## Godel's First Incompleteness theorem

What is the basis for reason? And mathematics?

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roydop
Posts: 325
Joined: Wed Jan 07, 2015 11:37 pm

### Godel's First Incompleteness theorem

If Godel's first incompleteness theorem can be described as: "Any sufficiently expressive math system must be either incomplete or inconsistent", what does "sufficiently expressive" mean? Able to be interpreted?
anne
Posts: 24
Joined: Thu Jan 03, 2019 1:26 am

### Re: Godel's First Incompleteness theorem

Gödel’s 1st theorem is meaningless and invalid

http://gamahucherpress.yellowgum.com/bo ... GODEL5.pdf

1) Gödel’s 1st theorem is meaningless

a) “Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory (Kleene 1967, p. 250)

note
"... there is an arithmetical statement that is true..."

In other words there are true mathematical statements which cant be proven
But the fact is Godel cant tell us what makes a mathematical statement true thus his theorem is meaningless

Godels 1st theorem is invalid as his G statement is banned by an axiom of the system he uses to prove his theorem

http://gamahucherpress.yellowgum.com/bo ... GODEL5.pdf

2) Godels theorm is invalid

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

Axiom of reduciblity

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

G statement impredicative

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
Logik
Posts: 4041
Joined: Tue Dec 04, 2018 12:48 pm

### Re: Godel's First Incompleteness theorem

Abandon the Godel universe. It is full of rats!

Go for decidability.

If I say "X = X is true" you should ask "HOW did you determine that? What WORK did you do to evaluate the result?
If you didn't do any work then you have only assumed the result to be true.

The difficulty of evaluating equality under complexity is conveyed in this experiment: viewtopic.php?f=26&t=26199

Godel's work is a subset of Turing/Rosser/Kleene's work and if you want a rapid catch-up on it all read Scott Aronson's blog: https://www.scottaaronson.com/blog/?p=710
when I teach Gödel to computer scientists, I like to sidestep the nasty details of how you formalize the concept of “provability in F.” (From a modern computer-science perspective, Gödel numbering is a barf-inducingly ugly hack!)

Instead, I simply observe Gödel’s Theorem as a trivial corollary of what I see as its conceptually-prior (even though historically-later) cousin: Turing’s Theorem on the unsolvability of the halting problem.
For those of you who’ve never seen the connection between these two triumphs of human thought: suppose we had a sound and complete (and recursively-axiomatizable, yadda yadda yadda) formal system F, which was powerful enough to reason about Turing machines. Then I claim that, using such an F, we could easily solve the halting problem. For suppose we’re given a Turing machine M, and we want to know whether it halts on a blank tape. Then we’d simply have to enumerate all possible proofs in F, until we found either a proof that M halts, or a proof that M runs forever. Because F is complete, we’d eventually find one or the other, and because F is sound, the proof’s conclusion would be true. So we’d decide whether M halts. But since we already know (thanks to Mr. T) that the halting problem is undecidable, we conclude that F can’t have existed.