On Wednesday, 8 December 2021 at 07:58:59 UTC-4, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 8. Dezember 2021 um 12:57:14 UTC+1:
> > On Wednesday, 8 December 2021 at 07:49:11 UTC-4, WM wrote:
> > > Gus Gassmann schrieb am Mittwoch, 8. Dezember 2021 um 12:43:34 UTC+1:
> > >
> > > > Every n ∈ ℕ is finite, and the set E(n) = {n, n+1, n+2, ...} has cardinality ℵo. So the set E is isomorphic to ℕ and hence has cardinality ℵo also.
> > > What is the first natnumber residing in an infinite endsegment but not being an index?
> > Why would you even think that there is such a thing?
> If all endsegments are infinite, then there is no exhaustion.

Whatever that means. Exhaustion is real; hell, I get exhausted some time. But there are infinitely many natural numbers, each of which defines a finite FISON and an infinite end segment. If your muckemythics doesn't allow for that, that is decidedly your problem, not anybody else's.

On Wednesday, 8 December 2021 at 08:01:42 UTC-4, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 8. Dezember 2021 um 12:52:01 UTC+1:
> > And the *SET* of all FISONs is actually infinite.
> No.
>
> But better this can be seen by the endsegments which have an infinite intersection.

Quit lying and deluding yourself. (Your intentional ambiguities don't help, either.)

On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:

> By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀

Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set of endsegments with |∩{E(1), E(2), ...., E(k)}| = ℵ₀ is
E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set. Trivially E has cardinality aleph_0. Potentially infinite set is nonsense.

--
William Hughes

onsdag 8 december 2021 kl. 11:16:10 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Dienstag, 7. Dezember 2021 um 18:06:04 UTC+1:
> > On Tuesday, December 7, 2021 at 6:00:59 PM UTC+1, WM wrote:
> >
> > > Exercise: Prove that the set E of endsegments E(n) with |E(n)| = ℵo cannot have |E| = ℵo.
> > What a silly "exercise".
> Very important for young students who have not yet comprehended the mistakes of set theory. This exercise shows once and for all that set theory is smashed to pieces by the fact that never two consecutive ℵo-sets can exist in the natural order of ℕ.
>
> Regards, WM

It is not important because it is FALSE. It is a FALSE statement!

You keep saying that there are "two consecutive" yet you have failed to DEMONSTRATE IT!

onsdag 8 december 2021 kl. 11:30:38 UTC+1 skrev WM:
> Gus Gassmann schrieb am Dienstag, 7. Dezember 2021 um 19:50:37 UTC+1:
> > On Tuesday, 7 December 2021 at 12:52:35 UTC-4, WM wrote:
>
> > > But it is not a finite subset because it has not upper end. It is not an infinite subset, because almost all endsegments are following.
> > Each end segment is infinite, but a finite set of end segments is just that: A finite set. There are, of course, (countably) infinitely many such sets.
> That is called potential infinity. The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is potentially infinite. However the sets which have an empty intersection |∩{E(1), E(2), ...}| = 0, like the set of all endsegments, are actually infinite, that is much more than all finite sets.
> > > Obviously it exists. All endsegments have an empty intersection. The potentially infinite collection of infinite endsegments has an infinite intersection.
> > The intersection over all possible finite sets of end segments is just as empty as the intersection over the set of all end segments.
> That is wrong. Try to understand logic, in particular the universal quantifier, that is the turned A:
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
> > Your hallucinations are getting weirder and more troublesome by the day..
> Chuckle. I use logic.
>
> Regards, WM

You do NOT use logic at all! Stop being such a fucking LIAR!

onsdag 8 december 2021 kl. 11:42:17 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 06:39:11 UTC+1:
>
> > There is no such thing as "potential infinite"
> You are wrong. By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite but less than the sets which have an empty intersection |∩{E(1), E(2), ...}| = 0, like the set of all endsegments. They are actually infinite. That is much more than all finite sets.
>
> Regards, WM

Nope, I am right because you cannot define what your "potential infinite" even fucking mathematically mean for sets.

>The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite

with cardinality aleph-0

onsdag 8 december 2021 kl. 11:44:07 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 06:39:46 UTC+1:
>
> > It does NOT show that. It shows only that FINITE collections has an infinite intersection.
> Finite without a largest set. That is potential infinity. Or can you find a largest set of endsegements that satisfies ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀?
>
> Regards, WM

Yet meaningless in set theory.

