CONTRADICTION IN ZFC: E(Y) Y C Z ^ (XeY <-> P(X,Y))

[Graham Cooper]

ZF just modified NAIVE SET THEORY

E(Y) XeY <-> P(X,Y)

so Russell's Paradox "doesn't hold".

E(Y) Y C Z ^ ( XeY <-> P(X,Y) )

=

E(Y) XeY <-> xeZ ^ P(X,Y)

=

A(z): A(p1,p2..pn):
E(y): A(x): x e y <-> (x e z & P(x,y,p1,p2..pn))
***AXIOM OF SPECIFICATION***

=

which is just
E(Y) XeY -> P(X,Y)

________________________________

So to find a CONTRADICTION IN ZFC

we need to formulate a Superset Z that Russell's Set can be a subset
of.

E(Y) Y C Z ^ ( XeY <-> P(X,Y) ) //ZFC AXIOM OF SPECIFICATION

LET P(X,Y) = X~eX

Y = RUSSELL'S SET

LET Z = { Y }

http://en.wikipedia.org/wiki/Axiom_of_regularity
Russell's paradox does not manifest in ZF because ZF does not prove
that the proposed paradoxical set actually exists (e.g., ZF's axiom of
separation only allows us to construct subsets of some existing set,
and thus cannot be used to construct the desired set).

But if Y contains all sets that don't contain themselves,
does Y contain Z?

Z does not contain itself, so ZeY
Also YeZ

No contradiction, so the singleton { RUSSELLSSET } could exist in ZFC.

This doesn't seem to be prevented by Axiom_Of_Regularity

Indeed AOR seems superfluous, if all it does is prevent:

XeXeXeX...

but cannot prevent

XeYeXeY...

it doesn't solve any class of problematic constructions.

So if RSeZ
Z = { RUSSELLSSET }

How does tacking on the subset_of_z constraint to Naive S.T.
E(Y) XeY <-> xeZ ^ P(X,Y)

how does SPECIFICATION (into a subset) prevent RUSSELL SETS?

Z itself has to be CONSTRUCTED as a subset of some further set.

Specify 2 RUSSELL SETS

RS1 = { RS2, ... }
RS2 = { RS1, ... }

then RS1 = { X | X e RS2 ^ P1(X,Y) }
and RS2 = { X | X e RS1 ^ P2(X,Y) }

BOTH RS1 and RS2 are RUSSELL SETS so Pn(X,Y)=X~eX

then RS1 = { X | X e RS2 ^ X~eX }
and RS2 = { X | X e RS1 ^ X~eX }

BOTH RUSSELL SETS specified in ZFC!

***************
*CONTRADICTION*
***************

Herc

[Tonico]

Idiot

[MoeBlee]

On Apr 25, 3:49 am, Graham Cooper <grahamcoop...@gmail.com> wrote:

> E(Y) Y C Z ^ ( XeY <-> P(X,Y) )
>

> LET P(X,Y) = X~eX

You're leaving off the quantification on 'X'.

What we do have:

AzEy(y subset of z & Ax(xey <-> ~xex))

> Y = RUSSELL'S SET

No, first order logic does not permit you to move from existentially
quantified 'y' that way. And your argument is circular anyway. You
have not proved that there is a set R (the "Russell set") such that
Ax(xeR<-> ~xex), so you can't set y = Russell set.

So there are two fallacies in your argument already:

(1) You improperly move from the existentially quantified 'y' to an
unwarranted claim about y.

(2) You are committing question begging by setting y equal to
something (the Russell set) you haven't proven to exist. You're
assuming the existence of the Russell set when the existence of the
Russell set is what you're trying to prove. That's circular.

MoeBlee

[Graham Cooper]

On Apr 26, 1:26 am, MoeBlee <modem...@gmail.com> wrote:
> On Apr 25, 3:49 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > E(Y) Y C Z ^ ( XeY <-> P(X,Y) )
>
> > LET P(X,Y) = X~eX
>
> You're leaving off the quantification on 'X'.
>
> What we do have:
>
> AzEy(y subset of z & Ax(xey <-> ~xex))

So if ZFC contains the 2 disjoint sets

Z1={1,2,3} and Z2={4,5,6}
how can Y be a subset of BOTH?

