I Bet $25 to your $1 (PayPal) That You Can’t Prove Naive Set Theory Inconsistent

[Charlie-Boo]

Agreement:

I, the owner of email account shymat...@aol.com, do hereby agree to
wager $25 against $1 from anyone, payable through PayPal, that they
cannot prove Naïve Set Theory inconsistent, subject to the condition
that the person states here that they enter into this wager within 24
hours after this offer appears and they are the first to give their
proof as part of this wager.

C-B

[Rupert]

I am happy to take you up on your offer, but you should first indicate what definition of Naive Set Theory you want to work with.

[Barb Knox]

In article
<47f759a0-b48c-42ce...@w14g2000vba.googlegroups.com>,

Do you understand Russell's Paradox?
E.g., <http://en.wikipedia.org/wiki/Russell's_paradox>.

If so, what trick do you have up your sleeve to avoid losing $25?
Maybe the "$" represents pesos?

> C-B

--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb |
-----------------------------

[George Greene]

On Feb 15, 4:56 pm, Charlie-Boo <shymath...@gmail.com> wrote:
> Agreement:
>

> I, the owner of email account shymathgu...@aol.com, do hereby agree to

> wager $25 against $1 from anyone, payable through PayPal, that they
> cannot prove Naïve Set Theory inconsistent,

The subject of your verb is wrong. "You" don't prove things.
The THEORY ITSELF proves things.
It has axioms and theorems logically follow from those axioms,
WHETHER YOU NOTICE IT OR NOT.
Naive set theory in particular has an axiom requiring sets of things-
satisfying-a-predicate
to exist. In particular it requires that a set of all sets not
containing themselves (as elements) exist.
So pay up. Though I guess you could pay the naive comprehension axiom
instead of me since
THAT AND NOT any PERSON is what proves it.

[George Greene]

On Feb 16, 3:28 am, Barb Knox <s...@sig.below> wrote:

> Do you understand Russell's Paradox?

I fear that the more relevant question may be whether he does or
doesn't understand
the whole notion of an axiom-schema in a first-order theory. Even
before that, as a part
of it and therefore as a pre-requisite for it, there is a prior notion
of a
"unary-predicate-schema" (inside the axiom-schema), and for some
people, it's just all too abstract.

[George Greene]

On Feb 15, 11:59 pm, Rupert <rupertmccal...@yahoo.com> wrote:
> I am happy to take you up on your offer, but you should first indicate what definition of Naive Set Theory you want to work with.

While that scores points for giving him enough rope to hang himself,
it unfortunately cuts the other way on the issue
of conventional communication. We cannot talk at all without prior
shared consensus on the meanings of SOME terms.
Moving "Naive Set Theory" OUTside that circle-of-terms-with-KNOWN-
definitions is conceding TOO much.
He will use whatever definition of Naive Set Theory THE REST OF US
use, because "Naive Set Theory"
IS ALREADY IN the dictionary.

[Charlie-Boo]

On Feb 15, 11:59 pm, Rupert <rupertmccal...@yahoo.com> wrote:

> On Friday, February 15, 2013 10:56:09 PM UTC+1, Charlie-Boo wrote:
> > Agreement:
>

> > I, the owner of email account shymathgu...@aol.com, do hereby agree to

>
> > wager $25 against $1 from anyone, payable through PayPal, that they
>
> > cannot prove Naïve Set Theory inconsistent, subject to the condition
>
> > that the person states here that they enter into this wager within 24
>
> > hours after this offer appears and they are the first to give their
>
> > proof as part of this wager.
>
> > C-B
>
> I am happy to take you up on your offer, but you should first indicate what definition of Naive Set Theory you want to work with.

Offer accepted.

I am not aware of that distinction. By definition do you mean "which
theory" or "which wording of this one theory"? What alternatives are
you thinking of?

C-B

[Charlie-Boo]

On Feb 16, 3:28 am, Barb Knox <s...@sig.below> wrote:

> In article
> <47f759a0-b48c-42ce-bbd2-eb23968ab...@w14g2000vba.googlegroups.com>,
>
>  Charlie-Boo <shymath...@gmail.com> wrote:
> > Agreement:
>
> > I, the owner of email account shymathgu...@aol.com, do hereby agree to

> > wager $25 against $1 from anyone, payable through PayPal, that they
> > cannot prove Naïve Set Theory inconsistent, subject to the condition
> > that the person states here that they enter into this wager within 24
> > hours after this offer appears and they are the first to give their
> > proof as part of this wager.
>
> Do you understand Russell's Paradox?
> E.g., <http://en.wikipedia.org/wiki/Russell's_paradox>.

Sure.

> If so, what trick do you have up your sleeve to avoid losing $25?

Can't show you my cards until you place your bet. :)

> Maybe the "$" represents pesos?

We can use Euros.

C-B

[Charlie-Boo]

On Feb 16, 10:43 am, George Greene <gree...@email.unc.edu> wrote:
> On Feb 15, 4:56 pm, Charlie-Boo <shymath...@gmail.com> wrote:
>
> > Agreement:
>
> > I, the owner of email account shymathgu...@aol.com, do hereby agree to
> > wager $25 against $1 from anyone, payable through PayPal, that they
> > cannot prove Naïve Set Theory inconsistent,
>
> The subject of your verb is wrong. "You" don't prove things.
> The THEORY ITSELF proves things.

There is also the 3-place relation x (as you say next) notices proof y
of theorem z.

> It has axioms and theorems logically follow from those axioms,
> WHETHER YOU NOTICE IT OR NOT.

Yes, the former is not the existential quantification of the latter.

> Naive set theory in particular has an axiom requiring sets of things-
> satisfying-a-predicate
> to exist.  In particular it requires that a set of all sets not
> containing themselves (as elements) exist.
> So pay up.  Though I guess you could pay the naive comprehension axiom

So you accept the wager? And if so, your proof?

> instead of me since
> THAT AND NOT any PERSON is what proves it.

The wager is that you will not notice a proof. (In particular, it
does not exist.)

C-B

[Charlie-Boo]

On Feb 16, 10:45 am, George Greene <gree...@email.unc.edu> wrote:
> On Feb 16, 3:28 am, Barb Knox <s...@sig.below> wrote:
>
> > Do you understand Russell's Paradox?
>
> I fear that the more relevant question may be whether he does or
> doesn't understand

The only real question is whether you can (notice a proof to) prove it
or not – if you put your money where your mouth is.

> the whole notion of an axiom-schema in a first-order theory.  Even
> before that, as a part
> of it and therefore as a pre-requisite for it, there is a prior notion
> of a
> "unary-predicate-schema" (inside the axiom-schema), and for some
> people, it's just all too abstract.

Those people apparently include all but 4 people. http://tinyurl.com/unarypredicateschema

C-B

[Charlie-Boo]

He's saying there are more than one, so choose one and Make Your Bet.

C-B

[George Greene]

On Feb 16, 5:31 pm, Charlie-Boo <shymath...@gmail.com> wrote:
> The only real question is whether you can (notice a proof to) prove it
> or not – if you put your money where your mouth is.

That is NOT a real questions. Real questions about math have exactly
zero to do with money.
More to the point, this is no more a "question" than "how much is
2+2?".
It may be stated in the FORM of a question (a la Jeopardy) but give
that the ANSWER
is KNOWN, there IS NO QUESTION *about* this topic.

And just for the record, tons of people already HAVE instantiated the
naive set theory comprehension-axiom-schema with the unary-predicate-
schema
~xex. So Shut Up.

As for whether I personally have produced a proof, see any number of
posts I have
made in this newsgroup going back 15 years or more. For merely 6
years, see, e.g.,
http://groups.google.com/group/sci.logic/msg/a90fcda397182786

[Charlie-Boo]

On Feb 16, 10:17 pm, George Greene <gree...@email.unc.edu> wrote:
> On Feb 16, 5:31 pm, Charlie-Boo <shymath...@gmail.com> wrote:
>
> > The only real question is whether you can (notice a proof to) prove it
> > or not – if you put your money where your mouth is.
>
> That is NOT a real questions.  Real questions about math have exactly
> zero to do with money.

But all questions about money have lots to do with math: debt,
interest rates - because, after all, money is the formalization of
work.

> More to the point, this is no more a "question" than "how much is
> 2+2?".
> It may be stated in the FORM of a question (a la Jeopardy) but give
> that the ANSWER
> is KNOWN, there IS NO QUESTION *about* this topic.

False assumption. You have no proof. The Frege-Russell argument is
flawed. If you actually produced an attempted proof here, you would
see. You are only adding more icing to the cake.

> And just for the record, tons of people already HAVE instantiated the
> naive set theory comprehension-axiom-schema with the unary-predicate-
> schema
> ~xex.  So Shut Up.
>
> As for whether I personally have produced a proof, see any number of
> posts I have
> made in this newsgroup going back 15 years or more.  For merely 6
> years, see, e.g.,http://groups.google.com/group/sci.logic/msg/a90fcda397182786

That is over 3 pages of text talking about various topics, and
contains no such proof. No citation shenanigans, please.

What is your proof that NST is inconsistent, or do you concede the $1
yet?

C-B

[Charlie-Boo]

On Feb 16, 10:17 pm, George Greene <gree...@email.unc.edu> wrote:

No. See David Hilbert, "Mathematical Problems", Göttinger
Nachrichten, 1900, pp. 253-297, and Archiv der Mathematik und Physik,
3dser., vol. 1 (1901), pp. 44-63, 213-237.

If you would actually present the purported proof, then I could refer
to your words to show you what the flaw is. Since nobody has offered
any attempted proof, I am now owed somewhere between $1 and $3 via
PayPal.

If anybody ever tries to pawn off the popular false proof, then I
would point out that there is no collection/class/concept of all sets
that do not contain themselves due to the same principle of
diagonalization that tells us there is no set containing that.

Send $1 PayPal payments to shymat...@aol.com please.

https://www.paypal.com/webapps/mpp/send-money-online

C-B

[Charlie-Boo]

> Send $1 PayPal payments to shymathgu...@aol.com please.
>
> https://www.paypal.com/webapps/mpp/send-money-online
>
> C-B

That's shy math guy 99 at aol dot com.

[fom]

On 2/16/2013 2:28 AM, Barb Knox wrote:
> In article
> <47f759a0-b48c-42ce...@w14g2000vba.googlegroups.com>,
> Charlie-Boo <shyma...@gmail.com> wrote:
>
>> Agreement:
>>
>> I, the owner of email account shymat...@aol.com, do hereby agree to
>> wager $25 against $1 from anyone, payable through PayPal, that they

>> cannot prove Na�ve Set Theory inconsistent, subject to the condition

>> that the person states here that they enter into this wager within 24
>> hours after this offer appears and they are the first to give their
>> proof as part of this wager.
>
> Do you understand Russell's Paradox?
> E.g., <http://en.wikipedia.org/wiki/Russell's_paradox>.

The following account ends with the
paragraph:

I have a very strong suspicion that no one will
ever derive a Russellian paradox in Lesniewskian
mereology. This is especially true if one considers
George Greene's explanation that the paradox arises
from grammatical form. As we have seen, Lesniewski
specifically devised his mereology to circumvent the
grammatical forms he thought would be problematic.




