Ping "Me" - On infinity and corresponding impossible-to-know arithmetic truths.

[Khong Dong]

Not 100% sure why I ping you - "Me" - perhaps partially because it strikes
me that you're technical, objective, deliberate, fair poster, in evaluating
technical matters, or perhaps because it's a shot in the dark I'm trying,
or perhaps because you said in the other thread:

"It's well known that we have to be careful if dealing with the notion
of /infinity/".

but whatever the reasons, I'm just wondering if you could comment on the
technical correctness of the three meta theorems below (which I posted in
MO last month or so) - basically proving it's logically impossible (there
being no meta proofs) to know, verify the truth value of:

cGC <-> "There being infinitely counter examples of Goldbach Conjecture".

Certainly the presentation isn't perfect given it's as long as only one post
but I'll try my best to answer any question you might have.

Thank you in advance,

Best Regards,

-Nam Nguyen

===========================================================================

===> Definitions

GC <-> Goldbach Conjecture
cGC <-> "There are infinitely many counter examples of Goldbach Conjecture"
SR <->(~GC /\ ~cGC)

knowable(S) <=> There's a FINITE meta mathematical proof that S is (non-vacuously) true.

undecidable(S) <=> neg(knowable(S)) and neg(knowable(neg(S))).

Caveat: This is meta level truth-"undecidable", _not_ FOL provability-"undecidable".

intrinsic(S) <=> TRUE(S), for all the meaning of S.

===> Assumptions

The definition of prime(p)is free of occurrences of the language constant successor function symbol - required for any proof by induction:

prime(p)<->((~p=0)/\~Az[p*z=z]/\((p=x*y)->(Az[x*z=z]\/Az[y*z=z])))

Let's let U be the set of Von Neumann's (or Zermelo's) finite ordinals, or some kind of (infinite) union between the two sets.

There exists an uncountable set K of language structures M's having these following properties:

1. All M's have the same domain - the universe of discourse U, mentioned
above.

2. A natural number n is interpreted by a term of the language - a numeral -
denoting an element e_n in U but is neither the element nor the term
(numeral)!

For the economy of typing, we might conveniently use natural numbers and
elements interchangeably but that does *not* take away the said
distinction.

3. The truth of a formula F in any of them can also be interpreted the truth
of F of (about) the natural numbers.

4. Any element in U that is interpreted as a non-odd prime natural number n
in any of the structures is interpreted as the same non-odd prime natural
number n in all of the structures.

5. Any element in U that is interpreted as an odd prime natural number in
any of the structures is also interpreted as an odd prime natural number
in all of the structures: but NOT necessarily the same odd prime natural
number!

===> Lemmas

Lemma 1: If p is a variable denoting an odd prime (element) in M, the
uniqueness of the numeral of p - numeral(p) - is unknown.

Lemma 2: If e is a free variable ranging over the set of even elements in U
of M, and p is a prime element bounded to e, the uniqueness of
numeral(p)is unknown.

These two lemmas are due mostly to the fact that the definition of prime(p)
mentioned above is free of occurrences of the language constant symbol S.

===> Meta Theorems

Meta Theorem 1 - If GC is true in an M of K, it's not knowable.

Proof:

Let E be the open interval { e | even(e) /\ (e >= E0) } where even(E0),
E0 >= 4, and is known to be a sum of two primes. It's sufficient to show
that if GC is true over E, it's not knowable. But indeed, we have to show
there exist primes p1, p2, for an even e in E, such that:

numeral(e) = numeral(p1) + numeral(p2)

but by Lemma 2, the uniqueness of numeral(p1) and numeral(p2) are unknown,
hence the expected numeral equality is not knowable. QED.

***

Meta Theorem 2 - If there are finitely many counter examples of Goldbach Conjecture it's not knowable.

Proof:

There being finitely many counter examples of Goldbach Conjecture means that
there exists an open interval E = { e | even(e) /\ (e >= E0) } previously
mentioned, in the complement of which finitely many counter-example
elements (numbers) would be found. But by Meta Theorem 1, the existence of E
is not knowable, hence so is its complement. QED.

***

Meta Theorem 3 - If cGC is true but GC is false, it's not knowable.

Proof:

In this case there are infinitely many even (counter-example) elements e's,
and infinitely many corresponding prime elements p's such that:

- p = maximum prime to the left a general counter example e
- numeral(e) = numeral(p) + numeral(o)

for some odd(o).

But by Lemma 1, numeral(p) is not knowable (not unique) hence the numeral
equality is not knowable. QED.

[Rupert]

I have a question about this.

You're talking about an infinite family of structures for the first-order language of arithmetic, right, all of them with the same domain of discourse U. Where U, if I understand correctly, is some fixed infinite subset of the union of the set of all finite von Neumann ordinals and the set of all
finite Zermelo ordinals.

And I assume "non-odd prime natural number n" means, a natural number n which is not an odd prime number.

So, what exactly does it mean to say that an element of U is interpreted as a natural number n in one of the structures? That means that, in that structure, the referent of the numeral for n is equal to the element of U in question, does it?

Also later on with your statement of Lemma 1.

"Lemma 1: If p is a variable denoting an odd prime (element) in M, the
uniqueness of the numeral of p - numeral(p) - is unknown."

So M is one of the structures in the collection, right? Now, are you assuming that every element of U is the referent of a numeral relative to M? And what does it mean to say the uniqueness of the numeral of p is unknown? What if I stipulate that the numeral of p is SSSSS0? Then surely I know the numeral of p, and that seems to be consistent with the hypotheses of your lemma. So what can this lemma really be saying?

[Khong Dong]

(Thank you for your post. I'll respond later.)

[Khong Dong]

On Thursday, 21 March 2019 04:22:16 UTC-6, Rupert wrote:

It means the natural number n here is either, Zero, One, Two, or a multiple.

>
> So, what exactly does it mean to say that an element of U is interpreted
> as a natural number n in one of the structures? That means that, in that
> structure, the referent of the numeral for n is equal to the element of U
> in question, does it?

Right. It means exactly as Assumption 2 states:

<quote>

2. A natural number n is interpreted by a term of the language - a numeral -
denoting an element e_n in U but is neither the element nor the term
(numeral)!

</quote>

It goes without saying that, in general, each different M in K would induce
a different interpretation on the individual element e_n in U.

> Also later on with your statement of Lemma 1.
>
> "Lemma 1: If p is a variable denoting an odd prime (element) in M, the
> uniqueness of the numeral of p - numeral(p) - is unknown."
>
> So M is one of the structures in the collection, right? Now, are you
> assuming that every element of U is the referent of a numeral relative to
> M?

Correct. For instance, the Zermelo element z_3 = {{{{}}} could be a referent
for the (prime) numeral SSSS0 for some M but could also be a referent for
the (prime) numeral SSSSSS0 for some different M'.

> And what does it mean to say the uniqueness of the numeral of p is
> unknown?

Because there's an 1-many relationship between an element e_n and the
numerals relative to which M of K we choose: in the example above you have
the element z_3 but it's invalid to say its numeral must necessarily be
SSSS0, or SSSSSS0, ... _without referring to a *particular* M_ in K.

> What if I stipulate that the numeral of p is SSSSS0? Then surely I know
> the numeral of p, and that seems to be consistent with the hypotheses
> of your lemma. So what can this lemma really be saying?

Here you've involved another issue: p as stated in Lemma 1 is a free
_free variable term of the language_ and all you'd know (would be given in
the Lemma) is that p is an odd prime element: but you can't sufficiently
know the specificity of M.

If you stipulate that the numeral of p be the particular numeral SSSSS0 then
you've sufficiently specified the underlying M, and in that case p as an
element (or as denoting an element) would have to be (or refer to) a
specific element. Iow, p would no longer be a free variable that Lemma 1 is
talking about: it'd be a language-defined symbol for an individual constant
element, a syntactical alias for SSSSS0.

[Rupert]

Okay. Well, I wouldn't mind seeing the proof of Lemma 2 then.

[Khong Dong]

Sure. Proof:

By Assumption 4, in a *general M* of K, the *general* even element e of U,
in Lemma 2, would be interpreted as the same natural even number, say, N_e.
Let's hypothesize a bit further that N_e is greater than the natural number
Two. Consequently, there must exist a prime element p in U and in M that's
a surrogate for (is interpreted to be) the natural prime number N_p next to
the left of N_e. In this sense, p is bound to the left of e.

But by Assumption 5 the bound (the less-than order binding) between e and p
is subject to (relative to) which specific M being the underlying structure:
the same prime element p might be bound that way to a different even element
e' being a surrogate for a different natural even number N_e', hence
numeral(p) can't be unique, by Assumption 2. Iow, the language numeral(p)
can't be interpreted to be both N_e and N_e'.

In summary, though e is being interpreted to be the identical natural even
number N_e in any M, e is general (free) hence Assumption 5 will ensure
the said bound between e and p can't be logically established.

Otoh, if e is specific, numeral(e) would be specified and consequently the
numeral S_p of the natural prime number N_p next to the left of that
specific N_e, would be known. And by Assumption 5 it wouldn't matter which
prime element p it is; numeral(p) = S_p.