On Wednesday, 8 December 2021 at 09:43:03 UTC-4, JVR wrote:
> On Wednesday, December 8, 2021 at 2:07:51 PM UTC+1, Gus Gassmann wrote:
> > On Wednesday, 8 December 2021 at 07:58:59 UTC-4, WM wrote:
[...]
> > > If all endsegments are infinite, then there is no exhaustion.
> > Whatever that means.
> I think it means the same thing that he used to refer to as 'ready', which he translated from German 'fertig'; now
> his new muddled term for that is 'complete' - he doesn't know, of course, what the technical term 'complete' means
> in mathematics. Remember he a Professor of General Studies teaching remedial math in a technical college, not a mathematician. So he gets to make up his own terminology in his slightly deficient but serviceable English.

Thanks for the translation help. I am still not sure what is supposed to get exhausted, but I will let the great perfessor answer that for himself. (This is a situation where his English is not serviceable, but woefully inadequate, although that may be unfair to the English language and instead a blatant exposure of his math deficiencies.)

zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:34:06 UTC+1:
> onsdag 8 december 2021 kl. 11:42:17 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 06:39:11 UTC+1:
> >
> > > There is no such thing as "potential infinite"
> > You are wrong. By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite but less than the sets which have an empty intersection |∩{E(1), E(2), ...}| = 0, like the set of all endsegments. They are actually infinite. That is much more than all finite sets.

> Nope, I am right because you cannot define what your "potential infinite" even fucking mathematically mean for sets.

It means finite but without a last element.

> >The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite
> with cardinality aleph-0

No, the ℵ₀-set is {E(1), E(2), ...} with |∩{E(1), E(2), ...}| = 0. It is much larger than the potentially infinite oo-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:31:41 UTC+1:
> onsdag 8 december 2021 kl. 11:16:10 UTC+1 skrev WM:
> > Fritz Feldhase schrieb am Dienstag, 7. Dezember 2021 um 18:06:04 UTC+1:
> > > On Tuesday, December 7, 2021 at 6:00:59 PM UTC+1, WM wrote:
> > >
> > > > Exercise: Prove that the set E of endsegments E(n) with |E(n)| = ℵo cannot have |E| = ℵo.
> > > What a silly "exercise".
> > Very important for young students who have not yet comprehended the mistakes of set theory. This exercise shows once and for all that set theory is smashed to pieces by the fact that never two consecutive ℵo-sets can exist in the natural order of ℕ.
> >
> It is not important because it is FALSE. It is a FALSE statement!
>
> You keep saying that there are "two consecutive" yet you have failed to DEMONSTRATE IT!

All infinite endsegments are infinite. Therefore they contain ℵo-sets of natnumbers which are larger than all indices. Therefore their indices have to be a potentially infinite set.

Regards, WM

How much light numbers are enough for the Riemann Hypothesis?
They went already to 10000000000000 (X. Gourdon, 2004).

The Riemann Hypothesis in Computer Science - Yu. V. Matiyasevich
https://www.researchgate.net/profile/Yuri-Matiyasevich/publication/326247349

Is this light enough, or should they conquer
another piece of darkness. Asking for a friend.

WM schrieb am Mittwoch, 8. Dezember 2021 um 18:18:33 UTC+1:
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:31:41 UTC+1:
> > onsdag 8 december 2021 kl. 11:16:10 UTC+1 skrev WM:
> > > Fritz Feldhase schrieb am Dienstag, 7. Dezember 2021 um 18:06:04 UTC+1:
> > > > On Tuesday, December 7, 2021 at 6:00:59 PM UTC+1, WM wrote:
> > > >
> > > > > Exercise: Prove that the set E of endsegments E(n) with |E(n)| = ℵo cannot have |E| = ℵo.
> > > > What a silly "exercise".
> > > Very important for young students who have not yet comprehended the mistakes of set theory. This exercise shows once and for all that set theory is smashed to pieces by the fact that never two consecutive ℵo-sets can exist in the natural order of ℕ.
> > >
> > It is not important because it is FALSE. It is a FALSE statement!
> >
> > You keep saying that there are "two consecutive" yet you have failed to DEMONSTRATE IT!
> All infinite endsegments are infinite. Therefore they contain ℵo-sets of natnumbers which are larger than all indices. Therefore their indices have to be a potentially infinite set.
>
> Regards, WM

William schrieb am Mittwoch, 8. Dezember 2021 um 14:25:38 UTC+1:
> On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:
>
> > By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀
> Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is
> E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set.