=======================
> >
> > Arguably, mereology as an investigation into foundational
> > mathematics is from Lesniewski
> >
> > Lesniewski wrote several papers criticizing Russell's
> > Principia as basically being incoherent
> >
> > He did his own investigations along the lines of logical
> > structure of sentences having existential import that
> > are different from both Frege and Russell
> >
> > Subsequently, he characterized a notion of class that was
> > different from Russell's
> >
> > At first, he tried to characterize his ideas in traditional
> > logical formats but ultimately began to pursue it
> > using formal syntax
> >
> > This actually gets dense and I have not really
> > examined it. But, for example he begins a system
> > he calls protothetic with
> >
> > A1. ((p <-> r) <-> (q <-> p)) <-> (r <-> q)
> >
> > A2. (p <-> (q <-> r)) <-> ((p <-> q) <-> r)
> >
> > and then lists 79 theorems about logical equivalence.
> >
> > He then switches notation to quantify over propositional
> > variables
> >
> > A1. ApAqAr(((p <-> r) <-> (q <-> p)) <-> (r <-> q))
> >
> > A2. ApAqAr((p <-> (q <-> r)) <-> ((p <-> q) <-> r))
> >
> > So that he can extend the system using variables ranging
> > over truth-functions
> >
> > A3. AGAp(AF(G(p,p) <->
> > ((Ar(F(r,r) <-> G(p,p))
> > <->
> > (Ar(F(r,r) <-> G(p <-> Aq(q),p))))
> > <-> Aq(G(q,p)))
> >
> > He then lists 422 theorems in order to obtain the
> > three logical axioms of Lukasiewicz grounding
> > the usual theory of deduction based on implication
> > and negation. His primary stated goal at the
> > outset for extending the original system was to
> > obtain
> >
> > ApAqAF((p <-> q) <-> (F(p) <-> F(q)))
> >
> > and the equivalent was obtained at step 381
> >
> > Next, he turns to his system of ontology
> > grounded upon the logical foundation of his
> > protothetic. His only axiom is
> >
> > A0. AZAz((Z class of z) <-> (
> > (-(AY(-(Y class of Z)))
> > /\
> > AYAX(((Y class of Z) /\ (X class of Z)) -> (Y class of X)))
> > /\
> > AY((Y class of Z) -> (Y class of z))
> > ))
> >
> >
> > In his exposition he comments on Russell's
> > paradox:
> >
> > "... can be strengthened in ontology by
> > means of the easily proved sentence which
> > says that:
> >
> > AZAz((Z class of z) -> (Z class of Z))
> >
> >
> > (I call this the 'ontological identity sentence';
> > it should be noticed that the yet stronger thesis
> >
> > AZ(Z class of Z)
> >
> > is not provable in ontology -- indeed, its
> > negation is provable.) In connection with this
> > sentence, I want to emphasize expressly that in
> > ontology there is always a very good possibility
> > of proving theses having a single component of
> > the type (Z class of Z) or (what is indifferently
> > the same in ontology) (z class of z). This does
> > not, however, lead to a contradiction via the
> > well-known schema of Principia Mathematica
> > because the definition directives of ontology
> > have been appropriately formulated so that
> > no thesis of the type
> >
> > AZ((Z class of x) <-> -(Z class of Z))
> >
> > can be obtained."
> >
> >
> > This assertion might best be viewed much like
> > the situation with general relativity. Philosophers
> > who have been looking at Lesniewski's systems
> > have not run into any contradictions such as
> > Russell's (at least, in so far as Peter Simons
> > has reported accurately)
> >
> > The character of his predicate in ontology
> > allows him to formulate terminological
> > explanations about ontology within the language
> > of ontology.
> >
> > The axiom of ontology, and the equivalent axioms
> > he discusses in his exposition derive from
> > his analysis of Russell's paradox.
> >
> > The Lesniewskian notion of class is based upon
> > a part relation and this is the formal mereology
> > associated with his investigations:
> >
> >
> > A1:
> > If P is a part of object Q, then Q is not a
> > part of object P
> >
> > A2:
> > If P is a part of object Q, and Q is a part of object R,
> > then P is a part of object R
> >
> > D1:
> > P is an ingredient of an object Q when and only when,
> > P is the same object as Q or is a part of object Q
> >
> > D2:
> > P is the class of objects p, when and only when the
> > following conditions are fulfilled:
> >
> > a)
> > P is an object
> >
> > b)
> > every p is an ingredient of object P
> >
> > c)
> > for any Q, if Q is an ingredient of object P, then
> > some ingredient of object Q is an ingredient of
> > some p
> >
> >
> >
> > These conditions formalize a statement Lesniewski
> > made in his analysis of Russell's paradox:
> >
> > "I use the expressions "the set of all objects m"
> > and "the class of all objects m" to denote every
> > object P which fulfills the two following
> > conditions:
> >
> > 1) every m is an ingredient of the object P
> >
> > 2) if I is an ingredient of object P, then
> > some ingredient of object I is an ingredient
> > of some m"
> >
> >
> >
> > Since ingredient is effectively the reflexive
> > subset relation in set theory by Zermelo's
> > 1908 language, you can see why Zuhair chose the
> > language he did to describe atoms. In an
> > atomistic theory, I and m must at least share
> > some atom of P
> >
> > I expressed this idea in a formal sense long
> > ago only to be flamed (singed by you and firebombed
> > by someone else)
> >
> > What actually makes mereology work is something that
> > is associated with the constructible universe. It is
> > called almost universality. Of course, that is not
> > how Lesniewski referred to it:
> >
> > "Lukasiewicz writes in his book as follows: 'we say
> > of objects belonging to a particular class, that
> > they are subordinated to that class'
> >
> > "It most often happens that a class is not subordinated
> > to itself, as being a collection of elements, it
> > generally possesses different features from each of
> > its elements separately. A collection of men is not
> > a man, a collection of triangles is not a triangle,
> > etc. In some cases, it happens in fact to be otherwise.
> > Let us consider e.g., the conception of a 'full class',
> > i.e., a class to which belong, in general, some
> > individuals. For not all classes are full, some
> > being empty; e.g., the classes: "mountain of pure
> > gold', 'perpetual motion machine', 'square circle',
> > are empty, because there are no individuals which
> > belong to those classes. One can then distinguish
> > among them those classes to which belong some
> > individuals, and form the conception of a 'full
> > class'. Under this conception fall, as individuals,
> > whose classes, e.g., the class of men, the class
> > of triangles, the class of first even number (which
> > contains only one element, the number 2), etc.
> > A collection of all those classes constitutes
> > a new class, namely 'the class of full classes'.
> > So that the class of full classes is also a full
> > class and therefore is subordinated to itself."
> >
> > In almost universal models of set theory, every subclass of
> > the universe is an element of the universe. Thus,
> > "is an element or is equal to" is the same as
> > "is a proper part or is equal to". So, the satisfaction
> > predicate can be reflexive containment... except for
> > one thing. The identity predicate in the theory can
> > not be based on extensionality. It must be based on
> > first-order object identity as described by Frege and
> > this cannot arise just because one invokes the
> > ontological position of a "theory of identity". The
> > reason that it must be based on object identity is that
> > reference to the universe can only be made if, as
> > Lesniewski has observed
> >
> > P is an object
> >
> > In an almost universal model, every proper part
> > of the universe is an element of a class of which
> > the universe is not an element. Thus every proper
> > part of the universe is distinguished from the
> > universe on the basis of object identity. Since
> > to be a 'full class' the universe can have no
> > other parts, it is unique and may be denoted
> > by a singular term.
> >
> > I have a very strong suspicion that no one will
> > ever derive a Russellian paradox in Lesniewskian
> > mereology. This is especially true if one considers
> > George Greene's explanation that the paradox arises
> > from grammatical form. As we have seen, Lesniewski
> > specifically devised his mereology to circumvent the
> > grammatical forms he thought would be problematic.
>
>
>

[Charlie-Boo]

On Feb 17, 7:49 am, fom <fomJ...@nyms.net> wrote:
> On 2/16/2013 2:28 AM, Barb Knox wrote:
>
> > In article

> > <47f759a0-b48c-42ce-bbd2-eb23968ab...@w14g2000vba.googlegroups.com>,
> >   Charlie-Boo <shymath...@gmail.com> wrote:
>
> >> Agreement:
>
> >> I, the owner of email account shymathgu...@aol.com, do hereby agree to

> >> wager $25 against $1 from anyone, payable through PayPal, that they

> >> cannot prove Naïve Set Theory inconsistent, subject to the condition

Interesting. Do you accept my wager?

C-B

[billh04]

Are you saying that the axioms of Naive Set Theory don't imply that
the set of all sets that do not contain themselves exists?

Or, are you saying that the axioms of Naive Set Theory do imply that
the set of all sets that do contain themselves exists, but that this
leads to a contradiction by the usual diagonal argument, and thus the
wff "x not in x" is not a wff at all since "x not in x" leads to a
contradiction? Hence, according to you, since "x not in x" is not a
wff, the axioms of Naive Set Theory are not inconsistent since "x not
in x" cannot appear validly in any statement.

[Graham Cooper]

On Feb 16, 7:56 am, Charlie-Boo <shymath...@gmail.com> wrote:
> Agreement:
>

> I, the owner of email account shymathgu...@aol.com, do hereby agree to

> wager $25 against $1 from anyone, payable through PayPal, that they
> cannot prove Naïve Set Theory inconsistent, subject to the condition
> that the person states here that they enter into this wager within 24
> hours after this offer appears and they are the first to give their
> proof as part of this wager.
>
> C-B

Here is a Small Depth Limited Formal Set Theory in PROLOG!


*** t( THEOREM , LEVEL ) ***


t(1,z(1)).
not(0).


*** PREDICATE CONSTRUCTION ***


if( and(X,Y) , or(X,Y) ).
if( and(not(X),Y) , or(X,Y) ).
if( and(X,not(Y)) , or(X,Y) ).
if( and(not(X),not(Y)) , not(or(X,Y)) ).

if( and(not(X),not(Y)) , not(and(X,Y)) ).
if( and(not(X),Y) , not(and(X,Y)) ).
if( and(X,not(Y)) , not(and(X,Y)) ).

if( and(X,Y) , if(X,Y) ).
if( and(not(X),not(Y)) , if(X,Y) ).
if( and(not(X),Y) , if(X,Y) ).
if( and(X,not(Y)) , not(if(X,Y)) ).

if( and(X,Y) , iff(X,Y) ).
if( and(not(X),not(Y)) , iff(X,Y) ).
if( and(not(X),Y) , not(iff(X,Y)) ).
if( and(X,not(Y)) , not(iff(X,Y)) ).


*** NEGATION ***

if( not(and(X,Y)) , or(not(X),not(Y)) ).
if( not(or(X,Y)) , and(not(X),not(Y)) ).
if( not(xor(X,Y)) , iff(X,Y) ).
if( not(not(X)) , X ).
if( X , not(not(X)) ).


*** TRANSITIVE RELATIONS ***


if( and(if(A,B),if(B,C)) , if(A,C) ).
if( and(or(A,B),if(B,C)) , or(A,C) ).
if( and(and(A,B),if(B,C)) , and(A,C) ).

*** ASSOCIATIVE RELATIONS ***

if( and(A,B) , and(B,A) ).
if( or(A,B) , or(B,A) ).

*** THEOREMHOOD ***

t(if(X,Y),z(1)) :- if(X,Y).
t(not(X),z(1)) :- not(X).
t(X,z(Z)) :- t(X,Z).


*** CARTESIAN JOIN ON THEOREM PAIRS ***

t( and(X,Y) , z(Z)) :- t(X,Z), t(Y,Z).
t( and(X,not(Y)) , z(Z)) :- t(X,Z), not(Y).
t( and(not(X),Y) , z(Z)) :- not(X), t(Y,Z).
t( and(not(X),not(Y)) , z(Z)) :- not(X), not(Y).



*** SETHOOD ***


t(e(A,B),z(1)) :- e(A,B).

*** DEMO SET ***


if( e(X,X) , e(X,selfish) ).
e( ideas, abstract ).
e( abstract, abstract ).
e( dog, animals ).
e( cat, animals ).



