On Thursday, May 5, 2022 at 4:44:09 PM UTC-4, Mostowski Collapse wrote:
> 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]
>
>

There would be a serious bug in my program if you could prove such nonsense.

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 explanation that you cannot prove the below:

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

That your Peano Axioms have models, where there
are pairs a,b with s(a)=b and ~(a e n) or ~(a e n).

One such model is for example:

s : Q -> Q, x |-> x+1

You can take a=-1/2 and b=1/2, we can verify:

s(a)=b
~(a e n)
~(b e n)

Easy exercise, based on the way you formulated
your Peano Axioms.

Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 22:47:49 UTC+2:
> On Thursday, May 5, 2022 at 4:44:09 PM UTC-4, Mostowski Collapse wrote:
> > 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]
> >
> >
> There would be a serious bug in my program if you could prove such nonsense.
> 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:
>
> 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
>

None of these supposed "successor functions" satisfy Peano's Axioms. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.

Dan

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

Well they all satisfy your Peano Axioms. You can go
through each axiom and check whether its satisfied.
Which Axiom is not satisfied. And what are the

instantiations of the Axiom that render it false?
For example if you set s = succQ, which Axiom is
not satisfied, and how is it not satisfied?

Just curious....

Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 23:30:20 UTC+2:
> On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:
> >
> > 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
> >
> None of these supposed "successor functions" satisfy Peano's Axioms. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

For example this axiom is satisfied:

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

by s = succQ. Which value for the variable "a" do
you take to make it false? The axiom gets false
if you have the following:

(a e n) True
~s(a)=0 False

Can you tell us? Here is material implication, this
the truth table of the => sign in your Axiom:

A B (A → B)
F F T
F T T
T F F
T T T
https://web.stanford.edu/class/cs103/tools/truth-table-tool/

To make the axiom false, you need to provoke row
Nr. 3 of the truth table. Thats the only way to make
material implication false. In row Nr 3. you find

A = T and B = F. So you need to find a value for
the variable "a" that makes (a e n) true and ~s(a)=0
false. How would you do that with s = succQ

Just curious....

Mostowski Collapse schrieb am Donnerstag, 5. Mai 2022 um 23:39:10 UTC+2:
> Well they all satisfy your Peano Axioms. You can go
> through each axiom and check whether its satisfied.
> Which Axiom is not satisfied. And what are the
>
> instantiations of the Axiom that render it false?
> For example if you set s = succQ, which Axiom is
> not satisfied, and how is it not satisfied?
>
> Just curious....
> Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 23:30:20 UTC+2:
> > On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:
> > >
> > > 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
> > >
> > None of these supposed "successor functions" satisfy Peano's Axioms. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.
> > 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 5:39:10 PM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 23:30:20 UTC+2:
> > On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:
> > >
> > > 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
> > >
> > None of these supposed "successor functions" satisfy Peano's Axioms. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.

> Well they all satisfy your Peano Axioms.

Wrong again, Jan Burse. Perhaps you are not familiar with Peano's Axioms. Some references:

https://mathworld.wolfram.com/PeanosAxioms.html
https://www.britannica.com/science/Peano-axioms
https://brilliant.org/wiki/peano-axioms/
https://encyclopediaofmath.org/wiki/Peano_axioms
https://proofwiki.org/wiki/Axiom:Peano%27s_Axioms

From DC Proof:

1. Set(n)
Axiom

2. 0 in n
Axiom

3. ALL(a):[a in n => s(a) in n]
Axiom

4. ALL(a):ALL(b):[a in n & b in n => [s(a)=s(b) => a=b]]
Axiom

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

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]]]
Axiom

Neither of your supposed "successor functions" above satisfies the Peano's Axioms. Each violates 2 of those axioms:

1. There does not exist x in the underlying set such that x + 1 = 0.

In each of the underlying sets, we have -1. And -1 + 1 = 0.

2. For all subsets P of the underlying set, if 0 in P and for all x in P, x+1 is also in P, then P includes all of the underlying set. (Induction)

N is a subset of the each underlying set. 0 in N. For all x in N, x+1 is also in N. And yet, N does NOT include all of the underlying set. It does NOT include -1, which an element of each of the underlying sets.

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, dumbo?

Whats wrong with you. For the milllionst time I do not
change n, I only change s. s = succQ perfectly well
satisfies your axioms. You are going bonkers.

I nowhere wrote to change n.