But again in Lemma 2, M and e aren't specific so S_p can't be known and by
Assumption 5 virtually any prime element can be that next-to-the-left prime
element. QED.

[Rupert]

I'm sorry, I don't understand this sentence.

[Khong Dong]

Perhaps you could say which part of it that would prevent you? Would you
understand all the antecedent (required) assumptions (particularly
Assumption 4) prior to Lemma 2?

[Rupert]

I think the main problem is that your use of the word "general" is unclear.

[Khong Dong]

In this context "general" would mean "non specific" or "free" (variable).

[Peter Percival]

Hardly. Neither M nor e are variables. You say yourself that M is a
structure and e a member of a structure's domain.

--
"He who will not reason is a bigot;
he who cannot is a fool;
he who dares not is a slave."
- Sir William Drummond

[Rupert]

I'm afraid that doesn't help me very much.

It's possible that with a bit of work I could come up with an attempted interpretation of what you mean. But usually when I try to do that you tell me that I shouldn't put words into your mouth and should just stick with the words you use. But the problem, you see, is that I find the words you use to be incomprehensible. So, you know, there you go.

[Khong Dong]

Why? You don't know what a free variable means? I could look up Shoenfield's and
explain it for you if you'd like.

>
> It's possible that with a bit of work I could come up with an attempted interpretation of what you
> mean.

Not really. If otoh you don't understand what free variable means then of course you wouldn't be
able to understand my meta proof but I can't do much to help. Like I said somewhere I don't explain the very trivial. Sorry.

[Peter Percival]

Ha ha ha! You really are a hoot!

[Khong Dong]

Seriously, I can't discuss mathematical logic matters with you (Rupert) if
you're unable to understand, as you seem to have been, what the phrase
"free variable" would mean, or would complain the phrase "doesn't help me
very much".

[Rupert]

I know what a free variable is.

But that is no particular help in understanding your sentence.

[Rupert]

Well, no-one's saying you have to. Do you want me to stop replying, do you?

[Khong Dong]

That's kind of a childish question which I won't answer. (It's fair though to say that I didn't specifically invite you in to this conversation).

[Me]

On Saturday, March 23, 2019 at 3:46:35 PM UTC+1, Khong Dong wrote:

> That's kind of a childish question which I won't answer.

Maybe, but maybe not.

> (It's fair though to say that I didn't specifically invite you in to this
> conversation).

Actually, Rupert is the expert here. I'm next to nothing in comparison. Hence you should be GLAD that he engaded in this discussion, seriously.

And I'm quite sure that he knows the meaning of the term "free variable".

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

[Khong Dong]

Then I don't think any mathematician or sci.logic poster in the right mind
would understand what you've complained on Lemma 2:

"Lemma 2: If e is a free variable ranging over the set of even elements
in U of M, and p is a prime element bounded to e, the uniqueness of
numeral(p)is unknown."

Rupert:

"Okay. Well, I wouldn't mind seeing the proof of Lemma 2 then."

NN:

"Sure. Proof:

By Assumption 4, in a *general M* of K, the *general* even element e of U,
in Lemma 2, would be interpreted as the same natural even number, say,
N_e."

Rupert:

"I'm sorry, I don't understand this sentence."

NN:

"Perhaps you could say which part of it that would prevent you? Would you
understand all the antecedent (required) assumptions (particularly
Assumption 4) prior to Lemma 2?"

Rupert:

"I think the main problem is that your use of the word "general" is
unclear."

NN:

"In this context "general" would mean "non specific" or "free" (variable)."

Rupert:

"I'm afraid that doesn't help me very much."

====================================================

Your "main problem" in understanding the Lemma 2 proof's is that the proof
would refer to the variable e (the Lemma 2's "e is a free variable ranging
over the set of even elements") as a _general_ variable which you complained

<quote>
the word "general" is unclear
</quote>

but which I has clearly explained "general" means "free": so you or any
student can replace e being a "general" even number here by e being a free variable denoting an even element in U in M. It's bizarre that you couldn't
even understand that much!

So either your complaint:

<quote>

the word "general" is unclear

</quote>

wasn't genuine or for some inexplicable reasons you forgot what a "free" variable would mean!

Btw, Rupert, the below inferrecne :

P(n) -> An[P(n)]

is possible by an inference rule named UG - universal - _GENERALIZATION_ !

Are you going to complaint TOO something like:

"I think the main problem is that the use of the word 'GENERALIZATION'
here is unclear"

?

Good grief!

[Khong Dong]

On Saturday, 23 March 2019 09:07:08 UTC-6, Me wrote:
> On Saturday, March 23, 2019 at 3:46:35 PM UTC+1, Khong Dong wrote:
>
> > That's kind of a childish question which I won't answer.
>
> Maybe, but maybe not.
>
> > (It's fair though to say that I didn't specifically invite you in to this
> > conversation).
>
> Actually, Rupert is the expert here. I'm next to nothing in comparison. Hence you should be GLAD that he engaded in this discussion, seriously.

Please don't take my being critical of his post here as a sign I didn't
appreciate his involving in the discussion. This sort of childish question
shouldn't;' have been thrown in at all. In any rate I couldn't and wouldn't
answer non-technical question as such.

If we all could Q/A in a manner of to-the-point that'd be great, I'd think.

[Khong Dong]

Anyway, the technical proof for Lemma 2 isn't that convoluted, Rupert.

Let M, M' be free variables over K. We don't know whether or not M = M' but
we do know by Assumption 4 the successions of even elements in both M and
M' are identical. But by Assumption 5 the same can't be logically said of
the successions of (odd) primes in M and M'.

That's all Lemma 2 really says.

[Khong Dong]

For instance, consider the below Zermelo's elements:

Even elements:

z_6 = {{{{{{{}}}}}}}
z_8 = {{{{{{{{{}}}}}}}}}

Prime elements:

z_3 = {{{{}}}}
z_11 = {{{{{{{{{{{{}}}}}}}}}}}}

Your M could have these 2-tuples in the binary relation symbolized by '<':

(z_6, z_8), (z_3, z_6), (z_3, z_11), (z_11, z_8), ...

But your M' otoh could have:

(z_6, z_8), (z_11, z_6), (z_11, z_3), (z_3, z_8), ...

Note the identical 2-tuple (z_6, z_8) exists in both cases - but the
remaining 2-tuples are different.

Incidentally, the numeral for z_3 would be different between M and M', ditto
for that of z_11, as implicated in Lemma 2!

Hope this has helped clarifying proof of Lemma 2 you've requested.

[Khong Dong]

Note that from Lemma 2, we can show it's invalid to _even assume_
Shoenfield's N8 [1] be true in any M of K (hence it'd be invalid to assume
it be true of the natural numbers) - independent of whether or not PA is
_syntactically_ consistent.

This in effect is one of the gateways into showing Completeness is invalid.

[1]

N8 <-> (x < Sy <-> (x < y \/ x = y))

[Khong Dong]

So you (Rupert) didn't undstand the proof of Lemma 2 because:

"the main problem is that your use of the word "general" is unclear. "

Ok, that "main problem" has been clearly explained. Would you now
understand Lemma 2? If not, what would be the _new main problem_
this time?

[Rupert]

You wrote "I can't discuss mathematical logic matters with you" and now you say that the response "Do you want me to stop replying" is childish. I think you need to re-think this.

[Rupert]

We now have a problem with your claim that it's invalid to assume N8 is true in any M of K.

[Peter Percival]

Khong Dong wrote:
> Not 100% sure why I ping you - "Me" - perhaps partially because it strikes

Has Franz responded?

[Khong Dong]

But, seriously, _showing_ that's true would require key dependent knowledge
- Lemma 2 - and if you still don't understand the proof of Lemma 2 (which
you requested and I responded) it wouldn't be a fruitful showing.

So, one thing at a time, are we done with your request:

"Okay. Well, I wouldn't mind seeing the proof of Lemma 2 then."

?

[Khong Dong]

On Sunday, 24 March 2019 09:37:33 UTC-6, Peter Percival wrote:
> Khong Dong wrote:
> > Not 100% sure why I ping you - "Me" - perhaps partially because it strikes
>
> Has Franz responded?

I'm busy so if you could ask him and file a fact-finding report to sci.logic
afterward that'd be great. Thanks.

[Khong Dong]

In principle, I could have added your now request on N8 as Meta Theorem 4,
but its proof still has to go through all the Assumptions 1-5, Lemmas 1-2,
not to mention a few more on some kind of logic non-logical symbol
polymorphism (akin to C++ abstract function polymorphism), and then of
course HP/Choice.

But one thing at a time: let's have a closure on Lemma 2 and even on cGC
(Meta Theorem 3) which has been quite outstanding (for 9 years or so).

[Peter Percival]

Peter Percival wrote:
> Khong Dong wrote:
>> Not 100% sure why I ping you - "Me" - perhaps partially because it
>> strikes
>
> Has Franz responded?
>
>

respond*ed*

[Peter Percival]

I ask because I don't see a reply from him, but I doubt that all Usenet
servers carry identical info.

