).
> e( dog, animals ).
> e( cat, animals ).
>
> *** ARITHMETIC ***
>
>
> t(bigger(N,M),z(1)) :- bigger(N,M).
> if(  bigger(N,M)       ,    not(bigger(M,N))   ).
> if(  not(bigger(N,M))  ,    bigger(M,N)        ).
> if(  add(M,N,SUM)      ,    bigger(SUM,M)      ).
> if(  add(M,N,SUM)      ,    bigger(SUM,N)      ).
>
> *** MODUS PONENS ***
>
> t(R,z(Z)) :- if(L,R) , t(L,Z).
>
> *** DEMO QUERIES ***
>
> ?- t( e(X,selfish) , z(z(1)) ).
>
> if( not(e(X,X)) , e(X,rusl) ).
> if( e(X,russell) , not(e(X,X)) ).
>
> not(e(rusl,rusl)).
>
> ?- t( not(e(rusl,rusl)) , z(1) ).
> ?- t( e(rusl,rusl) , z(z(1)) ).
>
> ************************
>
> The output is:
>
> abstract
> YES
> YES
>
> e.g. the last 2 queries
>
> ?- t( not(e(rusl,rusl)) , z(1) ).
>
> READS:  Is it a theorem that russells set is not an element of
> russells set with 1 deduction?
>
> YES
>
> ?- t( e(rusl,rusl) , z(z(1)) ).
>
> READS: Is it a theorem that russells set is an element of russells set
> with 2 or less deductions?
>
> YES
>
> Therefore
>
> |- rusl e rusl
> AND
> |- not( rusl e rusl)
>
> which satisifies the conditions of an Inconsistent System!
>
> Herc
> --
>
> Next Week :  Proving 1+1=4 in an Inconsistent System!www.BLoCKPROLOG,.com

Yes, there is no set of sets that don't contain themselves - the proof
is 3-4 lines long. But NST doesn't imply that there is.

C-B

Date Subject Author
2/17/13 Graham Cooper
2/17/13 Charlie-Boo
2/17/13 Graham Cooper
2/17/13 Charlie-Boo
2/17/13 Bernice Barnett
2/17/13 Charlie-Boo
2/18/13 fom