True, a Peano set, but not an ℵ₀ set. Therefore you should not use {E(1), E(2), E(3) ...}.

>Trivially E has cardinality aleph_0.

No, the ℵ₀-set is {E(1), E(2), E(3), ...} with |∩{E(1), E(2), ...}| = 0. It is much larger than the potentially infinite Peano-set {E(1), E(2), ..., E(k)} with |∩{E(1), E(2), ..., E(k)}| = ℵ₀.

All infinite endegments are infinite. All have an infinite set of natnumbers in common with all their predecessors. This set cannot be infinite without having a first element. What could it be if all indexes were an infinite set?

Reards, WM

FromTheRafters schrieb am Mittwoch, 8. Dezember 2021 um 13:49:43 UTC+1:
> WM used his keyboard to write :
> > FromTheRafters schrieb am Mittwoch, 8. Dezember 2021 um 11:39:19 UTC+1:
> >> WM brought next idea :
> >>>> (1) By ℵo-set you presumably mean a set of cardinality ℵo.
> >>>
> >>> In any case we can exclude that two consecutive ℵo-sets cannot exist in the
> >>> natural order of ℕ.
> >
> >> Endsegments are taken
> >> as infinite sets and your sequence of FISONs as sets are simply
> >> infinitely many finite individual sets. "THERE IS NO POTENTIAL
> >> INFINITY"
> >
> > Then name the last FISON.
> There is no last FISON.
>
Fine. On the other han every FISON has an infinite set of successors. This set could not have a first element if the FISONs would exhaust all natural numbers.

Regards, WM

On Wednesday, 8 December 2021 at 13:28:07 UTC-4, WM wrote:
> FromTheRafters schrieb am Mittwoch, 8. Dezember 2021 um 13:49:43 UTC+1:
[...]
> > There is no last FISON.
> >
> Fine. On the other han every FISON has an infinite set of successors. This set could not have a first element if the FISONs would exhaust all natural numbers.

You truly are insane. Of course for every FISON(n) the set of successors is {n+1, n+2, n+3, ...} (or {FISON(n+1), FISON(n+2), FISON(n+3),... }, depending on what you think the successor of a FISON is in this particular moon phase). In either interpretation, there is a natural order with a smallest element. What the fuck is wrong with you?

Gus Gassmann schrieb am Mittwoch, 8. Dezember 2021 um 18:40:29 UTC+1:
> On Wednesday, 8 December 2021 at 13:28:07 UTC-4, WM wrote:
> > FromTheRafters schrieb am Mittwoch, 8. Dezember 2021 um 13:49:43 UTC+1:
> [...]
> > > There is no last FISON.
> > >
> > Fine. On the other han every FISON has an infinite set of successors. This set could not have a first element if the FISONs would exhaust all natural numbers.
> Of course for every FISON(n) the set of successors is {n+1, n+2, n+3, ...}. In either interpretation, there is a natural order with a smallest element.

For all FISONs the set of successors is infinite. ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Note that finite is less than infinite. So we can state: Every FISON cover less than half of all natural numbers.

This means that there is an infinite intersection of all successors. Here quantifier exchange is correct.

Regards, WM

On 12/8/2021 11:18 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:31:41 UTC+1:
>> onsdag 8 december 2021 kl. 11:16:10 UTC+1 skrev WM:
>>> Fritz Feldhase schrieb am Dienstag, 7. Dezember 2021 um 18:06:04 UTC+1:
>>>> On Tuesday, December 7, 2021 at 6:00:59 PM UTC+1, WM wrote:
>>>>
>>>>> Exercise: Prove that the set E of endsegments E(n) with |E(n)| = ℵo cannot have |E| = ℵo.
>>>> What a silly "exercise".
>>> Very important for young students who have not yet comprehended the mistakes of set theory. This exercise shows once and for all that set theory is smashed to pieces by the fact that never two consecutive ℵo-sets can exist in the natural order of ℕ.
>>>
>> It is not important because it is FALSE. It is a FALSE statement!
>>
>> You keep saying that there are "two consecutive" yet you have failed to DEMONSTRATE IT!
>
> All infinite endsegments are infinite. Therefore they contain ℵo-sets of natnumbers which are larger than all indices. Therefore their indices have to be a potentially infinite set.

