ABC conjecture - Resolution.

[Khong Dong]

According to this Wiki link:

https://en.wikipedia.org/wiki/Abc_conjecture

the familiar statement of the ABC conjecture would involve functions of reals
(which of course aren't first order expressible). Let's call this Wiki link
version the canonical version of the ABC conjecture.

Let's consider below the, say, NN (Natural-Number) version of the conjecture
that does NOT refer to non-natural numbers (i.e. real numbers):

(NN) There are finitely many ABC numeral-triples ([a],[b],[c])'s where:


- In general, [n] is a numeral denoting a natural number n.

- [c] denotes an even number c which is a counter example of Goldbach
conjecture.

- rad([a][b][c]) < [c]

- Neither [a] nor [b] is S0

We'll prove the below two meta theorems MT-A, MT-B:

- MT-A: It's impossible to verify - hence invalid to claim - the NN version is true if it's so true.

- MT-B: It's impossible to verify - hence invalid to claim - the canonical version is true if it's so true.

Note: We'll make use the CGC-trichotomy proof in the link:

https://groups.google.com/d/msg/sci.logic/JTQDFTIfyv0/EhyblT-zBwAJ

which basically proves that it's impossible to logically verify the set, say
CGC, of counter example of Goldbach conjecture to be empty, finite, or infinite
- despite the corresponding language structure theoretical trichotomy being
true.

====> Proof of MT-A.

By syntactical form, an even [c] denoting a counter example of Goldbach
conjecture doesn't logically exclude the distinct possible existence of an ABC
numeral-triple ([a],[b],[c]) where both [a],[b] aren't S0 and one of [a],[b]
must necessarily not denote a prime, and rad([a][b][c]) < [c].

Let NN_S be the set of such c's of the above ABC numeral-triples. Certainly
NN_S is a subset of CGC. So by the CGC-trichotomy proof, it's impossible to
logically verify the set NN_S - as a subset of CGC - to be empty, finite, or
infinite. Hence it's invalid to claim the NN version is true, if it's so true.
QED.

====> Proof of MT-B.

Let Canonical _S be the set of c's of the ABC numeral-triples ([a],[b],[c])'s
where both [a],[b] aren't S0 and rad([a][b][c]) < [c]. We're assuming that
Canonical_S is the alluded finitute in the Wiki canonical version of the ABC
conjecture. Certainly NN_S is a subset of Canonical_S, but by MT-A it'd be
invalid to claim NN_S is finite, hence it'd also be invalid to claim
Canonical_S is finite. Hence it'd invalid to claim the canonical version is
true if it's so true. QED

[Khong Dong]

Note that the above meta proof is independent of IUTT (Inter-universal
Teichmüller theory) whatever IUTT might be about.

.

[Julio Di Egidio]

On Friday, 6 December 2019 06:20:05 UTC+1, Khong Dong wrote:
> According to this Wiki link:
>
> https://en.wikipedia.org/wiki/Abc_conjecture
>
> the familiar statement of the ABC conjecture would involve functions of reals
> (which of course aren't first order expressible). Let's call this Wiki link
> version the canonical version of the ABC conjecture.

We must be looking at a different WP:

<< It is stated in terms of three positive integers, a, b and c >>

<< The abc conjecture and its versions express, in concentrated form, some
fundamental feature of various problems in Diophantine geometry. >>

<< we introduce the notion of the radical of an integer: for a positive integer
n, the radical of n, denoted rad(n), is the product of the distinct prime
factors of n. >>

<< The abc conjecture is an integer analogue of the Mason–Stothers theorem for
polynomials. >>

> Let's consider below the, say, NN (Natural-Number) version of the conjecture
> that does NOT refer to non-natural numbers (i.e. real numbers):
>
> (NN) There are finitely many ABC numeral-triples ([a],[b],[c])'s where:

The "canonical" ABC conjecture is about numbers, not about numerals.

> - In general, [n] is a numeral denoting a natural number n.
>
> - [c] denotes an even number c which is a counter example of Goldbach
> conjecture.

In the "canonical" ABC conjecture there is no such assumption.

> - rad([a][b][c]) < [c]
>
> - Neither [a] nor [b] is S0

What does S0 mean, by the way?

> We'll prove the below two meta theorems MT-A, MT-B:
>
> - MT-A: It's impossible to verify - hence invalid to claim - the NN version is true if it's so true.
>
> - MT-B: It's impossible to verify - hence invalid to claim - the canonical version is true if it's so true.
>
> Note: We'll make use the CGC-trichotomy proof in the link:

<snip>

FWIW, as said, that is just not the ABC conjecture. Anyway, while I shall
leave to others getting into the formal details that I have snipped if they
like, it seems already apparent to me that we are yet again at your "it is
impossible to verify... because I say so", apparently adamant to the fact that
unknown/unproven is just NOT synonym with impossible to know/prove.

Julio

[Julio Di Egidio]

On Friday, 6 December 2019 07:10:40 UTC+1, Julio Di Egidio wrote:
> On Friday, 6 December 2019 06:20:05 UTC+1, Khong Dong wrote:

> > - Neither [a] nor [b] is S0
>
> What does S0 mean, by the way?

Never mind about that.

Julio

[Khong Dong]

On Thursday, 5 December 2019 23:10:40 UTC-7, Julio Di Egidio wrote:
> On Friday, 6 December 2019 06:20:05 UTC+1, Khong Dong wrote:
> > According to this Wiki link:
> >
> > https://en.wikipedia.org/wiki/Abc_conjecture
> >
> > the familiar statement of the ABC conjecture would involve functions of reals
> > (which of course aren't first order expressible). Let's call this Wiki link
> > version the canonical version of the ABC conjecture.
>
> We must be looking at a different WP:
>
> << It is stated in terms of three positive integers, a, b and c >>
>
> << The abc conjecture and its versions express, in concentrated form, some
> fundamental feature of various problems in Diophantine geometry. >>
>
> << we introduce the notion of the radical of an integer: for a positive integer
> n, the radical of n, denoted rad(n), is the product of the distinct prime
> factors of n. >>
>
> << The abc conjecture is an integer analogue of the Mason–Stothers theorem for
> polynomials. >>
>
> > Let's consider below the, say, NN (Natural-Number) version of the conjecture
> > that does NOT refer to non-natural numbers (i.e. real numbers):
> >
> > (NN) There are finitely many ABC numeral-triples ([a],[b],[c])'s where:
>
> The "canonical" ABC conjecture is about numbers, not about numerals.

Read the post more carefully. There's the Wiki "canonical" version with the
link and there is the NN (Natural-Number) version in my op.

There are two _different_ versions! They play different roles here!

[Julio Di Egidio]

You read more carefully, starting from the fact that the "canonical" version
is certainly not what you describe.

> and there is the NN (Natural-Number) version in my op.

Indeed also notice that the "canonical" version is already about natural
numbers.

> There are two _different_ versions! They play different roles here!

Sure, same result though: which you will simply keep not seeing, apparently...

Julio

[Khong Dong]

Good grief! Read the op again: the NN version is defined in such a way it has
something to do with Goldbach Conjecture, while the definition of the
"canonical" version doesn't depend on the Goldbach Conjecture at all. What has happened to your technical reading?

>
> Julio

[Khong Dong]

And the theme and the proof in the op is that:

- It's invalid to claim the NN version is true.
- The invalidity of claiming the NN version is true would logically entail the
invalidity of claiming the ABC conjecture is true.
- Hence, it's invalid to claim there's a proof for the ABC conjecture.

[Rupert]

On Friday, December 6, 2019 at 6:20:05 AM UTC+1, Khong Dong wrote:
> According to this Wiki link:
>
> https://en.wikipedia.org/wiki/Abc_conjecture
>
> the familiar statement of the ABC conjecture would involve functions of reals
> (which of course aren't first order expressible).

The use of reals in the statement of the ABC conjecture can be replaced with the use of rationals, which can be coded for by natural numbers. So there is, in fact, no problem with expressing this version of the ABC Conjecture in the language of first-order arithmetic.

[Rupert]

On Friday, December 6, 2019 at 6:20:05 AM UTC+1, Khong Dong wrote:

> According to this Wiki link:
>
> https://en.wikipedia.org/wiki/Abc_conjecture
>
> the familiar statement of the ABC conjecture would involve functions of reals
> (which of course aren't first order expressible). Let's call this Wiki link
> version the canonical version of the ABC conjecture.
>
> Let's consider below the, say, NN (Natural-Number) version of the conjecture
> that does NOT refer to non-natural numbers (i.e. real numbers):
>
> (NN) There are finitely many ABC numeral-triples ([a],[b],[c])'s where:
>
>
> - In general, [n] is a numeral denoting a natural number n.
>
> - [c] denotes an even number c which is a counter example of Goldbach
> conjecture.
>
> - rad([a][b][c]) < [c]

As I pointed out before, the Wikipedia article states that this is known to be false.

[Khong Dong]

Wrong. Because you snipped the last part of the NN version definition!
Perhaps you want to read that version _in full_ .

[Khong Dong]

So now you (Rupert) would understand what the NN version states, right?
Any rate, are meta statements MT-A, MT-B true to you? If not, what would be
the specific errors you'd see?

[Rupert]

In your "NN" version, you appear to attach importance to the fact that you are referring to numerals rather than natural numbers. I am not able to see how this would make any difference to the truth-value of your claims. Your "NN" version appears to me to be equivalent to the statement that Goldbach's conjecture is true.

I see no reason to believe MT-A or MT-B.

From your purported proof of MT-A, for example:

"By syntactical form, an even [c] denoting a counter example of Goldbach

conjecture doesn't logically exclude the distinct possible existence of an ABC numeral-triple ([a],[b],[c]) where both [a],[b] aren't S0 and one of [a],[b] must necessarily not denote a prime, and rad([a][b][c]) < [c]."

Well, how do you know that?

[Khong Dong]

Consider the even number c = 128 = 2^7, expressed in the below 4 equation-cases
involving some ABC triples:

1. 3 + 5^3 = 2^7 [ rad(abc)<c ]
2. 3^2 + (7*17) = 2^7 [ rad(abc)>c ]
3. 19 + 109 = 2^7 [ rad(abc)>c ]
4. 31 + 97 = 2^7 [ rad(abc)>c ]

First, 128 is not a counter example of Goldbach conjecture because of cases
3 and 4, in each of which *it's syntactically possible* to express the entire
equation with numerals and _without_ the language multiplication symbol '*'
(which of course must be required to define prime numbers).

