DC Proofs waterloo is Russells definite descriptions

[Mostowski Collapse]

DC Proof creator Dan-O-Matik suffers from some major
didactic deficiencies:

1) Errorneously translates "the X is Y" into "all X are Y".

2) Denies that ZFC is integral part of modern basic math.

Latest example:

Dan Christensen schrieb am Mittwoch, 24. November 2021 um 05:09:48 UTC+1:
> 17 ALL(x):ALL(f):ALL(y):ALL(z):[ALL(a):[a in x => f(a) in y]
> & EXIST(a):a in x
> & ALL(a):[a in z <=> a in y & EXIST(b):[b in x & f(b)=a]]
> => EXIST(a):a in z]
> Conclusion, 1
https://groups.google.com/g/sci.logic/c/tVUtBSQUhiE/m/YXt2f3OIBAAJ

Why the hell ALL(z)? How many images of the function f do you
expect to exist? Zero many images? More than one images?

Isn't the common notation f[a] from mathematics for the image by
the function f a definite description?

What does it take to show that it is a definite decription?

[Dan Christensen]

You seem to be getting more desperate by the hour, Jan Burse! See my reply to your identical posting at sci.logic just now.

Dan

[Mostowski Collapse]

f[A] notation is also on your beloved Wikipedia:

f[A] = {f(x):x in A}
https://en.wikipedia.org/wiki/Image_%28mathematics%29#Image_of_a_subset

You find it in usage everywhere in math textbooks.
You need an axiom schema from ZFC to have f[A] exist.
But the you can not only show that it exist, its also unique.

To translate "the X is Y" into "all X are Y" is a common beginner error.
Unfortunately in your case it leads to denial of ZFC.
Your childish behavior is kind of WM with extra steps:

FACEPALM MEME 🤣 HD 🤣
https://www.youtube.com/watch?v=g-bVEc8oZvk

[Mostowski Collapse]

Apropos pedagogical nightmare, here it is in your own words:

> > > & ALL(a):[a in z <=> a in y & EXIST(b):[b in x & f(b)=a]]

> z is the image of set x under f

Only if it z exists and if z is unique. Then you are allowed
to use the phrase "the X is Y". Otherwise z could be the king of
france, which doesn't exist. How can you show that z exists?

(Hint you need ZFC) I rather watch 2 1/2 hours garlic bread
sent to outer space, than the ever repeating nonsense of
Dan-O-Matik proofs spammed into sci.logic and sci.math.

2½ Hours of Unedited Garlic Bread Flight Footage
https://www.youtube.com/watch?v=YKAblynZYhI

BTW: You can fix your pedagogical nightmare if you rephrase:
> if z is an image of set x under f

After all "=>" translates to "if then" and "ALL(z)"
translates to the indeterminate article.

[Mostowski Collapse]

The co-domain of f is not given.
The subset axiom doesn't work for this here:

f[A] = {f(x):x in A}

Why do you think you can prove existence of f[A] with subset axiom?

Dan Christensen schrieb am Donnerstag, 25. November 2021 um 18:42:34 UTC+1:
>> f[A] notation is also on your beloved Wikipedia:
>>
>> f[A] = {f(x):x in A}
>> https://en.wikipedia.org/wiki/Image_%28mathematics%29#Image_of_a_subset
>>
>> You find it in usage everywhere in math textbooks.
>> You need an axiom schema from ZFC to have f[A] exist.

> See Subset Axiom on the Sets menu of DC Proof. It is documented in
the user manual.

[Dan Christensen]

See my reply today to your identical posting at sci.logic.

Dan

[Mostowski Collapse]

Not requested at all. My function f : V -> V doesn't
have a set like domain x and a set like codomain y.

> Mostowski Collapse schrieb am Mittwoch, 24. November 2021 um 02:56:37 UTC+1:
> > Or formally, s≠0 stands for EXIST(c):[c e s]:
> > ALL(a):[a≠0 => EXIST(b):[ALL(x):[x e a => f(x) e b] & b≠0]]
> https://groups.google.com/g/sci.logic/c/tVUtBSQUhiE/m/bLEb_i6BBAAJ

You can only put a Q.E.D. if you prove what was request.
Because Q.E.D. means that:

Inglese WWWWW Which Was What Was Wanted
Latino Q.E.D quod erat demonstrandum
https://it.wikipedia.org/wiki/Come_volevasi_dimostrare#In_altre_lingue

[Mostowski Collapse]

Here is a tableaux proof of the theorem
in FOL+ZF, using Exy for x in y:

∀a∃b∀x(Exb ↔ ∃y(Eya ∧ f(y)=x)) → ∀a(∃zEza → ∃b(∃zEzb ∧ ∀y(Eya → Ef(y)b))) is valid.
https://www.umsu.de/trees

The LHS ∀a∃b∀x(Exb ↔ ∃y(Eya ∧ f(y)=x)) is an instance of the axiom schema
of replacement from ZF. The RHS ∀a(∃zEza → ∃b(∃zEzb ∧ ∀y(Eya → Ef(y)b)))
is what was wanted.