[Rupert]

Well, I believe that the statement of Lemma 2 is not particularly coherent, but I think that probably with a bit of effort I could come up with a paraphrase of it (and a correct proof) which more or less captures the essence of what you mean, but you probably wouldn't agree that it was an accurate paraphrase of what you really mean. So I'm not sure where that leaves us, really.

[Khong Dong]

Tell you what. Fwiw, this "Ping Me" thread exists somewhere near the end of
my cGC time in sci.logic: my thread-op isn't perfect by any means but it's
sufficiently concrete and has some concrete illustrating examples for any
serious arguer to pursue an objective analysis. My cGC posting on MO too
has come near the same end (I've decided).

Basically I've used these fora for my own (technical) purposes: testing the
strength and weakness of my MR, cGC, physics mathematization, ... efforts.
I've got from these fora the "data" I wanted, and my time should be spent
to the next (and final) phase of my knowledge-formalized meta-mathematics
efforts.

With that then sure, this time I'm willing to hear from you suggested
rewording "which more or less captures the essence of what you [NN] mean" -
hopefully. In the end we might not still see eye-to-eye and I might still
disagree with your rewording but, you know, even for serious mathematics
effort in the real world out there, accomplished mathematicians have bitter
disagreed more that we might have here, imho.

But I'd thank in advance for your offer one way or the other: especially
if the paraphrases offering a correct reflection of my MR, cGC impossibility
effort or what not, "(and a correct proof)" of course.

So let's hear the paraphrase; the posting is yours now.

[Me]

On Sunday, March 24, 2019 at 6:08:29 PM UTC+1, Peter Percival wrote:

> I ask because I don't see a reply from him [...]

I'm glad that Rupert engaded in this discussion. This type of stuff is beyond my pay-grade. :-P

[Rupert]

Okay. So you wrote

"Lemma 2: If e is a free variable ranging over the set of even elements
in U of M, and p is a prime element bounded to e, the uniqueness of
numeral(p)is unknown."

The best I can do by way of making sense of this is as follows.

Suppose that we have a function from the set of elements of U with the property that they are the referent, for each structure M in K, of the numeral of an even number (necessarily the same numeral for each structure in the collection K), to the set of elements which are the referent, for each structure M in K, of the numeral of an odd prime number (not necessarily the same odd prime number for each structure in the collection K). Then there exist elements p of the range of this function and two structures M and M' in K with the property that the unique numeral denoting p relative to the structure M is not equal to the unique numeral denoting p relative to the structure M'.

That's the best I can do by way of making sense of what you mean.

[Khong Dong]

I've read your note above and with a small risk of misinterpreting its
content I'd say yes that's what Lemma 2 would say in different wording-
rendition.

Ultimately both versions say of the same fact: in this context, the length
of numeral(p) is not absolute but relative to a frame of reference: M,
M', etc...

This is akin to Lorentz length contraction which says that the uniqueness
of the length of the same object p is unknown in a frame of reference M
moving with the constant speed v - where v is a free variable.

Or in FOL we'd say the expression p=0 doesn't have a unique meaning: it has
multiple meanings, again p being free.

[Khong Dong]

In reality, my final "official" version of Lemma 2 would employ some meta
mathematics notations (set up as convention), such as my "undecidable(S)"
defined in the op post. So whether or not I completely use your phrasing
which has the word "function" and I have to see how easy it can be written
using these notations.

I.e., per se it's not a matter of phrasing-preference for preference sake.
The question though is whether or not you'd _understand_ the _essence_ of
Lemma 2, even with your version, toward the proof of Meta Theorems 1-3,
about cGC? Or do you think there's an irreversible logic flaw in all this?

[Khong Dong]

I think I now see the motivation why you've protested Lemma 2's wording as
being vague but that's neither due to my or your fault: it's due to Choice
when Choice is necessary.

Of course a choice function c has to be unique in its own existence
logically, but when Choice is necessary, there's another kind of (non
logical) uniqueness coming into question, in the following sense.

If c1, c2 are choice functions when Choice is necessary over the same
collection K, then it's unknown if they are uniquely the same or not
uniquely so (i.e. they're different). Iow, if Choice is necessary and you
say c is "unique", I'd ask "Really? _Which_ one have you claimed as
"unique", as you'd have c', c'', ... alongside?",

[Rupert]

Meh.

[Rupert]

c1, c2 are choice functions are they?

For which family of sets?

I'm pretty doubtful that you actually know what a choice function is.

[Khong Dong]

Instead of making vague (virtually random) complaining (and what
looks like innuendo or character assassination) can you just elaborate
_in details_ what you're trying to complain about?

[Rupert]

Well, I asked you choice functions for which family of sets, you find that unclear do you?

[Khong Dong]

Isn't that an aimless, non sequitur question, because in that _example_ about c1, c2
I didn't refer to any specific family?

[Khong Dong]

So, in summary, would you now understand Lemma 2 and we can move on to
Meta Theorem 1 (If GC is true in an M of K, it's not knowable)?

[Rupert]

So that illustrates my point that you don't know what a choice function is, doesn't it? Because a choice function is by definition a choice function for some family of sets.

As I've told you a number of times, I offer maths tutoring for 30 euros per hour. You definitely need it.

[Rupert]

Can I just clarify, all the structures in K are elementarily equivalent?

[Khong Dong]

On Monday, 25 March 2019 00:55:55 UTC-6, Rupert wrote:
> So that illustrates my point that you don't know what a choice function is, doesn't it? Because a choice function is by definition a choice function for some family of sets.

And a poster really .... really knowledgeable, competent about FOL wouldn't
ask "which family" upon hearing the word "Choice", "choice function". Why
did you ask "which family"?

>
> As I've told you a number of times, I offer maths tutoring for 30 euros per hour. You definitely need it.

Playing good old song "pay me" huh? Instead, you should have googled around
to find the word "family" when you saw the words "Choice", "choice
function".

[Khong Dong]

On Monday, 25 March 2019 00:58:10 UTC-6, Rupert wrote:
> Can I just clarify, all the structures in K are elementarily equivalent?

No need to add "elementarily equivalent" to your paraphrase. If you're unable
already to understand Lemma 2 as it was posted, no amount of paraphrasing from
you would help.

[Rupert]

My reason for asking the question about elementary equivalence, is not for the purpose of further clarifying Lemma 2 but to engage with your statement of Meta Theorem 1.

[Rupert]

On Monday, March 25, 2019 at 4:02:35 PM UTC+1, Khong Dong wrote:
> On Monday, 25 March 2019 00:55:55 UTC-6, Rupert wrote:
> > So that illustrates my point that you don't know what a choice function is, doesn't it? Because a choice function is by definition a choice function for some family of sets.
>
> And a poster really .... really knowledgeable, competent about FOL wouldn't
> ask "which family" upon hearing the word "Choice", "choice function". Why
> did you ask "which family"?

Because I wanted to know, and it's obviously a perfectly reasonable and indeed quite natural question?

[Khong Dong]

But since the proof of Meta Theorem 1 would make use of Lemma 2, is Lemma 2 correct to you?
(I think I've asked for your understanding confirmation on this Lemma).

[Rupert]

So I wrote an attempted paraphrase of your Lemma 2 and you were willing to accept that, so with that understanding of what Lemma 2 means, yes with the right hypotheses that statement is correct.

[Khong Dong]

Ok great: so we seem to have a common understanding of Lemma 2, it having been rendered in
two different phrasings.

Now, on the "elementarily equivalent" request: No there's _no need_ to rephrase K with it.

Given all the Assumptions and Lemmas on K such equivalence has to be proven to be true,
false, or invalid to claim one way or the other. In fact if you paraphrase K's dehinition as such
you'd be talking about a different meta theorem, not Meta Theorem 1.

[Rupert]

It's not a question of trying to re-phrase anything, it was just something I wanted to know.

[Khong Dong]

On Monday, 25 March 2019 13:43:42 UTC-6, Rupert wrote:
> It's not a question of trying to re-phrase anything, it was just something I wanted to know.

You're right; I read it too quick there Sorry about that.

[Khong Dong]

Any rate, would you agree Meta Theorem 1 is true?

[Rupert]

If you were to be so extraordinarily kind as to answer my query about whether all the structures in K are elementarily equivalent, then I might be in a position to answer that.

[Khong Dong]

The short answer is it's impossible to know this truth value one way or the
other, hence it's invalid to claim it one way or the other. All it'd take is
for a formula (e.g. cGC) to be of the nature of being impossible to know the
truth value in any M in K. The long answer is below.

Let:

- MT be the acronym for Meta Theorem
- M, M' be free over K
- EQ(M,M') <=> (M, M' are elementarily equivalent)

then the following is Meta Theorem 4:

Meta Theorem 4 (MT4):

For any M, M', M'' in K:

((MT1 and MT2 and MT3) => undecide(M |= cGC)) => undecide(EQ(M',M''))

So you'd have to know the truth of MT1 (Meta Theorem 1) first before being
able to know if "all the structures in K are elementarily equivalent"
could even have a truth value in meta level - which it does _not_ by MT4 above.

