On Wednesday, May 4, 2022 at 4:28:07 PM UTC-4, Mostowski Collapse wrote:
> s(1/2)=3/2 is possible.
>

You could certainly construct another successor on Q.

> It doesn't violate any of your axioms.
>

You wouldn't be able to apply the usual successor function. It's, for all elements in N, etc. Not for all elements in Q.

And the Induction Axiom wouldn't hold.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Dan Christensen wrote:

> You wouldn't be able to apply the usual successor function. It's, for
> all elements in N, etc. Not for all elements in Q.
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com Visit my
> Math Blog at http://www.dcproof.wordpress.com

you gave that "program" the initials of your name, are you insane? Who
does that. You look like a slavic nazi turned anglo-saxon overnight.

Dan Christensen wrote:

> On Wednesday, May 4, 2022 at 2:31:39 AM UTC-4, Mostowski Collapse (aka
>> We need this: dom(add) = NxN
>
> Nope. See my previous reply to you.

No. See this first.

String of suicides prompts relocation of US aircraft carrier sailors
https://cdni.russiatoday.com/files/2022.05/l/62728dce85f5407a436ac8a6.jpg
https://www.rt.com/news/554961-george-washington-sailors-suicide/
Over 200 sailors have been re-housed from the USS George Washington
aircraft carrier

US training Ukrainians at ex-Nazi Wehrmacht base
https://www.rt.com/russia/554975-training-ukraine-troops-germany/
The US has been training Ukrainian forces for eight years and is using a
major base in Germany to continue doing so, a top general said

'An act of genocide': A witness recalls the 2014 Odessa massacre
https://www.rt.com/russia/554946-tragedy-odessa-protests-ukraine/

On Wednesday, May 4, 2022 at 4:53:46 PM UTC-4, Lanny Torii wrote:
> Dan Christensen wrote:
>
> > On Wednesday, May 4, 2022 at 2:31:39 AM UTC-4, Mostowski Collapse (aka
> >> We need this: dom(add) = NxN
> >
> > Nope. See my previous reply to you.
> No. See this first.
>
>
> String of suicides prompts relocation of US aircraft carrier sailors
> https://cdni.russiatoday.com/files/2022.05/l/62728dce85f5407a436ac8a6.jpg
> https://www.rt.com/news/554961-george-washington-sailors-suicide/
> Over 200 sailors have been re-housed from the USS George Washington
> aircraft carrier
>
> US training Ukrainians at ex-Nazi Wehrmacht base
> https://www.rt.com/russia/554975-training-ukraine-troops-germany/
> The US has been training Ukrainian forces for eight years and is using a
> major base in Germany to continue doing so, a top general said
>
> 'An act of genocide': A witness recalls the 2014 Odessa massacre
> https://www.rt.com/russia/554946-tragedy-odessa-protests-ukraine/

WAR OF AGRESSION = NAZI

How's that Blitzkrieg going, Nazi Boy? (Hee, hee!)

Why would your induction axiom not hold if s(1/2)=3/2.

You only say:

> 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> => [0 in a & ALL(b):[b in a => s(b) in a]
> => ALL(b):[b in n => b in a]]]

So you talk about sets b which are subset of n.

Dan Christensen schrieb am Mittwoch, 4. Mai 2022 um 22:42:35 UTC+2:
> On Wednesday, May 4, 2022 at 4:28:07 PM UTC-4, Mostowski Collapse wrote:
> > s(1/2)=3/2 is possible.
> >
>
> You could certainly construct another successor on Q.
> > It doesn't violate any of your axioms.
> >
> You wouldn't be able to apply the usual successor function. It's, for all elements in N, etc. Not for all elements in Q.
>
> And the Induction Axiom wouldn't hold.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Corr.: Typo

So you talk about sets a which are subset of n.

