Does ZFC use an ORACLE to decide what is a Theoerem?

[Graham Cooper]

Unfortunately Moeblee only asks questions not answer them, but here he
seems to suggest ZFC DISRPOVES RUSSELL'S SET using some non-axiomatic
intelligence.

On Sep 25, 8:35 am, MoeBlee <modem...@gmail.com> wrote:
> But the point I've made here (and still pretty banal except you don't
> understand it) is that if a sentence S is a theorem by the pure first
> order calclus in a language with 'e' as a 2-place relation symbol,
> then S is a theorem of ZFC.


IF FOLtheorem(S) THEN ZFCtheorem(S)
is not algorithmic.

> > On Sep 18, 1:45 pm, MoeBlee <modem...@gmail.com> wrote:
> > > In such context as set theory, we don't have an algorithm to test
> > > whether something will result in contradiction,

Of course in PREDICATE CALCULUS the formula are not all true.

E(x)!(x->x)

so picking out the THEOREMS is .... the mathematical bit!

Herc

[MoeBlee]

On Sep 24, 7:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> Unfortunately Moeblee only asks questions not answer them,

That's a lie. I've answered literally thousands of questions in this
newsgroup.

> but here he
> seems to suggest ZFC DISRPOVES RUSSELL'S SET using some non-axiomatic
> intelligence.

It's a simple argument to see that ZFC (indeed the pure first order
calculus with a 2-place relation symbol, say 'e') proves

~ExAy(yex <-> ~yey)

MoeBlee

[Rupert]

On Sep 25, 2:32 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> Unfortunately Moeblee only asks questions not answer them, but here he
> seems to suggest ZFC DISRPOVES RUSSELL'S SET using some non-axiomatic
> intelligence.
>
> On Sep 25, 8:35 am, MoeBlee <modem...@gmail.com> wrote:
>
> > But the point I've made here (and still pretty banal except you don't
> > understand it) is that if a sentence S is a theorem by the pure first
> > order calclus in a language with 'e' as a 2-place relation symbol,
> > then S is a theorem of ZFC.
>
> IF FOLtheorem(S) THEN ZFCtheorem(S)
> is not algorithmic.
>

The set of theorems of ZFC is recursively enumerable.

[Graham Cooper]

On Sep 25, 1:55 pm, Rupert <rupertmccal...@yahoo.com> wrote:
> On Sep 25, 2:32 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > Unfortunately Moeblee only asks questions not answer them, but here he
> > seems to suggest ZFC DISRPOVES RUSSELL'S SET using some non-axiomatic
> > intelligence.
>
> > On Sep 25, 8:35 am, MoeBlee <modem...@gmail.com> wrote:
>
> > > But the point I've made here (and still pretty banal except you don't
> > > understand it) is that if a sentence S is a theorem by the pure first
> > > order calclus in a language with 'e' as a 2-place relation symbol,
> > > then S is a theorem of ZFC.
>
> > IF FOLtheorem(S) THEN ZFCtheorem(S)
> > is not algorithmic.
>
> The set of theorems of ZFC is recursively enumerable.

how can you systematically enumerate ~E(y) xey<->!(xex)

given
A(a,b,c,..)
A(y)E(z)A(x) xey<->p(x,y,a,b,c) ^ xez

is mutually exclusive w.r.t. y?

[Rupert]

On Sep 25, 6:22 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Sep 25, 1:55 pm, Rupert <rupertmccal...@yahoo.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Sep 25, 2:32 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > Unfortunately Moeblee only asks questions not answer them, but here he
> > > seems to suggest ZFC DISRPOVES RUSSELL'S SET using some non-axiomatic
> > > intelligence.
>
> > > On Sep 25, 8:35 am, MoeBlee <modem...@gmail.com> wrote:
>
> > > > But the point I've made here (and still pretty banal except you don't
> > > > understand it) is that if a sentence S is a theorem by the pure first
> > > > order calclus in a language with 'e' as a 2-place relation symbol,
> > > > then S is a theorem of ZFC.
>
> > > IF FOLtheorem(S) THEN ZFCtheorem(S)
> > > is not algorithmic.
>
> > The set of theorems of ZFC is recursively enumerable.
>
> how can you systematically enumerate ~E(y) xey<->!(xex)
>
> given
> A(a,b,c,..)
> A(y)E(z)A(x) xey<->p(x,y,a,b,c) ^ xez
>
> is mutually exclusive w.r.t. y?
>

I don't understand your question. The formula ~EyAx(xey<->~(xex)) is a
theorem of first-order logic. The set of theorems of first-order logic
is a recursively enumerable set.

[George Greene]

On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> Unfortunately Moeblee only asks questions not answer them, but here he
> seems to suggest ZFC DISRPOVES RUSSELL'S SET
> using some non-axiomatic intelligence.

NOBODY is suggesting THAT!
In the first place, EVERYbody FROM MoeBlee ON UP
who is telling you that "ZFC disproves Russell's set" IS LYING.
The proof that the Russell set cannot exist IS NOT from ZFC!
It is from NO AXIOMS *AT*ALL*, therefore it is NOT from the ZFC
axioms! IT ISN'T first-order ZFC that proves this! It's
JUST*PLAIN*LOGIC,
WITH NO axioms at all!

You DON'T NEED any AXIOMS to prove
~(P <--> ~P) !!
You therefore DON'T NEED any AXIOMS to prove
~( rRr <--> ~rRr ) !!
You can JUST HAVE a binary predicate R in your language JUST BY
SAYING YOU HAVE ONE! You DON'T NEED to say anything about
the meaning or behavior of R, and you'd BEST not say anything about it
in this case, because it is precisely the fact that NO axioms have
said ANYthing about R that makes THE PROOF UNIVERSALLY GENERALIZABLE
to ANY AND EVERY POSSIBLE binary relation R!

From the previous step, you DON'T NEED ANY AXIOMS to infer
Ex[ ~( xRr & ~xRx ) ] -- ALL that you DO need is a first-order
INFERENCE RULE. The rule in question here is existential
generalization.
From the above, you infer immediately, WITHOUT ANY AXIOMS, but just
from
THE DEFINITION of the denial/negation of a quantified statement,
~Ax[ xRr & ~xRx ]. To this you can then apply THE INFERENCE RULE --
NOT
any AXIOM, but THE FIRST-ORDER INFERENCE RULE,
universal generalization, to get
Ar[ ~Ax[ xRr & ~xRx ] ].
THAT IS THE THEOREM. With a further UG on the R it becomes a SECOND-
order validity called "Strawson's Theorem". This is the DENIAL of
Russell's Paradox. This says "A Russell object cannot exist because
it would be paradoxical". It says that for ANY object in your
universe, it does NOT bear relation R to all and only those objects
that do-not-bear-R-to-themselves.

> > But the point I've made here (and still pretty banal except you don't
> > understand it) is that if a sentence S is a theorem by the pure first
> > order calclus in a language with 'e' as a 2-place relation symbol,
> > then S is a theorem of ZFC.

This IS NOT A RELEVANT point.
Moe's insistence on trying to make this point just proves that
teaching is
not something he ought to be wasting his time trying.
Moe got it right that the fact that standard classical logic is
monotonic
IS BANAL.
*THIS* discussion, BY CONTRAST, is IMPORTANT.
The fact that Russell's Paradox IS NOT ABOUT sets is NOT banal.
The fact that Moe is being stupidly distracted by sheer egotism and
god-knows-what-else is NOT banal.

[George Greene]

On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> IF FOLtheorem(S) THEN ZFCtheorem(S)
> is not algorithmic.

It's MUCH BETTER than algorithmic, dumbass, it's AUTOMATIC.
It's automatic BECAUSE classical logic IS MONOTONIC.
It's automatic because (P-->Q) --> ((P&R)-->Q)
IS A TAUTOLOGY.
When we SAY that something is a theorem of pure/plain FOL,
what we MEAN is that we can derive it, using THE INFERENCE RULES
of first (and 0th) order logic, from NO AXIOMS AT ALL.
Obviously, ADDING axioms to the ORIGINAL NONE is not going
to make FEWER things derivable -- it's going to make MORE things
derivable. So this point is totally BANAL, SIMPLE, AND BASIC.

But you didn't know it because you just don't know jack about logic.

[George Greene]

Herc said this:

> > IF FOLtheorem(S) THEN ZFCtheorem(S)
> > is not algorithmic.

And THEN, you were STUPID enough to reply with THIS:

On Sep 24, 11:55 pm, Rupert <rupertmccal...@yahoo.com> wrote:
> The set of theorems of ZFC is recursively enumerable.

WHY ARE YOU SAYING THIS?!?? The mere fact that it is TRUE is NOT
a defense! "The sky is blue" is true! RELEVANCE matters!
HOW CAN YOU EXPECT anyone who doesn't even understand that
INFERENCE IS MONOTONIC to understand what "recursively enumerable"
means?!?? GET YOUR EXPECTATIONS *RATIONALIZED*!! *TODAY*!!

[Graham Cooper]

when you add the AXIOM

ALL(X) XeY <-> BLAH

then ANY LOGIC PRECURSOR TO THAT

IN ANY WAY REMOTELY INSTANTIABLE TO THAT FORM

XeY <-> BLAH

must COMPLY WITH THE UNIVERSAL QUANTIFER IN THE AXIOM.

THAT IS WHY AXIOMATIC LOGIC
SUPERCEDES OPEN COMPREHENSION

You HAVE NO CLUE HOW TO 'DERIVE' ANY OF THESE "TRUTHS"

You all TALK JACK ABOUT IMPOSSIBLE CONSISTENCY,
REQUIRING A SUPERTHEORY TO MODEL AN UNDERLYING THEORY
NO TRUTH PREDICATE

then you MAKE UP ALL THESE TAUTOLOGIES AND SNEAK A PEAK AT THEM IN THE
DARK WHEN IT SUITS

Herc

[MoeBlee]

On Sep 25, 6:56 am, George Greene <gree...@email.unc.edu> wrote:

> EVERYbody FROM MoeBlee ON UP
> who is telling you that "ZFC disproves Russell's set" IS LYING.

You're amazing.

MoeBlee

[George Greene]

On Sep 25, 10:57 am, Graham Cooper <grahamcoop...@gmail.com> wrote:

> when you add the AXIOM

This is ridiculous. We are talking about RUSSELL'S PARADOX.
We are NOT TALKING about adding any axioms. We are JUST
talking about LOGIC.

ZFC has axioms but there are lots of things you can prove
in FIRST-ORDER LOGIC, WITHOUT any axioms.
The things you DON'T need axioms for ACTUALLY SUBSUME EVERYthing
that you DO need axioms for, because every proof uses only a finite
number of the axioms and if a theorem follows from the axioms, then
{axioms}-->Theorem is a theorem OF LOGIC ALONE, WITHOUT
ANY axioms.

>
> ALL(X)  XeY <-> BLAH
>
> then ANY LOGIC PRECURSOR TO THAT
>
> IN ANY WAY REMOTELY INSTANTIABLE TO THAT FORM
>
> XeY <-> BLAH
>
> must COMPLY WITH THE UNIVERSAL QUANTIFER IN THE AXIOM.

This is an odd use of the verb "must comply".
If you have Ax[ xey <--> whatever ]
then you are not requiring anything to comply with anything.
you are just DEFINING y as a set.
Things that DO "comply with" whatever are IN y.
Things that DON'T "comply with whatever" are OUT of y.
NOTHING is REQUIRED in advance to comply!

[George Greene]

On Sep 25, 10:57 am, Graham Cooper <grahamcoop...@gmail.com> wrote:

> You HAVE NO CLUE HOW TO 'DERIVE' ANY OF THESE "TRUTHS"

Oh, stop LYING. We DO SO TOO know what a decent first-order
derivation looks like. We know what a propositional derivation looks
like, too.
And we DON'T derive TRUTHS!! The MERE fact that what we derive IS
DERIVABLE means that the things we derive are TAUTOLOGOUS or VALID,
NOT *merely* true!

It is AXIOMS that are "true", and we, for the moment, are talking
about things
that we can derive WITHOUT using any axioms.

> You all TALK JACK ABOUT IMPOSSIBLE CONSISTENCY,
> REQUIRING A SUPERTHEORY TO MODEL AN UNDERLYING THEORY
> NO TRUTH PREDICATE

That's right, DUMBASS, there IS NO truth predicate. That's PROVABLE.
It's BEEN PROVED. Who do you expect us to believe, Tarski or YOU??

> then you MAKE UP ALL THESE TAUTOLOGIES

Dumbass: we DON'T MAKE UP tautologies! WE PROVE them!
We DERIVE them! YOU CAN DO IT TOO!
ALL you have to do to check whether some proposition is a tautology
IS LOOK AT ITS TRUTH-TABLE! IF (and ONLY if) IT'S ALL T's, THEN
(and ONLY then) THE PROPOSITION IS A TAUTOLOGY.

[MoeBlee]

On Sep 25, 1:31 pm, George Greene <gree...@email.unc.edu> wrote:

> Dumbass: we DON'T MAKE UP tautologies!  WE PROVE them!
> We DERIVE them!   YOU CAN DO IT TOO!
> ALL you have to do to check whether some proposition is a tautology
> IS LOOK AT ITS TRUTH-TABLE!   IF (and ONLY if)  IT'S ALL T's, THEN
> (and ONLY then) THE PROPOSITION IS A TAUTOLOGY.

When there are more all-caps than lower case words, I guess it's the
lower case words that are the ones emphasized?

MoeBlee

[George Greene]

On Sep 25, 1:15 am, Rupert <rupertmccal...@yahoo.com> wrote:
> I don't understand your question. The formula ~EyAx(xey<->~(xex)) is a
> theorem of first-order logic.

OK, let me see if I can help you understand: you are talking to AN
IDIOT.
The idiot you are talking to does NOT BEGIN TO UNDERSTAND THE CONCEPT
of a "theorem of first-order logic". That is not even reasonable
usage on your part
in any case. AXIOMS produce theorems. YOU are talking about some
results that are produced from NO axioms, that are derived by applying
FOL *INFERENCE*RULES* to THE EMPTY set of axioms, to NO axioms.
The idiot you are talking to DOES NOT UNDERSTAND the difference
between a logic and a theory.
He DOES NOT UNDERSTAND the difference between an inference rule and an
axiom.
You MUST engage on a FAR MORE BASIC level if you hope to say anything
relevant.

> The set of theorems of first-order logic
> is a recursively enumerable set.

The theory implied BY ANY recursive axiom-set in FOL is recursively
enumerable (or simpler).
If the signature has a binary predicate and a constant then said
theory is NO simpler than
recursively enumerable (i.e. it is not as simple as totally
recursive).
So, since the empty set is a recursive set, the theory of nothing but
first-order VALIDITIES
(OVER THE RELEVANT SIGNATURE, IF IT is rich enough) is going to be
r.e.

BUT ALL That is WAY over herc's head.
You have to BEAT HIM AWAY from his MISuse of relevant TERMINOLOGY, ALL
OVER the place.
This is NOT a constructive use of your time, but it would be MUCH MORE
constructive than
trying to talk about anything as complex as recursive enumerability.

[George Greene]

On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> Of course in PREDICATE CALCULUS the formula are not all true.

This is just A STUPID thing for you to say.
You DON'T know what it MEANS for a formula to be *IN* "predicate
calculus".
First of all, you need to STOP CALLING it that. Look at the SIGN ON
THE DOOR.
THE ROOM IS NAMED sci.logic and in HERE we CALL it
first-order logic.

> E(x)!(x->x)


THAT is NOT a formula of first-order logic, DUMBASS!
You DON'T quantify over STATEMENTS or sentences in FIRST-order logic!
Doing THAT would make it SECOND-order!
Here is something that IS in first-order logic:
Ax[P(x)\/~P(x)].
You DON'T NEED ANY AXIOMS to prove THAT.