*** ARITHMETIC ***

add(1,2,3).


t(add(M,N,S),z(1)) :- add(M,N,S).
t(bigger(N,M),z(1)) :- bigger(N,M).
if( bigger(N,M) , not(bigger(M,N)) ).
if( not(bigger(N,M)) , bigger(M,N) ).
if( add(M,N,SUM) , bigger(SUM,M) ).
if( add(M,N,SUM) , bigger(SUM,N) ).
if( add(M,N,SUM) , add(N,M,SUM) ).


*** MODUS PONENS ***


t(R,z(Z)) :- if(L,R) , t(L,Z).





*** DEMO QUERIES ***

?- t( e(X,selfish) , z(z(1)) ).


if( not(e(X,X)) , e(X,rusl) ).
if( e(X,russell) , not(e(X,X)) ).

not(e(rusl,rusl)).

?- t( not(e(rusl,rusl)) , z(1) ).
?- t( e(rusl,rusl) , z(z(1)) ).


************************


The output is:

abstract
YES
YES

e.g. the last 2 queries

?- t( not(e(rusl,rusl)) , z(1) ).

READS: Is it a theorem that russells set is not an element of
russells set with 1 deduction?

YES

?- t( e(rusl,rusl) , z(z(1)) ).

READS: Is it a theorem that russells set is an element of russells set
with 2 or less deductions?

YES

Therefore

|- rusl e rusl
AND
|- not( rusl e rusl)


which satisifies the conditions of an Inconsistent System!


Herc
--

Next Week : Proving 1+1=4 in an Inconsistent System!
www.BLoCKPROLOG,.com

[Charlie-Boo]

Yes.

C-B

[Charlie-Boo]

Yes, there is no set of sets that don't contain themselves - the proof
is 3-4 lines long. But NST doesn't imply that there is.

C-B

[Charlie-Boo]

On Feb 17, 2:18 pm, billh04 <bill...@gmail.com> wrote:

Participation in this discussion constitutes acceptance of the wager.

C-B

[William Hale]

In article
<00d6d905-30e4-4096...@h9g2000vbk.googlegroups.com>,

I assume that you mean that you won't respond to my post unless I accept
your wager, which is ok. But, you can't put conditions on posting in a
public news group.

But, even if you could, you are not meeting the conditions that you
yourself posted in the starting post of this thread:

===========================================
Agreement:

I, the owner of email account shymathgu...@aol.com, do hereby agree to

wager $25 against $1 from anyone, payable through PayPal, that they
cannot prove Naïve Set Theory inconsistent, subject to the condition
that the person states here that they enter into this wager within 24
hours after this offer appears and they are the first to give their
proof as part of this wager.

C-B
============================================

This seems to require that one states that they enter into the wager
(which I didn't), that one accepts the wager within 24 hours after the
offer appears (where I posted after 24 hours), and that one be the first
to do so (where I am not the first to post).

Likewise, your claim that you are owed up to $3 via PayPal doesn't meet
the condition that you stated that only the first will be in the wager.

[William Hale]

In article
<b37aaf2f-ae08-4161...@y9g2000vbb.googlegroups.com>,

Isn't one of the axioms of Naive Set Theory the following:

Axiom. If P(x) is a predicate with one and only one free variable
x, then {x | P(x)} is a set.

If that is not one of the axioms of Naive Set Theory, then how do you
assert that the set, say, of even numbers exists?

[George Greene]

> > That is NOT a real question.  Real questions about math have exactly

> > zero to do with money.
>
> But all questions about money have lots to do with math:

SO WHAT? THAT IS NOT the issue under discussion!
I *SAID* that the questions about math had nothing to do with money!
I DID NOT SAY anything about questions about MONEY!

debt,
> interest rates - because, after all, money is the formalization of
> work.

> > More to the point, this is no more a "question" than "how much is
> > 2+2?".

> False assumption.

Liar. I HAVEN'T MADE any assumptions here.

> You have no proof.

Liar.
I CITED a proof. YOU CAN GOOGLE tons of proofs. If you were NOT
STUPID then
YOU could state a proof.

> The Frege-Russell argument is flawed.

So what??
The argument that no binary relation has an element in its domain that
bears the relation to
all and only those things that don't bear it to themselves IS NOT
flawed!
It just requires YOU -- YOU -- to answer the question, "OK, if we DO
have such an element,
does it bear the relation TO ITSELF, OR NOT?"

>  If you actually produced an attempted proof here, you would see.

Jeez; get over yourself.
I JUST PRESENTED a proof, YET YOU do not see.
Some forms of stupid apparently cannot be fixed.

[George Greene]

On Feb 17, 5:17 pm, Charlie-Boo <shymath...@gmail.com> wrote:
> Yes, there is no set of sets that don't contain themselves - the proof
> is 3-4 lines long.  But NST doesn't imply that there is.

Liar.
Nobody can help it if YOU don't know what Naive Set Theory IS.
Besides, given that, as I and many others have pointed out, the
non-existence of this set can be proved FROM NO AXIOMS AT ALL,
IF you are willing to use *SECOND* order logic, the fact that it CAN
ALSO
be proved from NST follows AUTOMATICALLY.

In the case of NST, though, there is a relevant SPEEDUP:
From the naive set comprehension schema, you can prove
Ex[Ay[yex<->~yey]], IN ONE LINE.
In one more line, you can then instantiate y to x,
which gives you a contradiction in one MORE line.

The 2nd-order logic proof is an indirect proof a validity, though,
as OPPOSED to a proof that the whole theory is inconsistent, since
in THAT proof, UNLIKE in the proof from NST, you are NOT priorly
committed (by an axiom)
to the provabiliity&truth of the first line.

[Graham Cooper]

Is any definable collection a set?

That is the usual meaning of Naive Set Theory.

EXIST(S) ALL(x) xeS <-> d(x)
for all WFF d

*** RUSSELLS SET ***

if( not(e(X,X)) , e(X,rusl) ).
if( e(X,russell) , not(e(X,X)) ).

Since not(e(X,X)) is a WFF rusl is a SET


Herc
--
www.BLoCKPROLOG.com

[George Greene]

On Feb 17, 5:14 pm, Charlie-Boo <shymath...@gmail.com> wrote:

>  >  Are you saying that the axioms of Naive Set Theory don't imply
> that
>  >  the set of all sets that do not contain themselves exists?
>
> Yes.

Wellm DAMN, then, you're just IGNORANT.
What MAKES Naive Set Theory naive is THE FACT THAT it has
THIS NAIVE SET COMPREHENSION AXIOM-schena, namely,
Ex[Ay[yex<->phi(y)] ],
EACH, ANY, AND EVERYinstance of this -- it can be instantiated with
EACH, ANY, AND EVERY appropriate phi(.) -- IS AN AXIOM of Naive Set
Theory.

And phi(x)<=df=>~xex really just does happen, unfortunately for you
and your $25, to be one such phi(.).
So Ex[Ay[yex<->~yey]] REALLY IS AN AXIOM of NST. The theory's
inconsistency follows logically (and quickly).

[Charlie-Boo]

On Feb 17, 5:38 pm, William Hale <bill...@yahoo.com> wrote:
> In article
> <00d6d905-30e4-4096-8f3f-38e644a6f...@h9g2000vbk.googlegroups.com>,

That's the first to post your particular proof. Meeting the deadline
is sufficient (by definition) but not necessarily necessary, subject
to the discretion of the judges.

> Likewise, your claim that you are owed up to $3 via PayPal doesn't meet
> the condition that you stated that only the first will be in the wager.

No duplicate proofs, that's all (unless the first copy has already
been shown to be flawed.)

"The neat thing about people saying you're wrong is that when you
prove you're right, they have all proved that they weren't aware of
the truth the whole time." - C-B

C-B

[Charlie-Boo]

On Feb 17, 5:48 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <b37aaf2f-ae08-4161-9735-31820c24e...@y9g2000vbb.googlegroups.com>,

Sure.

> If that is not one of the axioms of Naive Set Theory, then how do you
> assert that the set, say, of even numbers exists?

C-B

[Charlie-Boo]

If you present a valid proof here (before your cohorts) then you win
the $25. Otherwise you're going to have to cough up a buck.

C-B

[Charlie-Boo]

It very well may be.

> That is the usual meaning of Naive Set Theory.
>
> EXIST(S) ALL(x)  xeS   <->   d(x)
> for all WFF d
>
> *** RUSSELLS SET ***
>
>  if( not(e(X,X)) , e(X,rusl) ).
>  if( e(X,russell) , not(e(X,X)) ).
>
> Since not(e(X,X)) is a WFF  rusl is a SET

How do you know it's a collection? You cannot prove it is. If it
were a collection then set and collection would have to differ. Can
you prove that?

C-B

> Herc
> --www.BLoCKPROLOG.com

[Charlie-Boo]

But if having a phi is the same as being a set then there is no phi
due to diagonalization. Can you prove they are different?

C-B

[Graham Cooper]

You merely deny a machine parsed proof of exactly what you asked for.

> Axiom. If P(x) is a predicate with one and only one free variable
> x, then {x | P(x)} is a set.

[CB]
Sure.

1
LET
P(x) <-> not(X e X)

2
{x|P(X)} is a set

3
X e rusl <-> P(X)

4
X e rusl <-> not(X e X)

5
not(X e X) -> X e rusl
AND
X e rusl -> not(X e X)


6

*** RUSSELLS SET ***
if( not(e(X,X)) , e(X,rusl) ).

if( e(X,rusl) , not(e(X,X)) ).

7

?- t( not(e(rusl,rusl)) , z(1) ).

YES
?- t( e(rusl,rusl) , z(z(1)) ).

YES

8
CONTRADICTION!

So either PAYPAL $25
OR state which Step does not follow from the previous Steps!


Herc
--
NEVER TAKE A BET WITH THE ADJUDICATOR!

[William Hale]

In article
<dc74bcd4-e232-4a1c...@x13g2000vby.googlegroups.com>,

Isn't "x not in x" a predicate with one and only one free variable?

[Jeff Barnett]

Graham Cooper wrote, On 2/17/2013 4:07 PM:

> Is any definable collection a set?
>
> That is the usual meaning of Naive Set Theory.
>

I suggest we don't need your definition. The world (99%) tells the
student to consult "Naive Set Theory," Paul R. Halmos (1960) D. von
Nostrand. For the rest (1%) the term means ZF sans Choice and Continuum.
Why would you think of tossing your definition out here? And why Prolog
at all? Speak the common language.
--
Jeff Barnett

[Charlie-Boo]

Step 7 is no good because t and z are not defined.

Where's my $1?

C-B

[Charlie-Boo]

On Feb 17, 7:38 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <dc74bcd4-e232-4a1c-88db-8fbbd539e...@x13g2000vby.googlegroups.com>,

Not if predicate and set are the same thing. There is no predicate of
predicates that are not in themselves. Can you prove predicate and
set are not the same thing?

C-B

[Charlie-Boo]

That was 60 years later. Russell wrote to Gottlob Frege with news of
his paradox on June 16, 1902. The paradox was of significance to
Frege's logical work since, in effect, it showed that the axioms Frege
was using to formalize his logic were inconsistent. Specifically,
Frege's Rule V, which states that two sets are equal if and only if
their corresponding functions coincide in values for all possible
arguments, requires that an expression such as f(x) be considered both
a function of the argument x and a function of the argument f. In
effect, it was this ambiguity that allowed Russell to construct R in
such a way that it could both be and not be a member of itself.

C-B

[William Hale]

In article
<fc00c8f1-ee0d-48cb...@z4g2000vbz.googlegroups.com>,

Well, you can define things to be whatever you want them to be, so I
cannot prove "set" and "predicate" are the same or different until I
know what things are taken as primitives and what things are taken as
defined and what the axioms are.

Which comes first: predicate or set? Or, are they both primitives and
you have an axiom that they are the same?

