The old Google Groups will be going away soon, but your browser is incompatible with the new version.
Does ZFC use an ORACLE to decide what is a Theoerem?
 There are currently too many topics in this group that display first. To make this topic appear first, remove this option from another topic. There was an error processing your request. Please try again. Standard view   View as tree
 Messages 101 - 111 of 111 < Older

From:
To:
Cc:
Followup To:
Subject:
 Validation: For verification purposes please type the characters you see in the picture below or the numbers you hear by clicking the accessibility icon.

More options Oct 2 2012, 10:57 pm
Newsgroups: sci.logic, sci.math
From: Graham Cooper <grahamcoop...@gmail.com>
Date: Tue, 2 Oct 2012 19:57:04 -0700 (PDT)
Local: Tues, Oct 2 2012 10:57 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?
On Oct 3, 11:39 am, MoeBlee <modem...@gmail.com> wrote:

yes you did!

> If you wish to reference something I've said, then you'd do better to
> copy/paste what I actually posted.

> However, you might be referring to my point that

> ErAx(xer <-> (~x=r & ~xex))

> is inconsistent with Z set theory (which includes the axiom of
> regularity)

you said Naive Set Theory aswell.

.

> This is seen as, in Z set theory:

> ErAx(xer <->  (~x=r & ~xex))
> is equivalent to
> ErAx(xer <-> ~x=r)

> So take r u {r} (the union of r with the singleton whose only member
> is r) and we have EsAy yes, from which we derive a contradiction in
> the usual way.

no you can't.  r U {r} must be specified so it is NOT Russell's Set.

> But I don't claim that there are not systems in which ErAx(xer <->
> (~x=r & ~xex)) is consistent. Rather, I pointed out that ErAx(xer <->
> (~x=r & ~xex) is not consistent with certain ordinary set theory
> axioms, whether that's of concern to you or not.

Rubbish!  You rubbished my definition of a set.

> Also, you might have in mind my remark that if the variable y occurs
> free in the formula P, then a formula of the form:

> y = {x | P}

> is not a correct definition (i.e., is not a definition) since it is
> not in proper definitional form; it is circular.

For the SEVENTH TIME!!!!!!

that is how it is defined in ZFC SPECIFICATION AXIOM!!!

> We cannot define y in terms of a formula P that itself mentions y
> (i.e. has y free in it).

For the 8th time,  ZFC does.

Herc

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 3 2012, 4:45 am
Newsgroups: sci.logic, sci.math
From: Alan Smaill <sma...@SPAMinf.ed.ac.uk>
Date: Wed, 03 Oct 2012 09:44:47 +0100
Local: Wed, Oct 3 2012 4:44 am
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?

Graham Cooper <grahamcoop...@gmail.com> writes:
> On Oct 3, 11:39 am, MoeBlee <modem...@gmail.com> wrote:
>> This is seen as, in Z set theory:

>> ErAx(xer <->  (~x=r & ~xex))
>> is equivalent to
>> ErAx(xer <-> ~x=r)

>> So take r u {r} (the union of r with the singleton whose only member
>> is r) and we have EsAy yes, from which we derive a contradiction in
>> the usual way.

> no you can't.  r U {r} must be specified so it is NOT Russell's Set.

Yes, you can.

All this needs is that given any set z, z U {z} is also a set.
I'm sure you can work out the membership criterion for
r U {r} here.

> Herc

--
Alan Smaill

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 3 2012, 1:19 pm
Newsgroups: sci.logic, sci.math
From: MoeBlee <modem...@gmail.com>
Date: Wed, 3 Oct 2012 10:19:12 -0700 (PDT)
Local: Wed, Oct 3 2012 1:19 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?
On Oct 2, 9:57 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

I don't recall using the terminology "illegal" in any recent post on
the subject here discussed. But you claim I did, so I would appreciate
you showing a post in which I did.

> > If you wish to reference something I've said, then you'd do better to
> > copy/paste what I actually posted.

You ignored this again.

> > However, you might be referring to my point that

> > ErAx(xer <-> (~x=r & ~xex))

> > is inconsistent with Z set theory (which includes the axiom of
> > regularity)

> you said Naive Set Theory aswell.

If by 'naive set theory' you mean a theory with unrestricted
comprehension, then, since that theory is itself inconsistent,
perforce it is inconsistent with any other sentences in its language.

> > This is seen as, in Z set theory:

> > ErAx(xer <->  (~x=r & ~xex))
> > is equivalent to
> > ErAx(xer <-> ~x=r)

> > So take r u {r} (the union of r with the singleton whose only member
> > is r) and we have EsAy yes, from which we derive a contradiction in
> > the usual way.

