On Monday, September 20, 2021 at 7:42:20 PM UTC+2, Gus Gassmann wrote:
> On Monday, 20 September 2021 at 14:13:50 UTC-3, Greg Cunt wrote:
> > On Monday, September 20, 2021 at 6:54:35 PM UTC+2, Serg io wrote:
> > > On 9/20/2021 11:20 AM, WM wrote:
> > > >
> > > > Two consecutive infinite sets in 1, 2, 3, ... ?
> > > >
> > > {1, 2, 3, ...} = {1, 3, 5, 7, ...} u {2, 4, 6, 8, ...}
> > >
> > Sure. But _not_
> >
> > for Ax e {1, 3, 5, 7, ...} and Ay e {2, 4, 6, 8, ...}: x < y .
> >
> > I guess, WM is asking for two infinite sets A and B (with A c IN and B c IN) such that for all a e A and for all b e B: a < b.
> >
> This, of course, depends on the definition of '<'.

Right. Clearly, I was assuming the "standard order" on the set of natural numbers here. In Mückenheim's case things aren't that clear.

> If a < b is defined properly, e.g., for a, b in |N, a < b iff [ (a in A and b in B) OR ( (b-a)/2 in |N ], Serg io's sets fit the bill. (Of course, WM can't fathom that there could be more than one way to order a set, but that's another story.)

And right, the fact that such an order *is* possible might tell us something, except WM, of course.

Greg Cunt wrote:
> On Monday, September 20, 2021 at 6:54:35 PM UTC+2, Serg io wrote:
>> On 9/20/2021 11:20 AM, WM wrote:
>>>
>>> Two consecutive infinite sets in 1, 2, 3, ... ?
>>>
>> {1, 2, 3, ...} = {1, 3, 5, 7, ...} u {2, 4, 6, 8, ...}
>
> Sure. But _not_
>
> for Ax e {1, 3, 5, 7, ...} and Ay e {2, 4, 6, 8, ...}: x < y .
>
> I guess, WM is asking for two infinite sets A and B (with A c IN and B c IN) such that for all a e A and for all b e B: a < b.

Crank Wolfgang Mueckenheim is regularly confusing sets of numbers,
sequences of numbers and well-ordered set of numbers. Mainly on
purpose of supporting his idiotic sophistries and also because
he is a complete idiot incapable of any rigorous math discourse.

fredag 17 september 2021 kl. 22:10:38 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 17. September 2021 um 07:18:31 UTC+2:
> > torsdag 16 september 2021 kl. 12:36:02 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:19:37 UTC+2:
> > > > onsdag 15 september 2021 kl. 17:37:24 UTC+2 skrev WM:
> > > > > Gus Gassmann schrieb am Mittwoch, 15. September 2021 um 15:46:10 UTC+2:
> > > > > > On Wednesday, 15 September 2021 at 09:57:18 UTC-3, WM wrote:
> > > > > >
> > > > > > > Between every infinite endsegment and {ω} there are infinitely many endegments. They are dark. Otherwise we could find finite endsegments or the basic definition
> > > > > > >
> > > > > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k} (*)
> > > > > > >
> > > > > > > would be broken.
> > > > > >
> > > > > > you have no idea what your equation involving the set difference in (*) actually tells you.
> > > > > It tells me and every person able to read math text, that no empty set can result unless finite endsegments are passed by.
> > > > >
> > > > It means nothing of the sort!
> > > You don't like mathematics? It is this: ∀k ∈ ℕ: E(k+1) = E(k) \ {k}. Can you see the universal quantifer?
> > >
> > I can read it better than you
> But you don't accept it?
> > The fact that the intersection of all endsegments is empty does not imply finite endsegments.
> If a finite endsegment E(m) was the minimal endsegment, then the intersection of all would be E(m).
>
> But if no finite endsegment is among them, then the intersection is less than every finite set? Only real fools can accept that.
>
> All infinite endsegments have an infinite intersection with each other. This is proved by inclusion monotony! As long as an endsegment contains elements, it contains them with all preceding endsegments.
>
> Regards, WM

Nope, the intersection of endsegments is empty. only FINITE intersections of endsegments has infintie cardinality.