[Khong Dong]

On Monday, 25 March 2019 21:54:03 UTC-6, Khong Dong wrote:
> On Monday, 25 March 2019 19:36:05 UTC-6, Rupert wrote:
> > On Monday, March 25, 2019 at 9:45:04 PM UTC+1, Khong Dong wrote:
> > > On Monday, 25 March 2019 13:53:40 UTC-6, Khong Dong wrote:
> > > > On Monday, 25 March 2019 13:43:42 UTC-6, Rupert wrote:
> > > > > It's not a question of trying to re-phrase anything, it was just something I wanted to know.
> > > >
> > > > You're right; I read it too quick there Sorry about that.
> > >
> > > Any rate, would you agree Meta Theorem 1 is true?
> >
> > If you were to be so extraordinarily kind as to answer my query about
> > whether all the structures in K are elementarily equivalent, then I might
> > be in a position to answer that.
>
> The short answer is it's impossible to know this truth value one way or the
> other, hence it's invalid to claim it one way or the other. All it'd take is
> for a formula (e.g. cGC) to be of the nature of being impossible to know the
> truth value in any M in K. The long answer is below.
>
> Let:
>
> - MT be the acronym for Meta Theorem
> - M, M' be free over K
> - EQ(M,M') <=> (M, M' are elementarily equivalent)

Correction: just "M, M' be in K", _not_ "be free over K".

[Peter Percival]

Khong Dong wrote:
> On Monday, 25 March 2019 19:36:05 UTC-6, Rupert wrote:
>> On Monday, March 25, 2019 at 9:45:04 PM UTC+1, Khong Dong wrote:
>>> On Monday, 25 March 2019 13:53:40 UTC-6, Khong Dong wrote:
>>>> On Monday, 25 March 2019 13:43:42 UTC-6, Rupert wrote:
>>>>> It's not a question of trying to re-phrase anything, it was just something I wanted to know.
>>>>
>>>> You're right; I read it too quick there Sorry about that.
>>>
>>> Any rate, would you agree Meta Theorem 1 is true?
>>
>> If you were to be so extraordinarily kind as to answer my query about
>> whether all the structures in K are elementarily equivalent, then I might
>> be in a position to answer that.
>
> The short answer is it's impossible to know this truth value one way or the
> other, hence it's invalid to claim it one way or the other. All it'd take is
> for a formula (e.g. cGC) to be of the nature of being impossible to know the
> truth value in any M in K. The long answer is below.
>
> Let:
>
> - MT be the acronym for Meta Theorem
> - M, M' be free over K

What does M is free over K mean? I thought K was a set that had M as an
element. What does it mean to say of a structure that it is free over a
set that it is in?

> - EQ(M,M') <=> (M, M' are elementarily equivalent)
>
> then the following is Meta Theorem 4:
>
> Meta Theorem 4 (MT4):
>
> For any M, M', M'' in K:
>
> ((MT1 and MT2 and MT3) => undecide(M |= cGC)) => undecide(EQ(M',M''))
>
> So you'd have to know the truth of MT1 (Meta Theorem 1) first before being
> able to know if "all the structures in K are elementarily equivalent"
> could even have a truth value in meta level - which it does _not_ by MT4 above.
>
>

--
"He who will not reason is a bigot;
he who cannot is a fool;
he who dares not is a slave."
- Sir William Drummond

[Peter Percival]

Khong Dong wrote:
> On Monday, 25 March 2019 21:54:03 UTC-6, Khong Dong wrote:
>> On Monday, 25 March 2019 19:36:05 UTC-6, Rupert wrote:
>>> On Monday, March 25, 2019 at 9:45:04 PM UTC+1, Khong Dong wrote:
>>>> On Monday, 25 March 2019 13:53:40 UTC-6, Khong Dong wrote:
>>>>> On Monday, 25 March 2019 13:43:42 UTC-6, Rupert wrote:
>>>>>> It's not a question of trying to re-phrase anything, it was just something I wanted to know.
>>>>>
>>>>> You're right; I read it too quick there Sorry about that.
>>>>
>>>> Any rate, would you agree Meta Theorem 1 is true?
>>>
>>> If you were to be so extraordinarily kind as to answer my query about
>>> whether all the structures in K are elementarily equivalent, then I might
>>> be in a position to answer that.
>>
>> The short answer is it's impossible to know this truth value one way or the
>> other, hence it's invalid to claim it one way or the other. All it'd take is
>> for a formula (e.g. cGC) to be of the nature of being impossible to know the
>> truth value in any M in K. The long answer is below.
>>
>> Let:
>>
>> - MT be the acronym for Meta Theorem
>> - M, M' be free over K
>> - EQ(M,M') <=> (M, M' are elementarily equivalent)
>
> Correction: just "M, M' be in K", _not_ "be free over K".

Ok. Ignore my question about that. (You were probably going to anyway.)

>>
>> then the following is Meta Theorem 4:
>>
>> Meta Theorem 4 (MT4):
>>
>> For any M, M', M'' in K:
>>
>> ((MT1 and MT2 and MT3) => undecide(M |= cGC)) => undecide(EQ(M',M''))
>>
>> So you'd have to know the truth of MT1 (Meta Theorem 1) first before being
>> able to know if "all the structures in K are elementarily equivalent"
>> could even have a truth value in meta level - which it does _not_ by MT4 above.
>

[Rupert]

Okay well look. Getting back to your Meta Theorem 1. So the statement is "If GC is true in an M of K, it's not knowable", yeah? Well, knowable by whom? I for one don't even know what the definition of K is as yet.

[Khong Dong]

By _any_ mortal being existing in any universe in the realm of existence.
I'm almost certain you know what the English phrases " _NON_ transient ",
" _NON_ temporal" would mean.

Let me quote from the very first post of the thread:

<quote>

knowable(S) <=> There's a FINITE meta mathematical proof that S is (non-vacuously) true.

undecidable(S) <=> neg(knowable(S)) and neg(knowable(neg(S))).

Caveat: This is meta level truth-"undecidable", _not_ FOL provability-"undecidable".

</quote>

There being NO FINITE meta mathematical proof means just that: There is NO
FINITE meta mathematical proof. Period.

> I for one don't even know what the definition of K is as yet.

That's your (individual) transient not-knowable: one day going over the
thread you might know (as a non-transient distinct possibility).

So, would you now, after all these technical definitions, exemplifications,
elaborations, understand the truth of Meta Theorem 1?

Fwiw, face value, Meta Theorem 1 is a concrete technical resolution
on the century-old Goldbach Conjecture: next time if multi millions of
dollars/euros are offered as a prize to prove it _true_ you
(*general* "you") shouldn't even bother with.

And if anyone offers billion dollars/euros to prove cGC, a sick joke would be being played.

[Khong Dong]

On Tuesday, 26 March 2019 00:41:36 UTC-6, Peter Percival wrote:
> Khong Dong wrote:
> > On Monday, 25 March 2019 19:36:05 UTC-6, Rupert wrote:
> >> On Monday, March 25, 2019 at 9:45:04 PM UTC+1, Khong Dong wrote:
> >>> On Monday, 25 March 2019 13:53:40 UTC-6, Khong Dong wrote:
> >>>> On Monday, 25 March 2019 13:43:42 UTC-6, Rupert wrote:
> >>>>> It's not a question of trying to re-phrase anything, it was just something I wanted to know.
> >>>>
> >>>> You're right; I read it too quick there Sorry about that.
> >>>
> >>> Any rate, would you agree Meta Theorem 1 is true?
> >>
> >> If you were to be so extraordinarily kind as to answer my query about
> >> whether all the structures in K are elementarily equivalent, then I might
> >> be in a position to answer that.
> >
> > The short answer is it's impossible to know this truth value one way or the
> > other, hence it's invalid to claim it one way or the other. All it'd take is
> > for a formula (e.g. cGC) to be of the nature of being impossible to know the
> > truth value in any M in K. The long answer is below.
> >
> > Let:
> >
> > - MT be the acronym for Meta Theorem
> > - M, M' be free over K
>
> What does M is free over K mean? I thought K was a set that had M as an
> element. What does it mean to say of a structure that it is free over a
> set that it is in?

In this context it just means M is an individual term/variable denoting an element in K.

[Khong Dong]

I think at this juncture you should concede that Meta Theorem 1 is true, Rupert.
Even Torkel Franzen thought that in general the truth value of a Goldbach like statement
might be impossible to know.

[Peter Percival]

Khong Dong wrote:
> [...]

> Even Torkel Franzen thought that in general the truth value of a Goldbach like statement
> might be impossible to know.

Did he put his thought into writing? And, if yes, where?

[Khong Dong]

On Tuesday, 26 March 2019 10:48:19 UTC-6, Peter Percival wrote:
> Khong Dong wrote:
> > [...]
> > Even Torkel Franzen thought that in general the truth value of a Goldbach like statement
> > might be impossible to know.
>
> Did he put his thought into writing? And, if yes, where?

Don't have that Chapter PDF handy but it's in his "Incompleteness for the Dummy" or
something like that. But I'm confident Rupert wouldn't disagree with such an assessment.

[Rupert]

Okay, so what's the proof of Meta Theorem 1?

[Rupert]

Why on earth would I do that? Are you suggesting you've already posted a proof?

[Khong Dong]

Yes, I'm suggesting that: read the op of this thread.

[Khong Dong]

And it was "on earth" we agreed to understand the truth of Lemma 2, which would lead
right to the truth of Meta Theorem 1.