A(z): { {1,2,3}, {4,5,6} }
x e z

there is no value for x

>
> > Y = RUSSELL'S SET
>
> No, first order logic does not permit you to move from existentially
> quantified 'y' that way. And your argument is circular anyway. You
> have not proved that there is a set R (the "Russell set") such that
> Ax(xeR<-> ~xex), so you can't set y = Russell set.
>
> So there are two fallacies in your argument already:
>
> (1) You improperly move from the existentially quantified 'y' to an
> unwarranted claim about y.
>
> (2) You are committing question begging by setting y equal to
> something (the Russell set) you haven't proven to exist. You're
> assuming the existence of the Russell set when the existence of the
> Russell set is what you're trying to prove. That's circular.
>
> MoeBlee

Actually my error was more basic.

Specify 2 RUSSELL SETS
RS1 = { RS2, ... }
RS2 = { RS1, ... }

SHOULD BE:

RS1 C RS2
RS2 C RS1

but I cannot proceed while Axiom Of Specification is fundamentally
flawed to begin with.

Just using a standard RUSSELL SET

Y = { X | X~eX }

since ALL(Y) Y C Y

Y will stratify in ZFC. i.e. Russell Sets still exist

(once ALL(z) is fixed to EXIST(z), Y=/=Z is required also)

Herc

[MoeBlee]

On Apr 25, 10:44 am, Graham Cooper <grahamcoop...@gmail.com> wrote:

> Z1={1,2,3} and Z2={4,5,6}

> how can Y be a subset of BOTH?

WHICH y?

In the axiom schema, 'y' is an existentially bound variable.

In the formula we discussed, for every z there is a y. One is not
warranted to assert anything about that y other than what one has
proven about it. To assume anything else about y is to oblige oneself
to an undischarged assumption.

In any case, for ANY number of sets, there is a set (viz. the empty
set) that is a subset of them all.

But that has nothing to do with 'y' here, since the formula we're
discussing doesn't assert that there is a y for every z, but rather
that for every z there is a y.

For your edification in the basics of the first order predicate
calculus, I recommend to you 'Logic: Techniques Of Formal Reasoning'
by Kalish, Montague, and Mar.

MoeBlee

[Graham Cooper]

On Apr 26, 2:07 am, MoeBlee <modem...@gmail.com> wrote:

> On Apr 25, 10:44 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > Z1={1,2,3} and Z2={4,5,6}
> > how can Y be a subset of BOTH?
>
> WHICH y?
>
> In the axiom schema, 'y' is an existentially bound variable.
>
> In the formula we discussed, for every z there is a y. One is not
> warranted to assert anything about that y other than what one has
> proven about it. To assume anything else about y is to oblige oneself
> to an undischarged assumption.
>
> In any case, for ANY number of sets, there is a set (viz. the empty
> set) that is a subset of them all.
>
> But that has nothing to do with 'y' here, since the formula we're
> discussing doesn't assert that there is a y for every z, but rather
> that for every z there is a y.

close!

OK I mixed up E(y):A(z) and A(z):E(y)

If we take Specification as:

For any Well Formed Set there is a (possibly identical) subset sub-
defined via a predicate P.

then there is a trivial Z s.t. RS C Z

i.e. Z = RS

but Z would not be trivially constructable via the other axioms.

OK no contradiction!

Herc

[MoeBlee]

On Apr 25, 12:04 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> For any Well Formed Set there is a (possibly identical) subset sub-
> defined via a predicate P.

I gave you an EXACT statement of the axiom schema of specification:

If P is a formula in which y does not occur free, then all closures of
the following are axioms:

EyAx(xey <-> (xez & P)).

/

From that it does follow:

If P is a formula in which y does not occur free, then all closures of
the following are theorems:

AzEy(y subset of z & Ax(xey <-> P)).

/

More fundamentally, your confusions about first order logic would be
remedied by studying an introductory textbook in the subject.