The issue is you do not understand basic logic.

torsdag 16 september 2021 kl. 14:54:57 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:05:53 UTC+2:
>
> > All endsegments have the same cardinality and the set of all endsegments exists.
> Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ... and aleph_0 numbers beyond. Can you accept this?
>
> Regards, WM

Non-sequitor so no, I do not accept it.

Greg Cunt schrieb am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
> On Sunday, September 19, 2021 at 4:40:47 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Samstag, 18. September 2021 um 20:20:00 UTC+2:
>
> > > Let n e IN.
> > >
> > n does not belong to that set.
> To the set of all natural numbers, IN?

No. Every natural number has a unique prime decomposition. n has none.
>
> So we have n e IN, but

that is only an abbreviation for: A natural number is element of |N.

> > > So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?
> No answer?

Of course, since it is fact.

> > They are placeholders.
> So placeholders are elements in IN?

No. They are place holders for elements of |N.
>
>
> It's not a abbreviation of anything

It is well-known that matheologians claim counterfactual nonsense like:
- All fractions can be enumerated.
- There are uncountably many real numbers which can be well-ordered although only countably many can be distinguished. (This amounts to: there are many even prime numbers, but they cannot be constructed.)
- There are more paths in the Binary Tree than nodes. (This amounts to: there are more houses than bricks.)
- The bankruptcy of McDuck. (He earns 1000 $ every day and spends only one, but in the limit he will be bancrupt.)
- aleph_0 rational points separate uncountably many irrational points such that never two are existing without rational between them.
- All endsegments are infinite but their intersection is empty.

They are easily recognizable as fools. But your claim schlägt dem Fass die Krone ins Gesicht.

> But It might be the beginning of a proof.

Yes, but the proof shall be valid for all natural numbers like 1, 2, 3, ..., not only for n.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:11:54 UTC+2:
> fredag 17 september 2021 kl. 22:10:38 UTC+2 skrev WM:

> > All infinite endsegments have an infinite intersection with each other. This is proved by inclusion monotony! As long as an endsegment contains elements, it contains them with all preceding endsegments.
> >
> Nope, the intersection of endsegments is empty. only FINITE intersections of endsegments has infintie cardinality.
>
Every finite intersection. All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:18:09 UTC+2:
> torsdag 16 september 2021 kl. 14:54:57 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:05:53 UTC+2:
> >
> > > All endsegments have the same cardinality and the set of all endsegments exists.
> > Then there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ... and aleph_0 numbers beyond. Can you accept this?
> >
> Non-sequitor so no, I do not accept it.

What? All endsegments exhaust all indices? Or all endsegments have aleph_0 elements?

Regards, WM

Greg Cunt schrieb am Montag, 20. September 2021 um 19:13:50 UTC+2:
> On Monday, September 20, 2021 at 6:54:35 PM UTC+2, Serg io wrote:
> > On 9/20/2021 11:20 AM, WM wrote:
> > >
> > > Two consecutive infinite sets in 1, 2, 3, ... ?
> > >
> > {1, 2, 3, ...} = {1, 3, 5, 7, ...} u {2, 4, 6, 8, ...}
>
> Sure. But _not_
>
> for Ax e {1, 3, 5, 7, ...} and Ay e {2, 4, 6, 8, ...}: x < y .
>
> I guess, WM is asking for two infinite sets A and B (with A c IN and B c IN) such that for all a e A and for all b e B: a < b.

That is obvious, because the contents n, n+1, n+2, ... of endsegments follows upon their index n. Gassmann does not grasp that.

