ZFC IS INCONSISTENT.

[Foukzon]

Hа a level of metatheory ZFC is inconsistent. The proof of it
The unexpected fact, leans on that standard in the theory
Models the assumption, that set of all formulas of any theory
The first order, is countable ZFC-set.
The proof.
Let's designate a symbol x$Y a predicate: x there is an element of
set Y.
Let's designate a symbol x#Y a predicate: (1) x there is an element
countable sets Y which definable by means of some ZFC-formula F (x)
(2) Existence of set Y certain by means of formula F (x), together
with axioms ZFC.(3) (x$Y) it is demonstrable in ZFC.
Let's designate a symbol x~#Y a predicate: (1) x there is no element
countable sets Y which definable by means of some ZFC-formula F (x)
(2) Existence of set Y certain by means of formula F (x), together
with axioms ZFC. (3)(x$Y) it is demonstrable in ZFC.
Let's assume, that the set of all ZFC-formulas can be considered as
usual ZFC-set. Then by virtue of axioms of substitution in ZFC
existence is demonstrable set W which elements will be all ZFC-
definable sets. Then by virtue of an axiom of allocation in ZFC
existence so-called wild Rassel's set RW which is certain as follows
will be deduced: x$RW<->x~#x. For set RW by obvious image it is
received, that RW#RW<->RW~#RW.
Thus for RW in ZFC at a level of a metatheory it will be
demonstrable, that x$Y it is demonstrable in ZFC and it is
simultaneously indemonstrable in ZFC.
http://planetmath.org/?op=getobj&from=papers&id=329

[Rupert]

Foukzon wrote:
> Hа a level of metatheory ZFC is inconsistent. The proof of it
> The unexpected fact, leans on that standard in the theory
> Models the assumption, that set of all formulas of any theory
> The first order, is countable ZFC-set.

This is ungrammatical semi-coherent dribble. What's a countable
ZFC-set?

[Foukzon]

ZFC-set is a set which's existence can be proved in set theory ZFC.

[Rupert]

Foukzon wrote:
> ZFC-set is a set which's existence can be proved in set theory ZFC.

Okay, now the next bit is ungrammatical:

[JAYKOV]

Rupert писал(а):

x$Y if and only if x there is an element set Y.
x$Y ↔ x∈X

[JAYKOV]

x$Y if and only if x there is an element set Y.

x$Y ↔ x∈Y

[JAYKOV]

[Rupert]

That's ungrammatical. It doesn't make any sense.

[Rupert]

Okay, so you just meant the membership relation. Now for your predicate
x#Y. What makes you think this is definable in the first-order language
of set theory?

[Foukzon]

Dear Rupert. Predicate x#Y is definable in the first-order language of
set theory ZFC
from Godel's numbering of all ZFC-formulas. Comlete proof of this
fact will be posted later.

[JAYKOV]

[JAYKOV]

On Monday, February 13, 2006 2:51:30 PM UTC+2, Foukzon wrote:

complete proof

 

http://ru.scribd.com/doc/117651182/Inconsistent-Countable-Set

[Rupert McCallum]

You write "Let T be the collection of all sets X such that ZFC proves exists unique X psi(X), where psi(X) is a 1-place open formula." In the phrase "ZFC proves exists unique X psi(X)", X does not appear as a free variable. Are you assuming that every theorem of ZFC is true and letting X be the unique set such that psi(X)?

---

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Subject: Re: ZFC IS INCONSISTENT.

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[JAYKOV]

On Thursday, January 17, 2013 9:54:13 AM UTC+2, Rupert wrote:

You write "Let T be the collection of all sets X such that ZFC proves exists unique X psi(X), where psi(X) is a 1-place open formula." In the phrase "ZFC proves exists unique X psi(X)", X does not appear as a free variable. Are you assuming that every theorem of ZFC is true and letting X be the unique set such that psi(X)?

Yes.