Otoh, suppose 128 were a counter example of Goldbach conjecture then both cases
3, 4 can be ignored and only cases 1 and 2 would be considered as part of a
finitude both versions of the ABC conjecture are about.

But both of cases 1. and 2. are syntactically of the same language-symbol form
- where the language multiplication symbol '*' must by First Order _Logic_
formalism necessarily be present on the left side of the logical symbol '='
for the equation to be syntactically valid (i.e. well formed).

Consequently, if there is a counter example of Goldbach conjecture c (and at
least it's not known that we can logically exclude that distinct possibility)
being relevant to the ABC conjecture (of either version), then there would have
to be an ABC numeral-triple, hence ABC numeral-equation, like 1. and 2. (with
or without hypothetically assuming 128 were a counter example of Goldbach
conjecture): an (ABC-triple) equation of that restricted form, where '*' is
absolutely absent.

In summary, yes: we do you know _that_ - syntactically _that_ must be the case.

The important note here is that the common form of 3 and 4 here is ignored by
the alluded finitude in both versions of the ABC conjecture _anyway_ : since we
would have the undesired inequality rad(abc)>c.

It's an important note since it signifies we can examine the relationship
between Goldbach, cGC, ABC conjectures as a purely syntactical symbol game or
manipulation. And this examination would portend some disturbing facts about
the incoherent semantics-juxtaposition of the two FOL-with-equality logical
symbols - 'E' (Existence quantifier) and '=' (Equality), about it being invalid
to logically, semantically equate _general_ "equivalence" with "equality".

It would portend the collapse of the mathematical truth-reasoning envisioned
(philosophized) by Plato, built by Tarski, and invalidly codified by Gödel.

IUTT (Inter-Universal Teichmüller Theory) would be just an unfortunate and
accidental smokescreen masking of this collapse. How far have SME's on both
sides of the alleged ABC proof been able to agree even on such simple
(undergraduate level) issue of whether or not two isomorphic objects/structures
can categorically be treated the "same"?

Not that far at all apparently!

[Khong Dong]

> absolutely present.

Typo: was "absolutely absent", now "absolutely present".

[Khong Dong]

> absolutely present.

>
> In summary, yes: we do you know _that_ - syntactically _that_ must be the case.
>
> The important note here is that the common form of 3 and 4 here is ignored by
> the alluded finitude in both versions of the ABC conjecture _anyway_ : since we
> would have the undesired inequality rad(abc)>c.
>
> It's an important note since it signifies we can examine the relationship
> between Goldbach, cGC, ABC conjectures as a purely syntactical symbol game or
> manipulation. And this examination would portend some disturbing facts about
> the incoherent semantics-juxtaposition of the two FOL-with-equality logical
> symbols - 'E' (Existence quantifier) and '=' (Equality), about it being invalid
> to logically, semantically equate _general_ "equivalence" with "equality".

That is: about it being invalid to logically, semantically equate the
_general_ "equivalence" with a _specific_ "equality".

Two objects A and B might be existentially equivalent in term of probability of
existence but they might not be "the same" in term of existence. And merely
subjectively adding the string 'n=n' to any syntactical axiom-set won't make
what existentially might not be equal to be necessarily equal - logically.

[Jim Burns]

On 12/7/2019 2:00 PM, Khong Dong wrote:

> [...] and _without_ the language multiplication symbol '*'

> (which of course must be required to define prime numbers).

If '*' itself is a defined symbol, then the definition
of multiplication could be used in place of '*'.

For example,[1] along with addition '+', we might have
a square-operator '^2' instead of multiplication.
0^2 = 0
(x+1)^2 = x^2 + x + x + 1

Note that
(x+y)^2 = x^2 + 2*x*y + y^2

We could define multiplication '*' such that

x*y = z :<->
(x+y)^2 = x^2 + z + z + y^2

And then we could use that definition in place of 'x*y = z'
to define Prime() using '+' and '^2'.

[1]
https://math.stackexchange.com/questions/35988/can-multiplication-be-defined-in-terms-of-divisibility

[Khong Dong]

You've missed the point as always. Naturally for any numeral [x], you can
always syntactically replace it with [x]*S0. But that's _not_ what is being
talked about here!

What we talk about here is for a counter example of Goldbach conjecture, that
can't be the only way of syntactical replacement in which '*' would occur,
contrary to an even numeral denoting an even number being a sum of two primes:
[p1]+[p2]. Would you now understand that, JB?

And, please, if you could try to refrain from going off the topic talking about
"in-terms-of-divisibility" that would be much appreciated.

[Khong Dong]

On Saturday, 7 December 2019 18:43:51 UTC-7, Jim Burns wrote:

> On 12/7/2019 2:00 PM, Khong Dong wrote:
>
> > [...] and _without_ the language multiplication symbol '*'
> > (which of course must be required to define prime numbers).
>
> If '*' itself is a defined symbol

But it's *not*: it's a formal language symbol in the context.

[Jim Burns]

You mean '*' represents an undefined or primitive notion.
Defined notions are also formal.
https://en.wikipedia.org/wiki/Primitive_notion

But you didn't say so until now.

My point remains for cases in which the language
is expressive enough to _define_ the usual natural number
multiplication. '*' would not be _required_ to define
prime numbers. Even with '*' in the language, there would
be other ways to state the definition of Prime().

One example would be the language Goedel maps to numbers
in his Formally Undecidable Propositions. The basic signs
of KG's system P are _not_ '~', _or_ '\/', _forall_ 'A',
_nought_ '0', _successor_ 'S', _brackets_ '(' ')', and
variables of the first type, of the second type, of the
third type, and so on, for all natural numbers.

Clearly, he uses multiplication. But he must define it.
See? there is no basic sign for it. I think KG considers it
too obvious to waste space on in his paper.

In fact, he defines Prime(), too, in Definition 2.
And he _appears_ to use a dot (essentially '*') as a symbol
for multiplication in Definition 1. But it would necessarily
be defined without '*' or a dot, once all the definitions
are inserted for the defined terms.

[Jim Burns]

On 12/7/2019 9:30 PM, Khong Dong wrote:
> On Saturday, 7 December 2019 18:43:51 UTC-7,
> Jim Burns wrote:
>> On 12/7/2019 2:00 PM, Khong Dong wrote:

>>> [...] and _without_ the language multiplication symbol '*'
>>> (which of course must be required to define prime numbers).
>>
>> If '*' itself is a defined symbol, then the definition
>> of multiplication could be used in place of '*'.
>>
>> For example,[1] along with addition '+', we might have
>> a square-operator '^2' instead of multiplication.
>> 0^2 = 0
>> (x+1)^2 = x^2 + x + x + 1
>>
>> Note that
>> (x+y)^2 = x^2 + 2*x*y + y^2
>>
>> We could define multiplication '*' such that
>>
>> x*y = z :<->
>> (x+y)^2 = x^2 + z + z + y^2
>>
>> And then we could use that definition in place of 'x*y = z'
>> to define Prime() using '+' and '^2'.
>>
>> [1]
>> https://math.stackexchange.com/questions/35988/can-multiplication-be-defined-in-terms-of-divisibility
>
> You've missed the point as always. Naturally for any
> numeral [x], you can always syntactically replace it
> with [x]*S0.

Replacing [x] with [x]*S0
misses my point about
replacing x*y = z with (x+y)^2 = x^2 + z + z + y^2
-- which does not use '*'.

> But that's _not_ what is being talked about here!

Isn't it? What's

| [...] and _without_ the language multiplication symbol '*'
| (which of course must be required to define prime numbers).

? Techno-babble?

Whatever.

It's a point about foundations I posted because _I_ found it
interesting. _You_ aren't required to find foundations
interesting.

[...]

> And, please, if you could try to refrain from going off
> the topic talking about "in-terms-of-divisibility" that
> would be much appreciated.

You (NN) are notorious for hijacking the threads of
others. I'm not much interested in what _you_ would
appreciate, in that regard.

As for the rest of sci.logic, I imagine they might like
some actual mathematics inserted into your threads
occasionally. So, it's pretty close to a win all around.

[Khong Dong]

On Sunday, 8 December 2019 04:54:51 UTC-7, Jim Burns wrote:
> On 12/7/2019 10:52 PM, Khong Dong wrote:
> > On Saturday, 7 December 2019 18:43:51 UTC-7,
> > Jim Burns wrote:
> >> On 12/7/2019 2:00 PM, Khong Dong wrote:
>
> >>> [...] and _without_ the language multiplication symbol '*'
> >>> (which of course must be required to define prime numbers).
> >>
> >> If '*' itself is a defined symbol
> >
> > But it's *not*: it's a formal language symbol in the context.
>
> You mean '*' represents an undefined or primitive notion.
> Defined notions are also formal.
> https://en.wikipedia.org/wiki/Primitive_notion

Listen, crank Jim Burns. Instead of profoundly displaying your idiotic
ignorance in sci.logic, go and learn the below from Shoenfield's Mathematical
Logic (pg. 6):

"We emphasize that defined symbols are _not_ symbols of the language".

Too, in his book, '*' is not a defined symbol and you should not treat '*'
as a defined symbol in the language of Presburger arithmetic - as you've
ignorantly done. Crank.

[Khong Dong]

What a cheap, dishonest, and stupid retort from the crackpot Jim Burns.
What you typed above as a replacement is an equation - a full blown formula -
while what is expected here is a replacing _term_ - *only*!

Where's your "replacing" term for the term '[x]*S0'? Crank.

[Jim Burns]

'^2' is not a symbol in the language of Presburger
arithmetic.

'*' gets defined, but not in the language of Presburger
arithmetic.

| x*y = z :<->
| (x+y)^2 = x^2 + z + z + y^2

As you may recall, the recursive axioms for the
square-operator '^2" are

[Jim Burns]

You need to do a better job of expecting, then.

> Where's your "replacing" term for the term '[x]*S0'?

The replacement is for the formula 'x*y = z'.