Regards, WM

fredag 17 september 2021 kl. 21:44:43 UTC+2 skrev WM:
> Greg Cunt schrieb am Donnerstag, 16. September 2021 um 18:04:33 UTC+2:
>
> > No, n is a natural number, not a "place holder".
> What is its prime decomposition? How many factors has it?
>
> Regards, WM

It has as many as it does.

tisdag 21 september 2021 kl. 13:03:30 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:11:54 UTC+2:
> > fredag 17 september 2021 kl. 22:10:38 UTC+2 skrev WM:
>
> > > All infinite endsegments have an infinite intersection with each other. This is proved by inclusion monotony! As long as an endsegment contains elements, it contains them with all preceding endsegments.
> > >
> > Nope, the intersection of endsegments is empty. only FINITE intersections of endsegments has infintie cardinality.
> >
> Every finite intersection. All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.
>
> Regards, WM
nope, infinite intersection yields empty, but finite ones yield an infintie set.

WM wrote on 9/21/2021 :
> Greg Cunt schrieb am Montag, 20. September 2021 um 19:13:50 UTC+2:
>> On Monday, September 20, 2021 at 6:54:35 PM UTC+2, Serg io wrote:
>>> On 9/20/2021 11:20 AM, WM wrote:
>>>>
>>>> Two consecutive infinite sets in 1, 2, 3, ... ?
>>>>
>>> {1, 2, 3, ...} = {1, 3, 5, 7, ...} u {2, 4, 6, 8, ...}
>>
>> Sure. But _not_
>>
>> for Ax e {1, 3, 5, 7, ...} and Ay e {2, 4, 6, 8, ...}: x < y .
>>
>> I guess, WM is asking for two infinite sets A and B (with A c IN and B c IN)
>> such that for all a e A and for all b e B: a < b.
>
> That is obvious, because the contents n, n+1, n+2, ... of endsegments follows
> upon their index n. Gassmann does not grasp that.

Not obvious the way you wrote it. Your pipe-symbol partitioning has "n"
in both partitions so it is not even a set.

zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 14:02:08 UTC+2:
> fredag 17 september 2021 kl. 21:44:43 UTC+2 skrev WM:
> > Greg Cunt schrieb am Donnerstag, 16. September 2021 um 18:04:33 UTC+2:
> >
> > > No, n is a natural number, not a "place holder".
> > What is its prime decomposition? How many factors has it?
> >
> It has as many as it does.

It has none that anybody could find. It is matheology like all your ZF-nonsense.

Regards, WM

FromTheRafters schrieb am Dienstag, 21. September 2021 um 14:05:57 UTC+2:

> >> I guess, WM is asking for two infinite sets A and B (with A c IN and B c IN)
> >> such that for all a e A and for all b e B: a < b.
> >
> > That is obvious, because the contents n, n+1, n+2, ... of endsegments follows
> > upon their index n. Gassmann does not grasp that.
> Not obvious the way you wrote it. Your pipe-symbol partitioning has "n"
> in both partitions so it is not even a set.

Small wonder. 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not in both positions an infinite set can exist. Your claim is n infinite set of indices and an infinite set following every index. That is wrong.

Regards, WM

On 9/21/2021 5:57 AM, WM wrote:
> Greg Cunt schrieb am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
>> On Sunday, September 19, 2021 at 4:40:47 PM UTC+2, WM wrote:
>>> Greg Cunt schrieb am Samstag, 18. September 2021 um 20:20:00 UTC+2:
>>
>>>> Let n e IN.
>>>>
>>> n does not belong to that set.
>> To the set of all natural numbers, IN?
>
> No. Every natural number has a unique prime decomposition. n has none.
>>
>> So we have n e IN, but
>
> that is only an abbreviation for: A natural number is element of |N.
>
>>>> So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?
>> No answer?
>
> Of course, since it is fact.
>
>>> They are placeholders.
>> So placeholders are elements in IN?
>
> No. They are place holders for elements of |N.
>>
>>
>> It's not a abbreviation of anything
>
> It is well-known that matheologians claim counterfactual nonsense like:
> - All fractions can be enumerated.
> - There are uncountably many real numbers which can be well-ordered although only countably many can be distinguished. (This amounts to: there are many even prime numbers, but they cannot be constructed.)
> - There are more paths in the Binary Tree than nodes. (This amounts to: there are more houses than bricks.)
> - The bankruptcy of McDuck. (He earns 1000 $ every day and spends only one, but in the limit he will be bancrupt.)
> - aleph_0 rational points separate uncountably many irrational points such that never two are existing without rational between them.
> - All endsegments are infinite but their intersection is empty.
>
> They are easily recognizable as fools. But your claim schlägt dem Fass die Krone ins Gesicht.
>
>> But It might be the beginning of a proof.
>
> Yes, but the proof shall be valid for all natural numbers like 1, 2, 3, ..., not only for n.
>
> Regards, WM
>