> no you can't.  r U {r} must be specified so it is NOT Russell's Set.

r u {r} is a set in Z set theory.

> > But I don't claim that there are not systems in which ErAx(xer <->
> > (~x=r & ~xex)) is consistent. Rather, I pointed out that ErAx(xer <->
> > (~x=r & ~xex) is not consistent with certain ordinary set theory
> > axioms, whether that's of concern to you or not.

> Rubbish!  You rubbished my definition of a set.

What you gave is not a definition.

> > Also, you might have in mind my remark that if the variable y occurs
> > free in the formula P, then a formula of the form:

> > y = {x | P}

> > is not a correct definition (i.e., is not a definition) since it is
> > not in proper definitional form; it is circular.

> For the SEVENTH TIME!!!!!!

> that is how it is defined in ZFC SPECIFICATION AXIOM!!!

I'm impressed with your ability to count to seven, but, again, the
axiom schema of separation is:

If P is a formula in which y does not occur free, then all closures of

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

are axioms.

Since the schema requires that y not be free in P, there is no
"defined in the axiom schema of separation" where a set is defined by
a formula that mentions the set being defined.

> > We cannot define y in terms of a formula P that itself mentions y
> > (i.e. has y free in it).

> For the 8th time,  ZFC does.

Up to 8 now in the enumeration of your statements recently confirming

MoeBlee

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 3 2012, 3:56 pm
Newsgroups: sci.logic, sci.math
From: Graham Cooper <grahamcoop...@gmail.com>
Date: Wed, 3 Oct 2012 12:56:19 -0700 (PDT)
Local: Wed, Oct 3 2012 3:56 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?
On Oct 4, 3:19 am, MoeBlee <modem...@gmail.com> wrote:

MB you are 100% WRONG with EVERY STATEMENT IN THIS POST.

I'm not going to make HALF A DOZEN CORRECTIONS

since you IGNORE THEM ALL!

If you are INCAPABLE OF SOURCING THE FULL FORMULA for AXIOM OF
SPECIFICATION

that the name of the set Y does not appear in p(x,y,a1,a2,a3,..)
since it does!

Herc

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 3 2012, 4:13 pm
Newsgroups: sci.logic, sci.math
From: Graham Cooper <grahamcoop...@gmail.com>
Date: Wed, 3 Oct 2012 13:13:23 -0700 (PDT)
Local: Wed, Oct 3 2012 4:13 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?
SINCE YOU REFUSE TO STATE YOUR POSITION

YOU ARE JUST DUCKING AND WEAVING

AND HAVE NOTHING TO SAY

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 3 2012, 5:32 pm
Newsgroups: sci.logic, sci.math
From: Graham Cooper <grahamcoop...@gmail.com>
Date: Wed, 3 Oct 2012 14:32:02 -0700 (PDT)
Local: Wed, Oct 3 2012 5:32 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?
On Oct 3, 6:45 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

This is not a paradoxical set either.

xer  <->  x=r v !(xex)

although it's an arbitrary specification since it no longer means
*sets that don't contain themselves*.

Herc

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 4 2012, 5:10 am
Newsgroups: sci.logic, sci.math
From: Alan Smaill <sma...@SPAMinf.ed.ac.uk>
Date: Thu, 04 Oct 2012 10:07:05 +0100
Local: Thurs, Oct 4 2012 5:07 am
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?

Duh!!!

Combine this with *your* definition of r, and get back to us.

The only thing that matters with sets is what their elements are.

> Herc

--
Alan Smaill

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 5 2012, 7:22 am
Newsgroups: sci.logic
From: Frederick Williams <freddywilli...@btinternet.com>
Date: Fri, 05 Oct 2012 12:22:41 +0100
Local: Fri, Oct 5 2012 7:22 am
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?

George Greene wrote:

> On Oct 3, 5:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > This is not a paradoxical set either.

> > xer  <->  x=r v !(xex)

> This set doesn't exist in ZFC.
> Since, in ZFC, NO set contains itself, the right disjunct is just
> irrelevant.

I don't understand that.  I take it that ! means not, so in the language
of ZF !(xex) is well-formed and a theorem, so x=r v !(xex) is a theorem
also.

> This therefore becomes xer <-> x=r.
> ZFC has an axiom of foundation that prevents sets from containing
> themselves;
> since this set contains ONLY itself, it would be prohibited.