Woa! Thats quite amazing that trees can find it.
Pitty trees doesn't show some statistics about its search.
In a blink the website shows some numbers,

but then when it has found a proof it replaces
the HTML by a tableaux refutation method proof.

[Mostowski Collapse]

You can also use |= instead of the main →, it then shows:

∀a∃b∀x(Exb ↔ ∃y(Eya ∧ f(y)=x)) entails ∀a(∃zEza → ∃b(∃zEzb ∧ ∀y(Eya → Ef(y)b))).
https://www.umsu.de/trees

But I guess the distinction between |= and → is more
something for modal logic? Here in this example the
results are the same. Maybe the vdash |- would be a

better choice than the models |=, but anyway, nice tool.

[Brain Hubbs]

Mostowski Collapse wrote:

> You can also use |= instead of the main →, it then shows:
>
> ∀a∃b∀x(Exb ↔ ∃y(Eya ∧ f(y)=x)) entails ∀a(∃zEza → ∃b(∃zEzb ∧ ∀y(Eya →
> Ef(y)b))).
> https://www.umsu.de/trees

ohh my butt, they are even saying the capitalism destroyed eastern europe,
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capitalist china, saving the climate same time!!

PROJECT EAT HUMAN BABIES TO SOLVE WORLD HUNGER- CANNIBALISM TO GO
MAINSTREAM IN SCHOOLS (PURE EVIL)
https://153news.net/watch_video.php?v=Y6Y313YOYH79
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[Mostowski Collapse]

Micro penis tells his life:
> ohh my butt

[Mostowski Collapse]

I don't think using another sameness, as Dan-O-Matik
suggests changes anything. Dan-O-Matik suggest to
use something else than:

f ≈ g :<=> ∀x(f(x)=g(x))

but the main problem is not same-ness. The main
problems is Dan-O-Matik style functions, that pop
out of the function axiom.

For Dan-O-Matik style identification on the natural
function f : N -> N, we can prove that every
f_c : N u {c} -> N u {c} is also such a Dan-O-Matik

style function f_c : N -> N:

∀x((ExN ∨ x=c) → f(x)=x) → ∀x(ExN → f(x)=x) is valid.
https://www.umsu.de/trees/#~6x%28ExN~2x=c~5f%28x%29=x%29~5~6x%28ExN~5f%28x%29=x%29

I wonder what else definition of same-ness would
prevent the above theorem. The above theorem doesn't
make use of sameness, there is no f ≈ g in it.

[Mostowski Collapse]

We would only get a different result when the two
occerences of the function symbol f here:

∀x((ExN ∨ x=c) → f(x)=x) → ∀x(ExN → f(x)=x) is valid.
https://www.umsu.de/trees/#~6x%28ExN~2x=c~5f%28x%29=x%29~5~6x%28ExN~5f%28x%29=x%29

If we were forced to write it with two different function
symbols and a sameness between it:

f ≈ g ∧ ∀x((ExN ∨ x=c) → f(x)=x) → ∀x(ExN → g(x)=x) ?

Its then possibly that the above fallacy of Dan-O-Matiks
function spaces and function axiom does not anymore
happen, if additionally the sameness would violate the following:

The indiscernibility of identicals:
∀ x ∀ y [ x ≈ y → ∀ F ( F x ↔ F y ) ]
https://en.wikipedia.org/wiki/Identity_of_indiscernibles

But if the above is violated then the sameness is not
an identity anymore, and all the proofs that Dan-O-Matik
did claiming uniqueness are moot, because he

didn't use an identity, but some wonky sameness.