Mostowski Collapse schrieb am Mittwoch, 4. Mai 2022 um 23:29:19 UTC+2:
> Why would your induction axiom not hold if s(1/2)=3/2.
>
> You only say:
> > 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> > => [0 in a & ALL(b):[b in a => s(b) in a]
> > => ALL(b):[b in n => b in a]]]
> So you talk about sets b which are subset of n.
> Dan Christensen schrieb am Mittwoch, 4. Mai 2022 um 22:42:35 UTC+2:
> > On Wednesday, May 4, 2022 at 4:28:07 PM UTC-4, Mostowski Collapse wrote:
> > > s(1/2)=3/2 is possible.
> > >
> >
> > You could certainly construct another successor on Q.
> > > It doesn't violate any of your axioms.
> > >
> > You wouldn't be able to apply the usual successor function. It's, for all elements in N, etc. Not for all elements in Q.
> >
> > And the Induction Axiom wouldn't hold.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

Dan Christensen wrote:

>> US training Ukrainians at ex-Nazi Wehrmacht base
>> https://www.rt.com/russia/554975-training-ukraine-troops-germany/
>> The US has been training Ukrainian forces for eight years and is using
>> a major base in Germany to continue doing so, a top general said
>>
>> 'An act of genocide': A witness recalls the 2014 Odessa massacre
>> https://www.rt.com/russia/554946-tragedy-odessa-protests-ukraine/
>
> WAR OF AGRESSION = NAZI
> How's that Blitzkrieg going, Nazi Boy? (Hee, hee!)

no such thing, but the western europe and cacanada, did lost huge storage
of nazi_armament, a few generals and alot of nazi illegal combatants, in
the process. It stays by the mainstream media tv and newspapers. And they
want to lose more. The prices on food and fuel already exceed the double,
food shortage, heat shortage, detergent shortages and so on. How blitz
would be that??

On Wednesday, May 4, 2022 at 5:29:19 PM UTC-4, Mostowski Collapse wrote:
> Why would your induction axiom not hold if s(1/2)=3/2.
>

You would have to introduce quite a different successor function to capture all of the rationals.

> You only say:
> > 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> > => [0 in a & ALL(b):[b in a => s(b) in a]
> > => ALL(b):[b in n => b in a]]]
> So you talk about sets b which are subset of n.

Not subsets of Q.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

On Wednesday, May 4, 2022 at 5:37:19 PM UTC-4, Lanny Torii wrote:
> Dan Christensen wrote:
>
> >> US training Ukrainians at ex-Nazi Wehrmacht base
> >> https://www.rt.com/russia/554975-training-ukraine-troops-germany/
> >> The US has been training Ukrainian forces for eight years and is using
> >> a major base in Germany to continue doing so, a top general said
> >>
> >> 'An act of genocide': A witness recalls the 2014 Odessa massacre
> >> https://www.rt.com/russia/554946-tragedy-odessa-protests-ukraine/
> >
> > WAR OF AGRESSION = NAZI
> > How's that Blitzkrieg going, Nazi Boy? (Hee, hee!)
> no such thing, but the western europe and cacanada...

Just keep that sturdy length of rope handy, Nazi Boy. You may need it.

Dan Christensen wrote:

>> >> US training Ukrainians at ex-Nazi Wehrmacht base
>> >> https://www.rt.com/russia/554975-training-ukraine-troops-germany/
>> >> The US has been training Ukrainian forces for eight years and is
>> >> using a major base in Germany to continue doing so, a top general
>> >> said
>> >>
>> >> 'An act of genocide': A witness recalls the 2014 Odessa massacre
>> >> https://www.rt.com/russia/554946-tragedy-odessa-protests-ukraine/
>> >
>> > WAR OF AGRESSION = NAZI How's that Blitzkrieg going, Nazi Boy? (Hee,
>> > hee!)
>> no such thing, but the western europe and cacanada...
>
> Just keep that sturdy length of rope handy, Nazi Boy. You may need it.

What cacanada, you have to move from cacanada, immigrant. Provable you are
a nazi individual, born like that, in a family where nazism was built in
the system. Here some other proofs. Watch carefully the blitz, because you
don't know what you do.