Suppose one has a _formula_ F(x*y)
in which the _term_ 'x*y' occurs.

One can always introduce another variable in order
to re-write F(x*y) so that 'x*y = z' appears --
which allows us to insert the definition I gave as
I gave it.

F(x*y) <->
Az:( x*y = z -> F(z) ) <->
Az:( (x+y)^2 = x^2 + z + z + y^2 -> F(z) )

> Crank.

[Khong Dong]

On Sunday, 8 December 2019 14:16:08 UTC-7, Jim Burns wrote:
> On 12/8/2019 1:09 PM, Khong Dong wrote:
> > On Sunday, 8 December 2019 04:54:51 UTC-7,
> > Jim Burns wrote:
> >> On 12/7/2019 10:52 PM, Khong Dong wrote:
> >>> On Saturday, 7 December 2019 18:43:51 UTC-7,
> >>> Jim Burns wrote:
> >>>> On 12/7/2019 2:00 PM, Khong Dong wrote:
>
> >>>>> [...] and _without_ the language multiplication symbol '*'
> >>>>> (which of course must be required to define prime numbers).
> >>>>
> >>>> If '*' itself is a defined symbol
> >>>
> >>> But it's *not*: it's a formal language symbol in the context.
> >>
> >> You mean '*' represents an undefined or primitive notion.
> >> Defined notions are also formal.
> >> https://en.wikipedia.org/wiki/Primitive_notion
> >

> > Listen, crank Jim Burns. Instead of profoundlyu displaying

> > your idiotic ignorance in sci.logic, go and learn the below
> > from Shoenfield's Mathematical Logic (pg. 6):
> >
> > "We emphasize that defined symbols are _not_ symbols of
> > the language".
> >
> > Too, in his book, '*' is not a defined symbol and you should
> > not treat '*' as a defined symbol in the language of
> > Presburger arithmetic - as you've ignorantly done. Crank.
>
> '^2' is not a symbol in the language of Presburger
> arithmetic.

Idiotic trolling. Didn't you in the past refer to _your_ ~(2<2) as a formula
written in L(PA) - hence in the language of Presburger arithmetic?

When will you quit being both an idiotic troll and a dishonest Jim Burns?

In any rate, how do you think mathematicians prove 2+2=4 in PA
or in Presburger arithmetic? Idiot.

[Ross A. Finlayson]

Is that the only book you have?

[Khong Dong]

What's your point? Are you having a resolution for a version of the ABC
conjecture you'd like to argue for?

[Ross A. Finlayson]

"We already talked about Mochizuki here
and about Teichmuller spaces as about
Groethendieck universes here about
Frobenius and a "prime at infinity", and
Mochizuki's uniformization to the p-adics
from the integers with a prime at infinity.

It's got nothing to do with breaking or
changing the rules, it's about transforms
between systems, not the bait-and-switch.

That is to say, it's built way high above
standard numbers and the Archimedean,
and it's still there, not having inconstancy
in definition.

Inconstancy: kind of like inconsistency,
but in a variety of senses, worse. "

"I study Burns as a model of professionalism...".


"Mochizuki's group theoretic results
about abc conjecture up in Teichmuller,
here for example "so complicated",
don't allow HP to share "just because
it's so complicated that HP can't be
bothered to maintain consistency". "

"That's not to say nonstandard models
of the integers don't have reason for
their existence and properties, or that
various models of effectively the integers
don't have the varieties of their objects
in the nonstandard as about the quest of
the integer continuum for number theory,
and it's not necessarily to say that Cohen's
forcing models M and the other M aren't a
bait-and-switch, it's just that the above are
examples of illustrated fallacies, this forum
is about logic, there are right answers,
and those are not them.


Statements about arithmetic that would
establish consistency of arithmetic so might
be a Goedel sentence, may also simply enough
point to the extra-standard instead of the
anti-standard."


"Your "uncountably many", with
"almost all" of them wrong, is,
not sufficient that there's
anything to contradict, at all.

Because it's already wrong. "


"I appreciated "it's a pinata,
not a theory. And it's his party." "


""S" isn't just a notation that you can
purpose to your own device (or corrupt),
here it's the Peano's Successor function
and the only meaningful ("true") interpretation
of "substring_1" is Predecessor. "


https://groups.google.com/d/msg/sci.logic/vrr0ltxCi2U/XKky7w-fCwAJ

"(That also pretty well dispatched
"model structure language theory"
as a poor translation of a model-
theoretic discourse of Shoenfield.)"

https://groups.google.com/d/msg/sci.logic/XFR2K6rNunI/K3y5F3rhmvcJ

[Ross A. Finlayson]

https://en.wikipedia.org/wiki/Abc_conjecture

Where a, b, c are co-prime and a + b = c and d is the product
of distinct factors of abc, d <= abc, d/c <=ab, c/d >= 1/ab.

d is much smaller than abc when either a and b, or c,
are high powers of small primes.

With a and b being co-prime, it seems like a + b is
automatically co-prime to a and b. (Because neither
adds whole multiples of its existing primes already.)

There's an idea to re-write:

c - b = a; a, b > 0; c > b

Then there's the idea to figure whether and where
d > c or instead d > a or d > b. d=rad(abc) is usually
going to be smaller than a, b, and c. (Because "most"
numbers have powers in their prime factorizations.)

1/2 * 1/2
+
1/3 * 1/3
+
1/5 * 1/5
...

https://mathoverflow.net/questions/53443/sum-of-the-reciprocal-of-the-primes-squared
https://math.stackexchange.com/questions/2676173/series-of-reciprocals-of-primes-squared


Or, about .452 of numbers have powers in their integer factors,
so, d = rad(abc) is going to be smaller than abc, because
~.55 that that each isn't a power in factors, .55 * .55 *.55 < .5,
in three terms in the radical.

(One wonders then where the range begins that more numbers
have powers in their prime factors than don't, vis-a-vis usual
laws of small numbers. For example in [1,10] there's {4, 8, 9},
while in [1,20] there's {4, 8, 9, 12, 16, 18} and in [1,30] there's
{4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 28), i.e., there grows more.)

Nam, I'd be interested in solving this,
not quitting it and demanding others quit it too.

I.e., Nam doesn't know what undecideable means in model theory,
or rather conflates it roughly to a wrong reading of Shoenfield,
which besides very easily being reversed upon itself isn't conclusive,
or: questions about decideability in standard and nonstandard models
of integers are not relevant to Nam-speak, and, vice-versa.

Good luck, though, he could always be right, just not relevant.
(I.e., that's only "right" in an inconstant, inconsistent,
incomplete theory, i.e. that's wrong.)

[Khong Dong]

For sure (I'm almost certain) you're right that "We already talked about

Mochizuki here and about Teichmuller spaces as about Groethendieck universes

here" and I'm confident the mathematics community has talked out there about
IUUT in conjunction with the alleged ABC conjecture proof.

So where are we, RAF, with the status of the alleged ABC conjecture proof based
on IUUT? Or should we spend the rest of our precious time listening to JB
lecturing us that his own formula '~(2<2)' isn't considered to be written in
the forma language of arithmetic (PA or Presburger)?

Maybe you want to listen to JB's trolling "lecture", but I don't. Sorry.

[Khong Dong]

Then solve it, and present your solving here. Just stop your uncontrollable
blah blah blah ....

> I.e., Nam doesn't know what undecideable means in model theory,
> or rather conflates it roughly to a wrong reading of Shoenfield,
> which besides very easily being reversed upon itself isn't conclusive,
> or: questions about decideability in standard and nonstandard models
> of integers are not relevant to Nam-speak, and, vice-versa.

> Good luck, though, he could always be right, just not relevant.
> (I.e., that's only "right" in an inconstant, inconsistent,
> incomplete theory, i.e. that's wrong.)

Idiotic ranting of another sci.logic crank.

[Khong Dong]

Only a crank would say in the same sentence something like: "inconsistent, incomplete theory".

[Jim Burns]

I proved "my" formula ~(2<2) in Q, Robinson arithmetic,
so you (NN) would probably find all this less confusing if
you referred to ~(2<2) as a formula in L(Q).

However, since Q and PA have the same language, yes,
~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
'*' is an undefined symbol.

Suppose we were using a theory _other than_ Q or PA or
Presburger arithmetic, in which, instead of addition '+'
and multiplication '*', we had addition '+' and squaring
'^2'.

As well, instead of these axioms for '*',
x*0 = 0
x*Sy = x*y + x
we had these axioms for '^2'

0^2 = 0
(x+1)^2 = x^2 + x + x + 1

Even though that looks like a very different theory,
it would be _fundamentally_ the same as Q or PA
(depending on what other axioms there were), because

(i)
Obviously, we can define '^2' in terms of '*'
x^2 = y <->
x*x = y

(ii)
_Less obviously_ we can define * in terms of '^2'

x*y = z <->
(x+y)^2 = x^2 + z + z + y^2

> - hence in the language of Presburger arithmetic?

No, 'hence' does not belong there. Being a formula
of L(PA) = L(Q) does not imply being a formula of L(PrA),
since L(PrA) does not have '*'. Which you knew, of course.

"The language of Presburger arithmetic" could be taken
to have '<' as either (i)

an _undefined_ symbol, with axioms describing its use.
I think these would work:
~(x < 0)
(x < Sy) <-> ( x < y \/ x = y )
x < y \/ x = y \/ y < x

or (ii)

a _defined_ symbol
x < y :<->
Ez: x + Sz = y

I don't think defined/undefined is not generally seen as
an important distinction, unless one is making an argument
in which _which undefined symbols are in the language_
plays a role.

When you talked about requiring the use of the
symbol '*', it sounded as though you were making an
argument in which the defined/undefined distinction
mattered.

| [...] and _without_ the language multiplication
| symbol '*' (which of course must be required to
| define prime numbers).

I thought maybe you would prefer to argue from correct
information. It's kind of a tradition, here.

----
That quote reminds me. You have a lot of trouble with the
distinction between "I (NN) don't see how to do this" and
"This cannot be done". For example, that you (NN) don't
see how to define prime numbers without '*' doesn't mean
that it can't be done. And so on.

It takes a very determined person to continue to elide
that distinction, even after being given so many examples
of things they cannot do which nonetheless can be done.