[Rupert]

Ah yes okay.

The "proof" in the OP is dribble.

[Khong Dong]

_Honestly_ can you in technical details explain your utterance here?

[Rupert]

Well, I believe that if what you wrote in the OP is your idea of a proof, then you don't know what a proof is and you need remedial tutoring if there is to be any chance of successful communication at all. Which I don't feel inclined to give for free, since I think it'll probably be a thankless and futile task, and I doubt you'll be willing to pay. So there you go.

[Peter Percival]

Khong Dong wrote:
> [...]

> Yes, I'm suggesting that: read the op of this thread.

It might be useful to look in a few mathematics books to see how
mathematics is written. While style is not everything, but it does
count a lot. There may be books with such titles as "how to write
proofs". If so, they might be helpful. Certainly Halmos's /How to
write mathematics/ is available on the interweb. Amazon found only
this:
https://www.amazon.co.uk/Introduction-Writing-Mathematical-Proofs-Upper-Level/dp/1547033665/ref=sr_1_25?s=books&ie=UTF8&qid=1553626300&sr=1-25&keywords=how+to+write+proofs.
(I'm not endorsing Amazon.)

[Khong Dong]

Such a dishonest and inquisitional statement. You were very much part of this
_meta mathematics proof_ when agrrering to the truth of Lemma 2.
You now just lack courage to admit your failure to see it'd lead to Meta Theorem 1.
Not surprised at all!

[Khong Dong]

On Tuesday, 26 March 2019 12:54:23 UTC-6, Peter Percival wrote:
> Khong Dong wrote:
> > [...]
> > Yes, I'm suggesting that: read the op of this thread.
>
> It might be useful to look in a few mathematics books to see how
> mathematics is written. While style is not everything, but it does
> count a lot. There may be books with such titles as "how to write
> proofs". If so, they might be helpful. Certainly Halmos's /How to
> write mathematics/ is available on the interweb. Amazon found only
> this:
> https://www.amazon.co.uk/Introduction-Writing-Mathematical-Proofs-Upper-Level/dp/1547033665/ref=sr_1_25?s=books&ie=UTF8&qid=1553626300&sr=1-25&keywords=how+to+write+proofs.
> (I'm not endorsing Amazon.)

Impressive trolling, Peter. Do you know anything about _meta mathematics proof_ at all?

[Khong Dong]

After all you're the one who placed a long distance (oversea?) call to my
college trying to verify my having a Bachelor degree in Mathematics,
apparently with the hope that had I lied, my meta mathematical proof on cGC
would be _somehow_ successfully counter-argued by you.

Seriously, Rupert. You should concede that Meta Theorem 1 is true - if
you're true to yourself.

[Rupert]

I don't know if I recall doing that. If I did make an international call then I probably would have done it using Skype. But I really don't recall having done that. I recall that you sent me an alleged scanned copy of the testamur of your degree but for some strange reason wanted me to agree to incredibly strong restrictions on what I could say in public about the email conversation with no apparent point to them.

I don't recall any phone conversation with your college, although I do recall doing a bit of Googling about it, and I don't currently know that any tertiary institution ever awarded you a degree in mathematics, and if they ever did then either the degree is an utterly worthless one that you can pay for without having to show any real academic attainment, or else you suffered some kind of brain damage since the time you completed the degree, as Peter Percival has quite correctly observed.

You obviously haven't offered the slightest reason to think that Meta Theorem 1 is true, given that the attempted "proof" of your OP is complete dribble, so obviously there is no good reason why I should concede that it is true.

[Khong Dong]

On Wednesday, 27 March 2019 09:58:08 UTC-6, Rupert wrote:
> On Wednesday, March 27, 2019 at 3:42:02 PM UTC+1, Khong Dong wrote:
> > On Tuesday, 26 March 2019 13:09:21 UTC-6, Khong Dong wrote:

> > > You now just lack courage to admit your failure to see it'd lead to Meta Theorem 1.
> > > Not surprised at all!
> >
> > After all you're the one who placed a long distance (oversea?) call to my
> > college trying to verify my having a Bachelor degree in Mathematics,
> > apparently with the hope that had I lied, my meta mathematical proof on cGC
> > would be _somehow_ successfully counter-argued by you.
> >
> > Seriously, Rupert. You should concede that Meta Theorem 1 is true - if
> > you're true to yourself.
>
> I don't know if I recall doing that. If I did make an international call
> then I probably would have done it using Skype. But I really don't recall
> having done that.

Really that short memory? But see below ...

> I don't recall any phone conversation with your college, although I do
> recall doing a bit of Googling about it,

Would you, Dr. McCallum, share with sci.logic your opinion as to whether
it's a simple forgetfulness, or you've just lied through your teeth with
your above statement, given the _fact_ that then you said the relevant
below:

Rupert:

"I had a go at ringing up the college that you mentioned to me by email.
I didn't have any luck. The person at the switchboard didn't know what
the procedure was for confirming that someone has indeed been awarded
a degree."

?

_HOW_ , Dr. McCallum, would "doing a bit of Googling about it" of yours
enable you to have "a go at ringing up the college", then _talk at length_
to "The person at the switchboard"?

> and I don't currently know that any tertiary institution ever awarded you
> a degree in mathematics,

That's your knowledge problem.

> and if they ever did then either the degree is an
> utterly worthless one that you can pay for without having to show any real
> academic attainment, or else you suffered some kind of brain damage since
> the time you completed the degree, as Peter Percival has quite correctly
> observed.

You and Peter better have a legal certificate (from some doctors) "quite
correctly" (your phrase) proving I've suffered some kind of brain damage:
when I email my college's office, they might contact you to make sure you're
not accusing the college having taken bribe money and offer "degree [...]
utterly worthless".

Note:

Have you heard about some education institutes having taken bribes in the
news lately? If not, read:

https://www.vox.com/2019/3/15/18264399/college-admissions-scandal-lori-loughlin-cheating-huffman


Obviously if your mentioned "brain damage" was only some kind of joke,
insult to me and wasn't true, then apparently your accusing my college
"degree is an utterly worthless one that you can pay" would be true, and I
think the college might want to discuss with you about your accusing them.

You have three days from this post to apologize to my college, sci.logic,
and me that your statement above about my college's "utterly worthless"
degree and my mental health doesn't reflect the best of your honesty. And I
wouldn't want anything less than that in your apology.

If I don't hear the apology from you, I'll contact my college to ask if
indeed they were and/or have been the kind of institutes taking money from
students, state and federal governments, while deliberately offering degrees
"utterly worthless [...] that you can pay for without having to show any
real academic attainment". I'm interested to know for myself if they indeed
had cheated me taking my government-guaranteed student loan money in such a
dishonest manner.

> You obviously haven't offered the slightest reason to think that Meta
> Theorem 1 is true, given that the attempted "proof" of your OP is complete
> dribble, so obviously there is no good reason why I should concede that
> it is true.

Though you might not believe it, you don't have the competency-knowledge to
understand mathematical foundation issues - I've alerted you many times.
But you've always jumped into this kind of foundation talks (even if I
didn't directly invite you) and became inquisitional in knowledge exchange.
Not my fault.

[Rupert]

The problem with all this bluster is that you asked me not to mention the name of the college in public.

Sure, if that degree certificate you sent me by email was real, then I'll gladly publicly state that that either you suffered from brain damage since the award of the degree or else that particular college is not a reputable degree-awarding institution, and then do your worst and see if I care. Except the joke is that you won't allow me to publicly state which college it is, isn't it?

[Rupert]

Anyway, I just looked again at the testamur you sent me, it makes no mention of the word "mathematics". And the college is a liberal arts college. So no, I am not accusing that particular college of fraud. I am just saying you are lying that anyone ever awarded you a degree in mathematics.

[Khong Dong]

We'll see. So, did you _actually_ make the (oversea?) call to my college
through your Skype or you cell phone?

[Khong Dong]

Oh yes you did with your:

"if they ever did then either the degree is an utterly worthless one that

you can pay for without having to show any real academic attainment".

But let's hear from them, since they awarded me with a Bachelor Degree in
Mathematics as an accredited institution.

> I am just saying you are lying that anyone ever awarded you a degree
> in mathematics.

Really? I'm almost certain my college would know whether or _not_ they truthfully, honestly did give me a real Bachelor Degree in Mathematics.
We'll see. But you've accused and insulted the college on a _couple_ of
things. Bye.

[Rupert]

The testamur you showed me was in fact a Bachelor of Arts. You refuse to let me publicly say what the name of the college is, so I don't know that they ever really did give you a Bachelor of Arts degree. The testamur you gave me does not make any mention of the word "mathematics".

But just to be absolutely clear here. Any tertiary institution which ever did award you any kind of qualification suggesting that you had some kind of competence in mathematics, and you haven't had any accident suffering brain damage since that time, then yeah I'm more than happy to go public with saying that the credentials of that tertiary institution need to be called into question.

However, this tertiary institution we are talking about here, (1) they never, to my knowledge, gave you a degree in mathematics, and (2) you won't allow me to say publicly what the name of the institution is.