FALSE. Show a Proof.

>
> Regards, WM
>

On 12/8/2021 12:02 PM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 8. Dezember 2021 um 18:40:29 UTC+1:
>> On Wednesday, 8 December 2021 at 13:28:07 UTC-4, WM wrote:
>>> FromTheRafters schrieb am Mittwoch, 8. Dezember 2021 um 13:49:43 UTC+1:
>> [...]
>>>> There is no last FISON.
>>>>
>>> Fine. On the other han every FISON has an infinite set of successors. This set could not have a first element if the FISONs would exhaust all natural numbers.
>> Of course for every FISON(n) the set of successors is {n+1, n+2, n+3, ...}. In either interpretation, there is a natural order with a smallest element.
>
> For all FISONs the set of successors is infinite. ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

that is conflicting mush.

> Note that finite is less than infinite.

> >So we can state: Every FISON cover less than half of all natural numbers.

Fundamentally Wrong.

>
> This means that there is an infinite intersection of all successors.

Wrong.

> Here quantifier exchange is correct.

wrong again.

>
> Regards, WM
>

On Wednesday, December 8, 2021 at 1:25:49 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 8. Dezember 2021 um 14:25:38 UTC+1:
> > On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:
> >
> > > By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀
> > Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set of endsegments with |∩{E(1), E(2), ...., E(k)}| = ℵ₀ is
> > E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set.
> True, a Peano set, but not an ℵ₀ set

A Peano set has cardinality aleph_0.

> the potentially infinite Peano-set

"Potentially infinite set" is nonsense.

>{E(1), E(2), ..., E(k)}

Nope, this is not a Peano set. A Peano set does not have a last element

--
William Hughes

William schrieb am Mittwoch, 8. Dezember 2021 um 19:12:39 UTC+1:
> On Wednesday, December 8, 2021 at 1:25:49 PM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 8. Dezember 2021 um 14:25:38 UTC+1:
> > > On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:
> > >
> > > > By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀
> > > Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set of endsegments with |∩{E(1), E(2), ...., E(k)}| = ℵ₀ is
> > > E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set.
> > True, a Peano set, but not an ℵ₀ set
> A Peano set has cardinality aleph_0.

No. All FISONs are finite and therefore less than ℵ₀ . Hence all FISONs cover less than half of ℕ.
>
> > the potentially infinite Peano-set
> "Potentially infinite set" is nonsense.
> >{E(1), E(2), ..., E(k)}
>
> Nope, this is not a Peano set. A Peano set does not have a last element
>
{E(1), E(2), ..., E(k)} with not fixed k has no fixed last element.

Regards, WM

On 12/8/2021 12:37 PM, WM wrote:
> William schrieb am Mittwoch, 8. Dezember 2021 um 19:12:39 UTC+1:
>> On Wednesday, December 8, 2021 at 1:25:49 PM UTC-4, WM wrote:
>>> William schrieb am Mittwoch, 8. Dezember 2021 um 14:25:38 UTC+1:
>>>> On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:
>>>>
>>>>> By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀
>>>> Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is
>>>> E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set.
>>> True, a Peano set, but not an ℵ₀ set
>> A Peano set has cardinality aleph_0.
>
> No. All FISONs are finite and therefore less than ℵ₀ . Hence all FISONs cover less than half of ℕ.

wrong conclusion.

>>
>>> the potentially infinite Peano-set
>> "Potentially infinite set" is nonsense.
>>> {E(1), E(2), ..., E(k)}
>>
>> Nope, this is not a Peano set. A Peano set does not have a last element
>>
> {E(1), E(2), ..., E(k)} with not fixed k has no fixed last element.

you make mistake again, that is finite set because you stopped at k.

come on, this is easy to learn!

if you want no last element, write;

{E(1), E(2), ...} has no last element

a good Learning Lesson for you.

try to keep up.

(we know you are a slow learner, that's OK, Bashar was slow learner too, but not as slow as you.)

>
> Regards, WM
>

On Wednesday, December 8, 2021 at 2:37:54 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 8. Dezember 2021 um 19:12:39 UTC+1:
> > On Wednesday, December 8, 2021 at 1:25:49 PM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 8. Dezember 2021 um 14:25:38 UTC+1:
> > > > On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:
> > > >
> > > > > By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀
> > > > Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is
> > > > E={ E(k) | k in |N_P} ={E(1), E(2), E(3) ...} a Peano set.
> > > True, a Peano set, but not an ℵ₀ set
> > A Peano set has cardinality aleph_0.
> No.