But there is nothing prohibited about the formula !(xex).  When you
write 'This set doesn't exist in ZFC.' what do you mean?  The formula

xer  <->  x=r v !(xex)

is materially equivalent to

xer.

--
Where are the songs of Summer?--With the sun,
Oping the dusky eyelids of the south,
Till shade and silence waken up as one,
And morning sings with a warm odorous mouth.

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 5 2012, 11:33 am
Newsgroups: sci.logic
From: George Greene <gree...@email.unc.edu>
Date: Fri, 5 Oct 2012 08:33:43 -0700 (PDT)
Local: Fri, Oct 5 2012 11:33 am
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?

> > Since, in ZFC, NO set contains itself, the right disjunct is just
> > irrelevant.

> I don't understand that.  I take it that ! means not, so in the language
> of ZF !(xex) is well-formed and a theorem, so x=r v !(xex) is a theorem
> also.

I mis-parsed it.

>    xer  <->  x=r v !(xex)

> is materially equivalent to

>    xer.

Well, yes, but why would I PRE-intuit that he was saying something
THAT simple?
I just figured there had to be a mistake.

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 5 2012, 6:31 pm
Newsgroups: sci.logic, sci.math
From: Graham Cooper <grahamcoop...@gmail.com>
Date: Fri, 5 Oct 2012 15:31:09 -0700 (PDT)
Local: Fri, Oct 5 2012 6:31 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?
On Oct 6, 1:33 am, George Greene <gree...@email.unc.edu> wrote:

Will you 2 STOP USING AXIOM OF REGULARITY

during DISCUSSIONS OF NAIVE SET THEORY.

THERE ARE 2 SETS UNDER DISCUSSION HERE!!

XeR1 <-> !(XeX) ^ !(X=R1)

XeR2 <- !(XeX) v (X=R2)

One is "RUSSELLS SET" wihout R.
One if "RUSSELLS SET" with R.

You can't refute their definitions based on RUSSELLS SET.

SO IS:

is forcing a torrent of errors!

OPEN COMPREHENSION IS NOT A NAIVE THEORY
It just needs formula checks to stop the BAD FORMULA.

What you CALL  *NON_NAIVE_THEORY*
is really

YOU ELIMINATED FULL SPECIFCATION OF SETS
(regardless of a superset or not)
just because you can't handle the definition of sets with/without self
as a member.

Herc

To post a message you must first join this group.
You do not have the permission required to post.
More options Oct 8 2012, 5:06 pm
Newsgroups: sci.logic
From: George Greene <gree...@email.unc.edu>
Date: Mon, 8 Oct 2012 14:06:41 -0700 (PDT)
Local: Mon, Oct 8 2012 5:06 pm
Subject: Re: Does ZFC use an ORACLE to decide what is a Theoerem?

> > > xer  <->  x=r v !(xex)
> George Greene wrote:
> > This set doesn't exist in ZFC.
> > Since, in ZFC, NO set contains itself, the right disjunct is just
> > irrelevant.

On Oct 5, 7:22 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> I don't understand that.  I take it that ! means not, so in the language
> of ZF !(xex) is well-formed and a theorem, so x=r v !(xex) is a theorem
> also.

Of course, but the right disjunct is always true, so the point is,
THIS SIMPLIFIES.  I had mis-parsed it as always-false because
the simplification is even MORE trivial if it is always true.

> When you write 'This set doesn't exist in ZFC.' what do you mean?

The universal set doesn't exist in ZFC.

> The formula
>  xer  <->  x=r v !(xex)
> is materially equivalent to
>    xer.

Yeah, but you've lost context.  This was in the context of defining r
as a set and saying what it equals.

Herc wrote:
>> This is not a paradoxical set either.
>> xer  <->  x=r v !(xex)

This doesn't EXPLICITLY state its quantifiers, but the antecedent of
"this" is *r*.  And by way of specifying precisely which things are or
aren't
in r, Herc is talking about ALL x's and how to decide which x's are in
r and
which are out.  In other words, WITH the proper quantifiers IN
context,
Herc was saying, "The r whose existence is asserted below is not
ErAx[ xer <-> (x=r V *true*) ]  which simplifies to
ErAx[ xer <-> *true* ]  which simplifies,
AS YOU SAID (about the part INside the brackets), to
ErAx[ xer ].
But THAT is simply the assertion of the existence of a universal set.
Which, as *I* said, DOES NOT EXIST, not in ZFC anyway.

To post a message you must first join this group.
You do not have the permission required to post.
 End of messages < Older
 « Back to Discussions « Newer topic Older topic »