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Jan 16, 2006, 3:12:22 PM1/16/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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

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

Jan 16, 2006, 9:06:12 PM1/16/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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?

Feb 11, 2006, 12:06:45 PM2/11/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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

Feb 11, 2006, 7:00:26 PM2/11/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

Foukzon wrote:

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

Okay, now the next bit is ungrammatical:

Feb 12, 2006, 7:00:39 PM2/12/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

Rupert писал(а):

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

x$Y ↔ x∈X

Feb 12, 2006, 7:02:26 PM2/12/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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

x$Y ↔ x∈YFeb 12, 2006, 7:04:05 PM2/12/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

Feb 12, 2006, 10:25:48 PM2/12/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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

Feb 12, 2006, 10:27:05 PM2/12/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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?

Feb 13, 2006, 7:51:30 AM2/13/06

to Self-Reference Paradoxes & Incompleteness in Logic & Computer Science

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.

set theory ZFC

from Godel's numbering of all ZFC-formulas. Comlete proof of this

fact will be posted later.

Jan 17, 2013, 2:18:07 AM1/17/13

to sel...@googlegroups.com

Message has been deleted

Jan 17, 2013, 2:28:57 AM1/17/13

to sel...@googlegroups.com

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

complete proof

Jan 17, 2013, 2:54:13 AM1/17/13

to sel...@googlegroups.com

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)?

**From:** JAYKOV <advanced...@list.ru>

**To:** sel...@googlegroups.com

**Sent:**
Thursday, 17 January 2013 8:28 AM

**Subject:** Re: ZFC IS INCONSISTENT.

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Jan 17, 2013, 3:12:40 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 17, 2013, 3:19:58 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 17, 2013, 3:36:15 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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

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

Jan 17, 2013, 3:53:52 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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).

Jan 17, 2013, 4:28:55 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 17, 2013, 7:26:13 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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

Jan 17, 2013, 7:48:39 AM1/17/13

to sel...@googlegroups.com, Rupert McCallum

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.

Message has been deleted

Jan 19, 2013, 12:14:04 AM1/19/13

to sel...@googlegroups.com, Rupert McCallum

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

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.

Jan 19, 2013, 12:23:59 AM1/19/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 19, 2013, 1:07:13 AM1/19/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 19, 2013, 1:34:10 AM1/19/13

to sel...@googlegroups.com, Rupert McCallum

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

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?

Jan 23, 2013, 10:54:06 AM1/23/13

to sel...@googlegroups.com, Rupert McCallum

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.

Message has been deleted

Jan 25, 2013, 4:43:41 PM1/25/13

to sel...@googlegroups.com, Rupert McCallum

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

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

Message has been deleted

Jan 27, 2013, 5:15:30 PM1/27/13

to sel...@googlegroups.com, Rupert McCallum

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

If it so then your argument again the false

Jan 28, 2013, 7:35:38 AM1/28/13

to sel...@googlegroups.com, Rupert McCallum

No, it's not.

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

Jan 28, 2013, 9:28:06 AM1/28/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 28, 2013, 10:10:28 PM1/28/13

to sel...@googlegroups.com, Rupert McCallum

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

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

Jan 29, 2013, 12:30:05 PM1/29/13

to sel...@googlegroups.com, Rupert McCallum

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

Jan 29, 2013, 3:31:23 PM1/29/13

to sel...@googlegroups.com, Rupert McCallum

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

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.Jan 29, 2013, 3:36:24 PM1/29/13

to sel...@googlegroups.com, Rupert McCallum

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).

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).

Jan 29, 2013, 3:46:22 PM1/29/13

to sel...@googlegroups.com, Rupert McCallum

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).

Jan 29, 2013, 3:48:38 PM1/29/13

to sel...@googlegroups.com, Rupert McCallum

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.

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

~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.

Jan 30, 2013, 1:29:15 AM1/30/13

to sel...@googlegroups.com, Rupert McCallum

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.

Jan 30, 2013, 6:42:17 AM1/30/13

to sel...@googlegroups.com, Rupert McCallum

Yes. But this

_means_ that Th has omega model. Otherwise you cannot define predicate Pr_Th ([Phi]).Jan 31, 2013, 1:52:43 AM1/31/13

to sel...@googlegroups.com, Rupert McCallum

Why would that be?

Jan 31, 2013, 11:36:00 AM1/31/13