[George Greene]

On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> Of course in PREDICATE CALCULUS the formula are not all true.

If ALL you have is the pure calculus itself, WITH NO AXIOMS,
then NOthing is merely *TRUE*! SOME things are VALID and others
are CONTRADICTIONS. The validities aren't JUST true -- they're
PROVABLY
true and they STAY provable, NO MATTER HOW MANY AXIOMS YOU ADD LATER.
You NEVER NEED TO USE any of those added axioms in the proof, since
you ALREADY
HAD a proof EVEN WITH NO axioms.

> so picking out the THEOREMS is .... the mathematical bit!

You don't "pick out" these theorems: You DERIVE these theorems. You
PROVE these theorems.
That's what "theorem" MEANS -- a sentence that IS PROVED.
If you prove it FROM axioms then it is a theorem "of" whatever theory
that axiom-set is named for.
If you prove it from the axioms of ZFC then it is a theorem OF ZFC.
If you prove it from no axioms at all then it is a theorem of EVERY
theory over EVERY language including it,
but THAT IS NOT the point! The point IS that if you prove it from NO
axioms then it you are proving it
using *MERE*LOGIC* as OPPOSED to ANY axioms or ANY "theory"!

Ar[ ~Ax[ xRr <--> ~xRx ] ]
is a theorem OF LOGIC. It's a theorem that can be inferred, USING THE
INFERENCE RULES of first-
order logic, from NO AXIOMS AT ALL. You don't NEED axioms to prove
this, any more than you need
axioms to prove P-->P or P\/~P or ~(P/\~P) or P<-->P or ~(P<-->~P).

[MoeBlee]

So everybody just shut up so that without the nuisance of other people
posting, George alone can post exactly what he thinks are just the
right things to say to a crank. George knows. He knows exactly the
right context to presuppose, the right amount of formality, the right
amount of technical detail, the right level of terminology, and
exactly which considerations are relevant and which aren't, at any
point in a thread. George knows all this because ... well because ...
HE'S GEORGE, YOU STUPID FUCKING LYING SHITHEEL!!! It is George and
George alone who sets the agenda and context of any thread he wishes
to post in. It is George alone who determines what is appropriate and
how best mathematics can be communicated to cranks. And If you veer
from this program of George's in even the slightest way, then George
knows too that you need to be called A STUPID FUCKING LYING
SHITHEEL!!! And nevermind that too far often George is wrong,
confused, and utterly irrational about various matters in the subject
and will wear down even the most patient person trying to bring him
reason; and nevermind that what he says one day is likely to be the
opposite of what he said a few months ago or will say a few months
from now. That is the intellectual prerogative of George, which is a
prerogative that comes from his constant need to disagree for the sake
of disagreeing, to bark like a dog behind a fence at ANYTHING that
happens to pass by. Don't you EVER forget that that is the purpose of
this newsgroup, YOU IDIOTIC LYING SHITHEEL BENEATH THE CONTEMPT OF
GEORGE GREENE!!!

[Graham Cooper]

BUMP!

Herc

[MoeBlee]

P.S. I forgot to mention that this is all a MORAL issue. If your posts
do not follow precisely the agenda Greene has in mind then you are not
just unwise but you being a BAD PERSON. So don't be a BAD PERSON.
Refrain from posting. ALWAYS refrain from posting - because if Greene
has already spoken in a thread then it's his thread to say what is or
is not appropriate to post and by posting you're, in all likelihood,
posting what Greene doesn't WANT YOU TO POST, and if Greene has not
yet spoken in a thread then don't pollute the thread with your posts
since you can never know when Greene might decide to go into the
thread to do some of the barking and growling that is his own very
special "pedagogy". So just shut up EVERYBODY and let George Greene
have these threads to himself, which is EXACTLY THE WAY IT NEEDS TO
BE!!!

[Christopher Menzel]

On Tuesday, September 25, 2012 6:56:01 AM UTC-5, George Greene wrote:
> On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > Unfortunately Moeblee only asks questions not answer them, but here he
> > seems to suggest ZFC DISRPOVES RUSSELL'S SET
> > using some non-axiomatic intelligence.
>
> NOBODY is suggesting THAT!
>
> In the first place, EVERYbody FROM MoeBlee ON UP
> who is telling you that "ZFC disproves Russell's set" IS LYING.
>
> The proof that the Russell set cannot exist IS NOT from ZFC!
> It is from NO AXIOMS *AT*ALL*, therefore it is NOT from the ZFC
> axioms! IT ISN'T first-order ZFC that proves this! It's

> JUST*PLAIN*LOGIC.

>
> WITH NO axioms at all!

This is stupid on so many levels it's hard to know where to begin. First, the tone is stupid. There is a shadow of a point here, but it could be expressed objectively without the ludicrous insults, manufactured outrage and utterly unwarranted haughtiness.

Second, the shadow of a point is that the *general* proposition "there is a set of all non-self-membered sets", ∃x∀y(y∈x ↔ y∈y) is indeed provable in any first-order logic whose language includes the membership predicate. But, by definition of "proof" (check any textbook), every proof in a first-order logic in a given language L is a proof in any first-order theory T whose language is L, even if the proof makes no use of any of the axioms of T. So the proof that there is no Russell set is a proof in ZFC.

Third, what makes it relevant to say that ZFC proves there is no Russell set is that naive set theory proves that there *is* one. Thus, it is often illuminating to present the unprovability of the Russell set in ZFC by augmenting the language with class terms {x : φ}, supposing for reductio that {x : x ∉ x} exists, and then running the usual argument. (Doing so in fact *does* require the addition of a (non-logical) set theoretic abstraction schema y ∈ {x : φ} ↔ φ(y).) This helps people see more clearly that the comprehension schema of naive set theory is the culprit in the paradox and is far more illuminating than dryly pointing out that the nonexistence of the Russell set is a theorem of first-order logic.

Fourth, it is stupid (or, at the least, misleading) to say (let alone shout) that the nonexistence of the Russell set can be proved "with no axioms at all". What you mean, or should mean, is that it can be proved with no *non-logical* axioms. It can be proved from the axioms of first-order logic alone. Granted, in a natural deduction system, the theorem in question can be proved solely by means of inference rules for the various operators. But, in more advanced texts (Enderton, Mendelson, Shoenfield, etc), first-order logic is typically presented axiomatically with only Modus Ponens and Univ Generalization as rules of inference and in those systems the theorem in question relies upon logical axioms for its proof.

> The fact that Russell's Paradox IS NOT ABOUT sets is NOT banal.

It is true that at root the paradox is generated by assuming something that is logically false. What is interesting and important, conceptually and historically, is how this this logical falsehood was smuggled into set theory. This is why nearly every presentation of the paradox makes it about sets. For it shows in particular that an intuitively very appealing set theoretic principle — that, given any condition C, there is a set containing exactly the things that satisfy C — is false.

[Christopher Menzel]

On Wednesday, September 26, 2012 11:15:27 AM UTC-5, Christopher Menzel wrote:
> ...

> Second, the shadow of a point is that the *general* proposition
> "there is a set of all non-self-membered sets", ∃x∀y(y∈x ↔ y∈y)
> is indeed provable in any first-order logic whose language
> includes the membership predicate.

*sigh*

Of course, I meant to say that the general proposition "there is NO set of all non-self-membered sets" is indeed provable in any first-order logic whose language includes the membership predicate. And, of course, this is formalized: ~∃x∀y(y∈x ↔ y∉y).

[Frederick Williams]

Christopher Menzel wrote:
>
> [Excelent stuff snipped.]

> But, in more advanced texts (Enderton, Mendelson, Shoenfield, etc), first-order logic is typically presented axiomatically with only Modus Ponens and Univ Generalization as rules of inference and in those systems the theorem in question relies upon logical axioms for its proof.

Please sir, in Shoenfield MP and UG aren't the only rules, indeed,
they're not rules at all!

--
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.

[Christopher Menzel]

On Wednesday, September 26, 2012 11:35:22 AM UTC-5, Frederick Williams wrote:
> Christopher Menzel wrote:
>
> >
>
> > [Excelent stuff snipped.]
>
> > But, in more advanced texts (Enderton, Mendelson, Shoenfield, etc), first-order logic is typically presented axiomatically with only Modus Ponens and Univ Generalization as rules of inference and in those systems the theorem in question relies upon logical axioms for its proof.
>
> Please sir, in Shoenfield MP and UG aren't the only rules,

You are indeed correct, sir.

> indeed, they're not rules at all!

Well, they are *derived* rules, not primitive rules of the system, but a derived rule is still a rule. :-) Admittedly, though, my posts suggests they are primitive in Shoenfield's system. (And now that I think of it, I don't think UG is a primitive in Enderton's system either; I believe he, in effect, builds it into his statement of the axioms.)

[MoeBlee]

A reasonable exchange:

Kurt: Here's a ZFC proof of ~ExAy(yex <-> ~yey): [fill in proof here]. Notice that this proof makes no use of any non-logical axioms of ZFC but rather it is in the pure first order calculus. Indeed, every theorem of the pure first order calculus in a language whose sole non-logical symbol is the 2-place relation symbol 'e' is perforce a theorem of ZFC. Anyway, contrary to what another poster wrote, it is clear enough that ~ExAy(yex <-> ~yey) is a theorem of ZFC.

Alonzo: For sake of greater generality and for our pedagogical purposes, you should to mention that ~ExAy(yex <-> ~yey) is a theorem of logic and that 'e' could be any relation symbol R.

Kurt: Yes, I think I as much as said that in my post.

Alonzo: Quite so. I just wanted to reiterate the point.

Kurt: Thanks for that. By the way, did you read the article that just came out by this British chap Alan Turing?

/

An exchange with Greene:

MoeBlee: Here's a ZFC proof of ~ExAy(yex <-> ~yey): [fill in proof here]. Notice that this proof makes no use of any non-logical axioms of ZFC but rather it is in the pure first order calculus. Indeed, every theorem of the pure first order calculus in a language whose sole non-logical symbol is the 2-place relation symbol 'e' is perforce a theorem of ZFC. Anyway, contrary to what another poster wrote, it is clear enough that ~ExAy(yex <-> ~yey) is a theorem of ZFC.

Greene: That is IDIOTIC. You're LYING. ~ExAy(yex <-> ~yey) is NOT so a theorem of ZFC. It is CONTEMPTIBLE that you're saying that it is. What you DO NEED TO SAY is that 'e' could be any relation symbol R and that ~ExAy(yRx <-> ~yRy) is a theorem of LOGIC, you scumbag SHITHEEL of a human being.

MoeBlee: If you look through my posts, you'll see that I did mention that it is a theorem of logic, but then perforce a theorem of ZFC too, and that was in context of the other poster saying incorrectly that it is not a theorem of ZFC.

Greene: No, you LYING piece of garbage. It is NOT a theorem of ZFC. You are missing the WHOLE POINT that it is a theorem of LOGIC, you contemptible filthy worm. Now everybody else just SHUT UP while I am posting so that the crank doesn't get ALL THE WRONG IDEAS about this!!!

/

MoeBlee

[Frederick Williams]

Christopher Menzel wrote:
>
> On Wednesday, September 26, 2012 11:35:22 AM UTC-5, Frederick Williams wrote:
> > Christopher Menzel wrote:
> >
> > >
> >
> > > [Excelent stuff snipped.]
> >
> > > But, in more advanced texts (Enderton, Mendelson, Shoenfield, etc), first-order logic is typically presented axiomatically with only Modus Ponens and Univ Generalization as rules of inference and in those systems the theorem in question relies upon logical axioms for its proof.
> >
> > Please sir, in Shoenfield MP and UG aren't the only rules,
>
> You are indeed correct, sir.
>
> > indeed, they're not rules at all!
>
> Well, they are *derived* rules, not primitive rules of the system, but a derived rule is still a rule. :-)

I almost wrote 'not derived rules', but I decided to be more
argumentative. Since neither \forall nor -> is a symbol in Shoenfield's
language, neither rule is even statable. And one must be argumentative
in newsgroups, don't you think?

> Admittedly, though, my posts suggests they are primitive in Shoenfield's system. (And now that I think of it, I don't think UG is a primitive in Enderton's system either; I believe he, in effect, builds it into his statement of the axioms.)

[Graham Cooper]

> order logic, from NO AXIOMS AT ALL.  You don't NEED axioms to prove
> this, any more than you need
> axioms to prove P-->P  or  P\/~P or ~(P/\~P) or P<-->P or ~(P<-->~P).

What rot. These ARE the 5 Axioms Of Predicate Calculus when you write
them with quantifiers and predicate arguments.

Predicate Calculus was a dud! They coined together induction in it:

A(n) p(n) ^ p(n)->p(s(n)) -> A(n) p(n)

and thought it would prove things.

There is no transitive closure of this 'simple set of 1OL proofs'.
It's inconsistent, you're all switching to line by line Oracle mode
when it suits.

DCPROOF.com would fall apart if you could add theorems at will!

Herc

[George Greene]

> On Tuesday, September 25, 2012 6:56:01 AM UTC-5, George Greene wrote:
> > In the first place, EVERYbody FROM MoeBlee ON UP
> > who is telling you that "ZFC disproves Russell's set" IS LYING.
>
> > The proof that the Russell set cannot exist IS NOT from ZFC!
> > It is from NO AXIOMS *AT*ALL*, therefore it is NOT from the ZFC
> > axioms!  IT ISN'T first-order ZFC that proves this!  It's
> > JUST*PLAIN*LOGIC.
>
> > WITH NO axioms at all!

On Sep 26, 12:15 pm, Christopher Menzel <chris.men...@gmail.com>
wrote:

> This is stupid on so many levels it's hard to know where to begin.

I know exactly where to begin. YOU'RE stupid FOR PICKING THE WRONG
SIDE. I'm 52. You're considerably older. YOUR TIME IS SHORT. It is
therefore MORE important than usual For US to spend it saying
something CONSTRUCTIVE. I SAID something constructive. *I* was
defending the GENERALITY, the domain INdependence, of a 2nd-order
contradiction (or the validity of its denial). You BY CONTRAST are
belaboring a point that NOBODY IS EVEN CONTESTING ("classical
inference is monotonic" -- *DUH*) AND THAT I HAVE ALREADY CONCEDED
(not that IT NEEDED conceding SINCE NOBODY WAS ARGUING AGAINST IT).,
AS WELL AS ENDORSING a FALSE attack on MY character.

> First, the tone is stupid.

The tone IS ARGUMENTATIVE. This was not THE FIRST time I had tried to
make this point! The tone had been RATCHETED UP (down) to this level
of contempt BY THE PERSISTENT IDIOCY of the other party!

> There is a shadow of a point here,

It is NOT a shadow. Relevance DOES MATTER in inference from sets with
many axioms.

> but it could be expressed objectively

IT HAS BEEN, DUMBASS.

> without the ludicrous insults,

IT WAS, DUMBASS.

> manufactured outrage and utterly unwarranted haughtiness.

OK, guilty as charged on those two. I had my own personal reasons
for manufacturing outraged haughtiness as a tone. Your reaction is
charming.

> Second, the shadow of a point is that the *general* proposition
> "there is a set of all non-self-membered sets", ∃x∀y(y∈x ↔ y∈y)
> is indeed provable in any first-order logic whose
> language includes the membership predicate.

SHUT UP.
In the first place, YOU FORGOT A *NOT*. There IS NO set of non-self-
membered sets.

In the second place, that is NOT the point!
THE POINT is that there is NO barber who shaves all&only those barbers
who don't shave themselves EITHER (in ADDITION to there not being a
set of non-self-membered sets). The POINT is that there is NO wanker
who wanks all & only those who don't wank themselves! The POINT is
that ALL of these are non-existent FOR THE SAME reason! THE POINT is
that there is A SECOND-order validity that is a universal
generalization over ALL first-order binary predicates, REGARDLESS OF
WHAT THEY MEAN!
Meaning ANYthing about SETS is MISSING the forest FOR A TREE!
THAT was "the shadow of a" POINT!