Extra Credit Problems!!: Prove or show these statements are false.

1 - All fractions can be enumerated.

2 - There are uncountably many real numbers which can be well-ordered although only countably many can be distinguished. (This amounts to: there are
many even prime numbers, but they cannot be constructed.)

3 - There are more paths in the Binary Tree than nodes. (This amounts to: there are more houses than bricks.)

4 - The bankruptcy of McDuck. (He earns 1000 $ every day and spends only one, but in the limit he will be bancrupt.)

5 - aleph_0 rational points separate uncountably many irrational points such that never two are existing without rational between them.

6 - All endsegments are infinite but their intersection is empty.

On Tuesday, 21 September 2021 at 10:26:47 UTC-3, WM wrote:
> FromTheRafters schrieb am Dienstag, 21. September 2021 um 14:05:57 UTC+2:
>
> > >> I guess, WM is asking for two infinite sets A and B (with A c IN and B c IN)
> > >> such that for all a e A and for all b e B: a < b.
> > >
> > > That is obvious, because the contents n, n+1, n+2, ... of endsegments follows
> > > upon their index n. Gassmann does not grasp that.
> > Not obvious the way you wrote it. Your pipe-symbol partitioning has "n"
> > in both partitions so it is not even a set.
> Small wonder. 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not in both positions an infinite set can exist. Your claim is n infinite set of indices and an infinite set following every index. That is wrong.

Your stupidity boggles the mind. For every natural number n there are infinitely many others larger than it, which are elements of the end segment E(n). That's all there is to it. You are incredibly imaginative when it comes to deceiving yourself, but at the end it is still a simple unjustified quantifier switch. As Bob Dylan said "[your] mind has been mismanaged with great skill".

On Tuesday, September 21, 2021 at 1:03:30 PM UTC+2, WM wrote:

> All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.

Mumbo-jumbo. (Read: Nonsense)

WM expressed precisely :
> FromTheRafters schrieb am Dienstag, 21. September 2021 um 14:05:57 UTC+2:
>
>>>> I guess, WM is asking for two infinite sets A and B (with A c IN and B c
>>>> IN) such that for all a e A and for all b e B: a < b.
>>>
>>> That is obvious, because the contents n, n+1, n+2, ... of endsegments
>>> follows upon their index n. Gassmann does not grasp that.
>> Not obvious the way you wrote it. Your pipe-symbol partitioning has "n"
>> in both partitions so it is not even a set.
>
> Small wonder. 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not
> in both positions an infinite set can exist. Your claim is n infinite set of
> indices and an infinite set following every index. That is wrong.

Yet the non-negative integers {0,1,2,3,...}, all Aleph-zero of them,
get along just fine with the negative integers {-1,-2,-3,...}, all
Aleph-zero of them, in a set of cardinality Aleph-zero.

Are you denying the existence of the integers now? It wouldn't surprise
anyone here.

On 9/21/2021 9:53 AM, Greg Cunt wrote:
> On Tuesday, September 21, 2021 at 1:03:30 PM UTC+2, WM wrote:
>
>> All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.
>
> Mumbo-jumbo. (Read: Nonsense)
>

A = finite intersections
B = infinite intersection

All A amount to B because B is only over all finite sets.

On Tuesday, September 21, 2021 at 12:57:12 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
> > On Sunday, September 19, 2021 at 4:40:47 PM UTC+2, WM wrote:
> > > Greg Cunt schrieb am Samstag, 18. September 2021 um 20:20:00 UTC+2:

Let n e IN.