[Charlie-Boo]

On Feb 17, 10:26 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <fc00c8f1-ee0d-48cb-b82b-e8a58658a...@z4g2000vbz.googlegroups.com>,

Can you give an example of people being able to define NST as they
please? That would mean there is no such thing as NST in the first
place. Is that what you're saying?

> cannot prove "set" and "predicate" are the same or different until I
> know what things are taken as primitives and what things are taken as
> defined and what the axioms are.
>
> Which comes first: predicate or set? Or, are they both primitives and
> you have an axiom that they are the same?

What do you mean "comes first"? Why doesn't the definition of Naïve
Set Theory answer your questions?

If you don't know enough about NST to answer your questions, then it
sounds like you don't know enough to give a proof of inconsistency.
Is that what you are saying?

C-B

[William Hale]

In article
<f9603a2c-1dd5-4eb6...@w14g2000vba.googlegroups.com>,

> > > > > > > > > No. �See David Hilbert, "Mathematical Problems", G�ttinger

Yes: you. You are saying that predicate and set are the same thing.

> That would mean there is no such thing as NST in the first
> place. Is that what you're saying?

Yes, I am saying that there is no such thing as a standard NST. But,
that is not my point. Rather, I am trying to find out what you mean by
NST.

>
> > cannot prove "set" and "predicate" are the same or different until I
> > know what things are taken as primitives and what things are taken as
> > defined and what the axioms are.
> >
> > Which comes first: predicate or set? Or, are they both primitives and
> > you have an axiom that they are the same?
>

> What do you mean "comes first"? Why doesn't the definition of Na�ve

> Set Theory answer your questions?

I don't have the definition of Naive Set Theory. What is the definition
of Naive Set Theory?

In ZFC, predicate comes first since it is part of logic, even before (or
if ever) you begin ZFC. Then, in ZFC (for example), set is introduced as
an undefined object. Of course, you can some other axiom system other
than ZFC where a set is introduced as undefined or could even be defined
in terms of something else in that axiom system.

>
> If you don't know enough about NST to answer your questions, then it
> sounds like you don't know enough to give a proof of inconsistency.
> Is that what you are saying?

I am trying to find out what you mean by NST. In my understanding of
NST, NST would be inconsistent and "predicate" and "set" are not the
same thing.

[fom]

On 2/17/2013 8:31 AM, Charlie-Boo wrote:
>
> Interesting. Do you accept my wager?

Do you seriously believe I do not
understand my own posts?

Draw a rectangle.

Draw a circle in the rectangle.

Draw a point inside the circle.

Read the axiom of regularity.

Take the point inside the circle to
be the element with void intersection.

Draw a circle in the rectangle
disjoint from the first circle.

Label the second circle "use"

Label the point inside the first
circle "mention"

Define the power set for a given
set in terms of strict subsets instead
of the usual definition and see if
you do not notice anything strange.


I would really need to see your
axioms first. Russell and Frege
made mistakes not made by Cantor,
Lesniewski, and Zermelo. So, I do
not really know what the definition
of naive set theory is.

[Rupert]

On Saturday, February 16, 2013 11:11:52 PM UTC+1, Charlie-Boo wrote:
> On Feb 15, 11:59 pm, Rupert <rupertmccal...@yahoo.com> wrote:

>
> > On Friday, February 15, 2013 10:56:09 PM UTC+1, Charlie-Boo wrote:
>
> > > Agreement:
>
> >
>
> > > I, the owner of email account shymathgu...@aol.com, do hereby agree to
>
> >
>
> > > wager $25 against $1 from anyone, payable through PayPal, that they
>
> >
>
> > > cannot prove Naïve Set Theory inconsistent, subject to the condition
>
> >
>
> > > that the person states here that they enter into this wager within 24
>
> >
>
> > > hours after this offer appears and they are the first to give their
>
> >
>
> > > proof as part of this wager.
>
> >
>
> > > C-B
>
> >
>

> > I am happy to take you up on your offer, but you should first indicate what definition of Naive Set Theory you want to work with.
>
>
>
> Offer accepted.
>
>
>
> I am not aware of that distinction. By definition do you mean "which
>
> theory" or "which wording of this one theory"? What alternatives are
>
> you thinking of?
>
>
>
> C-B

I propose that by Naive Set Theory we meaning the following theory in the first-order language of set theory.

It has the axiom schema of comprehension, meaning that for any open formula phi(x,y_1,y_2,...y_n), we have the axiom Ay_1Ay_2...Ay_n EzAx(x e z <->phi(x,y_1,...,y_n)), where z is any variable not free in phi.

Do you accept this definition of Naive Set Theory?

[George Greene]

On Feb 18, 12:23 am, Charlie-Boo <shymath...@gmail.com> wrote:
> If you don't know enough about NST to answer your questions, then it
> sounds like you don't know enough to give a proof of inconsistency.

Jeezus, when will you EVER give it a REST?? This conversation IS TEN
YEARS OLD by now!!
groups.google.com/group/sci.math/browse_frm/thread/afebc0e3bfb839d/
1998229610689e6b

You were already participating in this thread in 2003! Why are YOU
PRETENDING that
YOU DON'T KNOW what Naive Set Theory *MEANS*?!?? IT *MEANS*
THIS AXIOM-SCHEMA:
Ex[Ay[yex<->phi(y)] ].

You should not be entertaining conversations with people under the
delusion that there
might be reasonable DOUBT about that.
And the relevant INDIRECT (or proof of contradiction) follows VERY
directly from
that axiom -- YOU JUST instantiate phi(y) to ~yey. It is also very
relevant that
EVEN WITHOUT this axiom, you would have a contradiction from
Ex[Ay[yRx<->~yRy]] FOR ANY BINARY RELATION R WHATSOEVER,
NOT JUST for set-membership! In other words, IN ADDITION to being
able
to prove Russell's Paradox "from Naive Set Theory", you CAN ALSO prove
the NON-
existence of the Russell set FROM NO AXIOMS AT ALL!

YOU HAVE BEEN THROUGH ALL OF THIS BEFORE.
Your decision to invoke money, PayPal, and "from Naive Set Theory"
WITHOUT *YOUR*SELF SAYING what "Naive Set Theory" *MEANS*
is grounds for excommunication, frankly.

[Charlie-Boo]

On Feb 17, 6:04 pm, George Greene <gree...@email.unc.edu> wrote:
> On Feb 17, 5:17 pm, Charlie-Boo <shymath...@gmail.com> wrote:
>
> > Yes, there is no set of sets that don't contain themselves - the proof
> > is 3-4 lines long.  But NST doesn't imply that there is.
>
> Liar.
> Nobody can help it if YOU don't know what Naive Set Theory IS.

How is one to learn, then? I don’t think too many people actually
do. What is the definitive account of Naïve Set Theory? Are we
talking about a particular early attempt or the result of evolution
over time?

> Besides, given that, as I and many others have pointed out, the
> non-existence of this set can be proved FROM NO AXIOMS AT ALL,
> IF you are willing to use *SECOND* order logic, the fact that it CAN
> ALSO
> be proved from NST follows AUTOMATICALLY.

You’d better be careful. That same mistake of not considering whether
or not truth and provability coincide that Godel used to catch Hilbert
is now being used to catch Russell and poor Frege, and may catch you
as well in more ways than one.

When we exhibit a sentence such as,

{ x | x ~e x } e { x | x ~e x }

(call it RSI for Russell’s Set Is In Itself), and note that,

RSI => ~RSI
~RSI => RSI
RSI = = ~RSI

and then say that cannot be, we are using a property of relations when
RSI is not a relation. We are assuming that every sentence is true or
false and conclude there is an inconsistency.

But always being true or false, and never both, while a property of
relations, is not a property of sentences such as RSI or “This is
false.” (or the various CBL-derived variations such as “ ‘It is true
of this.’ is true of ‘This is true of it.’” or the source of the Liar:
“ ‘It is not true of itself.’ is true of ‘It is not true of itself.’.
“)

> In the case of NST, though, there is a relevant SPEEDUP:
> From the naive set comprehension schema, you can prove
> Ex[Ay[yex<->~yey]], IN ONE LINE.
> In one more line, you can then instantiate y to x,
> which gives you a contradiction in one MORE line.

Sounds like it’s 3-4 lines after all.

> The 2nd-order logic proof is an indirect proof a validity, though,
> as OPPOSED to a proof that the whole theory is inconsistent, since
> in THAT proof, UNLIKE in the proof from NST, you are NOT priorly
> committed (by an axiom)
> to the provabiliity&truth of the first line.

Godel showed that if truth and provability coincide, then there is a
contradiction and Hilbert’s Programme cannot be attained.
http://www.cs.nyu.edu/pipermail/fom/2010-July/014890.html
http://www.cs.nyu.edu/pipermail/fom/2010-July/014918.html Now
consider what happens if class/concept/collection and set coincide and
the effect on the Russell Programme to thwart Frege.

C-B

[Charlie-Boo]

On Feb 18, 12:57 am, William Hale <bill...@yahoo.com> wrote:
> In article

> <f9603a2c-1dd5-4eb6-9268-bc70e80a2...@w14g2000vba.googlegroups.com>,

> > > > > > > > > > No.  See David Hilbert, "Mathematical Problems", Göttinger

Where did I say that? I did say this:

1. If predicate and set are the same thing, then there is no predicate
of sets that do not contain themselves.

2. (1) implies: If there is a predicate of sets that do not contain
themselves, then predicate and set are not the same thing (Modus
Tollens.)

3. A proof that relies on the assertion that there is a predicate of
sets that do not contain themselves will prove there is a predicate of
sets that do not contain themselves (definition of proof and
implication.)

4. (1) and (2) implies: A proof that there is a predicate of sets that
do not contain themselves will also prove that predicate and set are
not the same thing.

5. (3) and (4) implies: A proof that relies on the assertion that
there is a predicate of sets that do not contain themselves must also
prove that predicate and set are not the same thing.

6. (5) implies: If you cannot prove that predicate and set are not the
same thing then there is no proof that relies on the assertion that
there is a predicate of sets that do not contain themselves.

Which of these do you disagree with?

Formally (in CBL):

1: PRED , SE => - ~SE / PRED
2: ~SE / PRED => - PRED , SE
3: |- ( ( P=> Q ) => Q ) => |- P
4: |- ~SE / PRED => |- - PRED , SE
5: |- ( ( ~SE / PRED => P ) => P ) => |- - PRED , SE
6: ~ |- - PRED , SE => ~ |- ( ( ~SE / PRED => P ) => P )

I can explain and show formal proofs. The axiom is - ~P/P. I have
given CBL 101 many times on these pages. CBL can be used to generate C
++ programs to find the smallest common prime factor between 2 given
numbers or list all employees who make more than their manager,
generate theorems of recursion theory, theory of computation and
incompleteness in logic (see FOM July 2010 18 Word Proof) and
formalizations of paradoxes and logical arguments such as the Russell-
Frege conversation and other logical arguments displayed on these
pages. We take FOL, allow input and output variables in wffs, and
define a relation over sets of different cardinalities M # P / Q
program M calculates P in language Q and P / Q there is an M such that
M # P / Q.

> > That would mean there is no such thing as NST in the first
> > place.  Is that what you're saying?
>
> Yes, I am saying that there is no such thing as a standard NST. But,
> that is not my point. Rather, I am trying to find out what you mean by
> NST.

I am thinking about the fact that people say Russell’s Paradox proves
Naïve Set Theory inconsistent and Russell described his paradox to
Frege causing Frege to write that his system had problems.