You missed it because you, precisely as MoeBlee was intuiting,
were indulging a relevant moral failing. TALK about FAILING to make
a point objectively!

THE POINT is a 1962 theorem of J.F.Thomson! (well, this point is a
lot older than that, but this is the article that seems to get credit
for noticing the
generality in the modern context and reversing the polarity from a
paradox
to a validity).

This point is really old IN THIS NEWSGROUP, TOO.
I only found back to 2007 with Google yesterday, but other
archives have me on this in 2003, back when Franz Fritsche
was here:
George Greene:

"Russell's Paradox is WELL-expressed as the following
2nd-order tautology:

AR[~Er[Ax[(xRr) <-> ~(xRx)]]]

Short & natural: 'For any R, r does not exist'."

>
> [Indeed] one can understand Russell's Paradox better if, rather than
> thinking about what it is analogous to, develop a general theorem for
> which each is an instance.
>
Indeed. This theorem is called Thomson's theorem:

"Let S be any set and R any relation defined at least on S.
Then no element of S has R to all and only those elements
in S which do not R to themselves."

(J. F. Thomson, "On Some Paradoxes" in Analytical Philosophy,
ed. R. J. Butler, New York: Barnes & Noble, 1962, p. 104-119.)

[Graham Cooper]

OK here's a 1OL formula, true or false?

A(m) A(n) E(p) m>2, n>2, prime(m), prime(n), even(p), p=m+n

SIMPLE 1OL GEORGE! ADD IT TO ZFC

Herc

[MoeBlee]

On Sep 26, 12:46 pm, George Greene <gree...@email.unc.edu> wrote:

> I know exactly where to begin.

Let's start with the source of your uncontrollable rage.

> YOU'RE stupid FOR PICKING THE WRONG
> SIDE.

So, with six all-caps words there, I guess the emphasized word is the
one in lower case. Let's see how it works:

You're STUPID for picking the wrong side.

Yes, much better.

> I'm 52.

And the rage began when?

> You're considerably older.  YOUR TIME IS SHORT.

Good, let's rub it in some more. Chris, you're close to DEATH, you old
codger, you.

> It is
> therefore MORE important than usual For US to spend it saying
> something CONSTRUCTIVE.

Agreed, let the 20-somethings fritter their time being non-
constructive.

> I SAID something constructive.

Really, you SAID it?

> *I*

Damn those one-letter words you can't all-cap and have to use *
instead.

> was
> defending the GENERALITY,  the domain INdependence, of a 2nd-order
> contradiction (or the validity  of its denial).

Yeah, well *I* was defending the GENERALITY of the U.N. Declaration of
Human Rights and of the sanctity of God's Dominon over all live and
all the creatures fo the earth - real, mythical, and even logically
impossible.

> You BY CONTRAST are
> belaboring a point that NOBODY IS EVEN CONTESTING ("classical
> inference is monotonic" -- *DUH*) AND THAT  I HAVE ALREADY CONCEDED
> (not that IT NEEDED conceding SINCE NOBODY WAS ARGUING AGAINST IT).,
> AS WELL AS ENDORSING a FALSE attack on MY character.

This is my favorite part, when Greene says you're wrong about X but
then says he's already agreed with you about X so that you're wrong to
defend yourself from him saying you're wrong about somethign that
follows right from X.

Here, yes, he's agreed that the logic is monotonic but he's also said
that ~ExAy(yex <-> ~yey) is not a theorem of ZFC. But if you point
that since the logic is monotonic, ~ExAy(yex <-> ~yey) is a theorem of
ZFC, he yells at you that he's already agreed that the logic is
monotonic!

> > First, the tone is stupid.
>
> The tone IS ARGUMENTATIVE.

I love UNDERSTATEMENT.

> This was not THE FIRST time I had tried to
> make this point!  The tone had been RATCHETED UP (down) to this level
> of contempt BY THE PERSISTENT IDIOCY of the other party!

Said party being me. Saying that I'm idiotic is question begging here,
I think.

> > There is a shadow of a point here,
>
> It is NOT a shadow.  Relevance DOES MATTER in inference from sets with
> many axioms.
>
> > but it could be expressed objectively
>
> IT HAS BEEN, DUMBASS.

'Dumbass'. Another favorite among Greene's go-to insults.

> > without the ludicrous insults,
>
> IT WAS, DUMBASS.
>
> > manufactured outrage and utterly unwarranted haughtiness.
>
> OK, guilty as charged on those two.  I had my own personal reasons
> for manufacturing outraged haughtiness as a tone.  Your reaction is
> charming.

Another of my favorite Greeneisms. The momentary admission, often with
phrases something like, "I thank you for your patience with my
admittedly inelegant outburst" or something along those lines. But
then a post or paragraph or SENTENCE later he's right back at it.

> > >  Second, the shadow of a point is that the *general* proposition
> > "there is a set of all non-self-membered sets", ∃x∀y(y∈x ↔ y∈y)
> > is indeed provable in any first-order logic whose
> > language includes the membership predicate.
>
> SHUT UP.

See what I mean. Right back at it.

By the way, "SHUT UP" is Greene's own notation for "QED".

> In the first place, YOU FORGOT A *NOT*.  There IS NO set of non-self-
> membered sets.

On Menzel's behalf, I'll thank you for spotting the typo of omission.

> In the second place, that is NOT the point!

Ah, yes, the HEART of Greenery. What is the point in any discussion or
thread. With Greene there is always ONE POINT that is "THE POINT" and
only GREENE can properly correct everybody else about this POINT. No
matter what other considerations, no matter how correct you are about
them, no matter how reasonably balanced your post, if you are not
exactly addressing this POINT that Greene has determined to be THE
POINT, then you are LYING IDIOT, or worse, a DUMBASS.

> THE POINT is that there is NO barber who shaves all&only those barbers
> who don't shave themselves EITHER (in ADDITION to there not being a
> set of non-self-membered sets).   The POINT is that there is NO wanker
> who wanks all & only those who don't wank themselves!  The POINT is
> that ALL of these are non-existent FOR THE SAME reason!  THE POINT is
> that there is A SECOND-order validity that is a universal
> generalization over ALL first-order binary predicates, REGARDLESS OF
> WHAT THEY MEAN!

My response here is as Greene said earlier:

Greene is belaboring a point that NOBODY (at least not MoeBlee and
Menzel) is CONTESTING [all caps original]. A point that has been
AGREED upon over and over and over already. (Not the point as to what
is the POINT, but at least the point that the proof generalizes to any
2-place relation.)

> Meaning ANYthing about SETS is MISSING the forest FOR A TREE!
> THAT was "the shadow of a" POINT!

Again Greene ignores again my point that my response was to crank's
claim that it is not a theorem of ZFC. So I mentioned BOTH the tree
AND the forest when I said it's a ZFC theorem (the tree) and that it's
a first order validity (the forest). But Greene needs to rage, and so
he does.

> You missed it because you, precisely as MoeBlee was intuiting,
> were indulging a relevant moral failing.  TALK about FAILING to make
> a point objectively!

Yep, if you don't stick to THE POINT as Greene has determined THE
POINT then it is a MORAL failing. I'm sure glad I'm not a Catholic,
can you IMAGINE how embarrassing to have to confess to the priest that
I said that ~ExAy(yex <-> ~yey) is a theorem of ZFC!

> THE POINT is a 1962 theorem of J.F.Thomson!  (well, this point is a
> lot older than that, but this is the article that seems to get credit
> for noticing the
> generality in the modern context and reversing the polarity from a
> paradox
> to a validity).

Sounds interesting. (Not said sarcastically.)

> This point is really old IN THIS NEWSGROUP, TOO.
> I only found back to 2007 with Google yesterday, but other
> archives have me on this in 2003, back when Franz Fritsche
> was here:
> George Greene:
>
> "Russell's Paradox is WELL-expressed as the following
> 2nd-order tautology:
>
> AR[~Er[Ax[(xRr) <-> ~(xRx)]]]
>
> Short & natural: 'For any R, r does not exist'."

So it's stated there as second order. That doesn't make a person an
IDIOT for saying pretty much the same by saying that the proof works
for any 2-place relation symbol.

Moreover, the context was the crank saying that ZFC does not prove
~ExAy(yex <-> ~yey). In that context, with a crank who doesn't know
second order logic from second base in baseball, it is not required to
go into second order. It is sufficient to rebut the crank with a ZFC
prove and as bonus, as I DID, mention too that this works with any 2-
place relation symbol and as bonus (since the crank understands none
of this anyway) to mention it's a theorem of the pure first order
predicate calculus. (Though I do grant that in second order, it's even
more general since it ranges over relations and not just relation
symbols; though this is hardly the kind of thing that very much
matters in the purpose of rebutting the crank who claims that ZFC does
not prove ~ExAy(yex <-> ~yey)).

By the way, my purpose is not to convince or instruct the crank (which
is impossible) but rather to clearly rebut the crank, as a matter of
record. And that is exactly accomplished by giving the ZFC proof. It
is not required to include greatest generality and such things as
being able to state in second order. If making a techincal point in
mathematics always required stating in greatest generality and
mentioning the deepest possible mathematical and logical basis, then
making simple technical points would be continually burdensome.

> > [Indeed] one can understand Russell's Paradox better if, rather
than
> > thinking about what it is analogous to, develop a general theorem for
> > which each is an instance.
>
> Indeed.

Indeed.

Are we all indeedly happy now?

But again, my point (at that exact juncture; there were other points
at other junctures) was NOT to instruct the crank in the logical basis
of Russell's paradox, but specifically to rebut the crank's claim that
~ExAy(yex <-> ~yey) is not a theorem of ZFC.

> This theorem is called Thomson's theorem:
>
> "Let S be any set and R any relation defined at least on S.
> Then no element of S has R to all and only those elements
> in S which do not R to themselves."
>
> (J. F. Thomson, "On Some Paradoxes" in Analytical Philosophy,
> ed. R. J. Butler, New York: Barnes & Noble, 1962, p. 104-119.)

MoeBlee

[George Greene]

On Sep 26, 2:55 pm, MoeBlee <modem...@gmail.com> wrote:
> My response here is as Greene said earlier:
>
> Greene is belaboring a point that NOBODY (at least not MoeBlee and
> Menzel) is CONTESTING [all caps original]. A point that has been
> AGREED upon over and over and over already. (Not the point as to what
> is the POINT, but at least the point that the proof generalizes to any
> 2-place relation.)

YOU *LYING* idiot!
ACTIONS SPEAK LOUDER THAN WORDS!
IF you had agreed with my point THEN YOU WOULD HAVE STOPPED trying to
RE ASSERT the SKY-IS-BLUE-IRrelevance that a VALIDITY is "theorem of"
some PARTICULAR theory! As I continually reiterated this point, you
continually tried to accuse me, FALSELY, of NOT conceding that
inference is monotonic.

You now get to be guilty of lying about yourself as well.

[George Greene]

On Sep 26, 1:22 pm, MoeBlee <modem...@gmail.com> wrote:
> A reasonable exchange:
>
> Kurt: Here's a ZFC proof of ~ExAy(yex <-> ~yey): [fill in proof here]. Notice that
> this proof makes no use of any non-logical axioms of ZFC but rather
it is in the pure first order calculus. Indeed, every theorem of the
pure first order calculus in a language whose sole non-logical symbol
is the 2-place relation symbol 'e' is perforce a theorem of ZFC.
Anyway, contrary to what another poster wrote, it is clear enough that
~ExAy(yex <-> ~yey) is a theorem of ZFC.

There could not possibly be a reasonable exchange beginning with that
paragraph.
That paragraph IS NOT reasonable. The non-reasonability starts with
the sentence
beginning "Indeed". It MIGHT JUST BARELY be reasonable in a context
where
a hearer MIGHT NOT KNOW that classical inference is monotonic. BUT
EVERYBODY HERE *KNOWS* that.

[Graham Cooper]

MoeBlee, George G is best engaged in Logic Discussion only.

You're both good explainers for Pro Establishment,
MB talking top-down against alternate theories,
GG refuting bottom up against alternate theories.

But since both of you argue with 1 eye closed and a giant blind side,
because neither of you will ANSWER QUESTIONS...

the chance of you 2 resolving an issue logically is -->0

just my 2c

Herc

[MoeBlee]

On Sep 26, 2:03 pm, George Greene <gree...@email.unc.edu> wrote:
> On Sep 26, 1:22 pm, MoeBlee <modem...@gmail.com> wrote:> A reasonable exchange:
>
> > Kurt: Here's a ZFC proof of ~ExAy(yex <-> ~yey): [fill in proof here]. Notice that
>
>  > this proof makes no use of any non-logical axioms of ZFC but rather
> it is in the pure first order calculus. Indeed, every theorem of the
> pure first order calculus in a language whose sole non-logical symbol
> is the 2-place relation symbol 'e' is perforce a theorem of ZFC.
> Anyway, contrary to what another poster wrote, it is clear enough that
> ~ExAy(yex <-> ~yey) is a theorem of ZFC.
>
> There could not possibly be a reasonable exchange beginning with that
> paragraph.
> That paragraph IS NOT reasonable.
> The non-reasonability starts with
> the sentence
> beginning "Indeed".

Right, it's perfectly correct and to the point. herefore, not
reasonable.

> It MIGHT JUST BARELY be reasonable in a context
> where
> a hearer MIGHT NOT KNOW that classical inference is monotonic.  BUT
> EVERYBODY HERE *KNOWS* that.

Oh, for godsakes, ANOTHER Greene stipulation!

MoeBlee

[Graham Cooper]

Proof by resolution and restricted comprehension into subsets over all
sets are TWO DIFFERENT THEORIES you can't just ADD THEM TOGETHER.

Will you stop ignoring the POINTS AGAINST YOUR ARGUMENT.

Namely whether ZFC axioms are ENUMERABLE (ALGORITHMICALLY) RE!!

or NOT!

Herc

[George Greene]

This newsgroup is nearly unmoderated but I can't escalate
further without risking real-world consequences.
What you are saying about me below is fundamentally slanderous.
Whether calling someone a shitheel is or isn't fundamentally
slanderous is a question that my prior behavior begs, conceded,
but there really is a core difference. My epithets are clearly
directed
reactions to individual actions. AT NO TIME HAVE I ALLEGED
that anybody ALWAYS ROUTINELY gets it wrong because of THEIR
long-term/defining/deeply-ingrained character-flaws. That makes
YOURS DIFFERENT.

On Sep 26, 1:22 pm, MoeBlee <modem...@gmail.com> wrote:

> An exchange with Greene:
>
> MoeBlee: Here's a ZFC proof of ~ExAy(yex <-> ~yey): [fill in proof here]. Notice that this proof makes no use of any non-logical axioms of ZFC but rather it is in the pure first order calculus. Indeed, every theorem of the pure first order calculus in a language whose sole non-logical symbol is the 2-place relation symbol 'e' is perforce a theorem of ZFC. Anyway, contrary to what another poster wrote, it is clear enough that ~ExAy(yex <-> ~yey) is a theorem of ZFC.
>
> Greene: That is IDIOTIC. You're LYING. ~ExAy(yex <-> ~yey)  is NOT so a theorem of ZFC.

Accurate so far.

> It is CONTEMPTIBLE that you're saying that it is.

THAT, on the other hand, IS JUST LYING, AND IT'S CONTEMPTIBLE that
you're calling it analogous. The record of the progression of the
escalation IS THERE
until future archives censor it.