[Mostowski Collapse]

Although wikipedia thinks that principle 1 is undisputed
and lists some doubts for principle 2. We can use the
same list of doubts for principle 1 to challenge it:

They all challenge the F in the "conception of properties
used to define indiscernibility":
- pure vesus impure properties
- qualitative versus non-qualitative properties
- intrinsic versus extrinsic properties
https://en.wikipedia.org/wiki/Identity_of_indiscernibles#Indiscernibility_and_conceptions_of_properties

***********************************************************
You could add "transcending values versus non-transcending
values", where for Dan-O-Matik f(x) is transcending when
x is not element of a domain A. He clearly lives not in the
***********************************************************

world of first order logic (FOL), where any formula A(x) is
Ok for principle 1, its part of (FOL=), i.e. first order logic with
equality. He lives in a world different from first order logic

with equality, where certain formulas A(x) are meaning less,
even when they are wellformed, because they invoke transcend
values in some function application. This is in stark contrast to

first order logic with equality (FOL=) where a wellformed
formula has a truth value from {true,false} in a model.
There is no third truth value {true,false,meaningless}.

But did DC Proof succeed in providing a calculus for such a
logic that sees formulas A(x) from a 3-valued viewpoint? He
never defined some model theory for his DC Proof.

What would be a model theory that can do that?

[Mostowski Collapse]

Dan-O-Matik flash halucinated:
> You write "It is the subset of X where the limit in
question exists. Otherwise you could not formally
construct the required function f'. "

The limit in question is the function f'!

Look at the limit, it has a parameter x0, so it maps
x0 to L. Its one and the same function as f'. Where
do you think they differ?

[Mostowski Collapse]

If your shitty function axiom is so good, then please use
the derivative of the absolute value function f(x) = |x|,

which is the following function:

g(x) = x / |x|

And please show the following mathematical theorem,
that you can find in every math textbook:

g(x) is undefined at x = 0

[Dan Christensen]

To formally construct the real numbers and the required arithmetic operations is a massive project of literally tens of thousands of lines of formal proof. I will leave it to you as an exercise, Jan Burse.

Informally then...

f(x) = |x| = x for x>=0, -x for x <0. Compare that left and right-hand limits at x=0.

Right limit: Lim(h-->0+): ((f(0+h) - f(0))/h) = h/h = 1 (the slope of y=f(x) for x >= 0)

Left Limit: Lim(h-->0-): ((f(0+h) - f(0))/h) = -h/h = -1 (the slope of y=f(x) for x < 0)

The left and right-hand limits differ at x=0, so the limit does not exist there. Thus, the derivative of f(x)=|x| is said to be undefined at x=0.

So, we have: f'(x) is 1 if x>=0, -1 for x<0, and undefined for x=0.

I hope this helps sort out your wonky math, Jan Burse. (He likes to make "dark" inferences about functions outside of their domain in his wonky system. Really quite useless.)

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]

I don't see a proof by DC poop of:

g(x) is undefined at x = 0

LMAO!

[Mostowski Collapse]

As per your own decree, the Burse Paradox,
you cannot express it in DC poop, although you
insists that FOL function symbols and your

idiotic function axiom are necessary for
functions in DC poop. Congratulations you
shoot yourself in the foot.

[Dan Christensen]

On Thursday, February 10, 2022 at 12:09:58 PM UTC-5, Mostowski Collapse (aka Jan Burse) wrote:
> I don't see a proof by DC poop of:
> g(x) is undefined at x = 0

I guess that example was a bit complicated for you, Jan Burse. Here is simpler example from DC Proof. Only 3 lines.

Define the function f on set x such that:

1. ALL(a):[a in x => f(a) in x]
Axiom

Suppose...

2. ~t in x
Premise

Specifying that a=t in (1), we obtain:

3. t in x => f(t) in x
U Spec, 1

The definition of f (line 1) tells us nothing about elements NOT in x. In this case, we say that f(t) is UNDEFINED.

In summary: If f, x and t are such that ALL(a):[a in x => f(a) in x] & ~t in x, then, without any further information, we can say that f(t) is UNDEFINED.

Get it? Didn't think so. Oh, well...

[Mostowski Collapse]

And why can I prove in DC poop:

g(0) - g(0) = 0

When g(0) is indefined?

[Dan Christensen]

On Thursday, February 10, 2022 at 2:53:29 PM UTC-5, Mostowski Collapse wrote:
> And why can I prove in DC poop:
>
> g(0) - g(0) = 0
>

Let's see your proof.

Dan

[Mostowski Collapse]

I mean why can I not prove in DC poop:

g(0) - g(0) ≠ 0

Highschool math will tell me, when g(0)
is undefined, then undefined - undefined is
undefined and undefined is not zero.

[Mostowski Collapse]

What if I had defined g(x) as follows:

g(x) = 1 / (|x| / x)

Can you show:

g(x) = f'(x) where f(x) = |x|

LoL

[Dan Christensen]

On Thursday, February 10, 2022 at 3:11:22 PM UTC-5, Mostowski Collapse wrote:
> I mean why can I not prove in DC poop:
>
> g(0) - g(0) ≠ 0
>
> Highschool math will tell me, when g(0)
> is undefined, then undefined - undefined is
> undefined and undefined is not zero.