Yes. This is a fundamental property of a Peano set.

> {E(1), E(2), ..., E(k)} with not fixed k

is something that changes and is not a set.
Potentially infinite set is nonsense.

--
William Hughes

WM used his keyboard to write :
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:31:41 UTC+1:
>> onsdag 8 december 2021 kl. 11:16:10 UTC+1 skrev WM:
>>> Fritz Feldhase schrieb am Dienstag, 7. Dezember 2021 um 18:06:04 UTC+1:
>>>> On Tuesday, December 7, 2021 at 6:00:59 PM UTC+1, WM wrote:
>>>>
>>>>> Exercise: Prove that the set E of endsegments E(n) with |E(n)| = ℵo
>>>>> cannot have |E| = ℵo.
>>>> What a silly "exercise".
>>> Very important for young students who have not yet comprehended the
>>> mistakes of set theory. This exercise shows once and for all that set
>>> theory is smashed to pieces by the fact that never two consecutive ℵo-sets
>>> can exist in the natural order of ℕ.
>>>
>> It is not important because it is FALSE. It is a FALSE statement!
>>
>> You keep saying that there are "two consecutive" yet you have failed to
>> DEMONSTRATE IT!
>
> All infinite endsegments are infinite.

Wow! You're covering new ground here.

WM formulated the question :
> William schrieb am Mittwoch, 8. Dezember 2021 um 14:25:38 UTC+1:
>> On Wednesday, December 8, 2021 at 6:40:47 AM UTC-4, WM wrote:
>>
>>> By endsegments you see the difference between pot and act: The set of
>>> endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀
>> Note that if k in |N_P we have |∩{E(1), E(2), ..., E(k)}| = ℵ₀. So the set
>> of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is E={ E(k) | k in
>> |N_P} ={E(1), E(2), E(3) ...} a Peano set.
>
> True, a Peano set, but not an ℵ₀ set.

Liar. Prove it is either finite or the continuum.

On 12/8/2021 1:02 PM, WM wrote:

> For all FISONs the set of successors is infinite.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

ℕ = ∪{FISON'}

∀n ∈ ℕ: |ℕ\{0,...,n}| >= ℵ₀

> Note that finite is less than infinite.
> So we can state:
> Every FISON cover less than half of all natural numbers.

"Less than half than half of all natural numbers"
needs to be translated.

How about this?
∀n ∈ ℕ⁺ : |{0,...,n}| < |{0,...,n+n}| < |ℕ|

> This means that there is an infinite intersection of
> all successors.

This means that, for any natural number k,
there are ways that what follows k can be totally-ordered
(NOT steppable-and-two-ended ways)
and there are ways theycannot be totally-ordered
(steppable-and-two-ended ways).

> Here quantifier exchange is correct.

Here quantifier shift is incorrect.

A quantifier shift is ALWAYS an unreliable step.

Because we are looking for statement-paths for which
each step is reliable, a quantifier shift is always
incorrect for such a path.

That doesn't rule out some other reliable-step path that
doesn't have a quantifier shift. However, proclaiming
that here is correct doesn't make a path reliable.
Compare to:
Proclaiming that your checking account is not overdrawn
doesn't put money in your account.

On 12/8/2021 9:15 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 14:34:06 UTC+1:
>> onsdag 8 december 2021 kl. 11:42:17 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Mittwoch, 8. Dezember 2021 um 06:39:11 UTC+1:
>>>
>>>> There is no such thing as "potential infinite"
>>> You are wrong. By endsegments you see the difference between pot and act: The set of endsegments with |∩{E(1), E(2), ..., E(k)}| = ℵ₀ is infinite but less than the sets which have an empty intersection |∩{E(1), E(2), ...}| = 0, like the set of all endsegments. They are actually infinite. That is much more than all finite sets.
>
>> Nope, I am right because you cannot define what your "potential infinite" even fucking mathematically mean for sets.
>
> It means finite but without a last element.
[...]

Huh? That's infinite. Look...

{ 0 }
{ { 0 }, 1 }
{ { { 0 }, 1 }, 2 }
....

This is: infinity!