I did not and would not say that it was "contemptible" for you to say
that.
It's WRONG and it DETRACTS FROM THE DISCUSSION, but what is
CONTEMPTIBLE is your decision to REACT to my pointing out THAT YOU
WERE
WRONG by trying to belabor the point that you HAD TO BE RIGHT (because
inference is monotonic). *WE*ALL*KNOW* that inference is monotonic!
WE DON'T yet all know that SET THEORY AND EPSILON ARE *NOT* relevant
to the disproof of this contradiction! THAT still NEEDS to be said!
What YOU are saying DOES NOT, but you will LIE AND INSULT MY CHARACTER
in order to defend your right to keep saying it!

> What you DO NEED TO SAY is that 'e' could be any relation symbol R and that  ~ExAy(yRx <-> ~yRy) is a theorem of LOGIC,

Correct.

> you scumbag SHITHEEL of a human being.

I CALLED YOU A SHITHEEL (I do NOT call people scumbags) FOR CONTINUING
TO INSIST THAT *I* Was wrong and you were right, NOT IMMEDIATELY as
you are SLANDEROUSLY CLAIMING I would, here! THE RECORD IS CLEAR and
you are clearly mis-describing it. At least it OUGHT to be clear.
But when people as smart as MoeBlee and Chris Menzel can get it wrong,
clarification seems to be in order.

> MoeBlee: If you look through my posts, you'll see that I did mention that it is a theorem of logic,

No, I really don't see that.
More to the point,

> but then perforce a theorem of ZFC too,

THAT UNDERCUTS anything you may have said before.

> and that was in context of the other poster saying
> incorrectly that it is not a theorem of ZFC.

If he understands that it is a theorem OF PURE LOGIC then the ZFC
part WILL BE IRRELEVANT. YOU COULD HAVE DECIDED TO GO
with that. YOU COULD HAVE DECIDED TO EXCUSE ALL PARTIES
OF THE *BURDEN* of ZFC. HIS PROBLEMS WERE OBVIOUSLY MORE
basic than that. The fact that you didn't have sense enough to see &
try that
is a failing but it does NOT make you any sort of bad person; it does
NOT
constitute any sort of bad behavior. WHAT DID constitute insufferable
assholicity
was continuing to attack ME AFTER *I* DID redirect to the correct
level!

IF YOU UNDERSTAND that this a a theorem of pure logic then
I AM RIGHT AND YOU ARE WRONG! IT'S JUST THAT SIMPLE!
And it is equally simple to see that you, having stated something
that is technically correct, are not going to tolerate seeing it
blasted
as incorrect, NO MATTER HOW MANY LIES ABOUT ME you may have
to tell to defend YOUR intolerance!

This is not going away.

> Greene: No, you LYING piece of garbage.

Wrong. My reaction is NOT like that.

> It is NOT a theorem of ZFC.

Wrong again. You have NEVER seen that accented/emphasized that
way from me. That IS NOT like ANYthing I WOULD say.

> You are missing the WHOLE POINT that it is a theorem of LOGIC,

RIGHT.

> you contemptible filthy worm.

WRONG.
What's contemptible IS NOT missing that something is a theorem of
logic.
What's contemptible IS ACCUSING *ME* of NOT KNOWING that inference is
monotonic EVEN AFTER I POSTED the relevant tautology UNDER DURESS
("inference is monotonic" is NOT SUPPOSED TO NEED belaboring and
is NOT RELEVANT TO THIS DISCUSSION!).

> Now everybody else just SHUT UP while
> I am posting so that the crank doesn't get ALL THE WRONG IDEAS about this!!!

I was NOT asking you or anybody to shut up!
I had MADE A POINT! All YOU had to do WAS AGREE!
THE LOUDER THE BETTER!

[George Greene]

On Sep 26, 1:45 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > You don't NEED axioms to prove
> > this, any more than you need
> > axioms to prove P-->P  or  P\/~P or ~(P/\~P) or P<-->P or ~(P<-->~P).
>
> What rot.  These ARE the 5 Axioms Of Predicate Calculus when you write
> them with quantifiers and predicate arguments.

No, DUMBASS, THEY'RE NOT.
And IN ANY case, to the extent that first-order logic is A LOGIC,
we DON'T TALK about *ITS* A X I O M s.
We talk about ITS *RULES*OF*INFERENCE*.
An AXIOM is something in the OBJECT language FROM which you will try
to derive/prove THEOREMS, USING the rules of inference.

[Graham Cooper]

>
> Proof by resolution and restricted comprehension into subsets over all
> sets are TWO DIFFERENT THEORIES you can't just ADD THEM TOGETHER.

In fact, any theory with Proof By Resolution is 2OL, ranging of all
formula

and so is AXIOM OF SPECIFICATION, ranging of all predicate defined
sets.

20L + 2OL is not monotonic inference.

Herc

[MoeBlee]

On Sep 26, 2:15 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> whether ZFC axioms are ENUMERABLE (ALGORITHMICALLY)  RE!!

The axioms of ZFC are recursively enumerable.

The set of axioms of ZFC is recursive.

The theorems of ZFC are recursively enumerable.

The set of theorems of ZFC is not recursive.

This all established (either explicity or by easy inference by the
reader) in many a textbook in mathematical logic

MoeBlee

[Graham Cooper]

BRAVO! ANOTHER <INSERT SET THEORY>

FLAVOR OF THE MONTH!

NEW FLAVOR! OBJECT LANGUAGES!

You don't even know what P.C. AXIOMS I'm talking about.

Herc

[Graham Cooper]

On Sep 27, 5:20 am, MoeBlee <modem...@gmail.com> wrote:
> On Sep 26, 2:15 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > whether ZFC axioms are ENUMERABLE (ALGORITHMICALLY)  RE!!
>
> The axioms of ZFC are recursively enumerable.

???

>
> The set of axioms of ZFC is recursive.

???

>
> The theorems of ZFC are recursively enumerable.

???

>
> The set of theorems of ZFC is not recursive.

???


On Sep 25, 1:55 pm, Rupert <rupertmccal...@yahoo.com> wrote:
>
> The set of theorems of ZFC is recursively enumerable.
>

THAT MEANS ALGORITHMIC DUMBASS

Herc

[MoeBlee]

On Sep 26, 2:18 pm, George Greene <gree...@email.unc.edu> wrote:
> This newsgroup is nearly unmoderated but I can't escalate
> further without risking real-world consequences.

What does THAT mean?

> What you are saying about me below is fundamentally slanderous.

Bull.

> Whether calling someone a shitheel is or isn't fundamentally
> slanderous is a question that my prior behavior begs, conceded,
> but there really is a core difference.  My epithets are clearly
> directed
> reactions to individual actions.

Your insults are made so scattershot that they hardly have a
direction.

> AT NO TIME HAVE I ALLEGED
> that anybody ALWAYS ROUTINELY gets it wrong because of THEIR
> long-term/defining/deeply-ingrained character-flaws.  That makes
> YOURS DIFFERENT.

If it's not your need to ragingly disagree for the sake of
disagreement then what IS the reason for your massive record of saying
ridicuously incorrect things?

> On Sep 26, 1:22 pm, MoeBlee <modem...@gmail.com> wrote:
>
> > An exchange with Greene:
>
> > MoeBlee: Here's a ZFC proof of ~ExAy(yex <-> ~yey): [fill in proof here]. Notice that this proof makes no use of any non-logical axioms of ZFC but rather it is in the pure first order calculus. Indeed, every theorem of the pure first order calculus in a language whose sole non-logical symbol is the 2-place relation symbol 'e' is perforce a theorem of ZFC. Anyway, contrary to what another poster wrote, it is clear enough that ~ExAy(yex <-> ~yey) is a theorem of ZFC.
>
> > Greene: That is IDIOTIC. You're LYING. ~ExAy(yex <-> ~yey)  is NOT so a theorem of ZFC.
>
> Accurate so far.

Like I said.

> > It is CONTEMPTIBLE that you're saying that it is.
>
> THAT, on the other hand, IS JUST LYING, AND IT'S CONTEMPTIBLE that
> you're calling it analogous.  The record of the progression of the
> escalation IS THERE
> until future archives censor it.

Are the archives being censored?

(1) The conversation is obviously a fictionalized parody that
simplifies. (2) The escalations are often so RAPID that it barely
derseves being called a "progression".

> I did not and would not say that it was "contemptible" for you to say
> that.
> It's WRONG and it DETRACTS FROM THE DISCUSSION, but what is
> CONTEMPTIBLE is your decision to REACT to my pointing out THAT YOU
> WERE
> WRONG by trying to belabor the point that you HAD TO BE RIGHT (because
> inference is monotonic).   *WE*ALL*KNOW* that inference is monotonic!
> WE DON'T yet all know that SET THEORY AND EPSILON ARE *NOT* relevant
> to the disproof of this contradiction!  THAT still NEEDS to be said!
> What YOU are saying DOES NOT, but you will LIE AND INSULT MY CHARACTER
> in order to defend your right to keep saying it!

Your insults about being 'contemptible' and/or similar, over so many
instances, are ubiquitous. Whether the word is 'contemptible' or some
other close phrasing, or how soon they come, the point is that they
come pretty quickly and you spread them regularly.

>> >  What you DO NEED TO SAY is that 'e' could be any relation symbol R and that  ~ExAy(yRx <-> ~yRy) is a theorem of LOGIC,
>
> Correct.
>
> > you scumbag SHITHEEL of a human being.
>
> I CALLED YOU A SHITHEEL (I do NOT call people scumbags) FOR CONTINUING
> TO INSIST THAT *I* Was wrong and you were right, NOT IMMEDIATELY as
> you are SLANDEROUSLY CLAIMING I would, here!

Immediate or after a few exchanges. It comes soon enough. If I
gathered an actual sampling of some of your insults that you've
habitually spread over the years of posting, the parody I made just
now would be an even caricature that makes you even worse.

> THE RECORD IS CLEAR

It sure is. Over AT LEAST as DECADE of stupid and undeserved insults
from you, whether given farily quickly in a conversation or after some
back and forth first.

> and
> you are clearly mis-describing it.

It's a fair parody.

> At least it OUGHT to be clear.
> But when people as smart as MoeBlee and Chris Menzel can get it wrong,
> clarification seems to be in order.
>
> > MoeBlee: If you look through my posts, you'll see that I did mention that it is a theorem of logic,
>
> No, I really don't see that.

Then you must have a disorder in which sometimes you can read and
other times not.

> More to the point,
>
> > but then perforce a theorem of ZFC too,
>
> THAT UNDERCUTS anything you may have said before.

Bull. George Greene arbitrary bullshit.

> > and that was in context of the other poster saying
> >  incorrectly that it is not a theorem of ZFC.
>
> If he understands that it is a theorem OF PURE LOGIC then the ZFC
> part WILL BE IRRELEVANT.  YOU COULD HAVE DECIDED TO GO
> with that.  YOU COULD HAVE DECIDED TO EXCUSE ALL PARTIES
> OF THE *BURDEN* of ZFC.  HIS PROBLEMS WERE OBVIOUSLY MORE
> basic than that.  The fact that you didn't have sense enough to see &
> try that
> is a failing but it does NOT make you any sort of bad person; it does
> NOT
> constitute any sort of bad behavior.  WHAT DID constitute insufferable
> assholicity
> was continuing to attack ME AFTER *I* DID redirect to the correct
> level!
>
> IF YOU UNDERSTAND that this a a theorem of pure logic then
> I AM RIGHT AND YOU ARE WRONG!  IT'S JUST THAT SIMPLE!
> And it is equally simple to see that you, having stated something
> that is technically correct, are not going to tolerate seeing it
> blasted
> as incorrect, NO MATTER HOW MANY LIES ABOUT ME you may have
> to tell to defend YOUR intolerance!
>
> This is not going away.

What does "not going away" mean?

Anyway, the above is just more of the kind of crap from you I should
have tired from answering long ago, and same for the rest of your post
snipped.

MoeBlee

[MoeBlee]

On Sep 26, 2:27 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Sep 27, 5:20 am, MoeBlee <modem...@gmail.com> wrote:
>
> > On Sep 26, 2:15 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > whether ZFC axioms are ENUMERABLE (ALGORITHMICALLY)  RE!!
>
> > The axioms of ZFC are recursively enumerable.
>
> ???
>
>
>
> > The set of axioms of ZFC is recursive.
>
> ???
>
>
>
> > The theorems of ZFC are recursively enumerable.
>
> ???
>
>
>
> > The set of theorems of ZFC is not recursive.
>
> ???

When you write jsut "???" there's nothing concrete I can respond to.

> On Sep 25, 1:55 pm, Rupert <rupertmccal...@yahoo.com> wrote:

> > The set of theorems of ZFC is recursively enumerable.
>
> THAT MEANS ALGORITHMIC DUMBASS

It's funny-sad that you call Rupert 'dumbass' about the subject of
recursion and algorithms.

MoeBlee

[MoeBlee]

On Sep 26, 2:01 pm, George Greene <gree...@email.unc.edu> wrote:

> YOU *LYING* idiot!

For Greene, that's just a punctuation mark.

And the rest of his post is yet more of his bizarre ranting, to which
can't muster the interest to yet again respond.

MoeBlee

[MoeBlee]

On Sep 26, 2:12 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> MB talking top-down against alternate theories,

I am not "against" alternatives that are coherently presented.

MoeBlee

[Frederick Williams]

>
> > order logic, from NO AXIOMS AT ALL. You don't NEED axioms to prove
> > this, any more than you need
> > axioms to prove P-->P or P\/~P or ~(P/\~P) or P<-->P or ~(P<-->~P).

Graham Cooper wrote:
> What rot. These ARE the 5 Axioms Of Predicate Calculus when you write
> them with quantifiers and predicate arguments.

Give us a reference to just one formulation of predicate calculus which
has such axioms.

[Jesse F. Hughes]

Graham Cooper <graham...@gmail.com> writes:

> Predicate Calculus was a dud! They coined together induction in it:
>
> A(n) p(n) ^ p(n)->p(s(n)) -> A(n) p(n)

Er, not that this is actually the axiom of induction, but surely we
should all agree that the above formula is valid.

--
Jesse F. Hughes
"Basically there are two angry groups. I am a harsh force of
one. Against me is a society of mathematicians. So far it's been a
draw." -- JSH gives another display of keen insight.

[Jesse F. Hughes]

Graham Cooper <graham...@gmail.com> writes:

> OK here's a 1OL formula, true or false?
>
> A(m) A(n) E(p) m>2, n>2, prime(m), prime(n), even(p), p=m+n
>
> SIMPLE 1OL GEORGE! ADD IT TO ZFC

What's a comma? Does it mean "and"?

If so, the above statement is clearly false.

Does it mean "and" sometimes and "implies" other? (Most likely so!)
Then you need to learn to express yourself more clearly.

But, of course, you are literally mentally ill and we shouldn't expect
too much from you. I honestly hope you seek help.

--
Jesse F. Hughes
"Students said they wanted to make people feel more comfortable by not
having to choose a gender at the bathroom door."
-- Boston Globe article on gender-neutral bathrooms at universities.

[Graham Cooper]

On Sep 27, 6:28 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> Graham Cooper <grahamcoop...@gmail.com> writes:
> > OK here's a 1OL formula, true or false?
>
> > A(m) A(n) E(p) m>2, n>2, prime(m), prime(n), even(p), p=m+n
>
> > SIMPLE 1OL GEORGE!  ADD IT TO ZFC
>
> What's a comma?  Does it mean "and"?
>
> If so, the above statement is clearly false.
>
> Does it mean "and" sometimes and "implies" other?  (Most likely so!)
> Then you need to learn to express yourself more clearly.
>
> But, of course, you are literally mentally ill and we shouldn't expect
> too much from you.  I honestly hope you seek help.
>

OOOF! OOOF!

ULLRICH LAPDOG IS ON GO!