“Russell wrote to Gottlob Frege with news of his paradox on June 16,

1902. The paradox was of significance to Frege's logical work since,
in effect, it showed that the axioms Frege was using to formalize his

logic were inconsistent.” http://plato.stanford.edu/entries/russell-paradox/

People also refer to ZF replacing NST and that also occurred after the
Russell-Frege letters:
“Other responses to Russell's paradox have included . . . Ernst
Zermelo's 1908 axiomatization of set theory.“ - ibid

> > > cannot prove "set" and "predicate" are the same or different until I
> > > know what things are taken as primitives and what things are taken as
> > > defined and what the axioms are.
>
> > > Which comes first: predicate or set? Or, are they both primitives and
> > > you have an axiom that they are the same?
>

> > What do you mean "comes first"?  Why doesn't the definition of Naïve

> > Set Theory answer your questions?
>
> I don't have the definition of Naive Set Theory. What is the definition
> of Naive Set Theory?

http://groups.google.com/group/sci.logic/msg/04534f2e57bf8ed6?hl=en

> In ZFC, predicate comes first since it is part of logic, even before (or
> if ever) you begin ZFC. Then, in ZFC (for example), set is introduced as
> an undefined object. Of course, you can some other axiom system other
> than ZFC where a set is introduced as undefined or could even be defined
> in terms of something else in that axiom system.
>
>
>
> > If you don't know enough about NST to answer your questions, then it
> > sounds like you don't know enough to give a proof of inconsistency.
> > Is that what you are saying?
>
> I am trying to find out what you mean by NST. In my understanding of
> NST, NST would be inconsistent and "predicate" and "set" are not the
> same thing.

How are they defined so as to be different?

C-B

[fom]

>>>>>>>> No. See David Hilbert, "Mathematical Problems", G�ttinger

These are not his. But, they may inform as to the possibilities
with regard to his claim

news://news.giganews.com:119/5bidne6Ppslu1HzN...@giganews.com

[fom]

So you do pay attention!!!

I thought so from your remarks in another
thread. But, very often your remarks toward set
theory have seemed anti-mathematical. I see now that
you merely see the same problem with the same "urban
legend" from a different background.

[Charlie-Boo]

On Feb 18, 1:09 am, fom <fomJ...@nyms.net> wrote:
> On 2/17/2013 8:31 AM, Charlie-Boo wrote:
>
>
>
> > Interesting.  Do you accept my wager?
>
> Do you seriously believe I do not
> understand my own posts?

I’ve never really given that much thought, actually.

> Draw a rectangle.
>
> Draw a circle in the rectangle.
>
> Draw a point inside the circle.
>
> Read the axiom of regularity.
>
> Take the point inside the circle to
> be the element with void intersection.
>
> Draw a circle in the rectangle
> disjoint from the first circle.
>
> Label the second circle "use"
>
> Label the point inside the first
> circle "mention"
>
> Define the power set for a given
> set in terms of strict subsets instead
> of the usual definition and see if
> you do not notice anything strange.

Yes, definitely. All the axioms and theorems about power sets don’t
make sense anymore because you changed the definition of power set.
Don’t you think it’s not a good idea to change the rules in the middle
of a game, so to speak?

> I would really need to see your
> axioms first.  Russell and Frege

Which ones, for Program Synthesis, Theory of Computation, Recursion
Theory, Incompleteness in Logic or my Theory of Representability (an
amalgamation of all branches of Math/Logic/Computer Science)?

The most exciting latest development there is that I recently
developed computer program construction rules that are programming
language independent (save at most some literals in a simple table)!
And now I have expanded my branch relationships to include Program
Synthesis => Theory of Computation => Incompleteness in Logic (the
latter was the subject of a FOM discussion July 2010 18 Word Proof.)

> made mistakes not made by Cantor,
> Lesniewski, and Zermelo.  So, I do
> not really know what the definition
> of naive set theory is.

Turing made a lot of mistakes in his 1937 paper, but I still know what
his definitions are. Why does that cause a complete impasse?

C-B

[Charlie-Boo]

Sure. Thanks (for thinking instead of griping).

C-B

[William Hale]

In article
<33468096-5b28-41cf...@h11g2000vbf.googlegroups.com>,

> > > > > > > > > > > > > or not - if you put your money where your mouth is.

> > > > > > > > > > > G�ttinger

So, you are saying that one does not know (cannot prove) whether or not
predicate and set are the same things.

[Rupert]

Okay. So let phi(x) be the formula ¬(x e x). Then one special case of the axiom schema of comprehension is EzAx(x e z <-> ¬(x e x)). That gives you Ez(z e z <-> ¬(z e z)), which clearly leads to a contradiction.

Can I have my $25 now?

[Charlie-Boo]

On Feb 18, 5:20 am, George Greene <gree...@email.unc.edu> wrote:
> On Feb 18, 12:23 am, Charlie-Boo <shymath...@gmail.com> wrote:
>
> > If you don't know enough about NST to answer your questions, then it
> > sounds like you don't know enough to give a proof of inconsistency.
>
> Jeezus, when will you EVER give it a REST?? This conversation IS TEN
> YEARS OLD by now!!
> groups.google.com/group/sci.math/browse_frm/thread/afebc0e3bfb839d/
> 1998229610689e6b
>
> You were already participating in this thread in 2003!  Why are YOU
> PRETENDING that
> YOU DON'T KNOW what Naive Set Theory *MEANS*?!??  IT *MEANS*
> THIS AXIOM-SCHEMA:
> Ex[Ay[yex<->phi(y)] ].
>
> You should not be entertaining conversations with people under the
> delusion that there
> might be reasonable DOUBT about that.

Go to the source of those vicious rumors – I mean, circles.

> And the relevant INDIRECT (or proof of contradiction) follows VERY
> directly from
> that axiom -- YOU JUST instantiate phi(y) to ~yey.   It is also very

Are you sure you can do that?

> relevant that
> EVEN WITHOUT this axiom, you would have a contradiction from
> Ex[Ay[yRx<->~yRy]]  FOR ANY BINARY RELATION R WHATSOEVER,
> NOT JUST for set-membership!  In other words, IN ADDITION to being

You seem to be suggesting that set membership is a relation – no?

> able
> to prove Russell's Paradox "from Naive Set Theory", you CAN ALSO prove
> the NON-
> existence of the Russell set FROM NO AXIOMS AT ALL!
>
> YOU HAVE BEEN THROUGH ALL OF THIS BEFORE.
> Your decision to invoke money, PayPal, and "from Naive Set Theory"
> WITHOUT *YOUR*SELF SAYING what "Naive Set Theory" *MEANS*
> is grounds for excommunication, frankly.

Gee, I didn’t know I was even communicated in the first place. That
was fleeting news. Those were the days. But . . . the best is yet to
come. Wait till my team and I produce the world’s only actual program
synthesis system – based on mathematical logic. How does proof enable
program synthesis? By creating loops from proofs by induction (Manna/
Waldinger)? Nope. By proving (all x)P(x) => (exists y)Q(x,y) (Martin
Lof)? Naah. By proving the program you created satisfies the given
spec, silly! How else?

C-B

[Charlie-Boo]

On Feb 18, 2:34 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <33468096-5b28-41cf-8611-aa59c4532...@h11g2000vbf.googlegroups.com>,

> > > > > > > > > > > > Göttinger

Not necessarily. The purpose of the wager is in part to decide that.
But, again, for this explanation of why NST is inconsistent to be
valid, predicate and set cannot be the same. So one must assure that
to give validity to the proof.

C-B

[William Hale]

In article
<33468096-5b28-41cf...@h11g2000vbf.googlegroups.com>,
Charlie-Boo <shyma...@gmail.com> wrote:

> On Feb 18, 12:57�am, William Hale <bill...@yahoo.com> wrote:
> > In article
> > <f9603a2c-1dd5-4eb6-9268-bc70e80a2...@w14g2000vba.googlegroups.com>,

> > I am trying to find out what you mean by NST. In my understanding of
> > NST, NST would be inconsistent and "predicate" and "set" are not the
> > same thing.
>
> How are they defined so as to be different?

In logic, one defines what is meant by a predicate. It is defined
recursively. Here are some example.

*) If "P" is a predicate then "not P" is a predicate.
*) If "P" and "Q" are predicates then "P or Q" is a predicate.
*) If "P" is a predicate and "x" is an object, then "Ax.P" is a
predicate.
*) etc.

In (my understanding of) NST, the sets are undefined objects. There is
also an undefined two-place predicate symbol "in". If x and y are two
sets, then "x in y" is a predicate (with free variables x and y).

I don't define the sets as being different from the predicates. There is
no need to. The sets are undefined and the predicates are built up from
the two-place predicate symbol filled in with two objects by the
recursive rules given by logic (using the standard logical connectives).

[Charlie-Boo]

How do we know z e z has a truth value? What is the truth value of
"This is false."?

Is there a formula expressing the proposition that something is a
formula that does not apply to itself?

> Can I have my $25 now?

I certainly hope that you have all that is rightfully yours, no less,
no more.

C-B

[Charlie-Boo]

On Feb 18, 3:08 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <33468096-5b28-41cf-8611-aa59c4532...@h11g2000vbf.googlegroups.com>,

" 'x in y' is a predicate (with free variables x and y). I don't

define the sets as being different from the predicates."

That won't work. Diagonalize.

C-B

[Charlie-Boo]

On Feb 18, 3:08 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <33468096-5b28-41cf-8611-aa59c4532...@h11g2000vbf.googlegroups.com>,

>
>  Charlie-Boo <shymath...@gmail.com> wrote:
> > On Feb 18, 12:57 am, William Hale <bill...@yahoo.com> wrote:
> > > In article
> > > <f9603a2c-1dd5-4eb6-9268-bc70e80a2...@w14g2000vba.googlegroups.com>,
> > > I am trying to find out what you mean by NST. In my understanding of
> > > NST, NST would be inconsistent and "predicate" and "set" are not the
> > > same thing.
>
> > How are they defined so as to be different?
>
> In logic, one defines what is meant by a predicate. It is defined
> recursively. Here are some example.
>
>    *) If "P" is a predicate then "not P" is a predicate.
>    *) If "P" and "Q" are predicates then "P or Q" is a predicate.
>    *) If "P" is a predicate and "x" is an object, then "Ax.P" is a
> predicate.
>    *) etc.

That's actually close to CBL Set Theory, which fixes all of these
problems in a simple, unobjectionable manner.

C-B

[William Hale]

In article
<5a7686c5-375c-475d...@y9g2000vbb.googlegroups.com>,

I don't understand what you mean.

I was trying to answer your question "How are they defined so as to be
different". My answer is that I don't define the sets as being different
from the predicates. That is the answer to your question. However, I
don't define the sets as being not different from the predicates. That
is, sets are taken as undefined, and predicates (using "in" and the
logical connectives) are taken as defined recursively. I cannot now come
and define some or all sets to be some or all predicates. The sets are
(un)defined and the predicates are also defined: I cannot now redefine
something that is already defined.

[fom]

I changed no rules. There were differences of
opinion. I did not believe what I had been
taught. I analyzed, I wrote axioms, and I
discovered those different opinions.

Once a legitimate construct for "=" is
available, one needs only to add one
element to the "rubber band" power set.

1)
proper part
2)
proper part collection
3)
=
4)
subset collection

>> I would really need to see your
>> axioms first. Russell and Frege
>
> Which ones, for Program Synthesis, Theory of Computation, Recursion
> Theory, Incompleteness in Logic or my Theory of Representability (an
> amalgamation of all branches of Math/Logic/Computer Science)?
>

I meant for naive set theory.

Do you have a link for those others?