> > I am just saying you are lying that anyone ever awarded you a degree
> > in mathematics.
>
> Really? I'm almost certain my college would know whether or _not_ they truthfully, honestly did give me a real Bachelor Degree in Mathematics.
> We'll see. But you've accused and insulted the college on a _couple_ of
> things. Bye.

See you.

I have not, in fact, made any negative comments about this college whose name you refuse to let me say in public, and you are lying that the testamur that you sent me by email said "Bachelor Degree in Mathematics". It said "Bachelor of Arts", and the college named was a liberal arts college. And the most likely reason why you won't let me publicly name who the college is is that the testamur you sent me was a forgery and you are aware that that is a criminal matter.

[Rupert]

Looks like it, yes. Because I wanted to get some kind of confirmation from them that they really did award you a Bachelor of Arts degree and check whether you took any mathematics courses.

It was a liberal arts college.

I am not claiming to know whether they ever awarded you a Bachelor of Arts degree or whether you took any mathematics courses there, and you are lying when you say they gave you a Bachelor Degree in Mathematics.

If you did actually take a mathematics course there and got a passing grade, then yeah sure I'm happy to go public with saying their credentials as a tertiary institution need to be called into question, but of course you won't let me say what their name is. In any case it is not possible for them to sue me for defamation for making such public statements. If they go public saying that you have some competence in mathematics, (which they haven't done), then it is their role to thereby open themselves up to public criticism from other professional mathematicians who find your mathematical abilities wanting.

This conversation really is the height of absurdity.

[Peter Percival]

Khong Dong wrote:
> On Thursday, 28 March 2019 01:28:55 UTC-6, Rupert wrote:

>> [...]

>> So no, I am not accusing that particular college of fraud.
>
> Oh yes you did with your:
>
> "if they ever did then either the degree is an utterly worthless one that
> you can pay for without having to show any real academic attainment".

That starts with "if" and the "either" has lost its companion "or"s.

> But let's hear from them, since they awarded me with a Bachelor Degree in
> Mathematics as an accredited institution.

[Ben Bacarisse]

Rupert <rupertm...@yahoo.com> writes:

> The testamur you showed me was in fact a Bachelor of Arts. You refuse
> to let me publicly say what the name of the college is, so I don't
> know that they ever really did give you a Bachelor of Arts degree. The
> testamur you gave me does not make any mention of the word
> "mathematics".

Would you expect it to? I'm not so familiar with what US institutions
do, but my degree certificate makes no mention of the subject, just BA.
I've see examples on the web where there /is/ a space for the subject
studied to be written in, but it's not clear if that is the norm or if
the one you saw had this.

The whole thing is verging on the surreal. It just leaves me wondering
what it is that Nam wants to hide, which is surely not the result he was
aiming for. Why on Earth would you not say where you went and what
courses you took? Either that or stay private and tell people it's none
of their business. Both are perfectly reasonable stances to take, but
this "I'll tell one person who must keep it all secret" is bizarre.

> I have not, in fact, made any negative comments about this college
> whose name you refuse to let me say in public, and you are lying that
> the testamur that you sent me by email said "Bachelor Degree in
> Mathematics". It said "Bachelor of Arts", and the college named was a
> liberal arts college. And the most likely reason why you won't let me
> publicly name who the college is is that the testamur you sent me was
> a forgery and you are aware that that is a criminal matter.

I'd not thought of that. Despite Nam's public behaviour here, I never
thought he'd do anything like that.

--
Ben.

[Rupert]

On Thursday, March 28, 2019 at 3:20:17 PM UTC+1, Ben Bacarisse wrote:
> Rupert <rupertm...@yahoo.com> writes:
>
> > The testamur you showed me was in fact a Bachelor of Arts. You refuse
> > to let me publicly say what the name of the college is, so I don't
> > know that they ever really did give you a Bachelor of Arts degree. The
> > testamur you gave me does not make any mention of the word
> > "mathematics".
>
> Would you expect it to? I'm not so familiar with what US institutions
> do, but my degree certificate makes no mention of the subject, just BA.
> I've see examples on the web where there /is/ a space for the subject
> studied to be written in, but it's not clear if that is the norm or if
> the one you saw had this.

I don't really know. However I have a testamur from the University of New South Wales, Australia, saying "Bachelor of Science with Honours Class 1 in Pure Mathematics", and another one from the same university saying "Doctor of Philosophy" (without mentioning mathematics).

> The whole thing is verging on the surreal. It just leaves me wondering
> what it is that Nam wants to hide, which is surely not the result he was
> aiming for. Why on Earth would you not say where you went and what
> courses you took? Either that or stay private and tell people it's none
> of their business. Both are perfectly reasonable stances to take, but
> this "I'll tell one person who must keep it all secret" is bizarre.

Definitely is rather odd!

[Khong Dong]

On Saturday, 23 March 2019 11:14:41 UTC-6, Khong Dong wrote:
> On Saturday, 23 March 2019 10:53:05 UTC-6, Khong Dong wrote:
> > On Saturday, 23 March 2019 10:24:34 UTC-6, Khong Dong wrote:
> > > On Saturday, 23 March 2019 09:26:31 UTC-6, Khong Dong wrote:
> > > > On Saturday, 23 March 2019 04:45:46 UTC-6, Rupert wrote:
> > > > > On Saturday, March 23, 2019 at 12:05:57 AM UTC+1, Khong Dong wrote:
> > > > > > On Friday, 22 March 2019 15:22:22 UTC-6, Rupert wrote:
> > > > > > > On Friday, March 22, 2019 at 4:47:27 PM UTC+1, Khong Dong wrote:
> > > > > > > > On Friday, 22 March 2019 08:56:42 UTC-6, Rupert wrote:
> > > > > > > > > On Friday, March 22, 2019 at 3:47:48 PM UTC+1, Khong Dong wrote:
> > > > > > > > > > On Friday, 22 March 2019 08:41:33 UTC-6, Rupert wrote:
> > > > > > > > > > > On Friday, March 22, 2019 at 6:02:48 AM UTC+1, Khong Dong wrote:
> > > > > > > > > > > > On Thursday, 21 March 2019 20:35:22 UTC-6, Rupert wrote:
> > > > > > > > > > > > > On Friday, March 22, 2019 at 3:18:56 AM UTC+1, Khong Dong wrote:
> > > > > > > > > > > > > > On Thursday, 21 March 2019 04:22:16 UTC-6, Rupert wrote:
> > > > > > > > > > > > > > > On Thursday, March 21, 2019 at 7:49:07 AM UTC+1, Khong Dong wrote:
> > > > > > > > > > > > > > > > Not 100% sure why I ping you - "Me" - perhaps partially because it strikes
> > > > > > > > > > > > > > > > me that you're technical, objective, deliberate, fair poster, in evaluating
> > > > > > > > > > > > > > > > technical matters, or perhaps because it's a shot in the dark I'm trying,
> > > > > > > > > > > > > > > > or perhaps because you said in the other thread:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > "It's well known that we have to be careful if dealing with the notion
> > > > > > > > > > > > > > > > of /infinity/".
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > but whatever the reasons, I'm just wondering if you could comment on the
> > > > > > > > > > > > > > > > technical correctness of the three meta theorems below (which I posted in
> > > > > > > > > > > > > > > > MO last month or so) - basically proving it's logically impossible (there
> > > > > > > > > > > > > > > > being no meta proofs) to know, verify the truth value of:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > cGC <-> "There being infinitely counter examples of Goldbach Conjecture".
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Certainly the presentation isn't perfect given it's as long as only one post
> > > > > > > > > > > > > > > > but I'll try my best to answer any question you might have.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Thank you in advance,
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Best Regards,
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > -Nam Nguyen
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > ===========================================================================
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > ===> Definitions
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > GC <-> Goldbach Conjecture
> > > > > > > > > > > > > > > > cGC <-> "There are infinitely many counter examples of Goldbach Conjecture"
> > > > > > > > > > > > > > > > SR <->(~GC /\ ~cGC)

> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > knowable(S) <=> There's a FINITE meta mathematical proof that S is (non-vacuously) true.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > undecidable(S) <=> neg(knowable(S)) and neg(knowable(neg(S))).
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Caveat: This is meta level truth-"undecidable", _not_ FOL provability-"undecidable".
> > > > > > > > > > > > > > > >