You mean, you have something like:

1. 0 in n
Axiom

2. ALL(a):ALL(b):ALL(c):[a in n & b in n & c in n => [a-b=c <=> a=c+b]]
Axiom

3. ALL(a):[a in n & ~a=0 => g(a) in n]
Axiom

4. 0 in n & ~0=0 => g(0) in n
U Spec, 3

And you want to specify a=g(0), b=g(0), and c=0 in (2)?

The trouble is, you won't be able to prove that g(0) in n. As such, g(0) is undefined and you won't be able to apply (2).

[Mostowski Collapse]

You dont need ZFC for Highschool mathematics.
All that is needed is:

0 ≠ 1

Then by Russells Definite Descriptions, this
here is provable:

~∃z /(0,0,z)

[Dan Christensen]

On Friday, February 11, 2022 at 2:23:40 AM UTC-5, Mostowski Collapse (aka Jan Burse) wrote:

> Dan Christensen schrieb am Donnerstag, 10. Februar 2022 um 21:56:56 UTC+1:
> > On Thursday, February 10, 2022 at 3:11:22 PM UTC-5, Mostowski Collapse wrote:
> > > I mean why can I not prove in DC poop:
> > >
> > > g(0) - g(0) ≠ 0
> > >
> > > Highschool math will tell me, when g(0)
> > > is undefined, then undefined - undefined is
> > > undefined and undefined is not zero.
> > You mean, you have something like:
> >
> > 1. 0 in n
> > Axiom
> >
> > 2. ALL(a):ALL(b):ALL(c):[a in n & b in n & c in n => [a-b=c <=> a=c+b]]
> > Axiom
> >
> > 3. ALL(a):[a in n & ~a=0 => g(a) in n]
> > Axiom
> >
> > 4. 0 in n & ~0=0 => g(0) in n
> > U Spec, 3
> >
> > And you want to specify a=g(0), b=g(0), and c=0 in (2)?
> >
> > The trouble is, you won't be able to prove that g(0) in n. As such, g(0) is undefined and you won't be able to apply (2).

> You dont need ZFC for Highschool mathematics.

Huh???

> All that is needed is:
>
> 0 ≠ 1
>
> Then by Russells Definite Descriptions, this
> here is provable:
>
> ~∃z /(0,0,z)

Utter nonsense! Get back to us when you sober up, Jan Burse.

Dan

[Dan Christensen]

Utter nonsense. Guessing at what you MIGHT be getting at, consider the following proof:

1. 0 in n
Axiom

2. 1 in n
Axiom

3. ~1=0
Axiom


Define: f

4. ALL(a):[a in n => f(a) in n]
Axiom

5. f(0)=1
Axiom


Define: g

6. ALL(a):[a in n & ~a=0 => g(a) in n]
Axiom

7. g(1)=0
Axiom

8. f(g(1))=1
Substitute, 7, 5

9. ~f(g(1))=0
Substitute, 8, 3


Suppose...

10. x in n & ~x=0
Premise

11. x in n & ~x=0 => g(x) in n
U Spec, 6

12. g(x) in n
Detach, 11, 10

13. g(x) in n => f(g(x)) in n
U Spec, 4, 12

14. f(g(x)) in n
Detach, 13, 12

Define: f(g(a))

15. ALL(a):[a in n & ~a=0 => f(g(a)) in n]
Conclusion, 10

Specifying a=0, we see from the definition of f(g(a)) that f(g(0)) is undefined since ~0=0 is always false:

16. 0 in n & ~0=0 => f(g(0)) in n
U Spec, 15

[Mostowski Collapse]

About sober, what does ~0=0 mean? Isn't this
false? So that the implication is vacously true?

And why did you prove something with f o g?
I mentioned 0=\=1 because of zero divided by zero.

[Dan Christensen]

(Correction)

On Friday, February 11, 2022 at 6:59:16 PM UTC-5, Mostowski Collapse wrote:

> Dan Christensen schrieb am Freitag, 11. Februar 2022 um 22:25:27 UTC+1:
> > 16. 0 in n & ~0=0 => f(g(0)) in n
> > U Spec, 15

> About sober, what does ~0=0 mean? Isn't this
> false?

Yes. And since the antecedent is false, we cannot infer that f(g(0)) in n is true, or that it is false. In this case, f(g(0)) in n is simply UNDEFINED.