> The most exciting latest development there is that I recently
> developed computer program construction rules that are programming
> language independent (save at most some literals in a simple table)!
> And now I have expanded my branch relationships to include Program
> Synthesis => Theory of Computation => Incompleteness in Logic (the
> latter was the subject of a FOM discussion July 2010 18 Word Proof.)
>
>> made mistakes not made by Cantor,
>> Lesniewski, and Zermelo. So, I do
>> not really know what the definition
>> of naive set theory is.
>
> Turing made a lot of mistakes in his 1937 paper, but I still know what
> his definitions are. Why does that cause a complete impasse?

The difference between Frege and Cantor is highlighted
in Cantor's review of Frege. A Cantorian set is not
a Fregean extension. And, Zermelo's 1908 respected
Fregean description theory, whereas the "improved,"
subsequent versions did not.

Sometimes on a construction site, one can just throw
a bunch of guys together and get a wonderful job.

Other times, the same strategy leaves a mangled mess.

[fom]

It won't be that simple.

C-B permitted you the definition for which you asked. However,
you did not define your terms sufficiently well.

So, for example, the foundational system put forth by
Bertrand Russell and A. N. Whitehead in the first edition
of "Principia Mathematica" is based upon a "no-classes"
theory. This is derived from Russell's objections to
Meinong non-existent objects and Frege's description
theory.

In "On Denoting" Russell developed his own description
theory. In that theory, the word 'the' is treated as
a quantifier. The purpose behind that strategy is to
avoid "presupposition failure" that occurs when a sentence
that purports reference clearly has no objectively admitted
referent to which it refers. The classic example cited
in the literature is

"The present king of France is bald."

Is it true? Is it false?

After a significant analysis, Russell writes:

"Thus according to the meaning of denotation
lately explained, 'the difference between A
and B' has a denotation when A and B differ,
but not otherwise. This difference applies
to true and false propositions generally.
If 'aRb' stands for 'a has relation R to b'
then when aRb is true, there is such an entity
as the relation R between a and b; when aRb
is false, there is no such entity."


I do not want to ruin all of C-B's fun. You should
play the game and see where he takes you.

[George Greene]

On Feb 18, 2:53 pm, Charlie-Boo <shymath...@gmail.com> wrote:
> You seem to be suggesting that set membership is a relation – no?

NObody is saying ANYthing about what "set-membership" IS!
This is a THEORY! It's phrased IN A LANGUAGE!
The language HAS PREDICATES!
In THAT context, set-membership IS REPRESENTED AS a binary predicate!
You CAN ASK about ANY *two* things whether one is or is not a member
of
the other! The QUESTION is well-formed! IT ALWAYS HAS a yes-or-no
answer!
What is going on IN THE REAL WORLD (OUTside the theory) IS NOT
relevant!
Your question is about NAIVE SET *THEORY*! THAT does NOT have an
OPINION
on what set-membership IS! It is A FACT about the THEORY that the THE
THEORY HAS
a binary predicate where the thing on the right IS TREATED AS AND
EXPECTED TO BE a set!

NObody, ESPECIALLY NOT *YOUR* dumbass, gets to say whether
"set-membership IS" this, that, or the other! THAT is NOT a
theoretical question!
THAT is a METAPHYSICAL question and it is NOT relevant!
Obvioiusly there can be OTHER TREATMENTS of set membership
that need not use THIS linguistic paradigm! But when YOU SAY
"Naive Set Theory" then YOU BETTER KNOW what THAT means, withOUT
asking me OR ANYone irrelevant questions! AND WHERE'S MY DAMN
$25?!????????

[George Greene]

On Feb 18, 2:53 pm, Charlie-Boo <shymath...@gmail.com> wrote:

> > And the relevant INDIRECT (or proof of contradiction) follows VERY
> > directly from
> > that axiom -- YOU JUST instantiate phi(y) to ~yey.

> Are you sure you can do that?

Yes. More to the point, I'm sure YOU can.
Are you sure you can ask that question without BEING STUPID??

[George Greene]

On Feb 18, 2:53 pm, Charlie-Boo <shymath...@gmail.com> wrote:

> Gee, I didn’t know I was even communicated in the first place.  That
> was fleeting news.  Those were the days.  But . . . the best is yet to
> come.  Wait till my team and I produce the world’s only actual program
> synthesis system – based on mathematical logic.  How does proof enable
> program synthesis?

The QUESTION WAS ABOUT a proof of the inconsistency of naive set
theory
via Russell's Paradox, NOT PROGRAM SYNTHESIS! STOP CHANGING THE
SUBJECT!
AND PAY UP!!

[Nam Nguyen]

On 18/02/2013 5:28 PM, George Greene wrote:
> On Feb 18, 2:53 pm, Charlie-Boo <shymath...@gmail.com> wrote:

>> Gee, I didn�t know I was even communicated in the first place. That

>> was fleeting news. Those were the days. But . . . the best is yet to

>> come. Wait till my team and I produce the world�s only actual program
>> synthesis system � based on mathematical logic. How does proof enable

>> program synthesis?
>
> The QUESTION WAS ABOUT a proof of the inconsistency of naive set
> theory
> via Russell's Paradox, NOT PROGRAM SYNTHESIS! STOP CHANGING THE
> SUBJECT!
> AND PAY UP!!

Right. Fwiw, I have a bad feeling that C-B's arguing is rooted in
changing the subjects as often as he needs. Whatever the motivation
he might have, arguing with him is like wandering around a new area,
being carefree and _logicfree_ !

--
----------------------------------------------------
There is no remainder in the mathematics of infinity.

NYOGEN SENZAKI
----------------------------------------------------

[George Greene]

On Feb 16, 10:45 am, George Greene <gree...@email.unc.edu> wrote:
> I fear that the more relevant question may be whether he does or
> doesn't understand
> the whole notion of an axiom-schema in a first-order theory.

Indeed, precisely as I predicted, when I informed him that the phi(.)
in the naive comprehension schema,
Ex[Ay[yex<->phi(y)]] ,
could be instantiated as phi(y)= ~yey,
he asked, "are you sure you can do that?"

[Charlie-Boo]

On Feb 18, 3:38 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <5a7686c5-375c-475d-97b9-affffd811...@y9g2000vbb.googlegroups.com>,

I don't mean how do you explicitly define them so they will be
different.

You said, " "predicate" and "set" are not the same thing. "

I am asking, how can you tell from their definitions that they are
different? If x is defined by x+3=5 and y is defined by y+3=6 then x
is not equal to y because 5 is not equal to 6.

I am asking how you know they are not the same.

If they may or may not be the same, then there may or may not be a
predicate for x~ex and we cannot say that NST is definitely
inconsistent because there is a predicate viz x~ex that is not a set.

The problem is that we find out that x~ex is not a set and then try to
say x~ex is something different from a set but we still have to make
sure we don't declare the diagonal of the type of object as being an
instance of that object.

In formal Set Theory, they do the same thing: add "class" which
doesn't exist in math but they decide to introduce it and axiomatize
it.

1.If something is not a set then don't try to figure out what it is.
Just say it's not a set and leave it at that. Set is the most general
thing you can think of, so outside of this is beyond what you can
define.
2. It is what is called "kicking the can down the road." You will
eventually confront diagonalization of your new term and have the same
problem all over again. There is no class of classes that do not
contain themselves replaces Russell's Headache.
3. Axioms are for formalizing what we know informally or intuitively.
It is not for making up a property then formalizing it. That is just
looking at all the different systems we can define rather than what is
correct.

The best model of Set Theory is still the Theory of Computation.
Program = Set.

You can't prove a formal system inconsistent with vague informal
terms. You end up assuming things and I am pointing out the
responsibility for making assumptions.

The Mathematician's Oath:

"If you say it, you have to prove it. If you define it, you are
saying the definition is consistent."

C-B

[Charlie-Boo]

The only thing I know about NST is that it says wffs are sets and of
course they can be, and when someone says it is inconsistent, I say
not if you just say that sets begat sets and non-sets begat non-sets.
Whether wffs can represent non-sets depends on whether you want to
deal in non-sets or not. If not, then they aren't allowed in wffs and
all wffs are sets. But the question remains, what wffs are sets, for
which we work backwards from x~ex to generate what is not a set.

> Do you have a link for those others?

Start a thread asking for an explanation of Computationally Based
Logic (aka CBL) and I can explain it. Here is an early small part:
http://arxiv.org/html/cs/0003071

C-B

[William Hale]

In article
<65c848d5-792c-4021...@fn10g2000vbb.googlegroups.com>,

Take m to be the empty set, defined in different ways in ZFC (which is
not relevant here).

Take p to be the predicate "not x in x".

You are asking me, in this particular situation, to prove that m and p
are not the same.

To prove things in ZFC, I need to use the axioms of course.

But, even before I get to that point, I need to have a statement to
prove. You want me to prove the statement "not m = p". But, the
statement "not m = p" is not a syntactically correct statement in ZFC.
You want me to prove "not m = [not x in x]". A predicate like "not x in
x" is not a set. The statement "not m = [not x in x]" is not
syntactically valid.

If I have the statement "Ay. not y in m", where m is the empty set that
I mentioned before, then I cannot substitute for y the predicate "not x
in x". Or, are you saying that I can?

[Charlie-Boo]

You can't say there is a predicate of non self containing sets unless
you can show predicates and sets differ,

C-B

[William Hale]

In article
<6d1a80a4-20f3-40c4...@g8g2000vbf.googlegroups.com>,

If m is the empty set, are you saying "not m in m" is not a predicate
unless one can show predicates and sets differ?

If s is the set containing m and only m, are you saying "not s in s" is
not a predicate unless one can show predicates and sets differ?

[Rupert]

You asked me to derive a contradiction from the axiom schema of comprehension in first-order logic. I did this. I don't need to make any reference to the semantics of the first-order language of set theory. As far as the standard semantics goes, the formula z e z has a truth-value for every possible interpretation of the variable z.

>
>
> Is there a formula expressing the proposition that something is a
>
> formula that does not apply to itself?
>

Yes, if you add a truth predicate to the language, but that's neither here nor there.

>
>
> > Can I have my $25 now?
>
>
>
> I certainly hope that you have all that is rightfully yours, no less,
>
> no more.
>

You owe me $25.

>
>
> C-B

[Alan Smaill]

Rupert <rupertm...@yahoo.com> writes:

> You asked me to derive a contradiction from the axiom schema of
> comprehension in first-order logic. I did this. I don't need to make
> any reference to the semantics of the first-order language of set
> theory. As far as the standard semantics goes, the formula z e z has a
> truth-value for every possible interpretation of the variable z.

Of course.

>> > Can I have my $25 now?
>>
>>
>>
>> I certainly hope that you have all that is rightfully yours, no less,
>> no more.
>>
>
> You owe me $25.

C-B does not know what bad faith is.

>> C-B
>

--
Alan Smaill

[Rupert]

Russell and Whitehead's theory of descriptions has got nothing to do with it. I specified *exactly* in which theory I would derive a contradiction. I gave a completely precise definition, and C-B accepted it. Then I gave the proof that a contradiction can be derived in this system. The proof is easy. The only way that there can be any room for doubt is if you did not really understand my definition of my formal theory. But my definition was absolutely precise, using standard terminology.

C-B owes me $25.

[fom]

I won't argue with you. I missed the line above when
you specified the axiom of comprehension.

[Charlie-Boo]

You can't if phi(x) is the same thing as x being a set. There is no
phi of phis that are not in themselves. You can't prove that there is
such a phi, any more than can you say there is a set of x ~e x.

The lesson is, you can't prove a formal system inconsistent with
informal terms. And it's informal because it's trying to create
something bigger than a set without defining it. It uses "phi"
instead of "set" as if "Sets don't include x ~e x so we'll give it a
new name and say it DOES." Did someone say something about "stewpid"?