> > > > > > > > > > > > > > > > intrinsic(S) <=> TRUE(S), for all the meaning of S.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > ===> Assumptions
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The definition of prime(p)is free of occurrences of the language constant successor function symbol - required for any proof by induction:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > prime(p)<->((~p=0)/\~Az[p*z=z]/\((p=x*y)->(Az[x*z=z]\/Az[y*z=z])))
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Let's let U be the set of Von Neumann's (or Zermelo's) finite ordinals, or some kind of (infinite) union between the two sets.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > There exists an uncountable set K of language structures M's having these following properties:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > 1. All M's have the same domain - the universe of discourse U, mentioned
> > > > > > > > > > > > > > > > above.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > 2. A natural number n is interpreted by a term of the language - a numeral -
> > > > > > > > > > > > > > > > denoting an element e_n in U but is neither the element nor the term
> > > > > > > > > > > > > > > > (numeral)!
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > For the economy of typing, we might conveniently use natural numbers and
> > > > > > > > > > > > > > > > elements interchangeably but that does *not* take away the said
> > > > > > > > > > > > > > > > distinction.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > 3. The truth of a formula F in any of them can also be interpreted the truth
> > > > > > > > > > > > > > > > of F of (about) the natural numbers.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > 4. Any element in U that is interpreted as a non-odd prime natural number n
> > > > > > > > > > > > > > > > in any of the structures is interpreted as the same non-odd prime natural
> > > > > > > > > > > > > > > > number n in all of the structures.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > I have a question about this.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > You're talking about an infinite family of structures for the first-order language of arithmetic, right, all of them with the same domain of discourse U. Where U, if I understand correctly, is some fixed infinite subset of the union of the set of all finite von Neumann ordinals and the set of all
> > > > > > > > > > > > > > > finite Zermelo ordinals.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > And I assume "non-odd prime natural number n" means, a natural number n
> > > > > > > > > > > > > > > which is not an odd prime number.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > It means the natural number n here is either, Zero, One, Two, or a multiple.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > So, what exactly does it mean to say that an element of U is interpreted
> > > > > > > > > > > > > > > as a natural number n in one of the structures? That means that, in that
> > > > > > > > > > > > > > > structure, the referent of the numeral for n is equal to the element of U
> > > > > > > > > > > > > > > in question, does it?
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Right. It means exactly as Assumption 2 states:
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > <quote>
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > 2. A natural number n is interpreted by a term of the language - a numeral -
> > > > > > > > > > > > > > denoting an element e_n in U but is neither the element nor the term
> > > > > > > > > > > > > > (numeral)!
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > </quote>
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > It goes without saying that, in general, each different M in K would induce
> > > > > > > > > > > > > > a different interpretation on the individual element e_n in U.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > > Also later on with your statement of Lemma 1.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > "Lemma 1: If p is a variable denoting an odd prime (element) in M, the
> > > > > > > > > > > > > > > uniqueness of the numeral of p - numeral(p) - is unknown."
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > So M is one of the structures in the collection, right? Now, are you
> > > > > > > > > > > > > > > assuming that every element of U is the referent of a numeral relative to
> > > > > > > > > > > > > > > M?
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Correct. For instance, the Zermelo element z_3 = {{{{}}} could be a referent
> > > > > > > > > > > > > > for the (prime) numeral SSSS0 for some M but could also be a referent for
> > > > > > > > > > > > > > the (prime) numeral SSSSSS0 for some different M'.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > > And what does it mean to say the uniqueness of the numeral of p is
> > > > > > > > > > > > > > > unknown?
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Because there's an 1-many relationship between an element e_n and the
> > > > > > > > > > > > > > numerals relative to which M of K we choose: in the example above you have
> > > > > > > > > > > > > > the element z_3 but it's invalid to say its numeral must necessarily be
> > > > > > > > > > > > > > SSSS0, or SSSSSS0, ... _without referring to a *particular* M_ in K.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > > What if I stipulate that the numeral of p is SSSSS0? Then surely I know
> > > > > > > > > > > > > > > the numeral of p, and that seems to be consistent with the hypotheses
> > > > > > > > > > > > > > > of your lemma. So what can this lemma really be saying?
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Here you've involved another issue: p as stated in Lemma 1 is a free
> > > > > > > > > > > > > > _free variable term of the language_ and all you'd know (would be given in
> > > > > > > > > > > > > > the Lemma) is that p is an odd prime element: but you can't sufficiently
> > > > > > > > > > > > > > know the specificity of M.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > If you stipulate that the numeral of p be the particular numeral SSSSS0 then
> > > > > > > > > > > > > > you've sufficiently specified the underlying M, and in that case p as an
> > > > > > > > > > > > > > element (or as denoting an element) would have to be (or refer to) a
> > > > > > > > > > > > > > specific element. Iow, p would no longer be a free variable that Lemma 1 is
> > > > > > > > > > > > > > talking about: it'd be a language-defined symbol for an individual constant
> > > > > > > > > > > > > > element, a syntactical alias for SSSSS0.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Okay. Well, I wouldn't mind seeing the proof of Lemma 2 then.
> > > > > > > > > > > >
> > > > > > > > > > > > Sure. Proof:
> > > > > > > > > > > >
> > > > > > > > > > > > By Assumption 4, in a *general M* of K, the *general* even element e of U,
> > > > > > > > > > > > in Lemma 2, would be interpreted as the same natural even number, say, N_e.
> > > > > > > > > > >
> > > > > > > > > > > I'm sorry, I don't understand this sentence.
> > > > > > > > > >
> > > > > > > > > > Perhaps you could say which part of it that would prevent you? Would you
> > > > > > > > > > understand all the antecedent (required) assumptions (particularly
> > > > > > > > > > Assumption 4) prior to Lemma 2?
> > > > > > > > >
> > > > > > > > > I think the main problem is that your use of the word "general" is unclear.
> > > > > > > >
> > > > > > > > In this context "general" would mean "non specific" or "free" (variable).
> > > > > > >
> > > > > > > I'm afraid that doesn't help me very much.
> > > > > >
> > > > > > Why? You don't know what a free variable means? I could look up Shoenfield's and
> > > > > > explain it for you if you'd like.
> > > > >
> > > > > I know what a free variable is.
> > > > >
> > > > > But that is no particular help in understanding your sentence.
> > > >
> > > > Then I don't think any mathematician or sci.logic poster in the right mind
> > > > would understand what you've complained on Lemma 2:
> > > >
> > > > "Lemma 2: If e is a free variable ranging over the set of even elements
> > > > in U of M, and p is a prime element bounded to e, the uniqueness of
> > > > numeral(p)is unknown."
> > > >
> > > > Rupert:
> > > >
> > > > "Okay. Well, I wouldn't mind seeing the proof of Lemma 2 then."
> > > >
> > > > NN:
> > > >
> > > > "Sure. Proof:
> > > >
> > > > By Assumption 4, in a *general M* of K, the *general* even element e of U,
> > > > in Lemma 2, would be interpreted as the same natural even number, say,
> > > > N_e."
> > > >
> > > > Rupert:
> > > >
> > > > "I'm sorry, I don't understand this sentence."
> > > >
> > > > NN:
> > > >
> > > > "Perhaps you could say which part of it that would prevent you? Would you
> > > > understand all the antecedent (required) assumptions (particularly
> > > > Assumption 4) prior to Lemma 2?"
> > > >
> > > > Rupert:
> > > >
> > > > "I think the main problem is that your use of the word "general" is
> > > > unclear."
> > > >
> > > > NN:
> > > >
> > > > "In this context "general" would mean "non specific" or "free" (variable)."
> > > >
> > > > Rupert:
> > > >
> > > > "I'm afraid that doesn't help me very much."
> > > >
> > > > ====================================================
> > > >
> > > > Your "main problem" in understanding the Lemma 2 proof's is that the proof
> > > > would refer to the variable e (the Lemma 2's "e is a free variable ranging
> > > > over the set of even elements") as a _general_ variable which you complained
> > > >
> > > > <quote>
> > > > the word "general" is unclear
> > > > </quote>
> > > >
> > > > but which I has clearly explained "general" means "free": so you or any
> > > > student can replace e being a "general" even number here by e being a free variable denoting an even element in U in M. It's bizarre that you couldn't
> > > > even understand that much!
> > > >
> > > > So either your complaint:
> > > >
> > > > <quote>
> > > >
> > > > the word "general" is unclear
> > > >
> > > > </quote>
> > > >
> > > > wasn't genuine or for some inexplicable reasons you forgot what a "free" variable would mean!
> > > >
> > > > Btw, Rupert, the below inferrecne :
> > > >
> > > > P(n) -> An[P(n)]
> > > >
> > > > is possible by an inference rule named UG - universal - _GENERALIZATION_ !
> > > >
> > > > Are you going to complaint TOO something like:
> > > >
> > > > "I think the main problem is that the use of the word 'GENERALIZATION'
> > > > here is unclear"
> > > >
> > > > ?
> > >
> > > Anyway, the technical proof for Lemma 2 isn't that convoluted, Rupert.
> > >
> > > Let M, M' be free variables over K. We don't know whether or not M = M' but
> > > we do know by Assumption 4 the successions of even elements in both M and
> > > M' are identical. But by Assumption 5 the same can't be logically said of
> > > the successions of (odd) primes in M and M'.
> > >
> > > That's all Lemma 2 really says.
> >
> > For instance, consider the below Zermelo's elements:
> >
> > Even elements:
> >
> > z_6 = {{{{{{{}}}}}}}
> > z_8 = {{{{{{{{{}}}}}}}}}
> >
> > Prime elements:
> >
> > z_3 = {{{{}}}}
> > z_11 = {{{{{{{{{{{{}}}}}}}}}}}}
> >
> > Your M could have these 2-tuples in the binary relation symbolized by '<':
> >
> > (z_6, z_8), (z_3, z_6), (z_3, z_11), (z_11, z_8), ...
> >
> > But your M' otoh could have:
> >
> > (z_6, z_8), (z_11, z_6), (z_11, z_3), (z_3, z_8), ...
> >
> > Note the identical 2-tuple (z_6, z_8) exists in both cases - but the
> > remaining 2-tuples are different.
> >
> > Incidentally, the numeral for z_3 would be different between M and M', ditto
> > for that of z_11, as implicated in Lemma 2!
> >
> > Hope this has helped clarifying proof of Lemma 2 you've requested.
>
> Note that from Lemma 2, we can show it's invalid to _even assume_
> Shoenfield's N8 [1] be true in any M of K (hence it'd be invalid to assume
> it be true of the natural numbers) - independent of whether or not PA is
> _syntactically_ consistent.
>
> This in effect is one of the gateways into showing Completeness is invalid.
>
> [1]
>
> N8 <-> (x < Sy <-> (x < y \/ x = y))

It's an interesting coincidence (or may be not) that Rupert once said:

"There are mathematical propositions which can neither be proved true
nor proved false."