If I would want to change n I would write so. But I only
make a point that you can vary s considerably, and your
Peano Axioms are still satisfied.

Thats pretty easy to see. All these functions:

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

Satisfy your Peano Axioms when you set s = succX and
keep n what it is. Do you deny that? The symbol n doesnt change,
how do you want to falsify this axiom with succX:

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

You cannot falsify it with any of the succX. They all agree
on this fact. We do not have -1 e n.

Dan Christensen schrieb am Freitag, 6. Mai 2022 um 02:01:53 UTC+2:
> On Thursday, May 5, 2022 at 5:39:10 PM UTC-4, Mostowski Collapse wrote:
>
> > Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 23:30:20 UTC+2:
> > > On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:
> > > >
> > > > 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
> > > >
> > > None of these supposed "successor functions" satisfy Peano's Axioms. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.
> > Well they all satisfy your Peano Axioms.
> Wrong again, Jan Burse. Perhaps you are not familiar with Peano's Axioms. Some references:
>
> https://mathworld.wolfram.com/PeanosAxioms.html
> https://www.britannica.com/science/Peano-axioms
> https://brilliant.org/wiki/peano-axioms/
> https://encyclopediaofmath.org/wiki/Peano_axioms
> https://proofwiki.org/wiki/Axiom:Peano%27s_Axioms
>
> From DC Proof:
>
> 1. Set(n)
> Axiom
>
> 2. 0 in n
> Axiom
>
> 3. ALL(a):[a in n => s(a) in n]
> Axiom
>
> 4. ALL(a):ALL(b):[a in n & b in n => [s(a)=s(b) => a=b]]
> Axiom
>
> 5. ALL(a):[a in n => ~s(a)=0]
> Axiom
> 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]]]
> Axiom
>
>
> Neither of your supposed "successor functions" above satisfies the Peano's Axioms. Each violates 2 of those axioms:
>
> 1. There does not exist x in the underlying set such that x + 1 = 0.
>
> In each of the underlying sets, we have -1. And -1 + 1 = 0.
>
> 2. For all subsets P of the underlying set, if 0 in P and for all x in P, x+1 is also in P, then P includes all of the underlying set. (Induction)
>
> N is a subset of the each underlying set. 0 in N. For all x in N, x+1 is also in N. And yet, N does NOT include all of the underlying set. It does NOT include -1, which an element of each of the underlying sets.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Dumbo, you have everything right before your eyes, you write it yourself:

Dan Christensen schrieb am Freitag, 6. Mai 2022 um 02:01:53 UTC+2:
> It [you mean N] does NOT include -1
https://groups.google.com/g/sci.math/c/sYkWEHylBAA/m/yU7JqmzhAgAJ

You can still not see that this axiom stays true for succX?

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

Maybe see a doctor and order a new prescription of psychopharmaca.

LMAO!

Mostowski Collapse schrieb am Freitag, 6. Mai 2022 um 09:10:32 UTC+2:
> Since when is -1 e n, dumbo?
>
> Whats wrong with you. For the milllionst time I do not
> change n, I only change s. s = succQ perfectly well
> satisfies your axioms. You are going bonkers.
>
> I nowhere wrote to change n.
>
> If I would want to change n I would write so. But I only
> make a point that you can vary s considerably, and your
> Peano Axioms are still satisfied.
>
> Thats pretty easy to see. All these functions:
> succZ : Z -> Z, x |-> x+1
> succQ : Q -> Q, x |-> x+1
> succR : R -> R, x |-> x+1
> succC : C -> C, x |-> x+1
> Satisfy your Peano Axioms when you set s = succX and
> keep n what it is. Do you deny that? The symbol n doesnt change,
> how do you want to falsify this axiom with succX:
> ALL(a):[a e n => ~s(a)=0]
> You cannot falsify it with any of the succX. They all agree
> on this fact. We do not have -1 e n.
> Dan Christensen schrieb am Freitag, 6. Mai 2022 um 02:01:53 UTC+2:
> > On Thursday, May 5, 2022 at 5:39:10 PM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 23:30:20 UTC+2:
> > > > On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:
> > > > >
> > > > > 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
> > > > >
> > > > None of these supposed "successor functions" satisfy Peano's Axioms.. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.
> > > Well they all satisfy your Peano Axioms.
> > Wrong again, Jan Burse. Perhaps you are not familiar with Peano's Axioms. Some references:
> >
> > https://mathworld.wolfram.com/PeanosAxioms.html
> > https://www.britannica.com/science/Peano-axioms
> > https://brilliant.org/wiki/peano-axioms/
> > https://encyclopediaofmath.org/wiki/Peano_axioms
> > https://proofwiki.org/wiki/Axiom:Peano%27s_Axioms
> >
> > From DC Proof:
> >
> > 1. Set(n)
> > Axiom
> >
> > 2. 0 in n
> > Axiom
> >
> > 3. ALL(a):[a in n => s(a) in n]
> > Axiom
> >
> > 4. ALL(a):ALL(b):[a in n & b in n => [s(a)=s(b) => a=b]]
> > Axiom
> >
> > 5. ALL(a):[a in n => ~s(a)=0]
> > Axiom
> > 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]]]
> > Axiom
> >
> >
> > Neither of your supposed "successor functions" above satisfies the Peano's Axioms. Each violates 2 of those axioms:
> >
> > 1. There does not exist x in the underlying set such that x + 1 = 0.
> >
> > In each of the underlying sets, we have -1. And -1 + 1 = 0.
> >
> > 2. For all subsets P of the underlying set, if 0 in P and for all x in P, x+1 is also in P, then P includes all of the underlying set. (Induction)
> >
> > N is a subset of the each underlying set. 0 in N. For all x in N, x+1 is also in N. And yet, N does NOT include all of the underlying set. It does NOT include -1, which an element of each of the underlying sets.
> > 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:

> Wrong again, Jan Burse. Perhaps you are not familiar with Peano's
> Axioms. Some references:

sure, but the stupid *homepage_designer* nazi wanker Thomas
'PointedEars' Lahn <PointedEars@web.de> is *facepalm*ing himself now
repeatedly complaining about his own complains. What a fucking idiot. The
nazis of "ukraine" are calling his canceler for *a_sulky_liver_sausage*
and the wankers, the former largest economy of europe, still want to go to
a nazi shithole of a "country" for *kissing_the_ring* of the nazi khazar
zelenske. This is incredible.

On Friday, May 6, 2022 at 3:10:32 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:

> Dan Christensen schrieb am Freitag, 6. Mai 2022 um 02:01:53 UTC+2:
> > On Thursday, May 5, 2022 at 5:39:10 PM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Donnerstag, 5. Mai 2022 um 23:30:20 UTC+2:
> > > > On Thursday, May 5, 2022 at 2:51:37 PM UTC-4, Mostowski Collapse wrote:
> > > > >
> > > > > 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
> > > > >
> > > > None of these supposed "successor functions" satisfy Peano's Axioms.. In each case, there exists x in the respective set such that x+1=0. This effectively violates the 4th Peano Axiom that S(x)=/=0. Deal with it, Jan Burse. Just admit you were wrong.
> > > Well they all satisfy your Peano Axioms.
> > Wrong again, Jan Burse. Perhaps you are not familiar with Peano's Axioms. Some references:
> >
> > https://mathworld.wolfram.com/PeanosAxioms.html
> > https://www.britannica.com/science/Peano-axioms
> > https://brilliant.org/wiki/peano-axioms/
> > https://encyclopediaofmath.org/wiki/Peano_axioms
> > https://proofwiki.org/wiki/Axiom:Peano%27s_Axioms
> >
> > From DC Proof:
> >
> > 1. Set(n)
> > Axiom
> >
> > 2. 0 in n
> > Axiom
> >
> > 3. ALL(a):[a in n => s(a) in n]
> > Axiom
> >
> > 4. ALL(a):ALL(b):[a in n & b in n => [s(a)=s(b) => a=b]]
> > Axiom
> >
> > 5. ALL(a):[a in n => ~s(a)=0]
> > Axiom
> > 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]]]
> > Axiom
> >
> >
> > Neither of your supposed "successor functions" above satisfies the Peano's Axioms. Each violates 2 of those axioms:
> >
> > 1. [Peano Axiom 4] There does not exist x in the underlying set such that x + 1 = 0.
> >
> > In each of the underlying sets, we have -1. And -1 + 1 = 0.
> >
> > 2. [Peano Axiom 5] For all subsets P of the underlying set, if 0 in P and for all x in P, x+1 is also in P, then P includes all of the underlying set. (Induction)
> >
> > N is a subset of the each underlying set. 0 in N. For all x in N, x+1 is also in N. And yet, N does NOT include all of the underlying set. It does NOT include -1, which an element of each of the underlying sets.