A commna!?? Bwaha you can do better Jesse.

Herc

[Graham Cooper]

On Sep 27, 6:01 am, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> > > order logic, from NO AXIOMS AT ALL.  You don't NEED axioms to prove
> > > this, any more than you need
> > > axioms to prove P-->P  or  P\/~P or ~(P/\~P) or P<-->P or ~(P<-->~P).
> Graham Cooper wrote:
> > What rot.  These ARE the 5 Axioms Of Predicate Calculus when you write
> > them with quantifiers and predicate arguments.
>
> Give us a reference to just one formulation of predicate calculus which
> has such axioms.
>

None of you even KNOW what PREDICATE CALCULUS IS!

You are ALL OFF YOUR TOTAL ROCKERS!!!

!E(R)XeR<->!(XeX)

is monotonic inference! ? ? ? X X X

Nobody has DEDUCED THAT BY COMPUTER WITH A L->R GRAMMAR!

Not until I get onto it next week!

Herc

[|-| E R C]

On Sep 27, 7:20 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> !E(R)XeR<->!(XeX)
>

> is monotonic inference! X

T, P^!P |- !( T |- P)

add a contradictory formula P to a theory T
to prove that formula P is not part of T

is what??? 3OL

And you think all those theories just add onto

ALL(SETS) EXIST(SUPERSET) ALL(ELEMENTS)
ELEMENTS e SET <-> ELEMENTS e SUPERSET ^ phi

with what... HANDWAVING!

Herc

[Jesse F. Hughes]

My question was serious. What did that statement mean?

Does it mean

A(m) A(n) E(p) (m>2 & n>2 & prime(m) & prime(n) & even(p) & p=m+n) ?

If so, it's false. Duh.

So, you tell me. What did you mean, if not that?


--
Jesse F. Hughes

"You shouldn't hate Mother Mathematics."
-- James S. Harris

[Graham Cooper]

On Sep 27, 8:23 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> Graham Cooper <grahamcoop...@gmail.com> writes:
> > On Sep 27, 6:28 am, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >> Graham Cooper <grahamcoop...@gmail.com> writes:
> >> > OK here's a 1OL formula, true or false?
>
> >> > A(m) A(n) E(p) m>2, n>2, prime(m), prime(n), even(p), p=m+n
>
> >> > SIMPLE 1OL GEORGE!  ADD IT TO ZFC
>
> >> What's a comma?  Does it mean "and"?
>
> >> If so, the above statement is clearly false.
>
> >> Does it mean "and" sometimes and "implies" other?  (Most likely so!)
> >> Then you need to learn to express yourself more clearly.
>
> >> But, of course, you are literally mentally ill and we shouldn't expect
> >> too much from you.  I honestly hope you seek help.
>
> > OOOF! OOOF!
>
> > ULLRICH LAPDOG IS ON GO!
>
> > A commna!??   Bwaha you can do better Jesse.
>
> My question was serious.  What did that statement mean?
>
> Does it mean
>
>   A(m) A(n) E(p) (m>2 & n>2 & prime(m) & prime(n) & even(p) & p=m+n) ?
>
> If so, it's false.  Duh.
>
> So, you tell me.  What did you mean, if not that?
>

are you seriously that stupid? or just a sarcastic asshole?

Herc

[George Greene]

> On Sep 27, 5:19 am, George Greene <gree...@email.unc.edu> wrote:
> > to the extent that first-order logic is A LOGIC,
> > we DON'T TALK about *ITS*   A X I O M s.
> > We talk about ITS *RULES*OF*INFERENCE*.
> > An AXIOM is something in the OBJECT language FROM which you will try
> > to derive/prove THEOREMS, USING the rules of inference.

On Sep 26, 3:23 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> BRAVO!   ANOTHER  <INSERT SET THEORY>
>
> FLAVOR OF THE MONTH!

DAMN, you're stupid.
*I* am the one THROUGHOUT the thread who has been DENYING
the relevance of set theory! NOTHING that *I* say about this has
ANYTHING TO DO with set theory! If *I* said it then IT IS NOT a "set
theory" ANYthing, flavor of the month OR OTHERWISE!

> NEW FLAVOR!   OBJECT LANGUAGES!

DAMN, you're stupid.
ESPECIALLY if you don't know what an object language is.
This is NOT new. If you have ever written a compiler then you know
that it is possible
to write a program in one language that has output IN ANOTHER
language.
The language in which you describe THE LOGIC and THE RULES for
determining
"what follows FROM what" MAY BE DIFFERENT from the language that the
"whats"
ARE WRITTEN IN. This IS NOT ANY kind of NEW flavor! It's just new TO
YOU
because YOU'RE STUPID!

> You don't even know what P.C. AXIOMS I'm talking about.

Nor do I GIVE a shit. IDIOT: *I'm* the one with two degrees in
logic here.
You DON'T TELL me what axioms you're talking about! YOU ASK me what
axioms you GET to talk about!
And with respect to logic, you DON'T get to talk ABOUT ANY axioms
because THOSE are NOT on THIS level!

[George Greene]

On Sep 26, 12:15 pm, Christopher Menzel <chris.men...@gmail.com>
wrote:
> Fourth, it is stupid (or, at the least, misleading) to say (let alone shout) that the nonexistence of the Russell set can be proved "with no axioms at all".

You know, your basic problem is that you're just a damn liar.

> What you mean, or should mean, is that it can be proved with no *non-logical* axioms.

I hate to break this to you, but nobody who is fucking this up as
badly as you are has ANY kind of standing to tell ME what I *SHOULD*
or should not be doing. YOU SHOULD be taking MY side INSTEAD OF
ATTACKING me. The fact that you are TOO DAMN HATEFUL to see
THAT largely moots MOST other considerations. "Non-logical axioms" IS
REDUNDANT. YOU *SHOULD*KNOW* that. "Logical Axioms"is NOT
appropriate in THIS lecture BECAUSE THERE IS A PEDAGOGICAL GOAL of * S
E P A R A T I N G * the language from the logic!

> It can be proved from the axioms of first-order logic alone.

DIPSHIT, given that first-order logic IS A LOGIC, it's NOT GOING TO
HAVE *AXIOMS*!
It's going to have RULES OF INFERENCE *INSTEAD*! The THEORIES
DEVELOPED in/under the logic are going to be founded on axioms!
That's what axioms ARE *FOR*!!

> Granted, in a natural deduction system, the theorem in question can be proved solely by means of inference rules for the various operators.

Then why didn't you just say "George is right and everybody else is
wrong"??
Don't you REALIZE that if you grant THAT, you are entitled to say LESS
THAN EPSILON in this WHOLE discussion??
No, obviously, you don't realize that. But you SHOULD have.

> But, in more advanced texts (Enderton, Mendelson, Shoenfield, etc), first-order logic is typically presented axiomatically
> with only Modus Ponens and Univ Generalization as rules of inference

Bu THAT'S *MY* POINT! THOSE aer RULES OF INFERENCE and NOT axioms!
And the issue of what is more or less "advanced" is NOT germane.

> and in those systems the theorem in question relies upon logical axioms for its proof.
SEZ YOU. *JEEzus*. *I* would not CHARACTERIZE the lifting of a 0th-
order tautology as a "logical axiom", or
a definition of and in terms of not as a logical axiom, or a rule for
denying a quantified statement as a logical axiom.
But none of that is even the point. THE POINT is that LOGICAL axioms
are going to be at A COMPLETELY DIFFERENT *LEVEL*
from the axioms CHARACTERIZING THE THEORY being investigated, which
HAD been, IN THIS CASE, SET theory.
MY POINT WAS ALWAYS that the proof of "the theorem in question" (if
you even want to CALL it a theorem -- IT'S *VALID* -- it isn't proved
FROM anything)
did NOT depend on set theory! More to the point, JUST BECAUSE
SOMEbody decided to CALL some things "logical axioms" DOES NOT MAKE
*THAT*
treatment RELEVANT OR REASONABLE for THIS LESSON! *I* can *AND*DID*
characterize the machinery defining FOL in terms of rules of
inference.
That's MY Prerogative. The treatment that is relevant HERE is MINE!
*I*, *NOT* Mendelson, am teaching THIS course!


> It is true that at root the paradox is generated by assuming something that is logically false
>. What is interesting and important, conceptually and historically, is how this this logical falsehood was smuggled into set theory.

You are LYING, Chirs. THAT was interesting at THAT inflection point
in the history of consciousness. THAT was interesting from 1880-1950.
THIS IS A CRANK-THREAD IN 2012. THAT is NOT the point HERE! This is
NOT ABOUT sets! THIS IS about Herc NOT KNOWING ANY LOGIC!
And your insistence on using "Axiom" ON TWO DIFFERENT LEVELS is
HARMFUL to the project OF TEACHING him A LITTLE about THE BASICS
ehre!

> This is why nearly every presentation of the paradox makes it about sets.

That was then; this is now.

> For it shows in particular that an intuitively very appealing set theoretic
> principle — that, given any condition C, there is a set containing exactly the things that satisfy C — is false.

That naive set theory is inconsistent IS NOT the lesson Herc needed to
learn OR the point being relevantly belabored here!
We have had cranks saying A LOT of stupid things here -- "the halting
problem is based on an ill-formed question",
"Russell's Paradox is not a paradox", "Cantor's [anti-]diagonal
argument is bogus", AD NAUSEAM, but SO far we
have NOT had ANYone say, "Naive set theory REALLY IS consistent".
BEFORE that can even arise, SIMPLE BASIC
LOGIC *has* to be mastered SOMEwhat! If (as you have seen threads
titled here) "Russell's Paradox is not really a paradox",
then the fact that naive set theory leads to Russell's Paradox WILL
NOT be seen or understood as showing that
naive set theory is wrong!

FIRST THINGS FIRST!

I truly wonder who you think the target audience of this is.

How exactly do you think any rational reader WITH A LIFE in the field
is going to stoop to parsing a flame-war with Herc??

[George Greene]

On Sep 26, 3:27 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > The set of theorems of ZFC is recursively enumerable.
>
> THAT MEANS ALGORITHMIC DUMBASS

Dumbass: YOU DON'T HAVE a textbook, a dictionary, OR ANYTHING ELSE
that tells YOU what
"algorithmic" means. WE are talking about collections of axioms and
theorems.
WE know what recursively enumerable and [totally]recursive mean, in
that context.
YOU DON'T. Deciding to say "algorithmic" IS NOT going to prove that
you are so smart
that you know a word that we don't. GIVE IT UP. WE control the
terminology.

What is or isn't algorithmic is NOT all that important IN ANY case.
This thread really actually was ABOUT something. It was about
whether Russell's Paradox
really was or wasn't a paradox. If you agree that it is a paradox,
then the next question is,
HOW DO YOU KNOW it's a paradox? How was that PROVED? What do YOU
think a proof of it looks like??

[George Greene]

On Sep 26, 5:20 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> None of you even KNOW what PREDICATE CALCULUS IS!

I have two degrees in it. Lots of other people here have a third.

>
> You are ALL OFF YOUR TOTAL ROCKERS!!!
>
> !E(R)XeR<->!(XeX)
>
> is monotonic inference!

No, it isn't. ALL classical inference is monotonic.
You are so ignorant about logic that you don't even know what term
applies to what.
In propositional (0th-order) logic, one could say "inference is
monotonic" by saying
(P->Q)->((P&R)->Q).
The fact that that REALLY IS a tautology means that classical
inference REALLY IS monotonic.
That truly has not much to do with the fact that
~Er[Ax[ rRx <--> ~xRx ] ]
is valid. That is a very easy thing to infer, FROM NO axioms.
The fact that inference is monotonic means that you can ALSO infer
this from sets of axioms that
are BIGGER than empty, even though you won't need to actually USE any
of the axioms in the set to do it.
Put another way, if you have a car with a tank that holds 11 gallons,
and you can drive it 20 feet from A to B
ON FUMES, then, yes, you could also drive it from A to B if it had 5
gallons of gas in it, but that doesn't actually MATTER.
It's just the way mathematicians like to WRITE stuff. MoeBlee was
obsessed with that point, to no constructive end.

? ? ?  X X X
>
> Nobody has DEDUCED THAT BY COMPUTER WITH A L->R GRAMMAR!

People DON'T deduce things "with a grammar", DUMBASS!
That is NOT what DEDUCTION even MEANS!!

L-> R grammars were about COMPILERS for 3rd-gen source-code languages
FROM LONG AGO!
WE ARE NOT TALKING ABOUT ANY computer-programming language! NOT EVEN
Prolog!
We are talking about a FIRST-ORDER language with a simple SIGNATURE
including at least ONE binary predicate
(two if you count equality).

[Graham Cooper]

On Sep 27, 2:48 pm, George Greene <gree...@email.unc.edu> wrote:
> What is or isn't algorithmic is NOT all that important IN ANY case.
> This thread really actually was ABOUT something.   It was about
> whether Russell's Paradox
> really was or wasn't a paradox.  If you agree that it is a paradox,
> then the next question is,
> HOW DO YOU KNOW it's a paradox?  How was that PROVED?  What do YOU
> think a proof of it looks like??

<THREAD SAVER>

Let's not totally waste the thread on unventing here..

I think the main issue is YOU GUYS DON"T HAVE A THEORY!


******

SPECIFICATION:

A(SET) E(SUPERSET) A(X) XeSET <-> XeSUPERSET & p(X,SET)

SEE THAT p() ??

***THAT'S WHERE THE PREDICATE CALCULUS GOES***

*******

FOR INSTANCE (of SET)

p(X,SET) <-> (X mod 2 = 0)

E(SET) E(N) A(X) XeS <-> XeN & (X mod 2 = 0)

S = {0,2,4,6...}

SUPERSET = AXIOM OF INFINITY N

*******

FOR INSTANCE

p(X,SET) <-> !(X e X)

E(SET) E(SS) XeSET <-> XeU & !(X e X)

NO SUCH SUPERSET U ... NO SUCH SET!

******

But D.C. took 20 YEARS PROGRAMMING JUST TO CHECK THE GRAMMAR 1 LINE AT
A TIME!

MY THEORY WITH ONLY 1 AXIOM NOT() is ALREADY MORE POWERFUL!

num(0).
tru(t).
fal(f).

num(s(X)) <- num(X).
tru(not(X)) <- fal(X).
fal(not(X)) <- tru(X).

I can use UNIFY( formula1, formlua2 )

50 LINES OF CODE AND I CAN ALREADY DO THIS...

?- tru(not(not(not(not(f))))).

PROLOG: NO

?- tru(not(not(not(X)))).

PROLOG: X = f

********

LETS DO IMPLY WHILE I'M STILL HERE!

tru(if(X,t)).
tru(if(f,Y)).

Not going to take much work to get

DERIVE(THEOREM) <- DERIVE(A) ^ DERIVE(B) ^ TAUTOLOGY(A,B,THEOREM).

But even I don't know how to AUTOMATE derivations AFTER A
CONTRADICTION...

Whole new kettle of fish...


Herc

[Jesse F. Hughes]

I'm seriously stupid enough to think that you should know that a single
piece of punctuation (a comma) should be used to mean & in one instance
and -> in another instance of the same formula.

--
Jesse F. Hughes
"Well, I'm a pragmatist. I've been wrong MANY TIMES and it seems to
me that it would be simpler to be wrong with this paper."
--James S. Harris explains his latest paper

[George Greene]

On Sep 27, 2:01 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> Let's not totally waste the thread on unventing here..
>
> I think the main issue is YOU GUYS DON"T HAVE A THEORY!

You are LYING. We have FIVE DIFFERENT theories and they are NOT all
created
equal!

