> On Oct 2, 6:24 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > On Oct 3, 7:55 am, MoeBlee <modem...@gmail.com> wrote:
> > > On Oct 2, 2:43 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > > Those are only YOUR DEFINITIONS.
> > > They're correct and precise statements of the axiom schemata.
> > The TOPIC is that YOU SAID
> > E(RS) XeRS <-> !(X=RS) ^ !(XeX)
> > is ILLEGAL because
> I didn't use the terminology "illegal". I don't know precisely what
> you mean by "illegal".
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).
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.
> On Oct 3, 11:39 am, MoeBlee <modem...@gmail.com> wrote:
> > On Oct 2, 6:24 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > On Oct 3, 7:55 am, MoeBlee <modem...@gmail.com> wrote:
> > > > On Oct 2, 2:43 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > > > Those are only YOUR DEFINITIONS.
> > > > They're correct and precise statements of the axiom schemata.
> > > The TOPIC is that YOU SAID
> > > E(RS) XeRS <-> !(X=RS) ^ !(XeX)
> > > is ILLEGAL because
> > I didn't use the terminology "illegal". I don't know precisely what
> > you mean by "illegal".
> yes you did!
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
your most basic confusions about set theory and even basic logic.
> On Oct 2, 9:57 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > On Oct 3, 11:39 am, MoeBlee <modem...@gmail.com> wrote:
> > > On Oct 2, 6:24 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > > On Oct 3, 7:55 am, MoeBlee <modem...@gmail.com> wrote:
> > > > > On Oct 2, 2:43 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > > > > Those are only YOUR DEFINITIONS.
> > > > > They're correct and precise statements of the axiom schemata.
> > > > The TOPIC is that YOU SAID
> > > > E(RS) XeRS <-> !(X=RS) ^ !(XeX)
> > > > is ILLEGAL because
> > > I didn't use the terminology "illegal". I don't know precisely what
> > > you mean by "illegal".
> > yes you did!
> 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
> your most basic confusions about set theory and even basic logic.
> MoeBlee
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
then I cannot help you with your basic error of repeating 10 times now
that the name of the set Y does not appear in p(x,y,a1,a2,a3,..)
since it does!
> 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.
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*.
Graham Cooper <grahamcoop...@gmail.com> writes:
> On Oct 3, 6:45 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> 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.
> 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*.
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.
> 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.
> > 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.
> > > 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.
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.
xeR3 <-> !(XeX) **BAD**
R3 is just BAD!
SO IS:
E(Y) !(X=X) **BAD**
The fact your understanding of R3=BAD is more convuluted than !
(X=X)=BAD
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
*PARANOID_OF_PARADOXES_THEORY*
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.
> > > 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
paradoxical either":
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.