C-B

[Charlie-Boo]

On Feb 19, 12:26 am, William Hale <bill...@yahoo.com> wrote:
> In article

> <6d1a80a4-20f3-40c4-ba5c-fa9141143...@g8g2000vbf.googlegroups.com>,

No, I am only reminding people that there is no set of x ~e x so if
you want to use something else that includes x ~e x you'd better make
sure IT is not just sets as well. You have to prove set is not the
same as predicate for such a predicate to exit.

Diagonalization catches yet another victim! (Which is good for me,
being the Meister of Diagonalization.)

C-B

[Charlie-Boo]

The proof that x ~e x is not a set follows directly from the
definition of e and set. You can't stop that.

UO me $1,

C-B

[Charlie-Boo]

On Feb 19, 5:19 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

Do you know what good logic is? Is there a predicate of predicates
that don't apply to themselves?

C-B

> >> C-B
>
> --
> Alan Smaill

[Charlie-Boo]

On Feb 18, 2:34 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <33468096-5b28-41cf-8611-aa59c4532...@h11g2000vbf.googlegroups.com>,
>
>
>
>
>
>
>
>
>
>  Charlie-Boo <shymath...@gmail.com> wrote:

> > On Feb 18, 12:57 am, William Hale <bill...@yahoo.com> wrote:
> > > In article

> > > <f9603a2c-1dd5-4eb6-9268-bc70e80a2...@w14g2000vba.googlegroups.com>,
>
> > >  Charlie-Boo <shymath...@gmail.com> wrote:

> > > > On Feb 17, 10:26 pm, William Hale <bill...@yahoo.com> wrote:
> > > > > In article

> > > > > <fc00c8f1-ee0d-48cb-b82b-e8a58658a...@z4g2000vbz.googlegroups.com>,
>
> > > > >  Charlie-Boo <shymath...@gmail.com> wrote:

> > > > > > On Feb 17, 7:38 pm, William Hale <bill...@yahoo.com> wrote:
> > > > > > > In article

> > > > > > > <dc74bcd4-e232-4a1c-88db-8fbbd539e...@x13g2000vby.googlegroups.com>,
>
> > > > > > >  Charlie-Boo <shymath...@gmail.com> wrote:

> > > > > > > > On Feb 17, 5:48 pm, William Hale <bill...@yahoo.com> wrote:
> > > > > > > > > In article

> > > > > > > > > <b37aaf2f-ae08-4161-9735-31820c24e...@y9g2000vbb.googlegroups.co

> > > > > > > > > m>,
>
> > > > > > > > >  Charlie-Boo <shymath...@gmail.com> wrote:

> > > > > > > > > > On Feb 17, 2:18 pm, billh04 <bill...@gmail.com> wrote:
> > > > > > > > > > > On Feb 17, 3:18 am, Charlie-Boo <shymath...@gmail.com>
> > > > > > > > > > > wrote:
>
> > > > > > > > > > > > On Feb 16, 10:17 pm, George Greene
> > > > > > > > > > > > <gree...@email.unc.edu>
> > > > > > > > > > > > wrote:
>
> > > > > > > > > > > > > On Feb 16, 5:31 pm, Charlie-Boo <shymath...@gmail.com>
> > > > > > > > > > > > > wrote:
>
> > > > > > > > > > > > > > The only real question is whether you can (notice a
> > > > > > > > > > > > > > proof
> > > > > > > > > > > > > > to)
> > > > > > > > > > > > > > prove
> > > > > > > > > > > > > > it
> > > > > > > > > > > > > > or not - if you put your money where your mouth is.
>
> > > > > > > > > > > > > That is NOT a real questions.  Real questions about
> > > > > > > > > > > > > math
> > > > > > > > > > > > > have
> > > > > > > > > > > > > exactly
> > > > > > > > > > > > > zero to do with money.
> > > > > > > > > > > > > More to the point, this is no more a "question" than
> > > > > > > > > > > > > "how
> > > > > > > > > > > > > much
> > > > > > > > > > > > > is
> > > > > > > > > > > > > 2+2?".
> > > > > > > > > > > > > It may be stated in the FORM of a question (a la
> > > > > > > > > > > > > Jeopardy)
> > > > > > > > > > > > > but
> > > > > > > > > > > > > give
> > > > > > > > > > > > > that the ANSWER
> > > > > > > > > > > > > is KNOWN, there IS NO QUESTION *about* this topic.
>
> > > > > > > > > > > > > And just for the record, tons of people already HAVE
> > > > > > > > > > > > > instantiated
> > > > > > > > > > > > > the
> > > > > > > > > > > > > naive set theory comprehension-axiom-schema with the
> > > > > > > > > > > > > unary-predicate-
> > > > > > > > > > > > > schema
> > > > > > > > > > > > > ~xex.  So Shut Up.
>
> > > > > > > > > > > > > As for whether I personally have produced a proof, see
> > > > > > > > > > > > > any
> > > > > > > > > > > > > number
> > > > > > > > > > > > > of
> > > > > > > > > > > > > posts I have
> > > > > > > > > > > > > made in this newsgroup going back 15 years or more.
> > > > > > > > > > > > >  For
> > > > > > > > > > > > > merely
> > > > > > > > > > > > > 6
> > > > > > > > > > > > > years, see,
> > > > > > > > > > > > > e.g.,http://groups.google.com/group/sci.logic/msg/a90fcd
> > > > > > > > > > > > > a397
> > > > > > > > > > > > > 1827
> > > > > > > > > > > > > 86
>
> > > > > > > > > > > > No.  See David Hilbert, "Mathematical Problems",
> > > > > > > > > > > > Göttinger
> > > > > > > > > > > > Nachrichten, 1900, pp. 253-297, and Archiv der Mathematik
> > > > > > > > > > > > und
> > > > > > > > > > > > Physik,
> > > > > > > > > > > > 3dser., vol. 1 (1901), pp. 44-63, 213-237.
>
> > > > > > > > > > > > If you would actually present the purported proof, then I
> > > > > > > > > > > > could
> > > > > > > > > > > > refer
> > > > > > > > > > > > to your words to show you what the flaw is.  Since nobody
> > > > > > > > > > > > has
> > > > > > > > > > > > offered
> > > > > > > > > > > > any attempted proof, I am now owed somewhere between $1
> > > > > > > > > > > > and
> > > > > > > > > > > > $3
> > > > > > > > > > > > via
> > > > > > > > > > > > PayPal.
>
> > > > > > > > > > > > If anybody ever tries to pawn off the popular false
> > > > > > > > > > > > proof,
> > > > > > > > > > > > then I
> > > > > > > > > > > > would point out that there is no collection/class/concept
> > > > > > > > > > > > of
> > > > > > > > > > > > all
> > > > > > > > > > > > sets
> > > > > > > > > > > > that do not contain themselves due to the same principle
> > > > > > > > > > > > of
> > > > > > > > > > > > diagonalization that tells us there is no set containing
> > > > > > > > > > > > that.
>
> > > > > > > > > >  >  Are you saying that the axioms of Naive Set Theory don't
> > > > > > > > > > imply
> > > > > > > > > > that
> > > > > > > > > >  >  the set of all sets that do not contain themselves
> > > > > > > > > > exists?
>
> > > > > > > > > > Yes.
>
> > > > > > > > > Isn't one of the axioms of Naive Set Theory the following:
>
> > > > > > > > >      Axiom. If P(x) is a predicate with one and only one free
> > > > > > > > > variable
> > > > > > > > > x, then {x | P(x)} is a set.
>
> > > > > > > > Sure.
>
> > > > > >  >  Isn't "x not in x" a predicate with one and only one free
> > > > > > variable?
>
> > > > > > Not if predicate and set are the same thing.  There is no predicate
> > > > > > of
> > > > > > predicates that are not in themselves.  Can you prove predicate and
> > > > > > set are not the same thing?
>
> > > > > Well, you can define things to be whatever you want them to be, so I
>
> > > > Can you give an example of people being able to define NST as they
> > > > please?
>
> > > Yes: you. You are saying that predicate and set are the same thing.
>
> > Where did I say that?  I did say this:
>
> > 1. If predicate and set are the same thing, then there is no predicate
> > of sets that do not contain themselves.
>
> > 2. (1) implies: If there is a predicate of sets that do not contain
> > themselves, then predicate and set are not the same thing (Modus
> > Tollens.)
>
> > 3. A proof that relies on the assertion that there is a predicate of
> > sets that do not contain themselves will prove there is a predicate of
> > sets that do not contain themselves (definition of proof and
> > implication.)
>
> > 4. (1) and (2) implies: A proof that there is a predicate of sets that
> > do not contain themselves will also prove that predicate and set are
> > not the same thing.
>
> > 5. (3) and (4) implies: A proof that relies on the assertion that
> > there is a predicate of sets that do not contain themselves must also
> > prove that predicate and set are not the same thing.
>
> > 6. (5) implies: If you cannot prove that predicate and set are not the
> > same thing then there is no proof that relies on the assertion that
> > there is a predicate of sets that do not contain themselves.
>
> > Which of these do you disagree with?
>
> So, you are saying that one does not know (cannot prove) whether or not
> predicate and set are the same things.

You have to prove that predicate and set are different to know that
predicate x ~e x exists and use it in a proof.

C-B

[Charlie-Boo]

> Can I have my $25 now?

I can't let phi be set ¬(x e x) so how can you let phi be formula ¬(x
e x) when you haven't shown that formula is different from set?

C-B

[Charlie-Boo]

On Feb 18, 8:03 pm, Nam Nguyen <namducngu...@shaw.ca> wrote:
> On 18/02/2013 5:28 PM, George Greene wrote:
>
> > On Feb 18, 2:53 pm, Charlie-Boo <shymath...@gmail.com> wrote:

> >> Gee, I didn’t know I was even communicated in the first place.  That

> >> was fleeting news.  Those were the days.  But . . . the best is yet to

> >> come.  Wait till my team and I produce the world’s only actual program
> >> synthesis system – based on mathematical logic.  How does proof enable

> >> program synthesis?
>
> > The QUESTION WAS ABOUT a proof of the inconsistency of naive set
> > theory
> > via Russell's Paradox, NOT PROGRAM SYNTHESIS!  STOP CHANGING THE
> > SUBJECT!
> > AND PAY UP!!
>
> Right. Fwiw, I have a bad feeling that C-B's arguing is rooted in
> changing the subjects as often as he needs. Whatever the motivation
> he might have, arguing with him is like wandering around a new area,
> being carefree and _logicfree_ !

Actually, Russell's result is a corollary of Program Synthesis rules
of inference. Do you see why?

Program Synthesis has rules (formally expressed) like "There is a
recursive function not(x) such that if M is a program that decides set
P then not(M) is a program that decides set ~P." But then that proves
(again, formally) that the complement of every recursive set is
recursive, a theorem of Theory of Computation. It was shown on FOM in
July 2010 that Incompleteness in Logic is a corollary of Theory of
Computation theorems (and defended by Martin Davis himself.) So
studying Program Synthesis is a good prerequisite for learning Theory
of Computation, Incompleteness in Logic, and diagonalization in
general.

In fact, Theory of Computation provides a model for Set Theory, where
program replaces set. It deals with non-representable sets in a nice,
noncontroversial manner that could be applied to the question of which
sets exist.

C-B

[Charlie-Boo]

I was giving you a chance to realize on your own that there is no phi
of phis that are not in themselves, and you have to avoid saying there
is.

C-B

[Charlie-Boo]

On Feb 19, 12:12 am, William Hale <bill...@yahoo.com> wrote:
> In article

> <65c848d5-792c-4021-a070-6d1dfb4f7...@fn10g2000vbb.googlegroups.com>,

Actually, the determining factor is whether predicate and set are in
fact the same in every case. You said they aren't. You need that to
be true. How do you know it is true?