The ZEROth theory we have is ZEROth-order logic, or propositional
logic.
There are various treatments of this with "logical axioms" and "rules
of inference"
but the ONLY IMPORTANT thing you have to know is that every
proposition has
A TRUTH TABLE and SOME of them have ALL trues or ALL falses in the
payoff
column. That is a SOPHOMORE (IN HIGH SCHOOL) - level lesson.
There are logical axioms and rules of inference that can be used to
generate
families of these all-true propositions, but THEY ARE NOT important.
What is
MORE important is that you just be able to RECOGNIZE one when you see
it.
What C.Menzel IS WRONGLY telling you to treat as "logical axioms" in
first-
order logic can just be subsumed as "that's a lifted/first-order
instance of a [zeroth-
order] tautology (thought of as a schema; you have already noted that
you could
universally quantify over propositional variables).

[MoeBlee]

On Sep 26, 11:45 pm, George Greene <gree...@email.unc.edu> wrote:

> *I*, *NOT* Mendelson, am teaching THIS course!

The Alexander Haig approach to mathematical logic in action.

MoeBlee

[MoeBlee]

On Sep 26, 11:45 pm, George Greene <gree...@email.unc.edu> wrote:

> The fact that you are TOO DAMN HATEFUL to see
> THAT largely moots MOST other considerations.

Right, Chris is famously a real hateful guy.

MoeBlee

[MoeBlee]

Some highlights from Sep 26, 11:45 pm, George Greene
<gree...@email.unc.edu>:

> You know, your basic problem is that you're just a damn liar.

> The fact that you are TOO DAMN HATEFUL to see
> THAT largely moots MOST other considerations.

> DIPSHIT

> You are LYING

(But he's not a contemptibly hateful dipshit liar, I guess.)

All in response to Menzel's first post in the thread. That's a pretty
quick turnaround, I think, even if the actual word 'contempitble' was
not included.

MoeBlee

[George Greene]

On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> Unfortunately Moeblee only asks questions not answer them, but here he
> seems to suggest ZFC DISRPOVES RUSSELL'S SET
> using some non-axiomatic intelligence.

Truth-tables are, for PRACTICAL purposes, NON-axiomatic.
Prof.Menzel has done us all a great DISservice by intervening
to insist that since ~( P <--> ~P ) is a "theorem of" propositional
logic,
it must have been proved from "logical axioms".

You are NEW to this arena and treatments (such as Prof. Menzel cited
from Mendelson) that are approriate for intelligent undergrads who
have
already completed a course in propositional logic are NOT appropriate
for YOU. What YOU need to start by believing is that THEORIES grow
(or are derived from) AXIOMS and that LOGICs are defined via Rules Of
Inference. It truly almost DOES NOT EVEN MATTER what "logical axioms"
you use to "derive" ~( P <--> ~P ) as a "theorem". THE POINT about
this
proposition is that IT IS A TAUTOLOGY. IT IS NECESSARILY TRUE.
People have invented 3-axiom axiomatizations of propositional logic in
restricted languages but YOU need NOT be BOTHERED with any of that!
YOUR mission is to JUST LOOK AT it and NOTICE that it is a tautology.

As you yourself have already noticed, any propositional tautology can
be
lifted to a first-order validity just by replacing each letter with a
first-order formula (you just have to be careful to replace the SAME
letter with the SAME formula, over all occurrences).

The kind of "non-axiomatic intelligence" we WERE TALKING ABOUT is just
the intelligence of knowing what the inference rules are and knowing
that they are sound. Once you know how, you can USE them to infer
THEOREMS from AXIOMs.
You use first-order inference rules to derive theorems -- PHRASED in a
first-order LANGUAGE -- from axioms phrased in the SAME language.

[The denial of] Russell's Paradox does not even NEED any of those
axioms because it is fundamentally lifted (to 1st order AND THEN AGAIN
TO 2nd) from
~( P & ~P ).

[MoeBlee]

George Greene:

Since the crank you are talking to is ineducable, it's silly when you
protest that other people are interfering with your pedagogical
objectives for the crank. Some treatments of first order logic use
logical axioms and rules while other treatments use only rules. One is
not screwing up the education of the crank by referring to treatments
that use both axioms and rules. The crank will not be educated.
Posting to him and about him can't seriously be about educating him.

MoeBlee

[Graham Cooper]

We did 2 years of LOGIC subjects and THEOREM PROVERS

That why *I* am the only one here who knows how to use SKOLEM
FUNCTIONS, PREDICATE LOGIC AXIOMS, LOGIC CIRCUIT OPTIMISATION.

MB and GG for the 10th Time, if you REFUSE TO ANSWER ANY AND ALL
QUESTIONS on this topic

and you BOTH CONTINUE WITH YOUR TOP DOWN LECTURE POSTING STYLE

then NEITHER OF YOU WILL EVER KNOW YOUR MISTAKES.

****

I'm not even SAYING YOU GUYS ARE MISTAKEN HERE..

MERELY THAT YOUR REFUSAL TO ADDRESS ANY AND ALL POINTS AGAINST YOU

means it is IMPOSSIBLE for you to even ACKNOWLEDGE if you are in
error.

****

If you're BOTH HAPPY THAT the last 20 GIGABYTES OF LOGIC THEORY

supported by WORLDWIDE ACEDAMIA is 100% ACCURATE then you should have
no problems KEEPING THAT TO YOUR OWN TOPICS on SUPERINFINITY, META-
THEOREMS, NON-PROVABILITY, NON-ALGORITHMS.

****

COMPUTER SCIENTISTS USE THESE FORUMS TOO

WE HAVE NO INTEREST IN DEAD ENDS
- IIMPOSSIBLE CONSISTENCY
- IMPOSSIBLE SETS
- IMPOSSIBLE FUNCTIONS
- IMPOSSIBLE ENUMERATION
- IMPOSSIBLE COMPLEXITY
- IMPOSSIBLE SIZES
- IMPOSSIBLE GROUPS OF SETS


NONE OF THIS IS PART OF PROGRAMMING THEOREM PROVERS

WE THEORISE WHAT WE PROVE
WE PROVE WHAT WE THEORISE

****

I HAVE POSTED GAPING LOGICAL HOLES IN ALL OF YOUR IMPOSSIBLE BULLSHIT

THE FACT YOU BOTH FLATLY REFUSE TO ADDRESS THESE FLAWS
AND SAY 'I'M NOT GOING TO ANSWER ANY QUESTIONS ABOUT IT'

MEANS YOU FORFEIT YOUR RIGHT TO ARGUE TOP DOWN.

****

NEED I REMIND YOU BOTH OF THE UNWRITTEN RULES OF SCI.MATH

WHEN YOUR STATEMENT IN CHALLENGED YOU CAN EITHER
1/ SUPPORT IT
2/ RETRACT IT
3/ SHUT UP

****

I CHALLENGE *ALL* OF YOUR PROOFS MB AND GG.

YOU CAN START BY *ANSWERING* MY CHALLENGE TO *YOU*

CATEGORIES
****************

COUNTABLE UNCOUNTABLE


WHICH OF THESE GO IN WHICH CATEGORY?
******************************************************

1/ FUNCTIONS
2/ GODEL NUMBERS
3/ SETS

YOUR AXIOM OF CHOICE

1 FUNCTION PER 1 SET

IS BOGUS FROM THE START

Herc

[Graham Cooper]

it's just AND.

A(x) E(y) p(x) AND q(y)

->

E(x) p(x) -> E(y) q(y)

does that help?

I mixed up.... all primes have a sum..
vs all sums have a prime..

***

but the POINT is that SOME of these 1OL formulas are not monotonically
trivially solvable.

GEORGE GREEN REFUSES TO STATE WHAT

{ ZFC-FORMUILA } minus { FOL TRUTHS }

is even meant to be?

Herc

[Graham Cooper]

On Sep 27, 2:22 pm, George Greene <gree...@email.unc.edu> wrote:
> > On Sep 27, 5:19 am, George Greene <gree...@email.unc.edu> wrote:
> > >  to the extent that first-order logic is A LOGIC,
> > > we DON'T TALK about *ITS*   A X I O M s.
> > > We talk about ITS *RULES*OF*INFERENCE*.
> > > An AXIOM is something in the OBJECT language FROM which you will try
> > > to derive/prove THEOREMS, USING the rules of inference.

AHHH NOW I SEE WHY YOU'RE (ALL) SO CONFUSED!

you think AVOIDING MODUS PONENS and using

RULE1: LEFTHANDSIDE --> RIGHTHANDSIDE
RULE2: LEFTHANDSIDE --> RIGHTHANDSIDE
...

is something other than using a plain old grammar of axioms.

ANY AXIOM [+] MODUS PONENS

IS AN INFERENCE RULE.

(technically [+] UNIFY() to match sub arguments)

IF YOU CANNOT INFER ANYTHING FROM AXIOMS THEY MAY AS WELL BE A USELESS
PIECE OF PAPER LIKE GEORGE GREENES DEGREES.


****************************
GRAHAM COOPER (BINFTECH)
UNIVERSITY OF QUEENSLAND
--
http://tinyURL.com/BLUEPRINTS-MATHEMATICS
http://tinyURL.com/BLUEPRINTS-HYPERREALS
http://tinyURL.com/BLUEPRINTS-HALT-PROOF
http://tinyURL.com/BLUEPRINTS-QUESTIONS
http://tinyURL.com/BLUEPRINTS-POWERSET
http://tinyURL.com/BLUEPRINTS-THEOREM
http://tinyURL.com/BLUEPRINTS-FORALL
http://tinyURL.com/BLUEPRINTS-TURING
http://tinyURL.com/BLUEPRINTS-GODEL
http://tinyURL.com/BLUEPRINTS-TRUTH
http://tinyURL.com/BLUEPRINTS-PROOF
http://tinyURL.com/BLUEPRINTS-LOGIC
http://tinyURL.com/BLUEPRINTS-BRAIN
http://tinyURL.com/BLUEPRINTS-SETS
http://tinyURL.com/BLUEPRINTS-PERM
http://tinyURL.com/BLUEPRINTS-P-NP
http://tinyURL.com/BLUEPRINTS-LIAR
http://tinyURL.com/BLUEPRINTS-GUT
http://tinyURL.com/BLUEPRINTS-BB
http://tinyURL.com/BLUEPRINTS-AI
> WHICH CATEGORY DO THESE 2 FORMULA BELONG TO?

Whatever answer I give to that will not vitiate anything I've posted
here. And I really don't feel like playing answering games to
questions put to me ~ Moeblee (sci.logic)

[Christopher Menzel]

On Wednesday, September 26, 2012 12:33:18 PM UTC-5, Frederick Williams wrote:
> Christopher Menzel wrote:
...
> > > Please sir, in Shoenfield MP and UG aren't the only rules,
>>
> > You are indeed correct, sir.
> > >
> > > indeed, they're not rules at all!
> >
> > Well, they are *derived* rules, not primitive rules of the system, but a derived rule is still a rule. :-)
>
> I almost wrote 'not derived rules', but I decided to be more
> argumentative. Since neither \forall nor -> is a symbol in Shoenfield's
> language, neither rule is even statable.

Surely they are stateable and, indeed, they statED, albeit under the definitions of the two operators in question (though he calls MP "Detachment").

[Frederick Williams]

Ok, I'll let you win. Not because you're right--oh no--but because I'm
feeling generous :-).

[Christopher Menzel]

On Thursday, September 27, 2012 12:39:50 PM UTC-5, George Greene wrote:
> On Sep 24, 8:32 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > Unfortunately Moeblee only asks questions not answer them, but here he
> > seems to suggest ZFC DISRPOVES RUSSELL'S SET
> > using some non-axiomatic intelligence.
>
> Truth-tables are, for PRACTICAL purposes, NON-axiomatic.

They are also never presented as proof theories. Perhaps you have truth trees in mind.

> Prof.Menzel has done us all a great DISservice by intervening
> to insist that since ~( P <--> ~P ) is a "theorem of" propositional
> logic, it must have been proved from "logical axioms".

I did not so insist. My claim was that it was, at the least, misleading, to assert straight up that "you don't need any axioms" (childish shouting elided) to prove a logical theorem. And I stated the reason why this is misleading: Whether or not the assertion is true depends on the system you are talking about, the system relative to which the notion of "theorem" is defined. In natural deduction systems, as I explicitly granted, logical theorems are proved using only rules of inference (though in some cases, like the ND rules for the reflexivity of identity, the difference between axioms and rules is nil). But in the axiomatic systems that one finds in nearly every intermediate to advanced text on logic, every logical theorem depends on axioms. It is hardly a disservice to bring clarity to your assertion and, moreover, highlight the sense in which it is correct and the sense in which it is not. You're welcome.

> You are NEW to this arena and treatments (such as Prof. Menzel cited
> from Mendelson) that are approriate for intelligent undergrads who
> have already completed a course in propositional logic are NOT appropriate
> for YOU. What YOU need to start by believing is that THEORIES grow
> (or are derived from) AXIOMS and that LOGICs are defined via Rules Of
> Inference.

But it's just not true in general. As you note, it is true for the sorts of basic natural deduction systems one finds in typical Intro to Logic texts, but why not at least qualify your claim with something like "in most introductory presentations of logical systems"? People other than Herc are reading your posts, after all. Why mislead them?

[George Greene]

On Sep 27, 1:50 pm, MoeBlee <modem...@gmail.com> wrote:
> George Greene:
>
> Since the crank you are talking to is ineducable,

Moe, please just go to hell. Go DIRECTLY to hell; Do NOT pass Go;
Do NOT collect $200.

YOU do NOT get to say that the crank is ineducable SINCE YOU
ARE STILL HERE PRODUCING NOISE on the channel along which
*I* am trying to educate him. I HAVE called him STUPID but I have
NOT called him ineducable. The number of lies you are telling about
ME
really does call for a complete severing of diplomatic relations.
I have tossed some epithets at you but the precise fact that they ARE
interjections and epithets, AS OPPOSED to complete sentences about
you, necessarily makes what I have inflicted on you ON A FUNDAMENTALLY
LOWER ORDER than what you are saying about me. Your reaction is as
disproportionate as it is idiotic.

[George Greene]

On Sep 27, 3:29 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Sep 28, 3:50 am, MoeBlee <modem...@gmail.com> wrote:

> We did 2 years of LOGIC subjects and THEOREM PROVERS

"We" who? You Very Obviously Have Not Done ANY LOGIC AT ALL.
You DON'T grok BASIC concepts. You KEEP MISdescribing what is
happening here. You can't convince anybody you've done ONE year
of logic if you don't know the difference between an axiom and a rule
of inference, or what a first-order language is.

You may have TRIED to do some theorem-provers, but how is THAT
credible when you DON'T KNOW THE DEFINITION OF "theorem"???

[Frederick Williams]

George Greene wrote:
>
> [...] You can't convince anybody you've done ONE year

> of logic if you don't know the difference between an axiom and a rule

> of inference, or [...]

Since any axiom A can be replaced by the rule: from the empty set of
premises, A may be derived, what is the difference?

[George Greene]

On Sep 27, 3:40 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> GEORGE GREEN REFUSES TO STATE WHAT
>

> { ZFC-FORMULA } minus { FOL TRUTHS }

>
> is even meant to be?

I don't see why I need to state that -- YOU KNOW what "minus" means!
And it's not a question of what it's MEANT to be -- it's a question of
what it IS!