> When will you quit being both an idiotic troll and
> a dishonest Jim Burns?
>
> In any rate, how do you think mathematicians prove
> 2+2=4 in PA or in Presburger arithmetic?

Those look like two rather desperate attempts to get me
stop talking about defining '*'.

For what it's worth, mathematicians prove 2+2=4 from
appropriate axioms and definitions. For example,

1 = S0, 2=S1, 3=S2, 4=S3

x + 0 = x

x + Sy = S(x + y)

----
2 + 2 = 2 + S1 = S(2 + 1)

2 + 1 = 2 + S0 = S(2 + 0) = S2 = 3

S(2 + 1) = S3 = 4

Therefore 2 + 2 = 4

Apparently, there's a subtle connection between this
and a requirement to use '*' in defining primes.

> Idiot.

Or, well, there's another possible explanation.

[Khong Dong]

On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:

> However, since Q and PA have the same language, yes,
> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
> '*' is an undefined symbol.

'*' is a formal language symbol in those contexts, you were told.

> Suppose we were using a theory _other than_ Q or PA or
> Presburger arithmetic, in which, instead of addition '+'
> and multiplication '*', we had addition '+' and squaring
> '^2'.

But why does one need that stupid, idiotic, trolling "Suppose",
when investigating an argument about the difficult ABC
conjecture!

In the context of the ABC conjecture it's idiotic and wrong the
way you redefined '*' in term of '*'. Only s crank would utter
that redefinition and somehow bs about the new '*' would still
mean multiplication operation of the natural numbers! Crank.

[Khong Dong]

On Monday, 9 December 2019 13:51:34 UTC-7, Khong Dong wrote:
> On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:
>
> > However, since Q and PA have the same language, yes,
> > ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
> > '*' is an undefined symbol.
>
> '*' is a formal language symbol in those contexts, you were told.
>
> > Suppose we were using a theory _other than_ Q or PA or
> > Presburger arithmetic, in which, instead of addition '+'
> > and multiplication '*', we had addition '+' and squaring
> > '^2'.
>
> But why does one need that stupid, idiotic, trolling "Suppose",
> when investigating an argument about the difficult ABC
> conjecture!
>
> In the context of the ABC conjecture it's idiotic and wrong the
> way you redefined '*' in term of '*'.

That is: "in term of '+'".

[Peter Percival]

Khong Dong wrote:
> On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:
>
>> However, since Q and PA have the same language, yes,
>> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
>> '*' is an undefined symbol.
>
> '*' is a formal language symbol in those contexts, you were told.
>
>> Suppose we were using a theory _other than_ Q or PA or
>> Presburger arithmetic, in which, instead of addition '+'
>> and multiplication '*', we had addition '+' and squaring
>> '^2'.
>
> But why does one need that stupid, idiotic, trolling "Suppose",
> when investigating an argument about the difficult ABC
> conjecture!
>
> In the context of the ABC conjecture it's idiotic and wrong the
> way you redefined '*' in term of '*'. Only s crank would utter

Jim hasn't redefined '*' in terms of '*'. Note that his definition of
'^2' makes no reference to '*'.

[Khong Dong]

On Monday, 9 December 2019 13:58:24 UTC-7, Peter Percival wrote:
> Khong Dong wrote:
> > On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:
> >
> >> However, since Q and PA have the same language, yes,
> >> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
> >> '*' is an undefined symbol.
> >
> > '*' is a formal language symbol in those contexts, you were told.
> >
> >> Suppose we were using a theory _other than_ Q or PA or
> >> Presburger arithmetic, in which, instead of addition '+'
> >> and multiplication '*', we had addition '+' and squaring
> >> '^2'.
> >
> > But why does one need that stupid, idiotic, trolling "Suppose",
> > when investigating an argument about the difficult ABC
> > conjecture!
> >
> > In the context of the ABC conjecture it's idiotic and wrong the
> > way you redefined '*' in term of '*'. Only s crank would utter
>
> Jim hasn't redefined '*' in terms of '*'.

That's was just a typo which I've already made a caveat-correction.

But again it was idiotic of him to redefine ''*' in term of '+' in this context
and in crackpot manner tried to bs around his ill-advised redefinition
posts after posts.

[Peter Percival]

Khong Dong wrote:
> On Monday, 9 December 2019 13:58:24 UTC-7, Peter Percival wrote:
>> Khong Dong wrote:
>>> On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:
>>>
>>>> However, since Q and PA have the same language, yes,
>>>> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
>>>> '*' is an undefined symbol.
>>>
>>> '*' is a formal language symbol in those contexts, you were told.
>>>
>>>> Suppose we were using a theory _other than_ Q or PA or
>>>> Presburger arithmetic, in which, instead of addition '+'
>>>> and multiplication '*', we had addition '+' and squaring
>>>> '^2'.
>>>
>>> But why does one need that stupid, idiotic, trolling "Suppose",
>>> when investigating an argument about the difficult ABC
>>> conjecture!
>>>
>>> In the context of the ABC conjecture it's idiotic and wrong the
>>> way you redefined '*' in term of '*'. Only s crank would utter
>>
>> Jim hasn't redefined '*' in terms of '*'.
>
> That's was just a typo which I've already made a caveat-correction.
>
> But again it was idiotic of him to redefine ''*' in term of '+' in this context

What point was Jim making when he redefined '*' in term of '+'?

[Jim Burns]

On 12/9/2019 3:51 PM, Khong Dong wrote:
> On Monday, 9 December 2019 05:39:41 UTC-7,
> Jim Burns wrote:

>> However, since Q and PA have the same language, yes,
>> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
>> '*' is an undefined symbol.
>
> '*' is a formal language symbol in those contexts,
> you were told.

Both undefined and defined symbols are formal,

you were told.

>> Suppose we were using a theory _other than_ Q or PA or
>> Presburger arithmetic, in which, instead of addition '+'
>> and multiplication '*', we had addition '+' and squaring
>> '^2'.
>
> But why does one need that stupid, idiotic, trolling
> "Suppose", when investigating an argument about the
> difficult ABC conjecture!

I may not be able to explain that to you. Think about:
https://en.wiktionary.org/wiki/suppose
| 2. (transitive) To theorize or hypothesize.
| "I suppose we all agree that this is the best solution."

Note that my point only addresses your claim that
'*' is required to define prime numbers.

If that claim of yours is irrelevant to the ABC
conjecture, just say so.

> In the context of the ABC conjecture it's idiotic and
> wrong the way you redefined '*' in term of '*'.

'*' is defined by use of '+' and '^2'.

The axioms for '^2' use '+', '1', '0', and '^2'.
They're recursive, just as the axioms for '+'and '*'
are.

> Only s crank would utter that redefinition and somehow
> bs about the new '*' would still mean multiplication
> operation of the natural numbers! Crank.

Do you want the definition explained?

(x+y)^2 =
(x+y)*(x+y) =
x*x + y*x + x*y + y*y =

x^2 + 2*x*y + y^2

For x*y = z, we can therefore write
(x+y)^2 = x^2 + z + z + y^2.

If there were both positive and negative numbers,
there would be some ambiguity, because of the squares.
But there aren't.

(x*y = z) <->
((x+y)^2 = x^2 + z + z + y^2)

Note that the axioms for x^2 do not use multiplication.

[Jim Burns]

On 12/9/2019 4:14 PM, Peter Percival wrote:
> Khong Dong wrote:

>> But again it was idiotic of him to redefine ''*'
>> in term of '+' in this context
>
> What point was Jim making when he redefined '*'
> in term of '+'?

First, I need to correct Nam.
I defined '*' in terms of '+' and '^2'.

The point I was making was that it may not be required
to use the symbol '*' in the definition of prime numbers.

For example, if '*' were a defined symbol, the definition
could be used, instead of '*'.

Nam had said otherwise.

[Khong Dong]

On Monday, 9 December 2019 14:14:58 UTC-7, Peter Percival wrote:
> Khong Dong wrote:
> > On Monday, 9 December 2019 13:58:24 UTC-7, Peter Percival wrote:
> >> Khong Dong wrote:
> >>> On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:
> >>>
> >>>> However, since Q and PA have the same language, yes,
> >>>> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
> >>>> '*' is an undefined symbol.
> >>>
> >>> '*' is a formal language symbol in those contexts, you were told.
> >>>
> >>>> Suppose we were using a theory _other than_ Q or PA or
> >>>> Presburger arithmetic, in which, instead of addition '+'
> >>>> and multiplication '*', we had addition '+' and squaring
> >>>> '^2'.
> >>>
> >>> But why does one need that stupid, idiotic, trolling "Suppose",
> >>> when investigating an argument about the difficult ABC
> >>> conjecture!
> >>>
> >>> In the context of the ABC conjecture it's idiotic and wrong the
> >>> way you redefined '*' in term of '*'. Only s crank would utter
> >>
> >> Jim hasn't redefined '*' in terms of '*'.
> >
> > That's was just a typo which I've already made a caveat-correction.
> >
> > But again it was idiotic of him to redefine ''*' in term of '+' in this context
>
> What point was Jim making when he redefined '*' in term of '+'?

That's his point, so you can ask him to clarify for you.
But whatever it is it's idiotically trolling and technically incorrect in _this thread context_
of multiplication, primes, rad(abc), natural numbers, ABC conjecture.

[Khong Dong]

In so far as the concept of the natural numbers is concerned, what Jim Burns has been
consistently delusional about is somehow he believes he could redefine the concept
of natural numbers so that it would become just Presburger arithmetic having a lone binary operation namely addition!

What a crackpot delusion Jim Burns has shown posts after posts!

[Jim Burns]

On 12/9/2019 6:06 PM, Khong Dong wrote:

> In so far as the concept of the natural numbers is
> concerned, what Jim Burns has been consistently
> delusional about is somehow he believes he could
> redefine the concept of natural numbers so that it
> would become just Presburger arithmetic having a
> lone binary operation namely addition!

I use _two_ operators, '+' and '^2'.

Axioms:

x + 0 = x

x + Sy = S(x + y)

0^2 = 0

(x + 1)^2 = x^2 + x + x + 1

Definition:
(x*y = z) :<->
((x+y)^2 = x^2 + z + z + y^2)

Presburger arithmetic does not have a 1-ary
squaring operator '^2'.