C-B

[Rupert]

On Tuesday, February 19, 2013 10:31:13 PM UTC+1, Charlie-Boo wrote:
> On Feb 19, 5:19 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > Rupert <rupertmccal...@yahoo.com> writes:
>
> > > You asked me to derive a contradiction from the axiom schema of
>
> > > comprehension in first-order logic. I did this. I don't need to make
>
> > > any reference to the semantics of the first-order language of set
>
> > > theory. As far as the standard semantics goes, the formula z e z has a
>
> > > truth-value for every possible interpretation of the variable z.
>
> >
>
> > Of course.
>
> >
>
> > >> > Can I have my $25 now?
>
> >
>
> > >> I certainly hope that you have all that is rightfully yours, no less,
>
> > >> no more.
>
> >
>
> > > You owe me $25.
>
> >
>
> > C-B does not know what bad faith is.
>
>
>
> Do you know what good logic is?

Yes, I do. I gave you an impeccable proof in first-order logic. I asked if you accepted my definition of the formal theory, you accepted, and I gave you a proof in that formal theory which could be made machine-checkable. You owe me $25. Pay up.

> Is there a predicate of predicates
>
> that don't apply to themselves?
>

That's a change of topic. Predicates in what language?

[William Hale]

In article
<6eec2cb2-585b-449a...@r8g2000vbj.googlegroups.com>,

What is your answer to my question: Can one substitute the predicate
"not x in x" in the statement "Ay.not y in m"?

[William Hale]

In article
<2142ba24-78ff-472e...@cd3g2000vbb.googlegroups.com>,

More particularly, are you saying that "not m in m" is not a predicate?

[Rupert]

I have no idea at all what you are babbling on about.

>
>
> UO me $1,
>

No. I gave you a perfectly precise definition of a formal theory in which proofs are in principle machine-checkable and you agreed to the definition. Then I gave you a proof of a contradiction in that formal theory which could easily be made machine-checkable. You owe me $25. I'm not exactly holding my breath for the day you'll pay up, but your failure to see that you lost the bet just indicates that you don't understand the question you were asking.

[Charlie-Boo]

On Feb 19, 5:36 pm, Rupert <rupertmccal...@yahoo.com> wrote:
> On Tuesday, February 19, 2013 10:31:13 PM UTC+1, Charlie-Boo wrote:
> > On Feb 19, 5:19 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > > Rupert <rupertmccal...@yahoo.com> writes:
>
> > > > You asked me to derive a contradiction from the axiom schema of
>
> > > > comprehension in first-order logic. I did this. I don't need to make
>
> > > > any reference to the semantics of the first-order language of set
>
> > > > theory. As far as the standard semantics goes, the formula z e z has a
>
> > > > truth-value for every possible interpretation of the variable z.
>
> > > Of course.
>
> > > >> > Can I have my $25 now?
>
> > > >> I certainly hope that you have all that is rightfully yours, no less,
>
> > > >> no more.
>
> > > > You owe me $25.
>
> > > C-B does not know what bad faith is.
>
> > Do you know what good logic is?
>
> Yes, I do. I gave you an impeccable proof in first-order logic. I asked if you accepted my definition of the formal theory, you accepted, and I gave you a proof in that formal theory which could be made machine-checkable. You owe me $25. Pay up.

You did not prove there is a formula ¬(x e x). There is no formula of
formulas that are not in themselves. You did not establish that ¬(x e
x) is not the same thing.

In fact, I would argue that formula definitely is the same thing as
set and so there is no such formula. The axioms says that every
formula is a set, and since the purpose of a formula is to specify
sets that exist, we should be able to give a formula for any set that
does in fact exist.

If you referred to the set ¬(x e x) would that be a valid proof? Of
course not. Your proof has not been shown to be referring to the same
thing so you don't know if there is such a formula.

You can't use a mathematical object unless you can prove it exists
first.

C-B

[Charlie-Boo]

On Feb 19, 8:20 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <6eec2cb2-585b-449a-bda2-bae57ee4b...@r8g2000vbj.googlegroups.com>,

You mean substitute for y? Well, if "everything is a set", then "all
y" means "all sets", so no, "not x in x" is not a set and cannot be
substituted for y.

C-B

[Charlie-Boo]

There is no X of all X that do not satisfy X, for X being sets,
formulas, predicates, classes, collections etc. When you refer to
something existing you have to prove it exists which means your
definition is not of this form for some X. You have not done that.
You can't use "the set x ~e x" and except for a word change you are
doing the same thing, so the X must not be the same as set. You are
avoiding Russell Paradox in name only. The name of the term may be
different from "set" but you haven't show that it doesn't mean the
same thing as set.

You know the trick where you prove 1=2 by dividing a valid equation by
an expression that happens to equal 0? The author of that false proof
needs to show that the expression he is dividing by is not zero or
else he can't use the expression that way. He can use it for many
purposes but not for dividing into an equation. Likewise you need to
show your terms are not the same as "the set of sets that do not
contain themselves."

> > UO me $1,
>
> No. I gave you a perfectly precise definition of a formal theory in which proofs are in principle machine-checkable and you agreed to the definition. Then I gave you a proof of a contradiction in that formal theory which could easily be made machine-checkable. You owe me $25. I'm not exactly holding my breath for the day you'll pay up, but your failure to see that you lost the bet just indicates that you don't understand the question you were asking.

I don't understand it? Actually, I understand it very well. Who else
has ever pointed out that to have a class etc. of sets that do not
contain themselves you have to prove that class etc. is not the same
thing as set?

C-B

[William Hale]

In article
<28f6f4c9-4022-4161...@n6g2000vbf.googlegroups.com>,

What is your answer to the question: Can one substitute the predicate
"not m in x" in the statement "Ey.y = m"?

I am trying to clarify "substitution" not "existence".

[Charlie-Boo]

On Feb 19, 8:22 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <2142ba24-78ff-472e-b342-bdc6d4231...@cd3g2000vbb.googlegroups.com>,

Unless predicate is different from set the answer is yes, it is not.

C-B

[William Hale]

In article
<f11cff94-5fce-4c48...@z4g2000vbz.googlegroups.com>,

Can you answer the question without the qualification "unless predicate
is different from set"?

My answer is that "{ } not in { }" is a predicate and is even true.

[Charlie-Boo]

On Feb 19, 10:10 pm, William Hale <bill...@yahoo.com> wrote:
> In article

> <28f6f4c9-4022-4161-93d5-89e740f7a...@n6g2000vbf.googlegroups.com>,

Can I substitute the number 1/0 in identities such as N+N = 2*N ? How
about 1/M for unknown M?

The correct answers are: "no" and "not if M=0". Same as if we asked
if x ~e x is a set and whether or not there is a class for x ~e x:
"no" and "not if set = class".

C-B

[William Hale]

In article
<36f6c765-b78e-47fe...@h14g2000vbe.googlegroups.com>,

Your analogy doesn't match the situation I presented.

My predicate "not m in x" corresponds to your "1/m".
My statement "Ey.y = m" corresponds to your statement "An.n+n = 2*n".

Your conclusion for the second is "One cannot substitute 1/m for n if m
= 0".

Your conclusion for the first is "One cannot substitute <not m in x> for
y if set = class".

In the second case, you gave a condition (namely, m = 0 being true) for
1/m not being allowed to be substituted for n.

In the first case, you gave a condition (namely, set = class being true)
for "not m in x" not being allowed to be substituted for y.

In the second case, the condition "m = 0" says something about the
number "1/m".

In the first case, the condition "set = class" says nothing about the
predicate "not m in x".

[William Hale]

In article
<36f6c765-b78e-47fe...@h14g2000vbe.googlegroups.com>,
Charlie-Boo <shyma...@gmail.com> wrote:

But I didn't ask about "x not in x".

[Rupert]

On Wednesday, February 20, 2013 3:46:08 AM UTC+1, Charlie-Boo wrote:
> On Feb 19, 5:36 pm, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > On Tuesday, February 19, 2013 10:31:13 PM UTC+1, Charlie-Boo wrote:
>
> > > On Feb 19, 5:19 am, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> >
>
> > > > Rupert <rupertmccal...@yahoo.com> writes:
>
> >
>
> > > > > You asked me to derive a contradiction from the axiom schema of
>
> >
>
> > > > > comprehension in first-order logic. I did this. I don't need to make
>
> >
>
> > > > > any reference to the semantics of the first-order language of set
>
> >
>
> > > > > theory. As far as the standard semantics goes, the formula z e z has a
>
> >
>
> > > > > truth-value for every possible interpretation of the variable z.
>
> >
>
> > > > Of course.
>
> >
>
> > > > >> > Can I have my $25 now?
>
> >
>
> > > > >> I certainly hope that you have all that is rightfully yours, no less,
>
> >
>
> > > > >> no more.
>
> >
>
> > > > > You owe me $25.
>
> >
>
> > > > C-B does not know what bad faith is.
>
> >
>
> > > Do you know what good logic is?
>
> >
>
> > Yes, I do. I gave you an impeccable proof in first-order logic. I asked if you accepted my definition of the formal theory, you accepted, and I gave you a proof in that formal theory which could be made machine-checkable. You owe me $25. Pay up.
>
>
>
> You did not prove there is a formula ¬(x e x).

I said "the first-order language of set theory". I was assuming that you were familiar with the definition of that language. The formula ¬(x e x) is a well-formed formula in that language.

> There is no formula of
>
> formulas that are not in themselves. You did not establish that ¬(x e
>
> x) is not the same thing.
>

See any standard textbook for the definition of the first-order language of set theory. For example, see Machover's "Set Theory, Logic, and their Limitations".

>
>
> In fact, I would argue that formula definitely is the same thing as
>
> set and so there is no such formula. The axioms says that every
>
> formula is a set,

What nonsense.

> and since the purpose of a formula is to specify
>
> sets that exist, we should be able to give a formula for any set that
>
> does in fact exist.
>

Again, nonsense.

>
>
> If you referred to the set ¬(x e x) would that be a valid proof? Of
>
> course not. Your proof has not been shown to be referring to the same
>
> thing so you don't know if there is such a formula.
>

There's absolutely nothing wrong with my proof, and you owe me $25.

>
>
> You can't use a mathematical object unless you can prove it exists
>
> first.
>

I told you what axioms I was going to use. Apart from that I used nothing other than first-order logic.

[Rupert]

We agreed on what the axioms were that I was allowed to use. My proof proceeded in first-order logic using nothing but one such axiom.

> You have not done that.
>
> You can't use "the set x ~e x" and except for a word change you are
>
> doing the same thing, so the X must not be the same as set. You are
>
> avoiding Russell Paradox in name only.

What I am giving you *is* Russell's Paradox. Russell's paradox is the fact that naive set theory is inconsistent.

> The name of the term may be
>
> different from "set" but you haven't show that it doesn't mean the
>
> same thing as set.
>
>
>
> You know the trick where you prove 1=2 by dividing a valid equation by
>
> an expression that happens to equal 0? The author of that false proof
>
> needs to show that the expression he is dividing by is not zero or
>
> else he can't use the expression that way. He can use it for many
>
> purposes but not for dividing into an equation. Likewise you need to
>
> show your terms are not the same as "the set of sets that do not
>
> contain themselves."
>

What terms? There is a standard definition of a "term" in a first-order language. My first-order language has no constant or function symbols so the only terms are variables.

>
>
> > > UO me $1,
>
> >
>
> > No. I gave you a perfectly precise definition of a formal theory in which proofs are in principle machine-checkable and you agreed to the definition. Then I gave you a proof of a contradiction in that formal theory which could easily be made machine-checkable. You owe me $25. I'm not exactly holding my breath for the day you'll pay up, but your failure to see that you lost the bet just indicates that you don't understand the question you were asking.
>
>
>
> I don't understand it? Actually, I understand it very well.

No, your understanding of this topic is clearly extremely confused.