Do you know what a first-order LANGUAGE IS?
The axioms of ZFC are all phrased in terms one ONE predicate -- "e" or
epsilon
for set-membership. It's a binary predicate (it takes 2 terms from
the language
as arguments). You can state formulas in this language. You can also
state
all of the axioms of ZFC in this language. ANY FORMULA IN THIS
LANGUAGE is a "ZFC-formula", but WE wouldn't call it that. That's
YOUR
term because YOU'RE under-educated. What it REALLY is is a formula
over the first-order language with 1 binary predicate (and the
predicate-
name is spelled epsilon, written infix, i.e., between the operands).
Formulas can have parts that are formulas, but the issue here is not
about
"formulas" -- it is about SENTENCES. It is about STATEMENTS.
THOSE have TRUTH-VALUES.
We are "using" ZFC when we assume that some sentences in this language
-- THE AXIOMS of ZFC -- ZFC IS DEFINED BY ITS AXIOMS -- are true, AND
when we THEN use "naked logic" to determine what OTHER sentences
FOLLOW LOGICALLY FROM THE AXIOMS. The logical consequences of the
axioms -- the things that ARE PROVED USING LOGIC from the axioms --
are THEOREMS of ZFC -- *NOT* "ZFC-formulas"!

There are some sentences-in-this-language that you can prove WITHOUT
USING any of the axioms of ZFC as a premise.
~Er[ Ax[ xer <--> ~xex ] ] (which means "there is no Russell Set")
IS ONE OF THEM.
It is also the case that ZFC does NOT prove the denial of this
("there IS a Russell set"), and it's a good thing that it doesn't,
because every theory that DOES prove/force/require a Russell set to
exist IS INCONSISTENT (AND THEREFORE WORTHLESS). Naive set theory
unfortunately DOES require this set to exist -- THAT IS WHY IT IS
DISMISSED AS NAIVE.

[George Greene]

On Sep 27, 4:39 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> AHHH NOW I SEE WHY YOU'RE (ALL) SO CONFUSED!
>
> you think AVOIDING MODUS PONENS and using
>
> RULE1: LEFTHANDSIDE --> RIGHTHANDSIDE
> RULE2: LEFTHANDSIDE --> RIGHTHANDSIDE

It is counter-productive, in a forum like this one, to say what
ANYbody OTHER THAN YOURSELF thinks ABOUT ANYthing.
IF you want to impute an opinion to somebody else, QUOTE HER
STATING that opinion -- otherwise STAND PREPARED for the whole
discussion to be HEAT RATHER than light. It is not a constructive
use of anybody's time for anybody to ascribe (to anybody else)
opinions that the ascribEE does NOT even hold!

In the first place, WE DON'T avoid modus ponens.
You're just LYING about us. STOP it. Modus Ponens
IS an inference rule. It's not one I personally need much
(my school preferred Resolution, which if you are doing
Prolog with unification, you SHOULD KNOW about),
but you won't catch ME saying that there's
anything WRONG with MP. What I *did* call "wrong" in the
context of YOUR education was "logical axioms".

It is a TRIVIAL matter to STATE (or re-state) MOST rules of inference
as "logical axioms" -- there is A NATURAL duality between any "logical
axiom" whose main connective is "-->" and a rule of inference.
Some important rules of inference in FOL do not
fit easily into that paradigm, though. That paradigm works much
better
in 0th-order logic and as a "bridge" from 0th to 1st (you can lift 0th-
order
tautologies to 1st-order validities if you are careful about variable-
names).

> is something other than using a plain old grammar of axioms.

You're making a VERY odd use of "plain" here.
Axioms are ONE thing. Grammar is ANOTHER thing.
There is NOTHING PLAIN about MIXING the two!
A first-order language HAS a grammar but grammar is
only indirectly/tangentially relevant to inferring.
You do have to PARSE the premises FROM which you are inferring --
you have to match parts of them to roles/parts of the LHS of the
inference
rule -- and there is something grammar-related in doing that -- but it
is
hardly ABOUT "grammar" to the degree to which you are claiming.
The whole purpose of something like Prolog is that IT ALREADY handles
THAT!

> ANY AXIOM [+]  MODUS PONENS
>
> IS AN INFERENCE RULE.
>
> (technically [+] UNIFY() to match sub arguments)
>
> IF YOU CANNOT INFER ANYTHING FROM AXIOMS THEY MAY AS WELL BE A USELESS
> PIECE OF PAPER LIKE GEORGE GREENES DEGREES.
>
> ****************************
> GRAHAM COOPER (BINFTECH)
> UNIVERSITY OF QUEENSLAND

> --http://tinyURL.com/BLUEPRINTS-MATHEMATICShttp://tinyURL.com/BLUEPRINTS-HYPERREALShttp://tinyURL.com/BLUEPRINTS-HALT-PROOFhttp://tinyURL.com/BLUEPRINTS-QUESTIONShttp://tinyURL.com/BLUEPRINTS-POWERSEThttp://tinyURL.com/BLUEPRINTS-THEOREMhttp://tinyURL.com/BLUEPRINTS-FORALLhttp://tinyURL.com/BLUEPRINTS-TURINGhttp://tinyURL.com/BLUEPRINTS-GODELhttp://tinyURL.com/BLUEPRINTS-TRUTHhttp://tinyURL.com/BLUEPRINTS-PROOFhttp://tinyURL.com/BLUEPRINTS-LOGIChttp://tinyURL.com/BLUEPRINTS-BRAINhttp://tinyURL.com/BLUEPRINTS-SETShttp://tinyURL.com/BLUEPRINTS-PERMhttp://tinyURL.com/BLUEPRINTS-P-NPhttp://tinyURL.com/BLUEPRINTS-LIARhttp://tinyURL.com/BLUEPRINTS-GUThttp://tinyURL.com/BLUEPRINTS-BBhttp://tinyURL.com/BLUEPRINTS-AI

[George Greene]

On Sep 27, 3:29 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> That why *I* am the only one here who knows how to use SKOLEM
> FUNCTIONS,

No, you're not.
I don't personally think you know how to use them.
If you do, SKOLEMIZE THIS:
ErAx[ xRr <--> ~xRx ]. (that's Russell's Paradox).

Then, if you know so much about theorem-proving, Skolemize
ITS DENIAL, WHICH IS a theorem:
~ErAx[ xRr <--> ~xRx ].

Maybe then we'll ask you to prove it.
But the point is, IF you know how to prove that, then
what are we DISagreeing about??

> PREDICATE LOGIC AXIOMS,

Your problem is you know so little about axioms in general that
YOU CAN'T TELL THE DIFFERENCE between a "predicate logic axiom"
AND an axiom of something OTHER than logic.
ZFC is A COLLECTION OF AXIOMS (and set theory is the theory that
follows FROM THOSE axioms). THE PURPOSE of axioms in general
is to generate the theory of their own logical consequences.
Therefore, speaking of "logical axioms" IS BAD.
The people who TAUGHT you to CALL them that did you a DISservice.
Prof.Menzel coming back here and validating that as terminology
MADE YOU STUPIDER thereby.
The AXIOMS found/define/generate the THEORY.
The rules of inference define the LOGIC.
The logic comes BEFORE the theory.
The relevant first-order LANGUAGE comes "between" the two.

[Graham Cooper]

On Sep 29, 4:32 am, George Greene <gree...@email.unc.edu> wrote:
> On Sep 27, 3:29 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > That why *I* am the only one here who knows how to use SKOLEM
> > FUNCTIONS,
>
> No, you're not.
> I don't personally think you know how to use them.
> If you do, SKOLEMIZE THIS:
> ErAx[ xRr <--> ~xRx ].    (that's Russell's Paradox).

NO QUANTIFIERS
xRr() <-> ~ xRx

since the Exists is outside the Forall.

x depends on r, going left to right, but x is universal anyway so it
stays the same.

In pure relations or functions it's only different to this formula.

ArAx[ xRr <--> ~xRx ].

by making r into r(), a predicate, not a variable.

In PROLOG, atomic terms are treated like 0 argument predicates, both
lower case.

term
predicate( args )
VAR

r( arguments )
r( )
r

so r is just a term, a fixed value.
The Unify() pattern matcher doesn't need to tell whether it has
arguments or not straight away, first it checks the string 'r' is a
match, whether a term or predicate. Well *my* system does that, most
others probably check the arity or type first.

In LOGIC10.PRO
******************************
iff( e(X, r) , not(e(X, X)) ).
******************************

just using 'e' since caps R wont work here.

incidently,

eq(X,X).

will define an axiom for '=', via unification.

>
> Then, if you know so much about theorem-proving, Skolemize
> ITS DENIAL, WHICH IS a theorem:
> ~ErAx[ xRr <--> ~xRx ].

ArEx ~( xRr <-> ~xRx )

'for all sets r, exist an x
such that it never holds that
x is in r iff x is not a member of itself'

Strange reading, there is a 'breakout' element for all sets r.

NO QUANTIFIERS
~( x(r) R r <-> ~ x(r) R x(r) ).

x depends on r. (returns some element dependent on the value r)

Herc

[George Greene]

On Sep 24, 9:40 pm, MoeBlee <modem...@gmail.com> wrote:
> It's a simple argument to see that ZFC (indeed the pure first order
> calculus with a 2-place relation symbol, say 'e') proves
>
> ~ExAy(yex <-> ~yey)


OK, YOU HAD IT RIGHT FROM DAY 1.

I am not responsible for the fact that you chose to emphasize
the half before the parentheses more than the half inside them.
The choice of emphasis MATTERS.

[George Greene]

On Sep 25, 4:10 pm, MoeBlee <modem...@gmail.com> wrote:
> So everybody just shut up so that without the nuisance of other people
> posting, George alone can post exactly what he thinks are just the
> right things to say to a crank. George knows. He knows exactly the
> right context to presuppose, the right amount of formality, the right
> amount of technical detail, the right level of terminology, and
> exactly which considerations are relevant and which aren't, at any
> point in a thread. George knows all this because ... well because ...
> HE'S GEORGE, YOU STUPID FUCKING LYING SHITHEEL!!!

To the extent that this was a ever a flamewar, this is who started it
and when.

[MoeBlee]

On Sep 28, 6:25 pm, George Greene <gree...@email.unc.edu> wrote:
> On Sep 24, 9:40 pm, MoeBlee <modem...@gmail.com> wrote:
>
> > It's a simple argument to see that ZFC (indeed the pure first order
> > calculus with a 2-place relation symbol, say 'e') proves
>
> > ~ExAy(yex <-> ~yey)
>
> OK, YOU HAD IT RIGHT FROM DAY 1.

Now THERE I like youre all caps. ;-)

> I am not responsible for the fact that you chose to emphasize
> the half before the parentheses more than the half inside them.

Then rather than as much as DENOUNCE me for being "wrong" and "lying"
etc., all you had to do, as I illustrated with the spoof dialogue, was
say that you would rather emphasize the parenthetical part more.

> The choice of emphasis MATTERS.

I emphasized enough over the course of several posts.

Anyway, thanks.

MoeBlee

[MoeBlee]

Oh come on... ! You said I'm "lying" and "shitheel" (for my
persistence) and whatever else, and related crap against Menzel, and
you say my REPSONSE to that is the start of the flaming.

I don't understand you, George. When you're smart and studied in the
subject (said not patronizingly), I don't understand why you need to
play such a truly bizarre rage and arbitrary contrarian game virtually
all the time.

MoeBlee

[George Greene]

On Sep 28, 8:05 pm, MoeBlee <modem...@gmail.com> wrote:
> I emphasized enough over the course of several posts.

Obviously I disagree but I am not going to get strenuous about it.
More important things are now at stake.

> Anyway, thanks.

That was more charitable than I deserve and praiseworthy purely as de-
escalation.
Surely we will all be happier if we CAN keep talking. I have been on
this page for
20 years and most of the really brilliant people have retired in
disgust. I was just
about to encourage C.Menzel to go and do likewise. There is a
difference between
hurling an epithet in reaction to an individual error and launching a
full diatribe
against somebody's general overall character, but I am not currently
willing to
keep fighting THAT battle either.

It is not immediately obvious to everybody whether any individual
crank can or cannot
be reached; there are similarities and there are differences. I get
exasperated with
people for making their own new errors in reacting to error AND for
trying to flaunt their
own "correctness" AS OPPOSED to engaging the crank's INcorrectness.
Even if you are not interested in pedagogy as a hobby, you STILL have
to
have SOME empathy; you (if you are going to engage a crank at all; I
mean, you
know he's a crank; WHAT'S THE *POINT*; WHAT is in it for you??) have
to make
some token attempt to see how or why he could be THAT mistaken in THAT
way about THAT simple a point.

[George Greene]

On Sep 28, 8:05 pm, MoeBlee <modem...@gmail.com> wrote:

> Then rather than as much as DENOUNCE me for being "wrong" and "lying"

I *SAID*, "Everybody who has been" saying that the set theory piece
was more important...
That is NOT specific to you. I did mention you by name but I repeat,
I said EVERYbody from MoeBlee on up....
Look, you WERE wrong. Having merely stated something that merely
happens to be true does NOT make you RIGHT!
Stating a truth is a speech act WITH MORE implications THAN the
content of the statement! The content of the
statement has theoretical and logical consequences but THE ACT has
PRACTICAL FACTUAL consequences!

In arguing with a crank one will frequently have to assert truths that
NORMALLY go withOUT saying;
one will frequently have to say things that would NORMALLY be too
OBVIOUS to be BELABORED!
The fact that this kind of conversation is going to have a MUCH higher
than NORMAL density of
asserted really-true-truths makes "being a really-truly-indisputable-
truth" LESS valuable and LESS
invulnerable-to-attack THAN USUAL! SUPPLY AND DEMAND! Analytic
truths are going to be in
UNUSUALLY HIGH supply and therefore of LESS VALUE in this context!
DOING the RIGHT thing
means PICKING the RIGHT truth to STRESS! Things as BLATANTLY false
as the ones asserted
by cranks are going to have LOTS of absurd consequences in need of
refuting (from OUR viewpoint),
but from HIS, *ONE* of them is the KEY log in the JAM, and making a
contribution entails making
SOME effort to find the NEEDLE in the haystack!

[MoeBlee]

On Sep 28, 8:09 pm, George Greene <gree...@email.unc.edu> wrote:
> I have been on
> this page for
> 20 years and most of the really brilliant people have retired in
> disgust.

I've looked at the archives and I agree that sci.logic is a shadow of
what it was. For example, whatever happened Keith Ramsay?

But still I think we agree that there some good posters still, such as
Smaill, Menzel, Aatu, Jesse Hughes, Ullirich (will he be popping in
again sometime?) and others I would name...

Though, I still feel great loss in the absence of Franzen. I loved
him. He was a great soul.

>  I was just
> about to encourage C.Menzel to go and do likewise.

My time would be better spent not posting to cranks, I agree. I do it
as a kind of a minor addiction - like some people can't resist the
crossword puzzle each morning. That's basically what it is for me.

> There is a
> difference between
> hurling an epithet in reaction to an individual error and launching a
> full diatribe
> against somebody's general overall character, but I am not currently
> willing to
> keep fighting THAT battle either.

I wouldn't have made that kind of broadside if I didn't truly in my
heart think it was legit to make; it was from an accumulation of about
seven years of posting with you.

> It is not immediately obvious to everybody whether any individual
> crank can or cannot
> be reached;

We know by now that cranks can't be reached. What rare exceptions are
there? The only one I know is that for about the first year or two of
zuhair's postings I couldn't believe that he was not a put-on, himself
a parody of a crank; but it turned out that eventually he did figure
some stuff out and he's no crank for a while now.

> there are similarities and there are differences.  I get
> exasperated with
> people for making their own new errors in reacting to error AND for
> trying to flaunt their
> own "correctness" AS OPPOSED to engaging the crank's INcorrectness.

I can understand that. On the other hand, my purpose is basically just
to "work the crossword", which means filling in corrections to the
cranks, and I don't feel obligated to worry about some larger context
of pedagogical presentation for them, since that's a waste; they're
not receptive anyway.

> Even if you are not interested in pedagogy as a hobby, you STILL have
> to
> have SOME empathy;

I have very little empathy for cranks. I guess if I knew them
personally, then I would feel personal empathy for them, for whatever
their problem is that has driven them to being cranks. But as
interlocuters in a newsgroup, to me they're little more than agents of
misinformation to be corrected.