[Rupert]

On Thursday, January 17, 2013 9:12:40 AM UTC+1, JAYKOV wrote:

On Thursday, January 17, 2013 9:54:13 AM UTC+2, Rupert wrote:

You write "Let T be the collection of all sets X such that ZFC proves exists unique X psi(X), where psi(X) is a 1-place open formula." In the phrase "ZFC proves exists unique X psi(X)", X does not appear as a free variable. Are you assuming that every theorem of ZFC is true and letting X be the unique set such that psi(X)?

Yes.

Okay. So you are working in a metatheory which has a truth predicate, and you are taking it as an axiom that every theorem of ZFC is true. (This set is *not* definable in the first-order language of set theory.)

Then you go on to define the predicate "ZFC proves X is not a member of Y". The problem here is that this is really a predicate that should apply to open formulas, not sets. You would get different answers depending on which formula you used to define the set.

[JAYKOV]

----------------------------------------------------------------------------------------------------------------

   I work not with the formula, but with a collection of equivalent formulas which define unique set. Thus the answer   always the unique

 

[JAYKOV]

   I work not with the formula, but with a collection of equivalent formulas which define unique set. Thus the answer   always the unique,see formula (2.14).

 

[Rupert]

Your argument fails because your "Russell set" R is not definable in the first-order language of set theory, so you don't get the desired contradiction.

[JAYKOV]

And where in your opinion I used language of the second order?

 

[Rupert]

On Thursday, January 17, 2013 1:26:13 PM UTC+1, JAYKOV wrote:

On Thursday, January 17, 2013 11:28:55 AM UTC+2, Rupert wrote:

On Thursday, January 17, 2013 9:53:52 AM UTC+1, JAYKOV wrote:

On Thursday, January 17, 2013 10:36:15 AM UTC+2, JAYKOV wrote:

On Thursday, January 17, 2013 10:19:58 AM UTC+2, Rupert wrote:

On Thursday, January 17, 2013 9:12:40 AM UTC+1, JAYKOV wrote:

On Thursday, January 17, 2013 9:54:13 AM UTC+2, Rupert wrote:

You write "Let T be the collection of all sets X such that ZFC proves exists unique X psi(X), where psi(X) is a 1-place open formula." In the phrase "ZFC proves exists unique X psi(X)", X does not appear as a free variable. Are you assuming that every theorem of ZFC is true and letting X be the unique set such that psi(X)?

Yes.

Okay. So you are working in a metatheory which has a truth predicate, and you are taking it as an axiom that every theorem of ZFC is true. (This set is *not* definable in the first-order language of set theory.)

Then you go on to define the predicate "ZFC proves X is not a member of Y". The problem here is that this is really a predicate that should apply to open formulas, not sets. You would get different answers depending on which formula you used to define the set.
----------------------------------------------------------------------------------------------------------------

   I work not with the formula, but with a collection of equivalent formulas which define unique set. Thus the answer   always the unique,see formula (2.14).

The statement "ZFC proves X is not a member of Y" would have different truth-values depending on which formulas in the equivalence classes correponding to the two sets you used to define the two sets.
 

 

Your argument fails because your "Russell set" R is not definable in the first-order language of set theory, so you don't get the desired contradiction.

 

And where in your opinion I used language of the second order?

To define your Russell set, we need to define the notion of an open formula with exactly one free variable x having the property that there exists exactly one set S such that the formula becomes true under the interpretation which gives the variable x the value S. Then we need to define an an equivalence relation on the set of all such open formulas, whereby two formulas are equivalent if and only if the two sets corresponding to the formulas in question in this way are equal. In order to define this property and this equivalence relation, we need to use a truth predicate. It cannot be done in the first-order language of set theory alone.

[JAYKOV]

----------------------------------------------------------------------------------------------------------------------------------------------------

It would be true if I worked in ZFC, but I worked in ZFC#, but there your argument false. If you work in ZFC#, predicate Prov (x, y) already sufficient.

 

 

[JAYKOV]

It would be true if I worked in ZFC, but I worked in ZFC#, but there your argument false. If you work in ZFC#, predicate Prov (x, y) already sufficient. See Proposition 2.1.