Here is the truth table for A=>B:

A B A=>B
T T T
T F F
F T T <---
F F T <---

Notice that, from lines 3 and 4, the antecedent A is false, but the implication A=>B is true, and that the consequent B could be EITHER true (line 3) OR false (line 4).

> So that the implication is vacously true?
>

Yes, but we cannot use to infer the truth value of the consequent. (See truth table above)

> And why did you prove something with f o g?

I thought it was a simpler example that would help you understand.

It seem to me that your example was needlessly complicated by having to introduce the definition of subtraction and the + operator. IIUC you had a function g: N\{0} --> N and a wonky composition of g with subtraction on N, namely f(x) = g(x)-g(x). In your example, f(0) would be undefined since g(0) is undefined.

> I mentioned 0=\=1 because of zero divided by zero.

I needed ~1=0 on line 3 to indicate the existence of at least two distinct elements in n, namely 0 and 1.

[Mostowski Collapse]

Newest low in Dan-O-Matiks logic capabilities.
He doesn't see that Vacuous Truth means that

ALL(a):[a e null => Q]

Is logically equivalent to TRUE, i.e.

P v ~P

He is slower than the Snake Gary in Sponge Bob.

Gary the Snail's Most SAVAGE Moments!
https://www.youtube.com/watch?v=HvBgdUXis5Y

Dan Christensen schrieb am Samstag, 12. Februar 2022 um 06:45:59 UTC+1:
> I am stupid and I know it

[Dan Christensen]

On Wednesday, March 30, 2022 at 1:15:52 PM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> Newest low in Dan-O-Matiks logic capabilities.
> He doesn't see that Vacuous Truth means that
>
> ALL(a):[a e null => Q]
>
> Is logically equivalent to TRUE, i.e.
>
> P v ~P
>

Poor Jan Burse has never been the same since he found out that you can't make inferences about a function outside of its domain. His "dark elements" keep haunting him day and night.

BTW, vacuous truth is easily handled in DC Proof. Poor Jan Burse just can't seem to get the hang of it. Oh, well...

[Mostowski Collapse]

The error is on your side, not on Wikipedias side. Wikipedia
uses nowhere a Vacuously True only condition. They have
also the condition f ⊆ X x Y.

Its YOUR PROBLEM Dan-O-Matik. Scape goating Wikipedia
doesn't work. Its YOUR BLUNDER.

[Mostowski Collapse]

Also you don't need much logic to see that you proved
nonsense. Its only trivial high school logic:

1) Vacuously True <=> TRUE
∀a((Pa ∧ ¬Pa) → Q) ↔ (R∨¬R) is valid.
https://www.umsu.de/trees/#~6a%28P%28a%29~1~3P%28a%29~5Q%29~4R~2~3R

2) TRUE is Absorbed by Conjunction:
((R∨¬R) ∧ Q) ↔ Q is valid.
https://www.umsu.de/trees/#%28%28R~2~3R%29~1Q%29~4Q

3) Left Hand Side TRUE is Absorbed by Implication
((R∨¬R) → Q) ↔ Q is valid.
https://www.umsu.de/trees/#%28%28R~2~3R%29~5Q%29~4Q

Its then trivial to see that what you proved:

ALL(x):[Set(x)
=> EXIST(f):[ALL(a):[a in null => f(a) in x]
& ALL(g):[ALL(a):[a e null => g(a) e x] => g=f]]]
https://dcproof.com/EmptyFunctionsUniqueV2.htm

Is logical equivalent to:
EXIST(f):[ALL(g):[g=f]]]

But such a Theorem is obviously incorrect. Since there
is more then one function possible in mathematics, for
example the functions sinus sin() and cosinus cos(),

which have sin=\=cos, so your theorem is nonsense.
When you can prove an incorrect theorem in DC poop,
this means DC poop is inconsistent.

[Mostowski Collapse]

Maybe use this naming for what Dan-O-Matik hasnt
internalized yet about logic. He is such a baby:

1) Vacuously True <=> TRUE
∀a((Pa ∧ ¬Pa) → Q) ↔ (R∨¬R) is valid.
https://www.umsu.de/trees/#~6a%28P%28a%29~1~3P%28a%29~5Q%29~4R~2~3R

2) TRUE is Multiplicative Identity:

((R∨¬R) ∧ Q) ↔ Q is valid.
https://www.umsu.de/trees/#%28%28R~2~3R%29~1Q%29~4Q

3) FALSE is Additive Identity & Material Implication

((R∨¬R) → Q) ↔ Q is valid.
https://www.umsu.de/trees/#%28%28R~2~3R%29~5Q%29~4Q

4) Additive Associativity & Material Implication & De Morgan's laws
(P → (Q→R)) ↔ ((P∧Q) → R) is valid.
https://www.umsu.de/trees/#%28P~5%28Q~5R%29%29~4%28P~1Q~5R%29

Material implication: (A => B) <=> (~A | B)
https://en.wikipedia.org/wiki/Material_implication_%28rule_of_inference%29

De Morgan's laws: ~(A & B) <=> (~A | ~B)
https://en.wikipedia.org/wiki/De_Morgan's_laws

[Mostowski Collapse]

Propositional Logic is like Rubiks Cube. Some can do
it very quick, some are not that quick.