But that is only after SO MANY attempts to truly get through to them;
to address them eye to eye and without insult. It's a huge amount of
time, sincerely explaining and putting a lot of mental energy into
thinking up ways of saying things that will lead them to understand
even basic reason. But it never works. And it never works not just for
me but for NO ONE. I've never seen anyone, no matter how patient and
no matter how good a teacher, get through the hard concrete wall of a
crank.

> you (if you are going to engage a crank at all; I
> mean, you
> know he's a crank; WHAT'S THE *POINT*; WHAT is in it for you??) have
> to make
> some token attempt to see how or why he could be THAT mistaken in THAT
> way about THAT simple a point.

I have tried a lot before. Never works. JUST when you think the crank
is about to come around and see that your P->Q and ~Q entails ~P, just
as he seems on ther very verge and says conciliatory things and pays a
bit of lip service to your point, then in the very next paragraph he
belies it all and starts making a chimpanzee of himself again.

Anyway, at this exact moment I am enjoying having a chat with you, in
what seems like a timeout for you, a momentary break in the clouds
from your usual rage.

MoeBlee

[MoeBlee]

On Sep 28, 8:16 pm, George Greene <gree...@email.unc.edu> wrote:

> you WERE wrong.

I give up. You just said that I was right from the start.

> Having merely stated something that merely
> happens to be true does NOT make you RIGHT!

I understand the sense in which you mean that (as you explain the rest
of your post I haven't included here), but I don't see it the way you
do in that regard. And I'm too worn out now to elaborate. But, anyway,
I gave plenty of the context about the general principle too. My
address to the crank was fine. He's not receptive anyway and he NEVER
WILL BE. It's silly to hold posters who are correcting cranks to a
quite unreasonable (and with you, subjective) standard of proper
pedagogy.

You post as if you have a "control freak" need to have every anti-
crank poster present the material in just the exact way you think it
should be explained, right down to the very last nuance, to the point
that your objections become absurd and bizarre. And then you get
yourself so tangled up that your own posts turn out to be riddled with
outright and outlandish plain mathematical error. Or, you turn what
seems to be merely contrarian for no purpose.

I find it even impossible to talk with you most the time. But funny
thing is that I do share your frustration with the cranks and I do
feel a certain affinity for your sensiblity in regards the love of the
subject, and also I know that you have a liberal political/social
point of view that I am sympathetic with. But in the day-to-day
posting, I can't take you...to me, your demands such as that everybody
else should shut up in talking to the crank are bizarrely imperious,
among other things that make me wonder not just as to your sense of
fair play in conversation but (and this is not said sarcastically) as
to your mental health. Sometimes I wonder whether you're posting in
terrible physical or mental pain; If I knew you more personally, I'd
actively worry about you.

MoeBlee

[Graham Cooper]

On Sep 29, 11:43 am, MoeBlee <modem...@gmail.com> wrote:
> interlocuters in a newsgroup, to me they're little more than agents of
> misinformation to be corrected.

got it! CONDESCENDING IS IN!

"That was idiotic" is out!

>
> But that is only after SO MANY attempts to truly get through to them;
> to address them eye to eye and without insult. It's a huge amount of
> time, sincerely explaining and putting a lot of mental energy into
> thinking up ways of saying things that will lead them to understand
> even basic reason. But it never works. And it never works not just for

You repeated 3 times:

A(a,b,c..)
A(y) E(z) A(x) xey <-> xez & phi(x,y,a,b,c...)

is not cyclic

but
A(y) A(x) xey <-> p(x,y)

was ILLEGAL.

If I said IDIOTIC POINT

instead of YOU IDIOT

which was IN CONTEXT, you making all these idiotic arguments then
running away.

no one would have got upset!

Herc

[Nam Nguyen]

On 28/09/2012 7:43 PM, MoeBlee wrote:
> On Sep 28, 8:09 pm, George Greene <gree...@email.unc.edu> wrote:
>> I have been on
>> this page for
>> 20 years and most of the really brilliant people have retired in
>> disgust.
>
> I've looked at the archives and I agree that sci.logic is a shadow of
> what it was. For example, whatever happened Keith Ramsay?
>
> But still I think we agree that there some good posters still, such as
> Smaill, Menzel, Aatu, Jesse Hughes, Ullirich (will he be popping in
> again sometime?) and others I would name...

Except when certain foundational dialog is needed so that they
could share or offer their own knowledge on the new areas, helping
one to confront issues, they seem to be busy and not "being there". :-(

>
> Though, I still feel great loss in the absence of Franzen. I loved
> him. He was a great soul.

Indeed I miss Franzen quite a lot too. If you don't know certain things
and you'd genuinely like to know _Franzen would almost all the time_
_be helpful_ _explaining in details the issues_ . As well as his fine
professionalism manner is missed quiter a bit, given the nature of the
newsgroups such as this.

I was hoping Enderton would decide to play a role in sci.logic but
then ... that unfortunately didn't turn out to be the case, as we
all know why.

I always wonder, were Franzen and Enderton still with us today,
how would either of them perceive the issue of mathematical relativity
as has been forwarded, presented? (Fwiw, it was Franzen that I
"targeted" for confronting the issue of relativity of arithmetic
truth, since I had discovered in some of his writing a a few passages
that were pro Platonic-absolute-truth. This was shortly before
his leaving us).

Anyway your conversation with George reminded me similar conversation
in the past, which I excerpt below.


On 14/05/2011 10:05 PM, Nam Nguyen wrote:
> On 14/05/2011 5:28 PM, Nam Nguyen wrote:
>> On 14/05/2011 5:09 PM, Rupert wrote:
>>> On May 15, 2:01 am, Nam Nguyen<namducngu...@shaw.ca> wrote:
>>>> On 14/05/2011 9:55 AM, MoeBlee wrote:
>>>>
>>>>> On May 14, 8:50 am, Nam Nguyen<namducngu...@shaw.ca> wrote:
>>>>>> Franzen's being no longer with
>>>>>> us is a great loss
>>>>> There's no doubt of that at all. Franzen is beloved and greatly
>>>>> missed. But that doesn't diminish that we're lucky still to have
>>>>> Chris Menzel, Rupert, Aatu Koskensilta and several others.
>>>>
>>>> Really? So which _specific_ posts of some length have they
>>>> responded or contributed to the subjects of relativity of
>>>> mathematical reasoning and the 4 Principles, which I raised in
>>>> the past, for example?
>>>>
>>> What are these 4 Principles?
>>
>> I'll have to leave now, but I will respond.



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

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

[George Greene]

On Sep 28, 10:25 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> You repeated 3 times:
>
> A(a,b,c..)
> A(y) E(z) A(x)  xey  <->  xez & phi(x,y,a,b,c...)
>
> is not cyclic

Well, IT ISN'T. Why did anyone NEED to repeat it 3 times? Why didn't
you JUST GET IT the FIRST time?!??

>
> but
> A(y) A(x)  xey  <->  p(x,y)
>
> was ILLEGAL.

It's CONTRADICTORY.

>
> If I said IDIOTIC POINT

Then YOU WOULD BE AN IDIOT.

> instead of YOU IDIOT
>
> which was IN CONTEXT, you making all these idiotic arguments then
> running away.
>
> no one would have got upset!

That is NOT true. It is NOT the case that if you just insult the
point, rather than the person making it,
that makes it all better. Human nature being what it is, people ARE
offended when you attack "their"
points, NO MATTER HOW CORRECT your attack may be. Which in your
personal case is approximately Never,
around here.

Even making a point that is "technically" CORRECT will not help --
there are OTHER CONSIDERATIONS of MERIT
and you can STILL find your point attacked EVEN when your point is
provable, and if this happens, YOU WILL GET
UPSET, NO MATTER that the fact that "your technical point is BESIDE
the point and even CONTRARY TO THE
MISSION of the discussion" IS ALSO provable.

[Graham Cooper]

A(y) A(x)  xey  <->  p(x,y)

My axiom of naive specification stands,

and using it is trivial to make a consistent Russells Set.

RS = { x| p(x,y) }

p(x,y) = (x=/=y) ^ ~(xex)

The fact none of you realised it is the DEFINITION of Russell's Set,
not the SET OF NON-SELF-MEMBERS that is the problem speaks volumes
about you 2 putting the cart before the horse here.

What 'buffoons' would try to disprove an explicitly possible set?

Herc

[MoeBlee]

On Sep 28, 9:25 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Sep 29, 11:43 am, MoeBlee <modem...@gmail.com> wrote:
>
> > interlocuters in a newsgroup, to me they're little more than agents of
> > misinformation to be corrected.

The paragraph I wrote:

I have very little empathy for cranks. I guess if I knew them
personally, then I would feel personal empathy for them, for whatever
their problem is that has driven them to being cranks. But as

interlocuters in a newsgroup, to me they're little more than agents
of
misinformation to be corrected.

So, my comment was not that I take interlocuters in a newsgroup in
general to be little more than agents of misinformation, but rather
that I take CRANKS, in their role as interlocturers, to be little more
than agents of misinformation.

> You repeated 3 times:
>
> A(a,b,c..)
> A(y) E(z) A(x)  xey  <->  xez & phi(x,y,a,b,c...)
>
> is not cyclic

I did not once (let alone three times) use the rubric 'cyclic' in this
connection. Moreover, the above is your (not mine) notation that seems
to be a mangled, incorrect attempt at stating the axiom schema of
separation.

However, where 'P' is any formula in which 'z' does not occur free,
all closures of

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

are instances of the axiom schema of separation.

Moreover,

Z = {x | xey & P}

is a definition (thus not circular) of S, if 'S' is not free in 'P'.

> but
> A(y) A(x)  xey  <->  p(x,y)
>
> was ILLEGAL.

Again, I did not use the word "illegal" in this context. And I don't
know why you write a formula such as above.

What I said (possibly re-lettered here) is that if z occurs free in P,
then

EzAx(xez <-> (xey & P)) is not an instance of the axiom schema of
separation.

Also

z = {x | xey & P} is not a correct definition since it is circular by
defining z in terms of z itself (as z occurs free in P).

> then
> running away.

I don't "run away". Claiming that I "ran away" is childish.

MoeBlee

[MoeBlee]

On Sep 29, 3:08 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>  A(y) A(x)  xey  <->  p(x,y)
>
> My axiom of naive specification stands,

That's not naive (unrestricted comprehension). Unrestricted
comprehension is

EyAx(xey <-> P) where P is any formula in which 'y' does not occur
free.

The first quantifer is 'Ey' not, as you wrote, 'Ay'. And even
unrestricted comprehension requires that the variable for the
existential quantifier (in this case 'y') not occur free in P.

MoeBlee

[Graham Cooper]

I think you mean bound.

y occurs (bound) in ZFC axiom of specification phi(x,y,a,b,c..), for
the 4th time!!!!

same as here

A(y) A(x)  xey  <->  p(x,y)

That's why I write the formulas 1 under the other.

Try to argue this without mentioning z.

Naive Set Theory is *ANY* predicate-definable set.

According to you argument MB, this set is impossible.

abstract = {art,emotion,abstract}
curriculum = {teaching,discipline,curriculum}

RS = {abstract,curriculum}

It's just a REFLEXIVE "e" relation. NOTHING TO IT!


Herc

[MoeBlee]

On Oct 1, 2:08 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Oct 2, 4:18 am, MoeBlee <modem...@gmail.com> wrote:
>
>
>
>
>
> > On Sep 29, 3:08 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > >  A(y) A(x)  xey  <->  p(x,y)
>
> > > My axiom of naive specification stands,
>
> > That's not naive (unrestricted comprehension). Unrestricted
> > comprehension is
>
> > EyAx(xey <-> P) where P is any formula in which 'y' does not occur
> > free.
>
> > The first quantifer is 'Ey' not, as you wrote, 'Ay'. And even
> > unrestricted comprehension requires that the variable for the
> > existential quantifier (in this case 'y') not occur free in P.
>
> > MoeBlee
>
> I think you mean bound.

No, I mean what I wrote.

The axiom schema of unrestricted comprehension is:

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

EyAx(xey <-> P)

are axioms.

/

> y occurs (bound) in ZFC axiom of specification phi(x,y,a,b,c..),

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.

Now, of course y occurs bound in EyAx(xey <-> (xez & P) but, for the
axiom schema of separation, y must

not occur free in P.

> According to you argument MB,  this set is impossible.
>
> abstract = {art,emotion,abstract}

No, I did not say that it is impossible to have such a set. What I
said is that the above is not a DEFINITION of a set.

Whether a set may exist such that the set is a member of itself is a
separate matter (usually settled by whether we adopt the axiom of
regularity, merely eschew the axiom of regularity, or eschew the axiom
of regularity and adopt some axiom contradicting the axiom of
regularity).

The point I made is merely that

y = {x | P}

is not a definition of a set y if y occurs free in P; such a purported
defintion would be circular.

This is NOT saying that, for example, there does not exist a set y
such y = {x y z}. Without the axiom of regularity we cannot conclude
that there does not exist such a set. But I'm saying that while such
as set y might (without the axiom of regularity) exist such that y =
{x y z} it is still not the case that "y = {x y z}" is not DEFINITION
of a set.

Please recognize the difference between an existence statement:

(1) Ex x = {x y z}

and

(2) x =df {x y z}

as (1) can hold (without the axiom of regularity), while (2) is not a
correct definition (thus not actually a definition) since it is
circular.

MoeBlee

[Graham Cooper]

For the 5th time.

Y DOES OCCUR inside the predicate P in ZFC.

If you DUCKSPEAK NAIVE SET THEORY so that certain sets don't occurr
then it's no longer called NAIVE SET THEORY.

You are making NO SENSE.

...y = ....p(x,y,...) ZFC
...y = ....p(x) NST

why is y an allowed variable in ZFC and not NST????

Herc

[Graham Cooper]

> ...y = ....p(x,y,...)    ZFC LEGAL
> ...y = ....p(x)    NST ILLEGAL??

>
> why is y an allowed variable in ZFC and not NST????
>
> Herc

FORGET IT!

IN THIS SET THEORY NEWLY CALLED

"TRANS-NAIVE SET THEORY"

A(y)A(x) xey <-> p(a,b,c...,n)

where the arguments of p are arbitrary.

THEN RUSELL'S SET IS WELL DEFINED.

rs = { X | p(X,rs) }

p(X,rs) <-> (rs=/=X)^!(XeE)

You can define NAIVE SET UNRESTRICTED BULLSHIT ANYTHING YOU WANT NOW

HAPPY?


Herc

[MoeBlee]

On Oct 1, 8:37 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:

> For the 5th time.
>
> Y DOES OCCUR inside the predicate  P in ZFC.

I don't know what you mean by "the predicate P in ZFC". But, of
course, there are formulas P such that y occurs in P. That is not
contested.

Meanwhile, the point I made stands:

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.

And the axiom schema of unrestricted comprehension (which is not an
axiom schema of ZFC) is:

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

EyAx(xey <-> P)

are axioms.

MoeBlee

[Graham Cooper]

Those are only YOUR DEFINITIONS.

and bad ones.

for the 6TH TIME.

AXIOM OF SEPERATION

ALL(y) ..... PHI(y....)

^
|
y is the name of the set


GOT IT YET?

You cannot refute a theory exists where 'e' can be reflexive using old
definitions.

Herc

[MoeBlee]

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.

MoeBlee

[Graham Cooper]

The TOPIC is that YOU SAID

E(RS) XeRS <-> !(X=RS) ^ !(XeX)

is ILLEGAL because

A(Y) XeY <-> p(X,Y)

Y appears on the RHS inside p()

NO IFS, NO BUTTS. DO YOU AGREE SO FAR?


Herc

[MoeBlee]

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".

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).

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.

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.

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.

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

MoeBlee