 

 

[JAYKOV]

It would be true if I worked in ZFC, but I worked in ZFC#, but there your argument false. If you work in ZFC#, predicate Prov (x, y) already sufficient. See Proposition 2.1 and formula 2.17. 

 

 

[JAYKOV]

---------------------------------------------------------------------------------------------------------------------------------------------------------------------

In passing I want to pay your attention to following circumstance. A hypothesis consistency of ZFC is extremely unnatural. This hypothesis as it is known demands real existence of the corresponding model, satisfying to axioms of ZFC.And who being in senses, it can be charged, what something such is available in this remarkable world?

 

 

[Rupert]

So I was looking over your definition of ZFC#. In the proof of Prop. 2.1 you say that if Th proves Pr_Th([Phi]) then Phi must be in Ded(Th). This is false. For example, you could let Th=ZFC+~Con(ZFC), which is consistent if ZFC is. Then Th proves Pr_Th([0=1]) but 0=1 is not in Ded(Th). You fail to establish the existence of a consistent maximal nice extension of any consistent theory Th.

 

[JAYKOV]

------------------------------------------------------------------------------------------------------------------------------------------------------------------

What you means under formula ~Con(ZFC}? Is it like formula {2.6}?

 

 

 

[JAYKOV]

-------------------------------------------------------------------------------------------------------------

If it so then your argument again the false

 

 

 

[Rupert]

No, it's not.

You yourself use the notation ~Con(ZFC) earlier in the paper. It means the sentence asserting ZFC is inconsistent.

[Rupert]

On Sunday, January 27, 2013 11:15:30 PM UTC+1, JAYKOV wrote:

If it so then your argument again the false

Well, you are the one making an argument. And you have not convinced me, for reasons I have given.

[JAYKOV]

---------------------------------------------------------------------------------------------------------------------------------------------------

 

Well. But if you mean direct definition: ~Con (ZFC) <-->ZFC|-A&~A it cannot be done in the first-order language of set theory alone.Therefore yours contr example does not valid

 

[Rupert]

~Con(ZFC) can be formulated in the first-order language of set theory, in fact, the first-order language of number theory.

[JAYKOV]

--------------------------------------------------------------------------------------------------------------------------------------------------

 

Yes certainly. But in this case we deal with Th * = ZFC+E(standard model of ZFC). In standard model of ZFC

in fact arithmetic formula ~Con (ZFC) is fail.

 

 

 

[JAYKOV]

REMARK:In order to deduce "ZFC is inconsistent" from "ZFC |- ~con(ZFC)" one needs
something more than the consistency of ZFC, e.g., that ZFC has an
omega-model (i.e., a model in which the integers are the standard
integers).

http://www.cs.nyu.edu/pipermail/fom/2007-October/012015.html 

 

 

[JAYKOV]

On Tuesday, January 29, 2013 10:36:24 PM UTC+2, JAYKOV wrote:

On Tuesday, January 29, 2013 10:31:23 PM UTC+2, JAYKOV wrote:

On Tuesday, January 29, 2013 7:30:05 PM UTC+2, Rupert wrote:

On Tuesday, January 29, 2013 4:10:28 AM UTC+1, JAYKOV wrote:

On Monday, January 28, 2013 4:28:06 PM UTC+2, Rupert wrote:

On Sunday, January 27, 2013 11:15:30 PM UTC+1, JAYKOV wrote:

If it so then your argument again the false

Well, you are the one making an argument. And you have not convinced me, for reasons I have given.
---------------------------------------------------------------------------------------------------------------------------------------------------

 

Well. But if you mean direct definition: ~Con (ZFC) <-->ZFC|-A&~A it cannot be done in the first-order language of set theory alone.Therefore yours contr example does not valid

 

~Con(ZFC) can be formulated in the first-order language of set theory, in fact, the first-order language of number theory.
--------------------------------------------------------------------------------------------------------------------------------------------------

 

Yes certainly. But in this case we deal with Th * = ZFC+E(standard model of ZFC). In standard model of ZFC

in fact arithmetic formula ~Con (ZFC) is fail.