> What a crackpot delusion Jim Burns has shown posts
> after posts!

It must be so disturbing for you to encounter someone
who counts to _two_ correctly.

[Khong Dong]

Are you or are you not talking about multiplication and prime
numbers definitions of the natural numbers expressed in the
language of arithmetic? Why are you trolling here, Jim?

[Jim Burns]

Not expressed in this language: L(0,1,+,*).
Expressed in this language: L(0,1,+,^2).

> Why are you trolling here, Jim?

<JB<NN>>

|>> [...] and _without_ the language multiplication
|>> symbol '*' (which of course must be required to
|>> define prime numbers).
|>

[Peter Percival]

Khong Dong wrote:
> On Monday, 9 December 2019 14:14:58 UTC-7, Peter Percival wrote:
>> Khong Dong wrote:
>>> On Monday, 9 December 2019 13:58:24 UTC-7, Peter Percival wrote:
>>>> Khong Dong wrote:
>>>>> On Monday, 9 December 2019 05:39:41 UTC-7, Jim Burns wrote:
>>>>>
>>>>>> However, since Q and PA have the same language, yes,
>>>>>> ~(2<2) is also a formula in L(PA). In L(Q) = L(PA),
>>>>>> '*' is an undefined symbol.
>>>>>
>>>>> '*' is a formal language symbol in those contexts, you were told.
>>>>>
>>>>>> Suppose we were using a theory _other than_ Q or PA or
>>>>>> Presburger arithmetic, in which, instead of addition '+'
>>>>>> and multiplication '*', we had addition '+' and squaring
>>>>>> '^2'.
>>>>>
>>>>> But why does one need that stupid, idiotic, trolling "Suppose",
>>>>> when investigating an argument about the difficult ABC
>>>>> conjecture!
>>>>>
>>>>> In the context of the ABC conjecture it's idiotic and wrong the
>>>>> way you redefined '*' in term of '*'. Only s crank would utter
>>>>
>>>> Jim hasn't redefined '*' in terms of '*'.
>>>
>>> That's was just a typo which I've already made a caveat-correction.
>>>
>>> But again it was idiotic of him to redefine ''*' in term of '+' in this context
>>
>> What point was Jim making when he redefined '*' in term of '+'?
>
> That's his point, so you can ask him to clarify for you.

No need: I know why Jim was defining * in terms of +. I was wondering
if you do.

[Peter Percival]

^2 isn't a symbol of Presburger arithmetic, not even a definable one.

[Peter Percival]

Jim Burns wrote:
> On 12/9/2019 4:14 PM, Peter Percival wrote:
>> Khong Dong wrote:
>
>>> But again it was idiotic of him to redefine ''*'
>>> in term of '+' in  this context
>>
>> What point was Jim making when he redefined '*'
>> in term of '+'?
>
> First, I need to correct Nam.
> I defined '*' in terms of '+' and '^2'.
>
> The point I was making was that it may not be required
> to use the symbol '*' in the definition of prime numbers.

I got that and was wondering if Nam had!

[Jim Burns]

On 12/9/2019 8:09 PM, Peter Percival wrote:
> Jim Burns wrote:
>> On 12/9/2019 4:14 PM, Peter Percival wrote:
>>> Khong Dong wrote:

>>>> But again it was idiotic of him to redefine ''*'
>>>> in term of '+' in  this context
>>>
>>> What point was Jim making when he redefined '*'
>>> in term of '+'?
>>
>> First, I need to correct Nam.
>> I defined '*' in terms of '+' and '^2'.
>>
>> The point I was making was that it may not be required
>> to use the symbol '*' in the definition of prime numbers.
>
> I got that and was wondering if Nam had!

Ah. Sorry about my obtuseness.

In my posts, here at least, I have been striving mightily
to be as _obvious_ as is humanly possible. My goal is to
_sometimes_ be understood by _certain posters_ .
"You may say -- I'm a dreamer ... "

A side effect is that I get a little bit myopic with
respect to most subtlety. And some posters may get
the impression that I think they're a moron. Or that
I myself am a moron.

(sigh) It's a cost that I'm aware of, but that I accept,
in hopes of the possibility of maybe communicating,
someday.

So, as you can see, I have a plan.

[Jim Burns]

Defining Prime() in L(0,1,+,^2) and without '*':

Prime(x) :<->
~Ey,Ez:( ~(y=x) & ~(z=x) & (x+y)^2 = x^2 + z + z + y^2 )

The axioms for '^2' are

0^2 = 0

(x+1)^2 = x^2 + x + x + 1

[Khong Dong]

[You might as well _name_ it "nonPrime(x)" or "JB_Idiotic_Prime(x)" too - and there's zero difference in it being the most stupid abuse of the familiar
mathematical word "prime"?]

Yeah. With _your_ *new* language of "arithmetic", defining a *new* kind of
"primes" I feel you could shed some light on Mochizuki's alleged proof of
the ABC conjecture that the majority of real PhDs, mathematicians are still
baffled about. No?

Idiotic troll.

[Peter Percival]

They're not a new kind of primes. Jim's primes are 2, 3, 5, ..., as usual.

[Khong Dong]

On Monday, 9 December 2019 18:09:04 UTC-7, Peter Percival wrote:
> Jim Burns wrote:

> > The point I was making was that it may not be required
> > to use the symbol '*' in the definition of prime numbers.

So what's the point of idiotically making _that_ a point in talking about
ABC conjecture which is about the so-called the standard language structure for
the language of arithmetic L(0,S,+,*,<) - already including the language symbol '*'?

> I got that and was wondering if Nam had!

The song of two idiotic trolling birds.

[Khong Dong]

Idiotic, isn't it? When we talk about formalizing _linguistic DEFINITION_ of
"prime" we simply don't use axioms. Troll.

[Jim Burns]

I think this is another instance of you missing the
distinction between
"I (NN) don't see how this is true"
and
"This is not true".

>>> Yeah. With _your_ *new* language of "arithmetic",
>>> defining a *new* kind of
>>
>> They're not a new kind of primes.
>> Jim's primes are 2, 3, 5, ..., as usual.
>
> Idiotic, isn't it? When we talk about formalizing
> _linguistic DEFINITION_ of "prime" we simply don't
> use axioms.

Oh, "we" simply don't use axioms?
That would explain where your Mathematical Relativity
results come from.

Here is how we can define Prime() in L(0,1,+,*):
| Prime(x) :<->
| ~Ey,Ez:( ~(y=x) & ~(z=x) & x*y = z )

Do you think we _don't_ use the axioms below?
If you think we don't, how do we know '*' means 'multiply'?

x*0 = 0

x*(y+1) = x*y + x

This is how I tell you '^2' means 'squared'

0^2= 0

(x+1)^2 = x^2 + x + x + 1

> Troll.
>>> Idiotic troll.

[Khong Dong]

Inane babbling voice of a well known sci.logic crank.

>
> >>> Yeah. With _your_ *new* language of "arithmetic",
> >>> defining a *new* kind of
> >>
> >> They're not a new kind of primes.
> >> Jim's primes are 2, 3, 5, ..., as usual.
> >
> > Idiotic, isn't it? When we talk about formalizing
> > _linguistic DEFINITION_ of "prime" we simply don't
> > use axioms.
>
> Oh, "we" simply don't use axioms?
> That would explain where your Mathematical Relativity
> results come from.

Idiotic whining.

>
> Here is how we can define Prime() in L(0,1,+,*):
> | Prime(x) :<->
> | ~Ey,Ez:( ~(y=x) & ~(z=x) & x*y = z )
>
> Do you think we _don't_ use the axioms below?
> If you think we don't, how do we know '*' means 'multiply'?

You're simply grossly confused between syntactical part and
semantical part of a formalism investigation. Go and learn
some basics of FOL _first_ and in time you might be able to
make less mistakes.

[Peter Percival]

Of course we do. We use the properties of * which are encapsulated in
the axioms.

[Khong Dong]

Ignorant, clueless utterance. First, go and learn some basics about _syntactical_
formalism, then with some luck you might be educated on the matter. Troll.

[Peter Percival]

Try to be polite and argue your point rather than attacking people who
disagree with you.

[Khong Dong]

I did technically argue my point here many times but you and JB have kept being
ignorant of the technical matters and then went on trolling making idiotic babling
and attacks.

Anyway, if you want to dwell on technical mattered why don't you in a straightforward
manner acknowledge my meta theorems of this thread are correct, or disprove them.

Can you do that in no more than three posts?

[Khong Dong]

In fact, in _very early on_ here:

https://groups.google.com/d/msg/sci.logic/wsW51PKClQA/qmPvxTWRBAAJ
https://groups.google.com/d/msg/sci.logic/wsW51PKClQA/Xx6m2KmVBAAJ

I was polite and very helpful in saying, e.g:

"But it's *not*: it's a formal language symbol in the context."

but has this prevented JB and PP from trolling multiple times in my thread?

Of course _NOT_ !

[Peter Percival]

There you are then! If you can be polite _very early on_ why not keep
it up throughout the thread?

>
> "But it's *not*: it's a formal language symbol in the context."
>
> but has this prevented JB and PP from trolling multiple times in my thread?

Jim is always a model of good manners.

[Khong Dong]

To me that sounds very much an idiotic lie because this flamewar started out with JB's
idiotic "Techno-babble" attack on his opponent. Did you tell JB he should have
been polite there? If not, why not?

> > Of course _NOT_ !
> >
> >> Anyway, if you want to dwell on technical mattered why don't you in a straightforward
> >> manner acknowledge my meta theorems of this thread are correct, or disprove them.
> >>
> >> Can you do that in no more than three posts?

Well, are you going to disprove my MT-A and MT-B or are you going to troll again?

[Peter Percival]

A criticism of your posts is not an attack on you.

[Khong Dong]

Idiotic ranting, again. Troll.

>
> >>> Of course _NOT_ !
> >>>
> >>>> Anyway, if you want to dwell on technical mattered why don't you in a straightforward
> >>>> manner acknowledge my meta theorems of this thread are correct, or disprove them.
> >>>>
> >>>> Can you do that in no more than three posts?
> >
> > Well, are you going to disprove my MT-A and MT-B or are you going to troll again?