Dan-O-Matik on the other hand doesn't master propositional
logic at all, otherwise he would post one nonsense

after the other, without seeing that some of axioms
or axiom schemas are buggy.

[Mostowski Collapse]

Corr.:

Dan-O-Matik on the other hand doesn't master propositional

logic at all, otherwise he wouldn't post one nonsense

[Mostowski Collapse]

So Dan-O-Matik doesn't know this meta rule of replacement,
which can be meta derived in natural deduction(*)?

P <=> Q
--------------------
A(P) <=> A(Q)

This is left as an exercise. Or lookup W. V. Quine Mathematical Logic,
you find this rule also mentioned.

Mathematical Logic Revised Edition W. V. Quine
$40.00 - ISBN 9780674554511 - 11/16/1982 - 364 pages
https://www.hup.harvard.edu/catalog.php?isbn=9780674554511

(*) BTW: it works in any complete calculus, this meta
theorem. Need not be only natural deduction, can be
also another calculus.

[Mostowski Collapse]

Here is a proof by induction, over the implicative fragment:

Base Case:
A(P) = P, A(Q) = Q, trivial (P <=> Q) / (P <=> Q)
A(P) = R, A(Q) = R, trivial (P <=> Q) / (R <=> R)

Inductive Step:
A(P) = (A1(P) => A2(P)), A(Q) = (A1(Q) => A2(Q)),
By induction hypothessis we have:
(P <=> Q) / (A1(P) <=> A1(Q))
(P <=> Q) / (A2(P) <=> A2(Q))
All we have to show that we can now create a proof
in the calculus of choice:

A1(P) <=> A1(Q) A2(P) <=> A2(Q)
--------------------------------------------------------------------------
(A1(P) => A2(P)) <=> (A1(Q) => A2(Q))

Which should be possible:

((P↔Q) ∧ (R↔S)) → ((P→R) ↔ (Q→S)) is valid.
https://www.umsu.de/trees/#%28P~4Q%29~1%28R~4S%29~5%28%28P~5R%29~4%28Q~5S%29%29

[Dan Christensen]

On Sunday, April 3, 2022 at 1:16:15 PM UTC-4, Mostowski Collapse wrote:
> Here is a proof by induction, over the implicative fragment:
>
> Base Case:
> A(P) = P, A(Q) = Q, trivial (P <=> Q) / (P <=> Q)
> A(P) = R, A(Q) = R, trivial (P <=> Q) / (R <=> R)
>
> Inductive Step:
> A(P) = (A1(P) => A2(P)), A(Q) = (A1(Q) => A2(Q)),
> By induction hypothessis we have:
> (P <=> Q) / (A1(P) <=> A1(Q))
> (P <=> Q) / (A2(P) <=> A2(Q))
> All we have to show that we can now create a proof
> in the calculus of choice:
>
> A1(P) <=> A1(Q) A2(P) <=> A2(Q)
> --------------------------------------------------------------------------
> (A1(P) => A2(P)) <=> (A1(Q) => A2(Q))
>
> Which should be possible:
>
> ((P↔Q) ∧ (R↔S)) → ((P→R) ↔ (Q→S)) is valid.
> https://www.umsu.de/trees/#%28P~4Q%29~1%28R~4S%29~5%28%28P~5R%29~4%28Q~5S%29%29

Pay attention, Jan Burse! Recall that your assignment was to formally prove using some form of natural deduction that:

ALL(x):[Set(x)
=> EXIST(f):ALL(a):[a in null => f(a) in x]
& ALL(f1):ALL(f2):[ALL(a):[a in null => f1(a) in x] & ALL(a):[a in null => f2(a) e x] => f1=f2]]
=> ALL(x):[Set(x) =>ALL(f1): ALL(f2):[f1=f2]]

I take it, you were unable to do so. Oh, well...

[Mostowski Collapse]

Hey Uber Moron, you prove the same way like the other
formula. You can even derive a stronger statement,
which shows the full extend of your nonsense claim:

ALL(f1): ALL(f2):[f1=f2]]

Its extremly easy, even my grandmother can do it.