REMARK:In order to deduce "ZFC is inconsistent" from ZFC",i.e.,  ZFC |- ~con(ZFC), one needs
_something more_ than the consistency of ZFC, e.g., that ZFC has an

omega-model (i.e., a model in which the integers are the standard
integers).

http://www.cs.nyu.edu/pipermail/fom/2007-October/012015.html 

 

 

[JAYKOV]

On Tuesday, January 29, 2013 10:46:22 PM UTC+2, JAYKOV wrote:

On Tuesday, January 29, 2013 10:36:24 PM UTC+2, JAYKOV wrote:

On Tuesday, January 29, 2013 10:31:23 PM UTC+2, JAYKOV wrote:

On Tuesday, January 29, 2013 7:30:05 PM UTC+2, Rupert wrote:

On Tuesday, January 29, 2013 4:10:28 AM UTC+1, JAYKOV wrote:

On Monday, January 28, 2013 4:28:06 PM UTC+2, Rupert wrote:

On Sunday, January 27, 2013 11:15:30 PM UTC+1, JAYKOV wrote:

If it so then your argument again the false

Well, you are the one making an argument. And you have not convinced me, for reasons I have given.
---------------------------------------------------------------------------------------------------------------------------------------------------

 

Well. But if you mean direct definition: ~Con (ZFC) <-->ZFC|-A&~A it cannot be done in the first-order language of set theory alone.Therefore yours contr example does not valid

 

~Con(ZFC) can be formulated in the first-order language of set theory, in fact, the first-order language of number theory.
--------------------------------------------------------------------------------------------------------------------------------------------------

 

Yes certainly. But in this case we deal with Th * = ZFC+E(standard model of ZFC). In standard model of ZFC

in fact  arithmetic formula such that ~Con (ZFC) is fail.

[Rupert]

No. You said that if Th proves Pr_Th([Phi]), then Phi must be in Ded(Th), for an arbitrary starting Th. I showed this to be false.

[JAYKOV]

 

Yes. But this

_means_ that Th has omega model. Otherwise you cannot define predicate Pr_Th ([Phi]).

 

 

 

 

 

[Rupert]

Why would that be?

[JAYKOV]

It is in detail explained here

http://www.cs.nyu.edu/pipermail/fom/2007-October/012015.html

 

 

 

 

 

[Rupert]

This doesn't make me any the wiser about why you think your claim is true.

[JAYKOV]

About what claim it is concrete You speak above?

 

 

 

[Rupert]

I want to know why you think that if a theory Th has no omega-model, then the predicate Prov_Th cannot be defined.

[JAYKOV]

It is not of great importance. Condition

Е(omega model of ZFC) is assumed in this paper initially. See Abstract

Abstract In this article we derived an importent example of the inconsistentcountable set. Main result is: ~ (ZFC +Е (omega model of ZFC))

 

 

 

[Rupert]

I've refuted one of your assertions in the paper and you've given no satisfactory response. I don't believe that the argument in the paper is sound.

 

 

[JAYKOV]

No, because real assertion is

Th contains (ZFC +Е(omega model of ZFC))

 

 

 

 

 

[Rupert]

Ok, so then my argument works with the theory ZFC+E(omega model of ZFC)+~Con(ZFC+E(omega model of ZFC)).
 

 

 

 

 

[JAYKOV]

No. I assumed  Con(ZFC+E(omega model of ZFC)).But  Con(ZFC+E(omega model of ZFC)).and
~Con(ZFC+E(omega model of ZFC)).does not valid in omega model of ZFC simultaneously.

 

 

 

 

 

[Rupert]

You made the assertion that if Th proves Prov_Th(Phi) then Phi must be in Ded(Th). For which theories Th is this supposed to be true?

[JAYKOV]

No. Obviously there it is meant also that Phi must be true in omega model.