Where are your disproofs? Idiotic troll.

[Khong Dong]

On Tuesday, 10 December 2019 11:35:27 UTC-7, Khong Dong wrote:

> Anyway, if you want to dwell on technical mattered why don't you in a straightforward
> manner acknowledge my meta theorems of this thread are correct, or disprove them.
>
> Can you do that in no more than three posts?

Well, would you (PP, JB) be able to agree my MT-A, MT-B theorems in the op are correct?
If not, what would be the technical errors you could see?

[Jim Burns]

On 12/10/2019 4:58 PM, Khong Dong wrote:
> On Tuesday, 10 December 2019 11:35:27 UTC-7,
> Khong Dong wrote:

>> Anyway, if you want to dwell on technical mattered
>> why don't you in a straightforward manner acknowledge
>> my meta theorems of this thread are correct,
>> or disprove them.
>>
>> Can you do that in no more than three posts?

Surely, how many posts it would take would be determined
at least in part by how you would react to criticism.

For example, I recently made a straightforward point about
defining prime numbers without '*', which I didn't see as
controversial. Your reaction has been metaphorical
foaming-at-the-mouth rage, and we are far, far past three
posts in our discussion of that point.

And that was for a point which, at one point, you seemed
to want to shrug off as unimportant. Presumably there
is some higher Defcon Level to which you can go to,
in the event of an _important_ criticism. It strikes me
as highly unlikely that any such point would be dealt
with in only three posts -- because of your reaction.

(Anyway, why _three_ posts?)

> Well, would you (PP, JB) be able to agree my MT-A, MT-B
> theorems in the op are correct? If not, what would be
> the technical errors you could see?

The only technical error I noticed is the one I have
mentioned to you: if '*' is definable in the theory
you are using, it is not the case that it is required
to use '*' in defining prime numbers.

I haven't read your MT-A or your MT-B. From things
you have said, I gathered the impression that you
preferred it that way.

Was the impression I gathered incorrect? Are you asking
for my critique of MT-A and MT-B?

[Khong Dong]

On Tuesday, 10 December 2019 18:45:10 UTC-7, Jim Burns wrote:
> On 12/10/2019 4:58 PM, Khong Dong wrote:

> > Well, would you (PP, JB) be able to agree my MT-A, MT-B
> > theorems in the op are correct? If not, what would be
> > the technical errors you could see?
>
> The only technical error I noticed is the one I have
> mentioned to you: if '*' is definable in the theory
> you are using, it is not the case that it is required
> to use '*' in defining prime numbers.

Your answer here is nonsensical to me, since my question was about my MT-A,
MT-B but which you've just said right below you "haven't read your MT-A or
your MT-B"!

>
> I haven't read your MT-A or your MT-B. From things
> you have said, I gathered the impression that you
> preferred it that way.

Yes. The whole point of this thread is my two meta theorems MT-A or your MT-B
from the very first post.

> Was the impression I gathered incorrect? Are you asking
> for my critique of MT-A and MT-B?

Yes. Are they correct to you? If not, can you prove your _detailed_ disproofs?
(You certainly can respond that you don't have an answer).

[Khong Dong]

On Tuesday, 10 December 2019 20:45:29 UTC-7, Khong Dong wrote:
> On Tuesday, 10 December 2019 18:45:10 UTC-7, Jim Burns wrote:
> > On 12/10/2019 4:58 PM, Khong Dong wrote:
>
> > > Well, would you (PP, JB) be able to agree my MT-A, MT-B
> > > theorems in the op are correct? If not, what would be
> > > the technical errors you could see?
> >
> > The only technical error I noticed is the one I have
> > mentioned to you: if '*' is definable in the theory
> > you are using, it is not the case that it is required
> > to use '*' in defining prime numbers.
>
> Your answer here is nonsensical to me, since my question was about my MT-A,
> MT-B but which you've just said right below you "haven't read your MT-A or
> your MT-B"!
>
> >
> > I haven't read your MT-A or your MT-B. From things
> > you have said, I gathered the impression that you
> > preferred it that way.
>

> Yes. The whole point of this thread is my two meta theorems MT-A and MT-B

> from the very first post.

Edited for typo correction in the above.

[Jim Burns]

On 12/10/2019 10:45 PM, Khong Dong wrote:
> On Tuesday, 10 December 2019 18:45:10 UTC-7,
> Jim Burns wrote:
>> On 12/10/2019 4:58 PM, Khong Dong wrote:

>>> Well, would you (PP, JB) be able to agree my MT-A, MT-B
>>> theorems in the op are correct? If not, what would be
>>> the technical errors you could see?
>>
>> The only technical error I noticed is the one I have
>> mentioned to you: if '*' is definable in the theory
>> you are using, it is not the case that it is required
>> to use '*' in defining prime numbers.
>
> Your answer here is nonsensical to me, since my question
> was about my MT-A, MT-B but which you've just said right
> below you "haven't read your MT-A or your MT-B"!

I think it's a pretty solid prediction that you will
pick some reason, any reason to call whatever I say
"nonsense".

Your most recent reason for calling my definition
without '*' of Prime() nonsense[1] was that I included
in my post axioms for the squaring operator '^2'
which I used instead of using multiplication '*'.

Your attitude does not encourage me to do you any favors.
If I read your MT-A and MT-B, it won't be because you
asked me to.

[1]
Technically, "called it 'idiotic'" is more apt than
"called it 'nonsense'".

<NN<JB<<NN>>>>

|>>>>
|>>>> [You might as well _name_ it "nonPrime(x)" or
|>>>> "JB_Idiotic_Prime(x)" too - and there's zero difference
|>>>> in it being the most stupid abuse of the familiar
|>>>> mathematical word "prime"?]
|>

|> I think this is another instance of you missing the
|> distinction between
|> "I (NN) don't see how this is true"
|> and
|> "This is not true".
|
| Inane babbling voice of a well known sci.logic crank.
|

</NN<JB<<NN>>>>

>> I haven't read your MT-A or your MT-B. From things
>> you have said, I gathered the impression that you
>> preferred it that way.
>
> Yes. The whole point of this thread is my two meta
> theorems MT-A or your MT-B from the very first post.

What you're saying 'Yes" to is you preferring that I _not_
read MT-A or MT-B.

<sigh> It took several posts from me just to get you to
notice that I use a squaring operator '^2' and not only '+'.

It takes an unreasonable amount of effort to get you (NN)
to notice _anything_ you won't like noticing.

>> Was the impression I gathered incorrect? Are you asking
>> for my critique of MT-A and MT-B?
>
> Yes. Are they correct to you? If not, can you prove your
> _detailed_ disproofs? (You certainly can respond that
> you don't have an answer).

I can also just not do the favor you ask of me.
Anyway, why _three_ posts?

[Khong Dong]

On Wednesday, 11 December 2019 06:47:23 UTC-7, Jim Burns wrote:
> On 12/10/2019 10:45 PM, Khong Dong wrote:
> > On Tuesday, 10 December 2019 18:45:10 UTC-7,
> > Jim Burns wrote:

> >> Are you asking
> >> for my critique of MT-A and MT-B?
> >
> > Yes. Are they correct to you? If not, can you prove your
> > _detailed_ disproofs? (You certainly can respond that
> > you don't have an answer).
>
> I can also just not do the favor you ask of me.
> Anyway, why _three_ posts?

Pick up any number (Googleplex, Aleph_1, Aleph_2, ... Mahlo, ...): just go on
with your troll-dancing. You've been good at it.

[Khong Dong]

Or you, Jim Burns, can choose to be a decent sci.logic poster and present your
disproofs of MT-A and MT-B. (If you have any naturally.)

[Khong Dong]

Fwiw, I've been somewhat anxious on the possibility that you or sci.logic
would disprove MT-A, MT-B. But I think the possibility looks more and more
remote now.

[Peter Percival]

Khong Dong wrote:

> Well, are you going to disprove my MT-A and MT-B or are you going to troll again?

I don't think I'm obliged to, but if I were I'd want to know (at least)
what you mean by something being impossible to verify. There are,
aren't there, different kinds of impossibility and different kinds of
verification.

[Jim Burns]

On 12/11/2019 3:04 PM, Khong Dong wrote:
> On Wednesday, 11 December 2019 12:51:55 UTC-7,
> Khong Dong wrote:

>> Or you, Jim Burns, can choose to be a decent sci.logic
>> poster

I've read your posts, Nam. Do you consider yourself
to be a "decent poster"?

I'm satisfied to not be whatever you demonstrate
with your posts.

>> and present your disproofs of MT-A and MT-B.
>> (If you have any naturally.)
>
> Fwiw, I've been somewhat anxious on the possibility
> that you or sci.logic would disprove MT-A, MT-B.
> But I think the possibility looks more and more
> remote now.

Something that you need to keep in mind:
That you don't know of a disproof does not imply
the existence of a proof.

This is especially true when no one is looking for
a disproof.

[Khong Dong]

I did define that one way or the other many times so we've passed that
point and you've just happened to ask the question too late. Now I could
only give a hint below but that's it.

Let U be this singleton set: U = {{}}, and let S be the mets statement:
S = "It is possible to verify U has at least one element". The negation of S
is neg(S) = "It is not possible to verify U has at least one element".

If you understand what S means then you should understand what neg(S)
means, and what "impossible to verify" I've meant through these years.
If you still don't understand then sorry but you'd not be sufficiently strong,
competent enough on foundation matters for me to help. That's it.

[Peter Percival]

Why do you think that the primes as Jim Burns defines them are a "new"
(your word) kind of prime? The things that satisfy Jim's definition are
all and only the primes as usually understood.

[Khong Dong]

On Wednesday, 11 December 2019 14:22:42 UTC-7, Jim Burns wrote:
> On 12/11/2019 3:04 PM, Khong Dong wrote:
> > On Wednesday, 11 December 2019 12:51:55 UTC-7,
> > Khong Dong wrote:
>
> >> Or you, Jim Burns, can choose to be a decent sci.logic
> >> poster
>
> I've read your posts, Nam.

So, Jim, are my MT-A, MT-B. true to you?

