continuing fraud in mathematics

[ladyjane]

There is a continuing fraud in maths
every one knows the axiom of redudciblity is invalid since ramseys critique of it and every one knows godel used it to prove his theorem

But people still keep saying godel theorem this godel theorems that. Godel proved this and Godel proved that. They know AR is invalid They know godel used it -and thus his theorem must be invalid as it is proven by invalid axioms.

There is a fraud in that people keep talking as if godel proved something when in effect he did not as he used invalid principles. There seems to be an attempt to fraud the public by not pointing out that Godels theorem is invalid due to an invalid proof-

Note it is not about what others have proven since that is the fraud but the fraud is keeping from the publick that WHAT GODEL DID is invalid

Ramsey says

Such an axiom has no place in mathematics, and anything which cannot be
proved without using it cannot be regarded as proved at all.

This axiom there is no reason to suppose true; and if it were true, this
would be a happy accident and not a logical necessity, for it is not a
tautology. (THE FOUNDATIONS OF MATHEMATICS* (1925) by F. P. RAMSEY

Where does Godel specifically use it

b Godel uses the axiom of reducibility axiom 1V of his system is the axiom of reducibility "As Godel says “this axiom represents the axiom of reducibility (comprehension axiom of set theory)” [/b](K Godel , On formally undecidable propositions of principia mathematica and related systems in The undecidable , M, Davis, Raven Press, 1965,p.12-13)

. Godel uses axiom 1V the  axiom of reducibility in his formula 40 where   he states "x is a formula arising from the  axiom schema 1V.1  ((K Godel , On formally  undecidable propositions of principia mathematica and related systems in [i]The undecidable [/i], M, Davis, Raven Press,  1965,p.21



" [40. R-Ax(x) ≡ (∃u,v,y,n)[u, v, y, n <= x & n Var v &  (n+1) Var u & u Fr y & Form(y) & x = u ∃x {v Gen [[R(u)*E(R(v))] Aeq y]}]

x is a formula derived from the axiom-schema IV, 1 by substitution "

[Knox]

All axioms are invalid.

[Valfar_dreylen]

how is an axiom invalid? axioms are assumptions to build mathematical theory. assumptions are invalid or valid only if you choose them to be. you are trying to mixed two different mathematical systems. it is like being in euclid geometery and saying the parallel axiom is ‘invalid.’ when that axiom is invalid, it leads to another mathematical system, namely hyberbolic and eliptical. you saying AR is invalid leads to another system that godel is not talking about.

[Valfar_dreylen]

furthermore, AR was invented by russel and godel’s makes use of russel’s system he created to show that russel’s system is flawed. to not accept AR means not to accept russel’s system which was the entire point of godel’s work.

also, you really should type out everything you are typing to say. that copying and pasteing of that text makes it not only difficult to read, but whoever wrote it writes like a unprofessional child. he uses way too much bold and italics and changes font. no one serious would write like that because if his ideas were profound enough he would not need to call attention everywhere.

[Dh4lsim]

If in fact Godel did prove the the theorem… why is it that his formula is invalid? Simple. It’s the design. You can quote F.P. Ramsey all you want. That doesn’t change the fact that Godel came to the same conclusion. At the time, Godel knew that AR was invalid, that was his plan if you read more on Godel.

[ladyjane]

to not accept AR means not to accept russel’s system which was the entire point of godel’s work.

godel used the 2nd ed of PM
russell following wittgenstien took it out of the 2nd edition as it was invalid

AR was not part of russell revised system in 2 ed PM -which godel used along side of the rejected AR

godel states

A. Whitehead and B. Russell, Principia Mathematica, 2nd edition, Cambridge 1925. In particular, we also reckon among the axioms of PM the axiom of infinity (in the form: there exist denumerably many individuals), and the axioms of reducibility and of choice (for all types)”

ramsy says AR is invalid

Such an axiom has no place in mathematics, and anything which cannot be
proved without using it cannot be regarded as proved at all.

This axiom there is no reason to suppose true; and if it were true, this
would be a happy accident and not a logical necessity, for it is not a
tautology. (THE FOUNDATIONS OF MATHEMATICS* (1925) by F. P. RAMSEY

[Carleas]

LJ, the last time you started a thread on this topic, I tried to explain to you exactly what everyone here is trying to explain. Godel’s proof does not depend on the axiom of reducibility being true, therefore its falsity does not defeat the proof. If the AR is false, mathematics is incomplete, and that is consistent with a proof that math is either incomplete or inconsistent.

[ladyjane]

Godel’s proof does not depend on the axiom of reducibility being true,

you cannot use an invalid axiom to prove anything
if you could
then let me be your accountant and i will make up axioms and channell your money into my account - would you accept my proof with invalid axioms that your money is now my money -all mathematically correct by your views

[Carleas]

You can assume an invalid axiom to show that it is invalid. There’s nothing wrong with that. You can assume an axiom of questionable validity to examine the consequences of its truth or falsity. There’s nothing wrong with that, either.

[ladyjane]

You can assume an invalid axiom to show that it is invalid. There’s nothing wrong with that. You can assume an axiom of questionable validity to examine the consequences of its truth or falsity.

but you cant prove anything with an invalid axiom -as godel did
if you could then
let me be your accountant and i will make up axioms and channell your money into my account - would you accept my proof with invalid axioms that your money is now my money -all mathematically correct by your views

[Mr_Reasonable]

What’s the negative result of this fraud? Why are you so against it? How does it affect your life?