 

 

 

[Rupert]

Well, that's not what you said. There is no hope of communicating effectively unless you say what you mean.

So what exactly is the assertion that you want to make?

[JAYKOV]

http://ru.scribd.com/doc/124530914/Inconsistent-Countable-Set4

 

 

 

 

 

[Rupert]

But you've been telling me that you mean something different from what you actually say in that article. So that's no use to me.

 

 

 

 

[JAYKOV]

It is the corrected version 4 and corresponding corrections are brought.

 

 

 

 

[Rupert]

On Friday, February 8, 2013 5:44:29 PM UTC+1, JAYKOV wrote:

 

Well, that's not what you said. There is no hope of communicating effectively unless you say what you mean.

So what exactly is the assertion that you want to make?

 

http://ru.scribd.com/doc/124530914/Inconsistent-Countable-Set4

Can we talk about your Definition 2.4? Am I to understand that you are working in some fixed omega-model of Th?

[JAYKOV]

Yes of course.

 

 

 

 

 

[Rupert]

In your Equation (2.14), you are asserting that a certain formula is provable in Th. So this formula must be a formula in the first-order language of set theory. But you have a quantified variable n ranging over the natural numbers, and then later on n appears as a subscript rather than a free variable. So what you have is not a formula in the first-order language of set theory. If you think that it can be translated into it then you need to make it clear why.

[JAYKOV]

--------------------------------------------------------------------------------------------------------------------------------------

 

Okay. See revisited paper 5

 

http://ru.scribd.com/doc/125764961/Inconsistent-Countable-Set-5

 

 

[Rupert]

You haven't fixed the problem.

[JAYKOV]

On Monday, February 18, 2013 8:51:20 AM UTC+2, Rupert wrote:

On Saturday, February 16, 2013 5:55:48 PM UTC+1, JAYKOV wrote:

On Thursday, February 14, 2013 7:10:38 PM UTC+2, Rupert wrote:

On Thursday, February 14, 2013 5:28:27 PM UTC+1, JAYKOV wrote:

On Wednesday, February 13, 2013 12:59:39 PM UTC+2, Rupert wrote:

 

Explain in more details.Predicate (2.19) is done by first order language.

[JAYKOV]

Explain in more details

. Predicate (2.19) is done by first order languages.

 

 

 

 

 

[Rupert]

My objection was to (2.14).
 

 

[JAYKOV]

Okay. Explain in more details.

 

 

 

[JAYKOV]

 

But in the

formula (2.14) there is a typing error.

 

 

[JAYKOV]

[Rupert]

I gave you a pretty detailed explanation; which bit did you not understand?

[Rupert]

Correct it then.

[JAYKOV]

http://ru.scribd.com/doc/126086294/Inconsistent-Countable-Set-7

 

 The problem is fixed by Proposition 2.3. see formula (2.19)

 

 

[Rupert]

No. That doesn't fix the problem.

[JAYKOV]

Why you so think? Predicate (2.19) is done by first order languages.

 

[Rupert]

What is supposed to be the relationship between (2.19) and (2.14)?

[JAYKOV]

In that sense that definitions (2.20) and (2.16) is equivalent

http://fs23.formsite.com//viXra/files/f-2-2-7453659_rDLgPTH0_Inconsistentcountableset.8.pdf

 

 

 

 

[Rupert]

It is still not really clear to me how this has a bearing on the question of whether the formula (2.14) is expressible in the first-order language of set theory.

Do you think that there is any chance that you could write this up in a way that it is actually possible to read it and make sense of it?

Basically, I would be astonished if your claim was correct. But I really have no way of locating the error in the proof, if any, because it's just not written up well enough to make any sense of.

If you can't learn to write more clearly, then I doubt you're going to succeed in convincing anyone.

 

 

 

[JAYKOV]

Well. I will write this point with the maximum detail.

 

 

 

[JAYKOV]

[JAYKOV]

[JAYKOV]

I think that already there is no error, but as much as possible detailed statement is necessary.