UKRAINE WAR IN 17 SEC IT STARTS WHEN THE CAMERA IS ON AND IT STOPS WHEN
THE CAMERA IS OFF GET IT https://www.bitchute.com/video/CTYjHQsAj3VL/

On Wednesday, May 4, 2022 at 6:12:08 PM UTC-4, Lanny Torii wrote:
> Dan Christensen wrote:
>
> >> >> US training Ukrainians at ex-Nazi Wehrmacht base
> >> >> https://www.rt.com/russia/554975-training-ukraine-troops-germany/
> >> >> The US has been training Ukrainian forces for eight years and is
> >> >> using a major base in Germany to continue doing so, a top general
> >> >> said
> >> >>
> >> >> 'An act of genocide': A witness recalls the 2014 Odessa massacre
> >> >> https://www.rt.com/russia/554946-tragedy-odessa-protests-ukraine/
> >> >
> >> > WAR OF AGRESSION = NAZI How's that Blitzkrieg going, Nazi Boy? (Hee,
> >> > hee!)
> >> no such thing, but the western europe and cacanada...
> >
> > Just keep that sturdy length of rope handy, Nazi Boy. You may need it.
> What cacanada, you have to move from cacanada, immigrant. Provable you are
> a nazi individual, born like that, in a family where nazism was built in
> the system. Here some other proofs. Watch carefully the blitz, because you
> don't know what you do.
>

Looks like you might be turning in your keyboard for a spiffy SS uniform in the next week or so, Nazi Boy!

Nothing prevents your Peano axioms from being satisfied by for example:

s : Q -> Q

So they are nonsens…

Dan Christensen schrieb am Mittwoch, 4. Mai 2022 um 23:50:27 UTC+2:
> On Wednesday, May 4, 2022 at 5:29:19 PM UTC-4, Mostowski Collapse wrote:
> > Why would your induction axiom not hold if s(1/2)=3/2.
> >
> You would have to introduce quite a different successor function to capture all of the rationals.
> > You only say:
> > > 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> > > => [0 in a & ALL(b):[b in a => s(b) in a]
> > > => ALL(b):[b in n => b in a]]]
> > So you talk about sets b which are subset of n.
> Not subsets of Q.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

On Thursday, May 5, 2022 at 1:06:26 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Mittwoch, 4. Mai 2022 um 23:50:27 UTC+2:
> > On Wednesday, May 4, 2022 at 5:29:19 PM UTC-4, Mostowski Collapse wrote:
> > > Why would your induction axiom not hold if s(1/2)=3/2.
> > >
> > You would have to introduce quite a different successor function to capture all of the rationals.
> > > You only say:
> > > > 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> > > > => [0 in a & ALL(b):[b in a => s(b) in a]
> > > > => ALL(b):[b in n => b in a]]]

> > > So you talk about sets b which are subset of n.
> > Not subsets of Q.

> Nothing prevents your Peano axioms from being satisfied by for example:
>
> s : Q -> Q
>

s(-1) = ???

Can't be 0. That would contradict the 4th Axiom.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Since when is -1 e n ? Your axiom only says:

ALL(a):[a in n => ~s(a)=0]

So its possible to have s(-1)=0, which would be the case if s : Q -> Q.

Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 15:50:56 UTC+2:
> On Thursday, May 5, 2022 at 1:06:26 AM UTC-4, Mostowski Collapse wrote:
>
> > Dan Christensen schrieb am Mittwoch, 4. Mai 2022 um 23:50:27 UTC+2:
> > > On Wednesday, May 4, 2022 at 5:29:19 PM UTC-4, Mostowski Collapse wrote:
> > > > Why would your induction axiom not hold if s(1/2)=3/2.
> > > >
> > > You would have to introduce quite a different successor function to capture all of the rationals.
> > > > You only say:
> > > > > 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> > > > > => [0 in a & ALL(b):[b in a => s(b) in a]
> > > > > => ALL(b):[b in n => b in a]]]
>
> > > > So you talk about sets b which are subset of n.
> > > Not subsets of Q.
> > Nothing prevents your Peano axioms from being satisfied by for example:
> >
> > s : Q -> Q
> >
> s(-1) = ???
>
> Can't be 0. That would contradict the 4th Axiom.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Dan Christensen wrote:

> s(-1) = ??? Can't be 0. That would contradict the 4th Axiom. Dan
> Download my DC Proof 2.0 freeware at http://www.dcproof.com Visit my
> Math Blog at http://www.dcproof.wordpress.com

how could you ever be "programming" something like that.

the problem you have is that the nazi "ukrainians" may be nazis, educating
their children in being strong nazis, but if one came along saying it,
then *aioe.org* would exclud *sci.physics.relativity* from posting. I
believe a nazi bitch named *Thomas_Pointedhead* / *Big_Dog* did it. This
world is fucked up, my friend. Try again.