[Carleas]

“but you cant prove anything with an invalid axiom -as godel did”
Yes, you can. You can prove a conditional, or you can prove disjunction.

[ladyjane]

but you cant prove anything with an invalid axiom -as godel did"
Yes, you can. You can prove a conditional, or you can prove disjunction.

SO CAN I BE YOUR ACCOUNTANT

[Anthem]

@ladyjane

SO CAN I BE YOUR ACCOUNTANT

Good lord no. I won’t allow it. With your understanding of math I wouldn’t trust you to count pennies.

[ladyjane]

Good lord no. I won’t allow it. With your understanding of math I wouldn’t trust you to count pennie

hey going by carleas
i can just make up an invalid axiom then prove all his money is mine and hes happy with that
hey you be his accountant then based on that argument-u might get very rich

[Anthem]

According to you, axioms are meaningless anyway and not capable of determining anything.

And you really don’t understand what Carleas is saying.

If I were king for a day I’d have your knuckles sown to your lips.

[ladyjane]

And you really don’t understand what Carleas is saying

he is saying godel can use invalid axioms to make a proof
so you and me as his accountant can get very rich doing using invalid axioms to get his money

[ladyjane]

That AR is invalid I give you 2 references

From Kurt Godels collected works vol 3 p.119

books.google.com/books?id=gDzbuU … #PPA119,M1

“the axiom of reducibility is generally regarded as the grossest philosophical expediency “

the standford encyclopdeia of philosophy says of AR

plato.stanford.edu/entries/princ … thematica/

“many critics concluded that the axiom of reducibility was simply too ad hoc to be justified philosophically”

From Kurt Godels collected works vol 3 p.119

books.google.com/books?id=gDzbuU … #PPA119,M1

“the axiom of reducibility is generally regarded as the grossest philosophical expediency “

[Anthem]

@ladyjane

he is saying godel can use invalid axioms to make a proof

Yes, and clearly you don’t understand what this means, either.

[ladyjane]

he is saying godel can use invalid axioms to make a proof
Yes, and clearly you don’t understand what this means, either

I know exactly what he is saying
Godel did not prove his theorem on the basis that AR is invalid

Godel in his proof assumes AR IS invalid as he assumes peano axioms are valid

We proceed now to the rigorous development of the proof sketched above, and begin by giving an exact description of the formal system P, for which we seek to demonstrate the existence of undecidable propositions. P is essentially the system obtained by superimposing on the Peano axioms the logic of PM16

etc

1. ~(Sx1 = 0)

Zero is the successor of no number. Expanded into the basic signs, the axiom is: ~(a2 ∀ (~(a2(x1)) ∨ a2(0)))

This is the smallest axiom in the entire system (although there are smaller theorems, such as 0=0).

2. Sx1 = Sy1 ⊃ x1 = y1

If x+1 = y+1 then x=y. Expanding the ⊃ operator we get: ~(Sx1 = Sy1) ∨ (x1 = y1) And expanding the = operators we get: ~(a2 ∀ (~(a2(Sx1)) ∨ a2(Sy1))) ∨ (a2 ∀ (~(a2(x1)) ∨ a2(y1)))

3. x2(0).x1 ∀ (x2(x1) ⊃ x2(fx1)) ⊃ x1 ∀ (x2(x1))

The principle of mathematical induction: If something is true for x=0, and if you can show that whenever it is true for y it is also true for y+1, then it is true for all whole numbers x.

[178]II. Every formula derived from the following schemata by substitution of any formulae for p, q and r.

1. p ∨ p ⊃ p

2. p ⊃ p ∨ q

3. p ∨ q ⊃ q ∨ p

4. (p ⊃ q) ⊃ (r ∨ p ⊃ r ∨ q)

III. Every formula derived from the two schemata

1. v ∀ (a) ∨ Subst a(v|c)

2. v ∀ (b ⊃ a) ∨ b ⊃ v ∀ (a)

by making the following substitutions for a, v, b, c (and carrying out in I the operation denoted by “Subst”): for a any given formula, for v any variable, for b any formula in which v does not appear free, for c a sign of the same type as v, provided that c contains no variable which is bound in a at a place where v is free.23

IV. Every formula derived from the schema

1. (∃u)(v ∀ (u(v) ≡ a))

on substituting for v or u any variables of types n or n + 1 respectively, and for a a formula which does not contain u free. This axiom represents the axiom of reducibility (the axiom of comprehension of set theory).

V. Every formula derived from the following by type-lift (and this formula itself):

1. x1 ∀ (x2(x1) ≡ y2(x1)) ∨ x2 = y2.

This axiom states that a class is completely determined by its elements.

etc

In the proof of Proposition VI the only properties of the system P [which is Peano and PM ie axiom of reducibility employed were the following:

1. The class of axioms [which include AR] and the rules of inference (i.e. the relation “immediate consequence of”) are recursively definable (as soon as the basic signs are replaced in any fashion by natural numbers).

2. Every recursive relation is definable in the system P (in the sense of Proposition V).

Hence in every formal system that satisfies assumptions 1 and 2 [ which includes AR] and is ω-consistent, undecidable propositions exist of the form (x) F(x), where F is a recursively defined property of natural numbers, and so too in every extension of such

and as such his proof is invalid as AR is invalid-which you admitt
if godel can prove his theorem by assuming AR is valid-when every knows it is not

then let me be yours and his accountant and i will prove using invalid axioms that your money is mine