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Re: A HARD FLAW in Godel's Proof
Posted:
Nov 17, 2012 11:55 PM
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On Nov 18, 1:10 pm, "INFINITY POWER" <infin...@limited.com> wrote:
> THINKING CAPS ON!
> ARGUE LOGICALLY!
> ASSUME ANYTHING!
> ROLLBACK ASSUMPTIONS LATER ON <<!
>
> STEP 1: DEFINE a 2 parameter predicate DERIVE(THEOREM, DERIVATION)
>
> DERIVE(T,D) is TRUE IFF
> D contains a sequence of inference rules and substitutions
> and the final formula T in D is logically implied from the Axioms.
>
> - - - - - - - - - - - - - -
>
> STEP 2: DEFINE a Godel Statement.
>
> i.e. Godel Statement named G =
> ALL(M) ~DERIVE(G,M)
>
> - - - - - - - - - - - - - -
>
> STEP 3: IS G A THEOREM?
>
> ASSUME: YES G IS A THEOREM
> DERIVE( G:ALL(M)~DERIVE(G,M) , D )
>
> - - - - - - - - - - - - - -
>
> STEP 4: UNIFY THE QUERY TO THE AXIOMS TO GET THE ANSWER
>
> GOAL : DERIVE( G:ALL(M)~DERIVE(G,M) , D )
> SUBGOAL : G:ALL(M)~DERIVE(G,M)
>
> (SUBGOALs are a Derivation Process that calculate reverse D in the trace)
>
> - - - - - - - - - - - - - - -
>
> STEP 5: REMOVE THE QUANTIFIER
>
> G:~EXIST(M)DERIVE(G,M)
> G: ~DERIVE(G,M)
>
> M is a variable and Existential by Double Variable Instantiation Rule of
> UNIFY().
>
> - - - - - - - - - - - - - - -
>
INSERT A STEP:
STEP 6a
G: ~DERIVE(G, [G | M] )
G is the HEAD of M by definition. (either 1st or last element)
M are the REMAINING TAIL of deductions back to the axioms.
[G <- <M>]
[G <- N <- <O>]
...
[G <- N <- P <- ... <- AXIOMS ]
Now M is strictly FREE as it doesn't contain G as an element in it's
deduction list.
and 6a reduces to 6 below.
>
> STEP 6: M IS A FREE VARIABLE
>
> G: ~DERIVE(G,M)
>
> is a null statement that will return
>
> SUBGOAL: M?
>
> i.e. When parsed by a clever logic compiler, Godel's Statement will return a
> Query in response
>
> [PROVER]- "Why is sentence G not derivable?"
>
> Herc
>
> --
> if( if(t(S),f(R)) , if(t(R),f(S)) ).
> if it's sunny then it's not raining
> ergo
> if it's raining then it's not sunny
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