Another Ukraine Citizens About Ukraine Nazi Forces
https://www.bitchute.com/video/XzJR1vo222cT/

HITLER'S NEW AGE RELIGION 'CROSSDRESSER NAZI SOLDIERS'
https://www.bitchute.com/video/t8yOLyjNcAFB/
one more proof, the khazar zelenske is a nazi.

2022.05.04 Russian Sanctions On Europe
https://www.bitchute.com/video/i2e0EBj6PmiX/
you have to move away from fake_money western europe, now as you can. The
*fake_money* took their corrupt minds in a parallel world. They truly
believe in their fake_money, which is historically.

On Thursday, May 5, 2022 at 10:23:33 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 15:50:56 UTC+2:
> > On Thursday, May 5, 2022 at 1:06:26 AM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Mittwoch, 4. Mai 2022 um 23:50:27 UTC+2:
> > > > On Wednesday, May 4, 2022 at 5:29:19 PM UTC-4, Mostowski Collapse wrote:
> > > > > Why would your induction axiom not hold if s(1/2)=3/2.
> > > > >
> > > > You would have to introduce quite a different successor function to capture all of the rationals.
> > > > > You only say:
> > > > > > 6. ALL(a):[Set(a) & ALL(b):[b in a => b in n]
> > > > > > => [0 in a & ALL(b):[b in a => s(b) in a]
> > > > > > => ALL(b):[b in n => b in a]]]
> >
> > > > > So you talk about sets b which are subset of n.
> > > > Not subsets of Q.
> > > Nothing prevents your Peano axioms from being satisfied by for example:
> > >
> > > s : Q -> Q
> > >
> > s(-1) = ???
> >
> > Can't be 0. That would contradict the 4th Axiom.

> Since when is -1 e n ? Your axiom only says:
>
> ALL(a):[a in n => ~s(a)=0]

> So its possible to have s(-1)=0, which would be the case if s : Q -> Q.

What do you imagine is the next rational number after 0, Jan Burse? Can't be 1 since there are infinitely many rational numbers between 0 and 1. How will you get to 1/2 starting from 0?

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

On Thursday, May 5, 2022 at 11:03:10 AM UTC-4, Grant Shirasu wrote:

>
> the problem you have is that the nazi "ukrainians" may be nazis, educating

WAR OF AGRESSION = NAZI

Looks like you may soon be turning in your keyboard for a spiffy new SS uniform on your way to the Ukrainian front, Nazi Boy. Won't you have fun raping and murdering Ukrainian women and children!

I dont have to get to 1/2, I only said s : Q -> Q, does
also satisfy your axioms. I did nowhere say that your
axioms construct Q. But every s : A -> B, with:

s | N = succ

Satisfies your axioms, where succ : N -> N is the
usual successor function on the natural numbers.
f | A is a notation from set theory, it denotes

the function f restricted to the domain A.

Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 17:15:18 UTC+2:
> On Thursday, May 5, 2022 at 10:23:33 AM UTC-4, Mostowski Collapse wrote:
>
> > So its possible to have s(-1)=0, which would be the case if s : Q -> Q.
> What do you imagine is the next rational number after 0, Jan Burse? Can't be 1 since there are infinitely many rational numbers between 0 and 1. How will you get to 1/2 starting from 0?
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Mostowski Collapse wrote:

> I dont have to get to 1/2, I only said s : Q -> Q, does also satisfy
> your axioms. I did nowhere say that your axioms construct Q. But every s
> : A -> B, with: s | N = succ

yet another *Sulky_Liver_Sausage*.

https://youtu.be/R7MOsliXbtc

the problem you have is that the nazi "ukrainians" may be nazis, educating
their children in being strong nazis, but if one came along saying it,
then *aioe.org* would exclude *sci.physics.relativity* from posting. I
believe a nazi bitch named *Thomas_Pointedhead* / *Big_Dog* did it. This
world is fucked up, my friend. Try again.

