```Date: Dec 8, 2012 2:27 PM
Author: Graham Cooper
Subject: Re: A HARD FLAW in Godel's Proof

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 levelconsistency theory with my current forward chaining Modus PonensProvers!The PROOF is just an extra argument of MODUS PONENS listing theinferences used so far!how would you decide if the FORMULAwas 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 IFtrif( 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= 1LVL= 1THM= 2LVL= 1* 1 and 2 are THEOREMS *THM= if(1,3)LVL=1THM= if(3,4)LVL=1THM= if( and(2,4), 5)LVL = 1THM = if(5,6)LVL = 1THM= if(4,7)LVL = 1THM= if( and(6,7), 8)LVL = 1THM= if(8,9)LVL = 1THM= 3LVL = l(l(_))* 3 is a LEVEL 2 THEOREM *THM= 4LVL = l(l(l(_)))THM=7LVL = l(l(l(l(_))))Herc
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