> Since when is -1 e n, dumbo?
>

Geez, yer dumb, Jan Burse!

-1 is NOT an element N, but it is an element of each of the underlying sets of your supposed "successor functions" here. Perhaps you didn't know, but -1 + 1 = 0.

Not only do your supposed "successor functions" violate the axiom that 0 has no predecessor, but as I have shown here, they also violate the induction axiom.

Dan

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

I do not violate any axiom. Do you have a proof that
I violate an axiom. I dont think so. You are only loud
mouthing that something is violated.

But so far you didn't produce a proof. Ok, lets make
the problem dumbo compatible, so that crazy Dan
Christensen also understands it.

Can you prove the alleged axiom violation:

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

in set theory and natural deduction, from these axioms here:

1. Your "Peano Axioms"
2. New Axiom: ~(-1 e n)
3. New Axiom: s(-1) = 0

Just curious, how would you prove it? Please show
us a formal DC Proof, so that it is a computer
verified proof, and not some halucination.

Thanks in advance.

Dan Christensen schrieb am Freitag, 6. Mai 2022 um 16:25:28 UTC+2:
> On Friday, May 6, 2022 at 3:10:32 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> Geez, yer dumb, Jan Burse!
>
> -1 is NOT an element N, but it is an element of each of the underlying sets of your supposed "successor functions" here. Perhaps you didn't know, but -1 + 1 = 0.
>
> Not only do your supposed "successor functions" violate the axiom that 0 has no predecessor, but as I have shown here, they also violate the induction axiom.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

On Friday, May 6, 2022 at 10:40:13 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Freitag, 6. Mai 2022 um 16:25:28 UTC+2:
> > On Friday, May 6, 2022 at 3:10:32 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > Geez, yer dumb, Jan Burse!
> >
> > -1 is NOT an element N, but it is an element of each of the underlying sets of your supposed "successor functions" here. Perhaps you didn't know, but -1 + 1 = 0.
> >
> > Not only do your supposed "successor functions" violate the axiom that 0 has no predecessor, but as I have shown here, they also violate the induction axiom.

> I do not violate any axiom. Do you have a proof that
> I violate an axiom. I dont think so. You are only loud
> mouthing that something is violated.
>

Still in denial, I see. HA, HA, HA!!!!

Dan

No denial, I don't need to deny anything, because there
is nothing to deny. Only hot vapor. I am waiting for your
proof. Where is your proof? Can you prove the

alleged axiom violation:

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

in set theory and natural deduction, from these axioms here:

1. Your "Peano Axioms"
2. New Axiom: ~(-1 e n)
3. New Axiom: s(-1) = 0

Just curious, how would you prove it? Please show
us a formal DC Proof, so that it is a computer
verified proof, and not some halucination.

Thanks in advance.

Dan Christensen schrieb am Freitag, 6. Mai 2022 um 16:56:03 UTC+2:
> On Friday, May 6, 2022 at 10:40:13 AM UTC-4, Mostowski Collapse wrote:
>
> > Dan Christensen schrieb am Freitag, 6. Mai 2022 um 16:25:28 UTC+2:
> > > On Friday, May 6, 2022 at 3:10:32 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > > Geez, yer dumb, Jan Burse!
> > >
> > > -1 is NOT an element N, but it is an element of each of the underlying sets of your supposed "successor functions" here. Perhaps you didn't know, but -1 + 1 = 0.
> > >
> > > Not only do your supposed "successor functions" violate the axiom that 0 has no predecessor, but as I have shown here, they also violate the induction axiom.
> > I do not violate any axiom. Do you have a proof that
> > I violate an axiom. I dont think so. You are only loud
> > mouthing that something is violated.
> >
> Still in denial, I see. HA, HA, HA!!!!
>
> Dan

if you cannot write minus one -1 in DC Proof,
you can also use the constant m, maybe this
is easier for you:

Can you prove the alleged axiom violation:

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

in set theory and natural deduction, from these axioms here:

1. Your "Peano Axioms"
2. New Axiom: ~(m e n)
3. New Axiom: s(m) = 0

0 might be the first element in n. But what prevents
m outside of n, being the predecessor of 0. Can
you please tell us, via some proof?