> Every natural number has a unique prime decomposition. n has none.

Huh?! I mean... errr?

Since every natural number has a unique prime decomposition and n e IN (i.e.. n is a natural number), n has a unique prime decomposition.

What's wrong with you, Mückenheim?

> > So we have
> >
> > n e IN
> >
> that is only an abbreviation for: A natural number is element of IN.

No, it says/expresses: "n is a natural number" (well, actually it says/expresses "n is an element in IN").

> > > > So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?
> > > >
> > No answer?
> >
> Of course, since it is fact.

What is a fact?

That n isn't a natural number even though n e IN?

Are you SERIOUS?

> > > They are placeholders.
> > >
> > So placeholders are elements in IN?
> >
> No. They are place holders for elements of IN.

No, n and n+1 aren't "place holders for elements of IN", BUT _elements in IN_.

That's what the expressions

n e IN

and

n+1 e IN

"say" or "express", namely:

n is an element in IN

and

n+1 is an element in IN .

> > It is well-known that matheologians claim counterfactual nonsense

In this case you clearly qualify for a matheologian.

> They are easily recognizable as fools.

Indeed! :-)

Again: 'Let n e IN.'

> > might be the beginning of a proof.
> >
> Yes, but the proof shall be valid for all natural numbers like 1, 2, 3, ...., not only for n.

Holy shit! :-(

On Tuesday, September 21, 2021 at 5:11:27 PM UTC+2, Serg io wrote:
> On 9/21/2021 9:53 AM, Greg Cunt wrote:
> > On Tuesday, September 21, 2021 at 1:03:30 PM UTC+2, WM wrote:
> > >
> > > All finite intersections amount to the infinite intersection because the infinite intersection is only over all finite sets.
> > >
> > Mumbo-jumbo. (Read: Nonsense)
> >
> A = finite intersections
> B = infinite intersection
>
> All A amount to B because B is only over all finite sets.

Thanks. That helped a lot.

https://www.youtube.com/watch?v=xzGV9Bl6CGg

On Tuesday, September 21, 2021 at 1:13:05 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 21. September 2021 um 07:18:09 UTC+2:
> > torsdag 16 september 2021 kl. 14:54:57 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Donnerstag, 16. September 2021 um 07:05:53 UTC+2:
> > > >
> > > > All endsegments have the same cardinality and the set of all endsegments exists. (*)
> > > >
> > > Then

Nope. Seems you didn't get it, Mückenheim:

> > Non-sequitur

Your idiotic claim

there are aleph_0 numbers before the | in 1, 2 3, ..., n | n, n+1, n+2, ... and [bla]

does NOT follow from (*).

> > Can you accept this?
> >
> so no, I do not accept it.

Hint: All endsegments have the same cardinality and the set of all endsegments exists (at least in the context of set theory, say ZFC).

On Tuesday, September 21, 2021 at 3:26:47 PM UTC+2, WM wrote:

> 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not in both positions an infinite set can exist.

Why don't you express this using standard notation?

Say, An e IN: card({1, 2, 3, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 ,

or: An e IN: {1, 2, 3, ..., n} is finite & {n, n+1, n+2, ...} is infinite .

> Your claim is n infinite set of indices and an infinite set following every index. That is wrong.

Mumbo-Jumbo.

Yeah, there are infinitely many ("finite") numbers in IN and each one is followed by infinitely many natural numbers.

If the natural numbers in our system (i.e. our set theory) are defined due to von Neumann (or due to Zermelo), we actually have:

There are infinitely many numbers in IN [all of which are finite sets] and each one is followed by infinitely many natural numbers.

On Tuesday, September 21, 2021 at 3:48:53 PM UTC+2, Gus Gassmann wrote:
> On Tuesday, 21 September 2021 at 10:26:47 UTC-3, WM wrote:

> You are incredibly imaginative when it comes to deceiving yourself, but at the end it is still a simple unjustified quantifier switch. As Bob Dylan said "[your] mind has been mismanaged with great skill".