[Khong Dong]

On Wednesday, 11 December 2019 14:23:40 UTC-7, Peter Percival wrote:
> Khong Dong wrote:

> > Idiotic, isn't it? When we talk about formalizing _linguistic DEFINITION_ of
> > "prime" we simply don't use axioms. Troll.
>
> Why do you think that the primes as Jim Burns defines them are a "new"
> (your word) kind of prime?

Ah. Stupid trolling from PP again.

[Khong Dong]

If you don't know of your answer that's perfectly fine too
(you could just let us know in that case).

[Peter Percival]

Khong Dong wrote:
> On Wednesday, 11 December 2019 13:47:35 UTC-7, Peter Percival wrote:
>> Khong Dong wrote:
>>
>>> Well, are you going to disprove my MT-A and MT-B or are you going to troll again?
>>
>> I don't think I'm obliged to, but if I were I'd want to know (at least)
>> what you mean by something being impossible to verify. There are,
>> aren't there, different kinds of impossibility and different kinds of
>> verification.
>
> I did define that one way or the other many times so we've passed that
> point and you've just happened to ask the question too late. Now I could
> only give a hint below but that's it.
>
> Let U be this singleton set: U = {{}}, and let S be the mets statement:
> S = "It is possible to verify U has at least one element". The negation of S
> is neg(S) = "It is not possible to verify U has at least one element".
>
> If you understand what S means

I don't. There is more than one kind of possibility.

It just so happens that I have been thinking about this -

"Let there be given a topological space and, if X is a subset of it, let
Cl(X) be the closure of X. Let there be a set of propositions closed
under the possibility operator <>, and let the truth values of the
propositions be subsets of that topological space. Then

If Y is the truth value of proposition P,
then Cl(Y) is the truth value of <>P."

It would be a strange coincidence if your notion of possible was of that
kind. But even if it were, the precise notion of <> depends on what
kind of topological space it is. And then there are all the kinds of
possibility that aren't of that kind.

[Peter Percival]

What does your bad manners gain you?

Do you see that you calling me a stupid troll does not somehow nullify
my question?

[Khong Dong]

On Wednesday, 11 December 2019 15:38:24 UTC-7, Peter Percival wrote:
> Khong Dong wrote:
> >
> > Let U be this singleton set: U = {{}}, and let S be the mets statement:
> > S = "It is possible to verify U has at least one element". The negation of S
> > is neg(S) = "It is not possible to verify U has at least one element".
> >
> > If you understand what S means
>
> I don't.

Not surprised. Trolls usually don't have much knowledge competency.

[Khong Dong]

On Wednesday, 11 December 2019 15:40:21 UTC-7, Peter Percival wrote:
> Khong Dong wrote:
> > On Wednesday, 11 December 2019 14:23:40 UTC-7, Peter Percival wrote:
> >> Khong Dong wrote:
> >
> >>> Idiotic, isn't it? When we talk about formalizing _linguistic DEFINITION_ of
> >>> "prime" we simply don't use axioms. Troll.
> >>
> >> Why do you think that the primes as Jim Burns defines them are a "new"
> >> (your word) kind of prime?
> >
> > Ah. Stupid trolling from PP again.
> >
>
> What does your bad manners gain you?

Stop troll-whining.

> Do you see that you calling me a stupid troll does not somehow nullify
> my question?

Who cares a troll's question? Stupid troll (as you are) deserves being called such. Troll.

[Peter Percival]

Does anyone known to you in the real world read your posts here? Your
brother perhaps? If so, don't you feel a bit embarrassed?

[Khong Dong]

On Wednesday, 11 December 2019 18:41:21 UTC-7, Peter Percival wrote:
> Khong Dong wrote:
> > On Wednesday, 11 December 2019 15:40:21 UTC-7, Peter Percival wrote:
> >> Khong Dong wrote:
> >>> On Wednesday, 11 December 2019 14:23:40 UTC-7, Peter Percival wrote:
> >>>> Khong Dong wrote:
> >>>
> >>>>> Idiotic, isn't it? When we talk about formalizing _linguistic DEFINITION_ of
> >>>>> "prime" we simply don't use axioms. Troll.
> >>>>
> >>>> Why do you think that the primes as Jim Burns defines them are a "new"
> >>>> (your word) kind of prime?
> >>>
> >>> Ah. Stupid trolling from PP again.
> >>>
> >>
> >> What does your bad manners gain you?
> >
> > Stop troll-whining.
> >
> >> Do you see that you calling me a stupid troll does not somehow nullify
> >> my question?
> >
> > Who cares a troll's question? Stupid troll (as you are) deserves being called such. Troll.
> >
>
> Does anyone known to you in the real world read your posts here?

Hard to tell. Who knows, other fora might be aware of this thread. We'll see.

> If so, don't you feel a bit embarrassed?

I think the trolling cranks like you and JB are the only ones who should feel
embarrassed. Seriously.

(For me ... I feel glad that so far no errors have been found against my MT-A and MT-B, since you asked).

[Khong Dong]

Having read my op, Jim, are my MT-A, MT-B true?

[Jim Burns]

On 12/12/2019 12:42 AM, Khong Dong wrote:
> On Wednesday, 11 December 2019 14:28:00 UTC-7,
> Khong Dong wrote:
>> On Wednesday, 11 December 2019 14:22:42 UTC-7,
>> Jim Burns wrote:
>>> On 12/11/2019 3:04 PM, Khong Dong wrote:
>>>> On Wednesday, 11 December 2019 12:51:55 UTC-7,
>>>> Khong Dong wrote:

>>>>> Or you, Jim Burns, can choose to be a decent
>>>>> sci.logic poster
>>>
>>> I've read your posts, Nam.

||> I've read your posts, Nam. Do you consider yourself
||> to be a "decent poster"?
||>
||> I'm satisfied to not be whatever you demonstrate
||> with your posts.

>> So, Jim, are my MT-A, MT-B. true to you?
>
> Having read my op, Jim, are my MT-A, MT-B true?

Let me clarify: I have not read all of your posts.

I have read enough of your posts to have a good idea of
the style of interaction with which you meet any questioning
of your brilliance. If I had forgotten, you would have
reminded me recently, when I committed the apparently
horrible crime of pointing out something that -- judging
from your reaction -- you hadn't known.

I don't imagine you've forgotten your recent behavior,
either. I suspect that your asking for my comments on
MT-A and MT-B _in order to be outraged_ as you so often
are in response to my posts.

Is that what this is about? That's fine, then.
Be outraged about this:

This is a definition of prime number that does not
use multiplication '*'. Instead, it uses a squaring-
-operator '^2'.

Prime(z) :<->
~Ex,Ey:( ~(x=z) & ~(y=z) & (x+y)^2 = x^2 + z + z + y^2 )

The axioms for the squaring-operator are:

0^2 = 0

(x+1)^2 = x^2 + x + x + 1

You're welcome.

[Khong Dong]

On Thursday, 12 December 2019 17:07:02 UTC-7, Jim Burns wrote:
> On 12/12/2019 12:42 AM, Khong Dong wrote:
> > On Wednesday, 11 December 2019 14:28:00 UTC-7,
> > Khong Dong wrote:
> >> On Wednesday, 11 December 2019 14:22:42 UTC-7,
> >> Jim Burns wrote:
> >>> On 12/11/2019 3:04 PM, Khong Dong wrote:
> >>>> On Wednesday, 11 December 2019 12:51:55 UTC-7,
> >>>> Khong Dong wrote:
>
> >>>>> Or you, Jim Burns, can choose to be a decent
> >>>>> sci.logic poster
> >>>
> >>> I've read your posts, Nam.
>
> ||> I've read your posts, Nam. Do you consider yourself
> ||> to be a "decent poster"?
> ||>
> ||> I'm satisfied to not be whatever you demonstrate
> ||> with your posts.
>
> >> So, Jim, are my MT-A, MT-B. true to you?
> >
> > Having read my op, Jim, are my MT-A, MT-B true?
>
> Let me clarify: I have not read all of your posts.

But did you read the very first post where the meta theorems
MT-A, MT-B were written and proved?

If you did read, are my MT-A, MT-B true to you?

If you didn't read the op at all, why are you here in this thread?

[Khong Dong]

On Saturday, 7 December 2019 12:00:49 UTC-7, Khong Dong wrote:
> On Saturday, 7 December 2019 04:23:46 UTC-7, Rupert wrote:
> > On Saturday, December 7, 2019 at 10:30:18 AM UTC+1, Khong Dong wrote:
> > > On Friday, 6 December 2019 01:48:02 UTC-7, Khong Dong wrote:
> > > > On Friday, 6 December 2019 01:26:54 UTC-7, Rupert wrote:
> > > > > On Friday, December 6, 2019 at 6:20:05 AM UTC+1, Khong Dong wrote:
> > > > > > According to this Wiki link:
> > > > > >
> > > > > > https://en.wikipedia.org/wiki/Abc_conjecture
> > > > > >
> > > > > > the familiar statement of the ABC conjecture would involve functions of reals
> > > > > > (which of course aren't first order expressible). Let's call this Wiki link
> > > > > > version the canonical version of the ABC conjecture.
> > > > > >
> > > > > > Let's consider below the, say, NN (Natural-Number) version of the conjecture
> > > > > > that does NOT refer to non-natural numbers (i.e. real numbers):
> > > > > >
> > > > > > (NN) There are finitely many ABC numeral-triples ([a],[b],[c])'s where:
> > > > > >
> > > > > >
> > > > > > - In general, [n] is a numeral denoting a natural number n.
> > > > > >
> > > > > > - [c] denotes an even number c which is a counter example of Goldbach
> > > > > > conjecture.
> > > > > >
> > > > > > - rad([a][b][c]) < [c]
> > > > >
> > > > > As I pointed out before, the Wikipedia article states that this is known to be false.
> > > >
> > > > Wrong. Because you snipped the last part of the NN version definition!
> > > > Perhaps you want to read that version _in full_ .
> > >
> > > So now you (Rupert) would understand what the NN version states, right?
> > > Any rate, are meta statements MT-A, MT-B true to you? If not, what would be
> > > the specific errors you'd see?
> >
> > In your "NN" version, you appear to attach importance to the fact that you are referring to numerals rather than natural numbers. I am not able to see how this would make any difference to the truth-value of your claims. Your "NN" version appears to me to be equivalent to the statement that Goldbach's conjecture is true.
> >
> > I see no reason to believe MT-A or MT-B.
> >
> > From your purported proof of MT-A, for example:
> >
> > "By syntactical form, an even [c] denoting a counter example of Goldbach
> > conjecture doesn't logically exclude the distinct possible existence of an ABC numeral-triple ([a],[b],[c]) where both [a],[b] aren't S0 and one of [a],[b] must necessarily not denote a prime, and rad([a][b][c]) < [c]."
> >
> > Well, how do you know that?
>
> Consider the even number c = 128 = 2^7, expressed in the below 4 equation-cases
> involving some ABC triples:
>
> 1. 3 + 5^3 = 2^7 [ rad(abc)<c ]
> 2. 3^2 + (7*17) = 2^7 [ rad(abc)>c ]
> 3. 19 + 109 = 2^7 [ rad(abc)>c ]
> 4. 31 + 97 = 2^7 [ rad(abc)>c ]
>
> First, 128 is not a counter example of Goldbach conjecture because of cases
> 3 and 4, in each of which *it's syntactically possible* to express the entire
> equation with numerals and _without_ the language multiplication symbol '*'