Another Ukraine Citizens About Ukraine Nazi Forces
https://www.bitchute.com/video/XzJR1vo222cT/

HITLER'S NEW AGE RELIGION 'CROSSDRESSER NAZI SOLDIERS'
https://www.bitchute.com/video/t8yOLyjNcAFB/
one more proof, the khazar zelenske is a nazi.

2022.05.04 Russian Sanctions On Europe
https://www.bitchute.com/video/i2e0EBj6PmiX/
you have to move away from fake_money western europe, now as you can. The
*fake_money* took their corrupt minds in a parallel world. They truly
believe in their fake_money, which is historically.

The proof is pretty easy, just observe that
because s | N = succ we have of course:

/* Restriction Lemma */
ALL(a):[a e N => s(a) = succ(a)]

Now since all your axioms are of the form
ALL(a):[a e N => … ] and since the induction
axiom only talks about subsets of N,

we see it is enough for a function s : A -> B
to satisfy s | N = succ to then satisfy your
axioms. If I am not totally mistaken, the

converse can be also shown.

On Thursday, May 5, 2022 at 1:39:37 PM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 17:15:18 UTC+2:
> > On Thursday, May 5, 2022 at 10:23:33 AM UTC-4, Mostowski Collapse wrote:
> >
> > > So its possible to have s(-1)=0, which would be the case if s : Q -> Q.
> > What do you imagine is the next rational number after 0, Jan Burse? Can't be 1 since there are infinitely many rational numbers between 0 and 1. How will you get to 1/2 starting from 0?

> I dont have to get to 1/2,

Actually, you do, Jan Burse. If you claim to have a successor function on Q, you must be able to get from 0 to 1/2 by means of repeated application of that function. Otherwise, you don't really have a successor function on Q.

[snip]

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

Well the predecessor of 1/2 is -1/2. But I dont have to
reach 1/2 from some natural number as you wrote:

Dan Christensen after sniffing one kilogram cocaine:
> What do you imagine is the next rational number
after 0, Jan Burse? Can't be 1 since there are infinitely
many rational numbers between 0 and 1. How will
you get to 1/2 starting from 0.
https://groups.google.com/g/sci.math/c/sYkWEHylBAA/m/aJnhQLDEAgAJ

LoL

BTW: There are many many functions satisfying
your „Peano Axioms“, like for example:

succZ : Z -> Z, x |-> x+1
succQ : Q -> Q, x |-> x+1
succR : R -> R, x |-> x+1
succC : C -> C, x |-> x+1

C are the complex numbers. These functions all satisfy
your „Peano Axioms“ because these functions all satisfy

succZ | N = succQ | N = succR | N = succC | N = succN

and because of the way you wrote down your axioms.

Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 20:12:44 UTC+2:
> On Thursday, May 5, 2022 at 1:39:37 PM UTC-4, Mostowski Collapse wrote:
>
> > Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 17:15:18 UTC+2:
> > > On Thursday, May 5, 2022 at 10:23:33 AM UTC-4, Mostowski Collapse wrote:
> > >
> > > > So its possible to have s(-1)=0, which would be the case if s : Q -> Q.
> > > What do you imagine is the next rational number after 0, Jan Burse? Can't be 1 since there are infinitely many rational numbers between 0 and 1. How will you get to 1/2 starting from 0?
> > I dont have to get to 1/2,
> Actually, you do, Jan Burse. If you claim to have a successor function on Q, you must be able to get from 0 to 1/2 by means of repeated application of that function. Otherwise, you don't really have a successor function on Q.
>
> [snip]
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Because succQ(-1/2)=1/2 we have succQ^(-1)(1/2)=-1/2.
Too much math in one day, or whats the prololoblem?

LMAO!