Imho he's suffering from a delusion.

He will ALWAYS find ways to avoid to see "the facts".

You know, otherwise his idee fixe might be in danger. (This has to be avoided at all costs.)

On Tuesday, 21 September 2021 at 12:40:14 UTC-3, Greg Cunt wrote:
> On Tuesday, September 21, 2021 at 3:26:47 PM UTC+2, WM wrote:
>
> > 1, 2, 3, ..., n | n, n+1, n+2, ... is simply showing that not in both positions an infinite set can exist.
> Why don't you express this using standard notation?
>
> Say, An e IN: card({1, 2, 3, ..., n}) = n & card({n, n+1, n+2, ...}) = aleph_0 ,
>
> or: An e IN: {1, 2, 3, ..., n} is finite & {n, n+1, n+2, ...} is infinite .

Well, with all due respect, you know the reason. First, he is constitutionally unable to write proper mathematics. This has been well established. Even when just plagiarizing other people's work, he gets things wrong. But more importantly, how can he apply his usual misdirection and subterfuge without the inherent ambiguity of natural language? If he used the last line of yours that I quoted, his improper switching of quantifiers would be far too obvious, perhaps even to himself. His picture allows him to pretend at least that there are different limiting processes on the left and on the right of the vertical bar, namely that the left side has infinitely many elements, and the right side does as well, for every n. (Even writing this down sounds grating to me, so it is clear that there must be a flaw somewhere, but in addition to an incomprehension of mathematics that borders on the criminal, WM is also quite word-deaf.)

On 9/21/2021 10:13 AM, Greg Cunt wrote:
> On Tuesday, September 21, 2021 at 12:57:12 PM UTC+2, WM wrote:
>> Greg Cunt schrieb am Sonntag, 19. September 2021 um 19:18:09 UTC+2:
>>> On Sunday, September 19, 2021 at 4:40:47 PM UTC+2, WM wrote:
>>>> Greg Cunt schrieb am Samstag, 18. September 2021 um 20:20:00 UTC+2:
>
> Let n e IN.
>
>> Every natural number has a unique prime decomposition. n has none.
>
> Huh?! I mean... errr?
>
> Since every natural number has a unique prime decomposition and n e IN (i.e. n is a natural number), n has a unique prime decomposition.
>
> What's wrong with you, Mückenheim?
>
>>> So we have
>>>
>>> n e IN
>>>
>> that is only an abbreviation for: A natural number is element of IN.
>
> No, it says/expresses: "n is a natural number" (well, actually it says/expresses "n is an element in IN").
>
>>>>> So n is in IN and n+1 is in IN, but neither n nor n+1 are natural numbers? Is that your claim?
>>>>>
>>> No answer?
>>>
>> Of course, since it is fact.
>
> What is a fact?
>
> That n isn't a natural number even though n e IN?
>
> Are you SERIOUS?
>
>>>> They are placeholders.
>>>>
>>> So placeholders are elements in IN?
>>>
>> No. They are place holders for elements of IN.
>
> No, n and n+1 aren't "place holders for elements of IN", BUT _elements in IN_.
>
> That's what the expressions
>
> n e IN
>
> and
>
> n+1 e IN
>
> "say" or "express", namely:
>
> n is an element in IN
>
> and
>
> n+1 is an element in IN .
>
>>> It is well-known that matheologians claim counterfactual nonsense
>
> In this case you clearly qualify for a matheologian.
>
>> They are easily recognizable as fools.
>
> Indeed! :-)
>
> Again: 'Let n e IN.'
>
>>> might be the beginning of a proof.
>>>
>> Yes, but the proof shall be valid for all natural numbers like 1, 2, 3, ..., not only for n.
>
> Holy shit! :-(
>

whoa, that last one is a keeper!

Let n e IN
n = 1 + 2
therefore n = 3

however, "proof shall be valid for all natural numbers like 1, 2, 3, ..., not only for n"