> (which of course must be required to define prime numbers).
>

> Otoh, suppose 128 were a counter example of Goldbach conjecture then both cases
> 3, 4 can be ignored and only cases 1 and 2 would be considered as part of a
> finitude both versions of the ABC conjecture are about.
>
> But both of cases 1. and 2. are syntactically of the same language-symbol form
> - where the language multiplication symbol '*' must by First Order _Logic_
> formalism necessarily be present on the left side of the logical symbol '='
> for the equation to be syntactically valid (i.e. well formed).
>
> Consequently, if there is a counter example of Goldbach conjecture c (and at
> least it's not known that we can logically exclude that distinct possibility)
> being relevant to the ABC conjecture (of either version), then there would have
> to be an ABC numeral-triple, hence ABC numeral-equation, like 1. and 2. (with
> or without hypothetically assuming 128 were a counter example of Goldbach
> conjecture): an (ABC-triple) equation of that restricted form, where '*' is
> absolutely absent.
>
> In summary, yes: we do you know _that_ - syntactically _that_ must be the case.
>
> The important note here is that the common form of 3 and 4 here is ignored by
> the alluded finitude in both versions of the ABC conjecture _anyway_ : since we
> would have the undesired inequality rad(abc)>c.
>
> It's an important note since it signifies we can examine the relationship
> between Goldbach, cGC, ABC conjectures as a purely syntactical symbol game or
> manipulation. And this examination would portend some disturbing facts about
> the incoherent semantics-juxtaposition of the two FOL-with-equality logical
> symbols - 'E' (Existence quantifier) and '=' (Equality), about it being invalid
> to logically, semantically equate _general_ "equivalence" with "equality".
>
> It would portend the collapse of the mathematical truth-reasoning envisioned
> (philosophized) by Plato, built by Tarski, and invalidly codified by Gödel.
>
> IUTT (Inter-Universal Teichmüller Theory) would be just an unfortunate and
> accidental smokescreen masking of this collapse. How far have SME's on both
> sides of the alleged ABC proof been able to agree even on such simple
> (undergraduate level) issue of whether or not two isomorphic objects/structures
> can categorically be treated the "same"?
>
> Not that far at all apparently!

To Rupert et al.:

I've looked and looked again many times at MT-A and MT-B, and NN_S being a
subset of Canonical_S is the final nail in a coffin of any hope that the
current alleged proof of the ABC conjecture is valid, independent of whether
or not IUUT is properly understood.

The only way I'd be wrong is if the CGC-trichotomy proof turns out to be wrong
or invalid. But one shouldn't bet on that unless one could find a blatant
counter example of the law of thought named here as HP (an impossibility in and
of itself), or unless one could prove the phrase "when Choice is necessary" is
a logically meaningless notion (another impossibility).

[Julio Di Egidio]

On Friday, 13 December 2019 07:39:31 UTC+1, Khong Dong wrote:

> The only way I'd be wrong is if the CGC-trichotomy proof turns out to be wrong
> or invalid.

It's plain gibberish, as you have been told already and repeatedly.

You are simply unteachable and unreasonable, and your absurd English, and
your nonsensical syntactical constructions, and your semantic messing up
of all abstraction levels (you should have written [1], not S0) and so on:
you are one of the few genuine cranks around here...

Julio

[Jim Burns]

I answered this last question upthread.
Maybe you didn't read the answer, so maybe I should
repeat it over and over until you do. Or something.

Would this help?
Try to remember what you were thinking when you posted
this in WM's thread "Three incompatible facts require
dark numbers":

<NN<JB>>
|> What allows us to talk about induction on finite
|> vN ordinals is a set-theoretic definition of finite
|> vN ordinals.
|>
|> Sy = (y U {y})
|
| Wrong. 'S' isn't a set-theoretic language symbol for
| your "set-theoretic definition" to be valid.

Which reminds me.

You have questions you would like answered.
Maybe, if you answered mine, I would look more favorably
on yours. It's worth a shot, isn't it?

As I asked before,

S is a _defined_ set-theoretic symbol. You quoted the
definition.

'U' and '{ ... }' are also _defined_ set-theoretic symbols.
They're defined in terms of 'e' (membership).

Are 'U' and '{ ... }' also invalid?

[Khong Dong]

Ah. So you're in this thread as a troll-dancer, not a decent
sci.logic poster trying to respond about the two theorems
of the thread. No. surprise.

[Khong Dong]

Listen carefully child Julio. It's true we don't teach about numeral S0 to
kindergarten children as you are with innocent mind full of precious
blankness. When you grow up a little more - if that ever happens - I'll
teach you one or two things about the formal Language of Arithmetics
where you'd learn there exist infinitely many numerals like S0.

For now just go home before your parents worrying too much about
you hanging around with the residential troll-dancers (like JB, etc...).

[Julio Di Egidio]

On Friday, 13 December 2019 21:44:30 UTC+1, Khong Dong wrote:
> On Friday, 13 December 2019 01:33:55 UTC-7, Julio Di Egidio wrote:
> > On Friday, 13 December 2019 07:39:31 UTC+1, Khong Dong wrote:
> >
> > > The only way I'd be wrong is if the CGC-trichotomy proof turns out to be wrong
> > > or invalid.
> >
> > It's plain gibberish, as you have been told already and repeatedly.
> >
> > You are simply unteachable and unreasonable, and your absurd English, and
> > your nonsensical syntactical constructions, and your semantic messing up
> > of all abstraction levels (you should have written [1], not S0) and so on:
> > you are one of the few genuine cranks around here...
>

> Listen carefully child Julio.

You are indeed as stupidly abusive as you are totally clueless.

> It's true we don't teach about numeral S0

The point, you utter moron, I'll say it again, was not messing up with the
levels abstraction, where you should have written [1] instead of S0: S0 is
just one possible encoding, i.e. an implementation detail, not even anything
canonical about it.

And that was the minor of all the points I had raised, you useless MORON.

EOD.

Julio

[Khong Dong]

The point is you failed to recognize the 2 versions of the ABC conjecture and
even failed to know what S0 means, and now are trying to dishonestly save
your face by making character assassinations against your opponent.

Where's your courage of a man?

> And that was the minor of all the points I had raised, you useless MORON.

The only point here is, courage wise, you're not half of the half-man you used
to be at birth.

> Julio

[Julio Di Egidio]

On Friday, 13 December 2019 23:05:41 UTC+1, Khong Dong wrote:
> On Friday, 13 December 2019 14:02:13 UTC-7, Julio Di Egidio wrote:
> > On Friday, 13 December 2019 21:44:30 UTC+1, Khong Dong wrote:
> > > On Friday, 13 December 2019 01:33:55 UTC-7, Julio Di Egidio wrote:
> > > > On Friday, 13 December 2019 07:39:31 UTC+1, Khong Dong wrote:
> > > >
> > > > > The only way I'd be wrong is if the CGC-trichotomy proof turns out to be wrong
> > > > > or invalid.
> > > >
> > > > It's plain gibberish, as you have been told already and repeatedly.
> > > >
> > > > You are simply unteachable and unreasonable, and your absurd English, and
> > > > your nonsensical syntactical constructions, and your semantic messing up
> > > > of all abstraction levels (you should have written [1], not S0) and so on:
> > > > you are one of the few genuine cranks around here...
> > >
> > > Listen carefully child Julio.
> >
> > You are indeed as stupidly abusive as you are totally clueless.
> >
> > > It's true we don't teach about numeral S0
> >
> > The point, you utter moron, I'll say it again, was not messing up with the
> > levels abstraction, where you should have written [1] instead of S0: S0 is
> > just one possible encoding, i.e. an implementation detail, not even anything
> > canonical about it.
>
> The point is you failed to recognize the 2 versions of the ABC conjecture

I have never said they weren't two version: you are just too retarded to admit
that you have rather been clearly told, and ot just by me, that NEITHER is the
ABC conjecture. But indeed you are too fucked up in the main brain to even
allow yourself to read.

> and even failed to know what S0 means

Initially I thought you meant group theory, LOL, for how out of place that was.
Initially, you stupid fuck, I corrected myself 2 minutes later: but of course
you delusional moron will keep pretending another version of reality... where I
didn't explain how fucked up even that is: [a][b][c][and the plain denial then
the ad hominem, that is indeed the only game that you retard can play.]

>, and now are trying to dishonestly save
> your face by making character assassinations against your opponent.

Denial, then ad hominem: that is you stupid fuck, and your 3 year old retorts.

> Where's your courage of a man?

Moron, delusional, abusive, stupid, spamming CRANK.

Too retarded to even appreciate the difference between unknown and unknowable...

*Plonk*

Julio

[Peter Percival]

You do seem to think that posters to sci.logic are obliged to answer
your questions while you avoid answering theirs. We are not circus
animals here to jump through hoops on your command.

>

[Khong Dong]

No known sci.logic posters (including the late TF whom we'd always
respect) would answer all questions, but that doesn't you or Jim Burns
less of a dancing troll (and crank). Troll.