LoL

Mostowski Collapse schrieb am Freitag, 6. Mai 2022 um 16:59:17 UTC+2:
> No denial, I don't need to deny anything, because there
> is nothing to deny. Only hot vapor. I am waiting for your
> proof. Where is your proof? Can you prove the
> alleged axiom violation:
>
> ~ALL(a):[a e n => ~s(a)=0]
>
> in set theory and natural deduction, from these axioms here:
>
> 1. Your "Peano Axioms"
> 2. New Axiom: ~(-1 e n)
> 3. New Axiom: s(-1) = 0
>
> Just curious, how would you prove it? Please show
> us a formal DC Proof, so that it is a computer
> verified proof, and not some halucination.
>
> Thanks in advance.
> Dan Christensen schrieb am Freitag, 6. Mai 2022 um 16:56:03 UTC+2:
> > On Friday, May 6, 2022 at 10:40:13 AM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Freitag, 6. Mai 2022 um 16:25:28 UTC+2:
> > > > On Friday, May 6, 2022 at 3:10:32 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > > > Geez, yer dumb, Jan Burse!
> > > >
> > > > -1 is NOT an element N, but it is an element of each of the underlying sets of your supposed "successor functions" here. Perhaps you didn't know, but -1 + 1 = 0.
> > > >
> > > > Not only do your supposed "successor functions" violate the axiom that 0 has no predecessor, but as I have shown here, they also violate the induction axiom.
> > > I do not violate any axiom. Do you have a proof that
> > > I violate an axiom. I dont think so. You are only loud
> > > mouthing that something is violated.
> > >
> > Still in denial, I see. HA, HA, HA!!!!
> >
> > Dan

On Friday, May 6, 2022 at 10:59:17 AM UTC-4, Mostowski Collapse wrote:
> No denial, I don't need to deny anything, because there

<yawn>

EOD

So you cannot prove the axiom violation:

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

Congratulations! Well s=succQ satisfies your
Peano Axioms. There is no proof that s=succQ
doesn't satisfy the Peano Axioms, not from

your side so far, against all your claims.

LMAO!

Dan Christensen schrieb am Freitag, 6. Mai 2022 um 17:05:08 UTC+2:
> On Friday, May 6, 2022 at 10:59:17 AM UTC-4, Mostowski Collapse wrote:
> > No denial, I don't need to deny anything, because there
> <yawn>
>
> EOD

On Friday, May 6, 2022 at 11:07:33 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> So you cannot prove the axiom violation:
> ~ALL(a):[a e n => ~s(a)=0]
> Congratulations! Well s=succQ satisfies your
> Peano Axioms.

What I said about the integers also applies for the rational numbers. Yesterday, you wrote "Because succQ(-1/2)=1/2 ..."

So, you also have succQ(-1) = 0, right?

Oooopsie... that would violate Peano's 4th Axiom. Likewise, Peano's 5th Axiom (Induction) is violated. Sorry, but (Q, succQ, 0) simply does NOT satisfy Peano Axioms.

Dan

I have of course succQ(-1)=0. But I have also ~(-1 e N).
Whats your point? Do you still believe this axiom is
violated for s = succQ and n = N?

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

By which value of a? Or do you believe this axiom is
violated for s = succQ and n = N?

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]]]

By which value of a? Can you please show us the witnesses
of these violations?

Thanks in advance.

Dan Christensen schrieb am Samstag, 7. Mai 2022 um 07:46:21 UTC+2:
> On Friday, May 6, 2022 at 11:07:33 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > So you cannot prove the axiom violation:
> > ~ALL(a):[a e n => ~s(a)=0]
> > Congratulations! Well s=succQ satisfies your
> > Peano Axioms.
> What I said about the integers also applies for the rational numbers. Yesterday, you wrote "Because succQ(-1/2)=1/2 ..."
>
> So, you also have succQ(-1) = 0, right?
>
> Oooopsie... that would violate Peano's 4th Axiom. Likewise, Peano's 5th Axiom (Induction) is violated. Sorry, but (Q, succQ, 0) simply does NOT satisfy Peano Axioms.
>
> Dan

The set Q is the set of rational numbers, you might write
down rational numbers as a/b, in as far 2/3 and 4/6

are the same rational numbers. The rational numbers
contain the natural numbers, like 10/2 is the same as 5.