Shoenfield's N8 seems to fit right in, whether or not Rupert himself
realizes it.

[Khong Dong]

Note, however, that while N8 is a valid specific wff, Godel's general
sentence G(T) isn't valid, well-formed, because "magnitude" of prime
can't be adequately defined in general.

[Khong Dong]

> 5. Any element in U that is interpreted as an odd prime natural number in
> any of the structures is also interpreted as an odd prime natural number
> in all of the structures: but NOT necessarily the same odd prime natural
> number!
>
> ===> Lemmas

>
> Lemma 1: If p is a variable denoting an odd prime (element) in M, the
> uniqueness of the numeral of p - numeral(p) - is unknown.
>

> Lemma 2: If e is a free variable ranging over the set of even elements in U
> of M, and p is a prime element bounded to e, the uniqueness of
> numeral(p)is unknown.
>

> These two lemmas are due mostly to the fact that the definition of prime(p)
> mentioned above is free of occurrences of the language constant symbol S.
>
> ===> Meta Theorems
>
> Meta Theorem 1 - If GC is true in an M of K, it's not knowable.
>
> Proof:
>
> Let E be the open interval { e | even(e) /\ (e >= E0) } where even(E0),
> E0 >= 4, and is known to be a sum of two primes. It's sufficient to show
> that if GC is true over E, it's not knowable. But indeed, we have to show
> there exist primes p1, p2, for an even e in E, such that:
>
> numeral(e) = numeral(p1) + numeral(p2)
>
> but by Lemma 2, the uniqueness of numeral(p1) and numeral(p2) are unknown,
> hence the expected numeral equality is not knowable. QED.
>
> ***
>
> Meta Theorem 2 - If there are finitely many counter examples of Goldbach Conjecture it's not knowable.
>
> Proof:
>
> There being finitely many counter examples of Goldbach Conjecture means that
> there exists an open interval E = { e | even(e) /\ (e >= E0) } previously
> mentioned, in the complement of which finitely many counter-example
> elements (numbers) would be found. But by Meta Theorem 1, the existence of E
> is not knowable, hence so is its complement. QED.
>
> ***
>
> Meta Theorem 3 - If cGC is true but GC is false, it's not knowable.
>
> Proof:
>
> In this case there are infinitely many even (counter-example) elements e's,
> and infinitely many corresponding prime elements p's such that:
>
> - p = maximum prime to the left a general counter example e
> - numeral(e) = numeral(p) + numeral(o)
>
> for some odd(o).
>
> But by Lemma 1, numeral(p) is not knowable (not unique) hence the numeral
> equality is not knowable. QED.

To "Me" et al.,

I think the meta theorems in the thread is straightforward, easy to
understand. But let me have a closure on the thread by pointing out
some notions about FOL framework reasoning many mathematicians would
tend to not recognize.

==========================> Interpretation - Sui Generis, Non

Consider the (LSTS) language structure theoretical set M object that
contains the ordered-pair "snipplet":

<'+',<e1,e2>>

which of the two distinct formulas of two different languages would be true in M:

e1 < e2

+e1 = e2

?

The answer: it depends on which _specific_ language one's _interpretation_
would be in. Iow, mathematical truth is not sui generis - it depends on individual subjective interpretation.

==========================> Detonation - Choice Necessity: Impossibility

Let F, F' be two unary function symbols _denoting_ two Choice functions,
where Choice is necessary, would the below formula:

Ax[F(x) = F'(x)]

have a truth value in the underlying LSTS?

The answer is of course NO.

***********

==========================> KF (Karma Fourier) Summation.

In summary, per each individual mortal being, his/her formalized
mathematical knowledge (truth) would be a summation (sigma):

Sigma(IKi + DKi)

where:

IK = interpretation-knowledge
DK = detonation-knowledge

as mentioned above.

[Khong Dong]

To further illustrate this KF summation, consider the optical illusion in:

https://www.womansworld.com/posts/my-wife-and-my-mother-in-law-166601

The Lady's fur-coat, hair, headscarf, decorating feather, are analogous to
(say) induction-able truth of the natural numbers: everyone would see the
same truth.

For instance, the distance between any point on the headscarf to any point
on the fur-coat would be the absolute.

Otoh, the distance between her left eyelash and her nostril isn't absolute:
it subjectively, relatively, depends on *which* lady one wold interpret
"the" lady be.

Consequently, it's invalid to claim the distance between the left eyelash
and the nostril is n-pixel long as true or false: it's meaningless until
ones fixes - interprets - which lady is the underlying one.

The concept of the standard language structure N is very much akin to the
concept of "Lady" in the picture: but instead of only two ladies we could
see, we'd have uncountably many successor functions and uncountably many
less-than relations that are of identical domain but not (non-logically)
isomorphic, and yet N is still purported to convey - to be a surrogate
concept for - the notion of the natural numbers!

Incompleteness does _not_ pertain to the realm of syntactical provability
only: it also pertains to the realm of language structures and concepts.

Godel should have known better.

[Khong Dong]

On Friday, 19 April 2019 11:52:23 UTC-6, Khong Dong wrote:

A typo: should have been "e1 + e2".

[Khong Dong]

Typo correction: in all of this it's meant "denotation" and not "detonation", of course.

[Khong Dong]

On Friday, 22 March 2019 08:41:33 UTC-6, Rupert wrote:
> On Friday, March 22, 2019 at 6:02:48 AM UTC+1, Khong Dong wrote:
> > On Thursday, 21 March 2019 20:35:22 UTC-6, Rupert wrote:
> > > On Friday, March 22, 2019 at 3:18:56 AM UTC+1, Khong Dong wrote:
> > > > On Thursday, 21 March 2019 04:22:16 UTC-6, Rupert wrote:

> > > > > I have a question about this.
> > > > >
> > > > > You're talking about an infinite family of structures for the first-order language of arithmetic, right, all of them with the same domain of discourse U. Where U, if I understand correctly, is some fixed infinite subset of the union of the set of all finite von Neumann ordinals and the set of all
> > > > > finite Zermelo ordinals.
> > > > >
> > > > > And I assume "non-odd prime natural number n" means, a natural number n
> > > > > which is not an odd prime number.
> > > >
> > > > It means the natural number n here is either, Zero, One, Two, or a multiple.
> > > >
> > > > >
> > > > > So, what exactly does it mean to say that an element of U is interpreted
> > > > > as a natural number n in one of the structures? That means that, in that
> > > > > structure, the referent of the numeral for n is equal to the element of U
> > > > > in question, does it?
> > > >
> > > > Right. It means exactly as Assumption 2 states:
> > > >
> > > > <quote>
> > > >

> > > > 2. A natural number n is interpreted by a term of the language - a numeral -
> > > > denoting an element e_n in U but is neither the element nor the term
> > > > (numeral)!
> > > >

> > > > </quote>
> > > >
> > > > It goes without saying that, in general, each different M in K would induce
> > > > a different interpretation on the individual element e_n in U.
> > > >
> > > > > Also later on with your statement of Lemma 1.
> > > > >

> > > > > "Lemma 1: If p is a variable denoting an odd prime (element) in M, the
> > > > > uniqueness of the numeral of p - numeral(p) - is unknown."
> > > > >

Ok. So by now you (Rupert) might have understood Lemma 2 (the proof of), I'm
hoping. If not, the below category presentation from Dr. Cheng would show that
my Lemma 2 is correct (though was differently worded of course).

"Category Theory in Life" - Eugenia Cheng, (16:01 - 17:13):

https://www.youtube.com/watch?v=ho7oagHeqNc

Basically from her presentation, one could see that if 6 is a Von Neumann's
surrogate element for the natural number Six, 5 for Five, and 7 for Seven, then
there are two distinct orders regarding 5 and 7, simply because the two prime
surrogates 5, 7 can be interchanged around the even surrogate 6:

(1) 5 < 7
(2) 7 < 5

Therefore the numeral(5) and numeral(7) are _subjectively determined_ despite
intuition, since language expressions are agnostic about subjective intuition.

And that's the fixed even surrogate 6. If you replace 6 by a general even
surrogate e, then for sure the numeral of the surrogate prime next less than
e can't be fixed (inductively determined), and neither can the surrogate prime
next greater than e be. Which is what Lemma 2 is about.

So hopefully now the three meta theorems 1 - 3 would be understood by you.