Mostowski Collapse schrieb am Donnerstag, 5. Mai 2022 um 20:51:37 UTC+2:
> Well the predecessor of 1/2 is -1/2. But I dont have to
> reach 1/2 from some natural number as you wrote:
>
> Dan Christensen after sniffing one kilogram cocaine:
> > What do you imagine is the next rational number
> after 0, Jan Burse? Can't be 1 since there are infinitely
> many rational numbers between 0 and 1. How will
> you get to 1/2 starting from 0.
> https://groups.google.com/g/sci.math/c/sYkWEHylBAA/m/aJnhQLDEAgAJ
>
> LoL
>
> BTW: There are many many functions satisfying
> your „Peano Axioms“, like for example:
>
> succZ : Z -> Z, x |-> x+1
> succQ : Q -> Q, x |-> x+1
> succR : R -> R, x |-> x+1
> succC : C -> C, x |-> x+1
>
> C are the complex numbers. These functions all satisfy
> your „Peano Axioms“ because these functions all satisfy
>
> succZ | N = succQ | N = succR | N = succC | N = succN
>
> and because of the way you wrote down your axioms.
> Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 20:12:44 UTC+2:
> > On Thursday, May 5, 2022 at 1:39:37 PM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 17:15:18 UTC+2:
> > > > On Thursday, May 5, 2022 at 10:23:33 AM UTC-4, Mostowski Collapse wrote:
> > > >
> > > > > So its possible to have s(-1)=0, which would be the case if s : Q -> Q.
> > > > What do you imagine is the next rational number after 0, Jan Burse? Can't be 1 since there are infinitely many rational numbers between 0 and 1. How will you get to 1/2 starting from 0?
> > > I dont have to get to 1/2,
> > Actually, you do, Jan Burse. If you claim to have a successor function on Q, you must be able to get from 0 to 1/2 by means of repeated application of that function. Otherwise, you don't really have a successor function on Q.
> >
> > [snip]
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 20:12:44 UTC+2:
> > On Thursday, May 5, 2022 at 1:39:37 PM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 17:15:18 UTC+2:
> > > > On Thursday, May 5, 2022 at 10:23:33 AM UTC-4, Mostowski Collapse wrote:
> > > >
> > > > > So its possible to have s(-1)=0, which would be the case if s : Q -> Q.
> > > > What do you imagine is the next rational number after 0, Jan Burse? Can't be 1 since there are infinitely many rational numbers between 0 and 1. How will you get to 1/2 starting from 0?
> > > I dont have to get to 1/2,
> > Actually, you do, Jan Burse. If you claim to have a successor function on Q, you must be able to get from 0 to 1/2 by means of repeated application of that function. Otherwise, you don't really have a successor function on Q.
> >

> Well the predecessor of 1/2 is -1/2. But I dont have to
> reach 1/2 from some natural number as you wrote:
>

Otherwise, induction doesn't work on the rational numbers. (It doesn't.) Deal with it, Jan Burse.

> Dan Christensen after sniffing one kilogram cocaine:
> > What do you imagine is the next rational number
> after 0, Jan Burse? Can't be 1 since there are infinitely
> many rational numbers between 0 and 1. How will
> you get to 1/2 starting from 0.

So you can't do it. Thought so.

> LoL
>
> BTW: There are many many functions satisfying
> your „Peano Axioms“, like for example:
>
> succZ : Z -> Z, x |-> x+1
> succQ : Q -> Q, x |-> x+1
> succR : R -> R, x |-> x+1
> succC : C -> C, x |-> x+1
>

Sadly for you, none of these supposed "successor functions" can be used as a basis for proof by induction on any of these sets. Just admit you are wrong, Jan Burse. You are looking like a complete idiot here.

Dan

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Nothing magical. Its really a no brainer.
They only explain why you cannot prove the following:

ALL(a):ALL(b):[s(a)=b => a e n & b e n]

From your "Peano Axioms".

Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 22:39:42 UTC+2:
> > succZ : Z -> Z, x |-> x+1
> > succQ : Q -> Q, x |-> x+1
> > succR : R -> R, x |-> x+1
> > succC : C -> C, x |-> x+1
> >
> Sadly for you, none of these supposed "successor functions" can be used as a basis for proof by induction on any of these sets. Just admit you are wrong, Jan Burse. You are looking like a complete idiot here.