+Q is usually defined in school as, assuming positive denominator:

a/b +Q c/d := (a *Z d +Z c *Z b) / (b *N d)

You can make this more precise because an equivalence
relation is involved. This is only a sketch. You need
operations on Z, the positive and negative numbers,

since the numerator can be negative. Now you can show:

x e N & y e N => x +Q y = x +N y

Another way to express this is to say:

+Q | NxN = +N

Mostowski Collapse schrieb am Samstag, 7. Mai 2022 um 11:23:13 UTC+2:
> I have of course succQ(-1)=0. But I have also ~(-1 e N).
> Whats your point? Do you still believe this axiom is
> violated for s = succQ and n = N?
> ALL(a):[a e n => ~s(a)=0]
> By which value of a? Or do you believe this axiom is
> violated for s = succQ and n = N?
> 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]]]
> By which value of a? Can you please show us the witnesses
> of these violations?
>
> Thanks in advance.
> Dan Christensen schrieb am Samstag, 7. Mai 2022 um 07:46:21 UTC+2:
> > On Friday, May 6, 2022 at 11:07:33 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > > So you cannot prove the axiom violation:
> > > ~ALL(a):[a e n => ~s(a)=0]
> > > Congratulations! Well s=succQ satisfies your
> > > Peano Axioms.
> > What I said about the integers also applies for the rational numbers. Yesterday, you wrote "Because succQ(-1/2)=1/2 ..."
> >
> > So, you also have succQ(-1) = 0, right?
> >
> > Oooopsie... that would violate Peano's 4th Axiom. Likewise, Peano's 5th Axiom (Induction) is violated. Sorry, but (Q, succQ, 0) simply does NOT satisfy Peano Axioms.
> >
> > Dan

Since +Q agrees on NxN with +N. You also have that
succQ agrees on N with succN. Just observe that
succQ can be defined as follows:

succQ(x) := x +Q 1.

Now since we had this lemma:

/* +Q agrees on NxN with +N */
> x e N & y e N => x +Q y = x +N y

We can also prove this lemma:

x e N => succQ(x) = succN(x)

And therefore succQ satisfies all Peano Axioms
because the particular way they are written
in the form ALL(v):[v e N => ...] and because

the induction axiom only talks about subsets
of N. Of course the Peano Axioms are not change
in my claim, they still have n in it, we do not

replace n by Q. This would violate Occams
Razor, since I nowhere said change n. Changing
n as well is an idee fix of Dan Christensen.

I nowhere shared the idea to change n as well.
This violates Occams Razor to also assume n
is changed, since I didn't state change n:

Occam's razor
is the problem-solving principle that "entities should
not be multiplied beyond necessity".
https://en.wikipedia.org/wiki/Occam's_razor

Mostowski Collapse schrieb am Samstag, 7. Mai 2022 um 11:25:40 UTC+2:
> The set Q is the set of rational numbers, you might write
> down rational numbers as a/b, in as far 2/3 and 4/6
>
> are the same rational numbers. The rational numbers
> contain the natural numbers, like 10/2 is the same as 5.
>
> +Q is usually defined in school as, assuming positive denominator:
>
> a/b +Q c/d := (a *Z d +Z c *Z b) / (b *N d)
>
> You can make this more precise because an equivalence
> relation is involved. This is only a sketch. You need
> operations on Z, the positive and negative numbers,
>
> since the numerator can be negative. Now you can show:
>
> x e N & y e N => x +Q y = x +N y
>
> Another way to express this is to say:
>
> +Q | NxN = +N
> Mostowski Collapse schrieb am Samstag, 7. Mai 2022 um 11:23:13 UTC+2:
> > I have of course succQ(-1)=0. But I have also ~(-1 e N).
> > Whats your point? Do you still believe this axiom is
> > violated for s = succQ and n = N?
> > ALL(a):[a e n => ~s(a)=0]
> > By which value of a? Or do you believe this axiom is
> > violated for s = succQ and n = N?
> > 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]]]
> > By which value of a? Can you please show us the witnesses
> > of these violations?
> >
> > Thanks in advance.
> > Dan Christensen schrieb am Samstag, 7. Mai 2022 um 07:46:21 UTC+2:
> > > On Friday, May 6, 2022 at 11:07:33 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > > > So you cannot prove the axiom violation:
> > > > ~ALL(a):[a e n => ~s(a)=0]
> > > > Congratulations! Well s=succQ satisfies your
> > > > Peano Axioms.
> > > What I said about the integers also applies for the rational numbers. Yesterday, you wrote "Because succQ(-1/2)=1/2 ..."
> > >
> > > So, you also have succQ(-1) = 0, right?
> > >
> > > Oooopsie... that would violate Peano's 4th Axiom. Likewise, Peano's 5th Axiom (Induction) is violated. Sorry, but (Q, succQ, 0) simply does NOT satisfy Peano Axioms.
> > >
> > > Dan