MoeBlee

[Graham Cooper]

On Apr 26, 3:17 am, MoeBlee <modem...@gmail.com> wrote:
> On Apr 25, 12:04 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > For any Well Formed Set there is a (possibly identical) subset sub-
> > defined via a predicate P.
>
> I gave you an EXACT statement of the axiom schema of specification:
>

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

> More fundamentally, your confusions about first order logic would be
> remedied by studying an introductory textbook in the subject.
>
> MoeBlee

You do not comprehend the difference between mathematics and English.

There is no EXACT STATEMENT in English of ANY FORMULA.

Herc

[MoeBlee]

On Apr 25, 12:20 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> There is no EXACT STATEMENT in English of ANY FORMULA.

The axiom schema of specification is not a formula. Rather, the axiom
schema of specification is a statement that all formulas of a certain
kind are axioms. All such formulas are instances of the axiom schema.
Thus, the axiom schema determines a certain set of formulas, all of
which are stipulated to be axioms of Z set theory.

I gave you an exact a statement of the axiom schema of specification.

If P is a formula in which y does not occur free, then all closures
of
the following are axioms:

EyAx(xey <-> (xez & P)).

MoeBlee

[Graham Cooper]

On Apr 26, 4:10 am, MoeBlee <modem...@gmail.com> wrote:
> On Apr 25, 12:20 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > There is no EXACT STATEMENT in English of ANY FORMULA.
>
> The axiom schema of specification is not a formula. Rather, the axiom
> schema of specification is a statement that all formulas of a certain
> kind are axioms.

King of like a recipe then? Or a formula you could call it?

Herc

[MoeBlee]

On Apr 25, 1:39 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> King of like a recipe then?  Or a formula you could call it?

The axiom schema of specification is not a formula in the object
language that is the language for set theory. Rather, it is a
statetment in the meta-language, and may be rendered as a formula in
the meta-language when we are using a formalized meta-language.

In any case, I gave you an exact statement of the axiom schema of
specification. Any supposed derviation of a supposed contradiction you
claim to derive would be ultimately checked to see whether you used
only some precise equivalent of the statement I gave along with
correct first order logic.

MoeBlee

[Graham Cooper]

On Apr 26, 5:27 am, MoeBlee <modem...@gmail.com> wrote:
> On Apr 25, 1:39 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > King of like a recipe then?  Or a formula you could call it?
>
> The axiom schema of specification is not a formula in the object
> language that is the language for set theory. Rather, it is a
> statetment in the meta-language, and may be rendered as a formula in
> the meta-language when we are using a formalized meta-language.
>
>

So you lied?

[MoeBlee]

On Apr 25, 3:23 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> So you lied?

Of course not.

And you're not listening.

I'll say it even more explictly for you:

As ordinarily stated, the axiom schema of specifcation is not itself a
formula in the sense of being a formula of the language of set theory,
especially of the set theory in which each instance of the schema is
an axiom. On the other hand, while the axiom schema of specification
is, as ordinarily stated, conveyed in English (or whatever natural
language) and not itself as a formula, we may formalize the meta-
language for set theory and thus we may formalize the axiom schema of
separation as a formula in such a meta-language.

MoeBlee

[Jesse F. Hughes]

MoeBlee <mode...@gmail.com> writes:

[To Graham Cooper:]

> More fundamentally, your confusions about first order logic would be
> remedied by studying an introductory textbook in the subject.

No, they wouldn't.

--
"But you people are scum of the earth who pretend to be something that
is clearly beyond you--real mathematicians. I wouldn't be having
these problems with Gauss or Euler. I wouldn't be having these
problems with Fermat or Archimedes." -- James S. Harris on pretending

[MoeBlee]

On Apr 26, 6:46 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> MoeBlee <modem...@gmail.com> writes:
>
> [To Graham Cooper:]
>
> > More fundamentally, your confusions about first order logic would be
> > remedied by studying an introductory textbook in the subject.
>
> No, they wouldn't.

Of course I considered that my comment would not apply to the
particular person, but I made it anyway in a spririt of generosity
that one day he might find himself in a lucid state.

MoeBlee