## Godels theorem is invalid as his G statement is banned

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### Godels theorem is invalid as his G statement is banned

Godels theorem is invalid as his G statement is banned by an axiom of the system he uses to prove his theorem

a flaw in theorem Godels sentence G is outlawed by the very axiom he uses to prove his theorem
ie the axiom of reducibiilty AR -thus his proof is invalid

[url]
http://www.enotes.com/topic/Axiom_of_reducibility [/url]

russells axiom of reducibility was formed such that impredicative statements were banned

but godels uses this AR axiom in his incompleteness proof ie axiom 1v
and formular 40

and as godel states he is useing the logic of PM ie AR

"P is essentially the system obtained by superimposing on the Peano axioms the logic of PM [ie AR axiom of reducibility]"

now godel constructs an impredicative statement G which AR was meant
to ban

The impredicative statement Godel constructs is
http://en.wikipedia.org/wiki/G%C3%B6del ... eorems#F...

the corresponding Gödel sentence G asserts: G cannot be proved to be true within the theory T

now godels use of AR bans godels G statement

thus godel cannot then go on to give a proof by useing a statement his own axiom bans
but in doing so he invalidates his whole proof
bas
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### Re: Godels theorem is invalid as his G statement is banned

Gödel was writing proofs about a deductive system by "standing outside" of it and looking in. A lot of his work was considering how deduction works inside of simple systems, some of which were so reduced that they could not even properly describe arithmetic. Some textbooks and other mathematicians refer to these as "reduxes". So you have , for example, "reduxes of Peano arithmetic".

To some degree (and in a strange way) Gödel's Incompleteness program was to first be sympathetic to an Axiom of Reducibility. But then to eventually cave in and show that AR cannot be obtained even if you really wanted it.

now godel constructs an impredicative statement G which AR was meant to ban

Right, because the actual act of banning G would require a procedure. Such a procedure would look like a decision procedure on all possible arithmetical expressions. Denote this procedure PAR

Does PAR exist?

At this point, your acceptance of AR looks like "PAR exists because I said so.". Well, I would personally love to have a procedure that solves the Halting Problem, but some good proofs show I couldn't have such a thing. Not because no smart person has come up with one. It can't be had because it doesn't exist.

hyksos
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