The used rules of replacement are:

1) Vacuously True <=> TRUE
∀a((Pa ∧ ¬Pa) → Q) ↔ (R∨¬R) is valid.
https://www.umsu.de/trees/#~6a%28P%28a%29~1~3P%28a%29~5Q%29~4R~2~3R

2) TRUE is Multiplicative Identity:
((R∨¬R) ∧ Q) ↔ Q is valid.
https://www.umsu.de/trees/#%28%28R~2~3R%29~1Q%29~4Q

3) FALSE is Additive Identity & Material Implication
((R∨¬R) → Q) ↔ Q is valid.
https://www.umsu.de/trees/#%28%28R~2~3R%29~5Q%29~4Q

The used replacement steps are:

ALL(x):[Set(x)
=> EXIST(f):ALL(a):[a in null => f(a) in x]
& ALL(f1):ALL(f2):[ALL(a):[a in null => f1(a) in x] & ALL(a):[a in null => f2(a) e x] => f1=f2]]

Apply 1) to get:

ALL(x):[Set(x)
=> EXIST(f):(R∨¬R)

& ALL(f1):ALL(f2):[ALL(a):[a in null => f1(a) in x] & ALL(a):[a in null => f2(a) e x] => f1=f2]]

Apply 1) to get:

ALL(x):[Set(x)
=> EXIST(f):(R∨¬R)
& ALL(f1):ALL(f2):[(R∨¬R) & ALL(a):[a in null => f2(a) e x] => f1=f2]]

Apply 1) to get:

ALL(x):[Set(x)
=> EXIST(f):(R∨¬R)
& ALL(f1):ALL(f2):[(R∨¬R) & (R∨¬R) => f1=f2]]

Apply 2) to get:

ALL(x):[Set(x)
=> EXIST(f):(R∨¬R)
& ALL(f1):ALL(f2):[(R∨¬R) => f1=f2]]

Apply 3) to get:

ALL(x):[Set(x)
=> EXIST(f):(R∨¬R)
& ALL(f1):ALL(f2):[f1=f2]]

Apply EXIST(v):A when v not in A is the same as A:

ALL(x):[Set(x)
=> (R∨¬R)
& ALL(f1):ALL(f2):[f1=f2]]

Apply 2) to get:

ALL(x):[Set(x)
=> ALL(f1):ALL(f2):[f1=f2]]

Apply the Forall Instantiation rule of Natural Deduction:

[Set(null)
=> ALL(f1):ALL(f2):[f1=f2]]

Apply Modus Ponens of Natural Deduction with your Set(null) Axiom:

ALL(f1):ALL(f2):[f1=f2]]

[Dan Christensen]

My previous reply here still applies. Deal with it Jan Burse. We are STILL waiting for that FORMAL proof. Maybe you need to familiarize yourself with what is meant by formal proof. See https://en.wikipedia.org/wiki/Formal_system

Dan

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

On Sunday, April 3, 2022 at 5:20:48 PM UTC-4, Mostowski Collapse wrote:
> Hey Uber Moron, you prove the same way like the other
> formula. You can even derive a stronger statement,

> which shows the full extend of your nonsense claim: ...

[Mostowski Collapse]

I showed a formal proof. Only you cannot grasp it.

LoL

[Dan Christensen]

On Sunday, April 3, 2022 at 7:44:12 PM UTC-4, Mostowski Collapse wrote:
> I showed a formal proof. Only you cannot grasp it.
>

You listed several claims, each without a formal proof.

[Mostowski Collapse]

No I listed step by step proofs.

We have now reduced two proofs from here:
https://dcproof.com/EmptyFunctionsUniqueV2.htm

To utter nonsense:

What Dan-O-Matik calls: "On every set, there is exists a unique empty function (lines 1-91)"
EXIST(f):ALL(g):[g=f]
https://groups.google.com/g/sci.logic/c/qdh5sCBDPJo/m/T8HochqvCAAJ

What Dan-O-Matik calls: "The function space (fs) of all functions mapping the empty set to any set x has only a single element (lines 92-137)"
ALL(f1):ALL(f2):[f1=f2]]
https://groups.google.com/g/sci.logic/c/3CrCpBI8I2E/m/jGjj7n69CAAJ

Your are only incapable to grasp it.

[Mostowski Collapse]

Dan-O-Matik braught up:
> non-negative real number x such the x =/= |x|.

Can we agree about the following new axiom for DC Proof?

