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Re: A HARD FLAW in Godel's Proof
Posted:
Dec 8, 2012 2:27 PM
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On Nov 18, 6:46 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> OK so the T/F PREDICATE
> DERIVES(T,<t1, t2, t3, t4,,,,T>)
>
> is easy to program!
> ...As long as D is a given argument, for now.
>
> STEP 2!
>
> - - - - - - - - - - - - - -
> STEP 2: DEFINE a Godel Statement.
> i.e. Godel Statement named G =
> ALL(M) ~DERIVE(G,M)
> - - - - - - - - - - - - - -
>
Thanks for bumping this thread, this ties together some high level
consistency theory with my current forward chaining Modus Ponens
Provers!
The PROOF is just an extra argument of MODUS PONENS listing the
inferences used so far!
how would you decide if the FORMULA
was a THEOREM of those AXIOMS?
AXIOM --> THM ---> THM ---> THM ---> FORMULA?
***************************************
A SIMPLY WAY TO [LIST] A THEORY!
***************************************
t(NEW,l(L)) :- trif(AXIOM,NEW,L), t(AXIOM,1).
TRANSITIVE IF
trif( OLD , NEW , l(L) ) :- if( OLD , NEW ).
trif( OLD , NEW , l(L) ) :- if( OLD , MID ) , trif( MID, NEW , L ).
NOW we can ask PROLOG what are all the THEOREMS?
..............................
?- t( THM , LVL ). <<<<<<< LIST THEORY!
THM= 1
LVL= 1
THM= 2
LVL= 1
* 1 and 2 are THEOREMS *
THM= if(1,3)
LVL=1
THM= if(3,4)
LVL=1
THM= if( and(2,4), 5)
LVL = 1
THM = if(5,6)
LVL = 1
THM= if(4,7)
LVL = 1
THM= if( and(6,7), 8)
LVL = 1
THM= if(8,9)
LVL = 1
THM= 3
LVL = l(l(_))
* 3 is a LEVEL 2 THEOREM *
THM= 4
LVL = l(l(l(_)))
THM=7
LVL = l(l(l(l(_))))
Herc
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