 

 

[JAYKOV]

I confirm that Godel's completeness theorem does not leave any chances for ZFC to be consistent

 

 

[Rupert]

Well, I find your text badly written and incomprehensible. Best of luck with trying to convince other people. 

 

[JAYKOV]

 

Well.Let con(ZFC) be a sentence in ZFC asserting that ZFC has an omega-model M. Let A_M be an wff over M. Let S be the theory ZFC+con(ZFC). Is the reflection for S: Bew_{S}(A_M) --> A_M is satisfied?

[Rupert]

What do you mean by a wff over M? Do you mean a wff with all quantifiers relativised to M? Are we adding a constant symbol for M to the language?
 

Let S be the theory ZFC+con(ZFC). Is the reflection for S: Bew_{S}(A_M) --> A_M is satisfied?

What do you mean by satisfied? Provable in some theory? Which theory? Or do you just mean true? I would say that most set theorists believe it is true.

[JAYKOV]

Yes.

What do you mean by satisfied?

S|- Bew_{S}(A_M) -->S|- A_M?

 

 

 

 

 

 

 

[JAYKOV]

[Rupert]

Could I just ask one more question. This wff A, is it allowed to have free variables? What does the predicate Bew_S mean when applied to a formula with free variables, is that the same as applying the predicate Bew_S to the universe closure of the formula?

Anyway, my opinion would be that every instance of this schema is in fact true, at least if you restrict A to sentences, but the question of in which theory you can prove it is very different.

 

 

 

 

 

 

[JAYKOV]

Ok. See complete def.2.4-2.6 

http://ru.scribd.com/doc/129443535/Lobs-Theorem4 

 

 

 

 

[JAYKOV]

But then to you also it should

be obvious that definitions (2.16) and (2.20) see http://vixra.org/pdf/1302.0048v3.pdf

is equivalent

 

 

 

 

 

 

 

 

 

[JAYKOV]

Ok. See def 2.4-2.6 
http://ru.scribd.com/doc/129443535/Lobs-Theorem4

[Rupert]

What's the definition of the doubly indexed family Psi_n,k ?
 

 

 

 

 

 

 

 

 

 

[JAYKOV]

Short:

Let W={q_n(x)} be the set (countable) of the all one-place open wff q(x) such that conditions ( * ) or ( ** ) is satisfied. Let Y be the quotient set (countable) of W by <--->,i.e. Y=W/<--->. Let q_n,n=1,2,... be the element of W and [q_n]_k,k =1,2,...be the respective element of Y={[q_n]_k},n=1,2,....,k =1,2,... . {Psi_n,k}=Y.
   

 

 

 

 

 

 

 

 

[Rupert]

What are conditions (*) and (**)? I can't seem to find them anywhere in the paper.
 

[JAYKOV]

 

(*) + (**)<----->Th#|-E!x_n[\Phy_(x_n)] see Definition 2.2.

 

 

 

 

 

 

[Rupert]

Is Y the quotient of W by equivalence of formulas in Th, Th#, or in the omega-model?
 

[JAYKOV]

Y the quotient of W by equivalence of formulas: in (1) Th#, or (2) in the omega-model.

 

 

 

 

 

 

[Rupert]

Well, which one?
 

 

 

 

 

 

 

[JAYKOV]

Well, which one?
 

Take at first (2)

 

 

 

 

 

 

[Rupert]

I don't get how the quotient of an indexed family of formulas by an equivalence relation is supposed to become a doubly indexed family of formulas.
 

 

 

 

 

 

 

[JAYKOV]

I am sorry.

There a typing error. Correctly so

Let Y be the quotient set (countable) of W by <--->,i.e. Y=W/<--->. Let q_n,n=1,2,... be the element of W and [q_n]_k,k =1,2,...be the respective element of Y={[q_n]_k},n=1,2,....,k =1,2,... ._Thus {{Psi_n,k]}=Y_where Psi_n,k is an .element of
[q_n]_k