Dan Christensen wrote:

> On Friday, May 6, 2022 at 11:07:33 AM UTC-4, Mostowski Collapse (aka Jan
> Burse) wrote:
>> So you cannot prove the axiom violation: ~ALL(a):[a e n => ~s(a)=0]
>> Congratulations! Well s=succQ satisfies your Peano Axioms.
>
> What I said about the integers also applies for the rational numbers.
> Yesterday, you wrote "Because succQ(-1/2)=1/2 ..."

Danenko Christeskij puts the foot in his mouth.

nevermind, this american citizen is saying you slavic nazis turned anglo-
saxons overnight, have to prepare for no water, no food, no heat and
probably also for atomic war. Not my fault. Deal with the american.

Germany and Switzerland suddenly urge citizens to become PREPPERS
https://www.bitchute.com/video/SNUZmNoQ7OH5/

Natalia Usmanova Woman From Azovstal
https://www.bitchute.com/video/Y2VYQgQ5G0jr/

A captured militant of the Armed Forces of Ukraine about the crimes of the
Ukrainian command. https://www.bitchute.com/video/69xwcmk3acmf/

Mostowski Collapse wrote:

> 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]]]
>
> By which value of a? Can you please show us the witnesses of these
> violations? Thanks in advance.
>
> Dan Christensen schrieb am Samstag, 7. Mai 2022 um 07:46:21 UTC+2:

nevermind, you Janeko Burseski, this american citizen is saying you slavic
nazis turned anglo-saxons overnight, have to prepare for no water, no
food, no heat and probably also for atomic war. Not my fault. Deal with
the american.

Germany and Switzerland suddenly urge citizens to become PREPPERS
https://www.bitchute.com/video/SNUZmNoQ7OH5/

Natalia Usmanova Woman From Azovstal
https://www.bitchute.com/video/Y2VYQgQ5G0jr/

A captured militant of the Armed Forces of Ukraine about the crimes of the
Ukrainian command. https://www.bitchute.com/video/69xwcmk3acmf/

Is that all you found today in your bitchute cocaine delirium?

Emmet Shibanuma schrieb am Samstag, 7. Mai 2022 um 13:28:30 UTC+2:
> I dont want to live in this world anymore,
> I am an italian comrad and my ass hurts
> from beeing fucked by Putin.
>

Mostowski Collapse wrote:

> Is that all you found today in your bitchute cocaine delirium?
>
> Emmet Shibanuma schrieb am Samstag, 7. Mai 2022 um 13:28:30 UTC+2:
>> I dont want to live in this world anymore,
>> I am an italian comrad and my ass hurts from beeing fucked by Putin.
>>

the man is american, you dementia hit brainless idiot. You are going to starve in cold big time. Not my words. Not my fault. You disgusting fake_money nazi Janeko Burseskij.

lol, slavic fake_money nazi switzerland turned anglo-saxon, starving in cold, unable to steal gas and oil on fake_money anymore!!

LIVE UPDATES: US Authorities Should Be Held Accountable for Kiev's Crimes in Ukraine - Duma Speaker
https://sputniknews.com/20220506/live-updates-biden-announces-more-security-aid-to-ukraine-including-artillery-munitions--radar-1095322099.html
"For the crimes committed in Ukraine by the Kyiv Nazi regime, the leadership of the United States should also be held accountable, joining the list of war criminals", the speaker of the Russian State Duma Vyacheslav Volodin said.
He noted that Washington has confessed that the US is coordinating the Ukrainian military, and therefore de-facto participating in the conflict, fighting against Russia.

On Saturday, May 7, 2022 at 5:23:13 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> I have of course succQ(-1)=0. But I have also ~(-1 e N).

Of course (N, S, 0) satisfies PA, where S is the usual successor function on N. Sadly for you, and contrary to your outrageous claim, (Q, succQ, 0) where succQ(x)=x+1 on Q, does NOT satisfy PA since succQ(-1)=0. When will you learn, Jan Burse?

Dan

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