ALL(f):ALL(g):[EXIST(a):[f(a)=/=g(a)] => f=/=g]

[Dan Christensen]

On Monday, April 4, 2022 at 2:54:02 AM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> No I listed step by step proofs.
>

Some "steps!" You really don't have clue, do you, Jan Burse?

> We have now reduced two proofs from here:
> https://dcproof.com/EmptyFunctionsUniqueV2.htm
>
> To utter nonsense:
>

Only in your wet dreams, Jan Burse.

You are the one claiming that x =/= |x| for some non-negative real x. And that we can make inferences about a function outside of its domain of definition--your "dark elements."

You should team up with WM and AP as a junior partner. Dark elements meets dark numbers meets one atom universe. They have been pushing shit like this for decades here. The sci.math Dream Team. (HA, HA, HA!)

Dan

[Mostowski Collapse]

I do not believe that there is non-negative real
number x such the x =/= |x|. Thats your crazy invention.

I was asking in this thread whether you agree:

"If two functions differ at some argument, then they differ"

ALL(f):ALL(g):[EXIST(a):[f(a)=/=g(a)] => f=/=g]

[Willie Dukes]

Mostowski Collapse wrote:

> I was asking in this thread whether you agree:
> "If two functions differ at some argument, then they differ"
> ALL(f):ALL(g):[EXIST(a):[f(a)=/=g(a)] => f=/=g]
> Dan Christensen schrieb am Dienstag, 5. April 2022 um 22:29:03 UTC+2:
>> You are the one claiming that x =/= |x| for some non-negative real x.

He can't answer. But the nazis of "ukraine", the nazis of "polakia", the nazis of the switzarland and the nazis of america, helping them, will be concentrated in forced labour camps, building highways, bridges etc in russia. You nazis can't escape. We write 2022 and the nazis still making wars?? Get out of here.
That's the problem with the russians, they are not evil enough. The will never
bomb infrastructure, water, electricity, etc, staging fals flags as the nazis does.
And btw, without mother Russia, many of you wankers not even would be born. Neither you and your parents. Think about it.

Disgusting: How the Ukrainian Nazi/Nationalist Movement Post WW2 Was Bought by the CIA
https://www.veteranstoday.com/2022/04/05/disgusting-how-the-ukrainian-nazi-nationalist-movement-post-ww2-was-bought-by-the-cia/

Russia calls for international probe into US-led biolabs in Ukraine
https://www.veteranstoday.com/2022/04/06/russia-calls-for-international-probe-into-us-led-biolabs-in-ukraine/

“Fakes emerge throughout Ukraine crisis just to be debunked later”
https://www.veteranstoday.com/2022/04/06/fakes-emerge-throughout-ukraine-crisis-just-to-be-debunked-later/

German chemical behemoth warns of country’s “total collapse” if Russian oil, gas is cut off
https://www.thetruthseeker.co.uk/?p=251808

“Vaccines” are Genocide Bio Weapons
https://www.thetruthseeker.co.uk/?p=251830

[Dan Christensen]

On Wednesday, April 6, 2022 at 7:12:52 AM UTC-4, Willie Dukes wrote:
> Mostowski Collapse wrote:
>
> > I was asking in this thread whether you agree:
> > "If two functions differ at some argument, then they differ"
> > ALL(f):ALL(g):[EXIST(a):[f(a)=/=g(a)] => f=/=g]
> > Dan Christensen schrieb am Dienstag, 5. April 2022 um 22:29:03 UTC+2:
> >> You are the one claiming that x =/= |x| for some non-negative real x.

> He can't answer. But the nazis of "ukraine"...

You are forgetting: WAR OF AGRESSION = NAZI

Looks like there will be a good chance you and your boss will be appearing before the Nuremberg War Crimes Tribunal II. Just hope they don't send you to a prison in the USA or Russia. I'm told conditions are absolutely appalling there.

[Jabe Jukado]

Dan Christensen wrote:

>> He can't answer. But the nazis of "ukraine"...
>
> You are forgetting: WAR OF AGRESSION = NAZI

Idiot, wars ARE aggression, but the nazi is you.

> Looks like there will be a good chance you and your boss will be
> appearing

Meanwhile a general or more caught in Marionopol if this is true.

Numerous Reports: U.S. Army General CAPTURED in Mariupol with AZOV Nazis!
https://www.thetruthseeker.co.uk/?p=251848

You are strange, for a nazi to be.

[Dan Christensen]

On Wednesday, April 6, 2022 at 2:09:54 PM UTC-4, Jabe Jukado wrote:
> Dan Christensen wrote:
>
> >> He can't answer. But the nazis of "ukraine"...
> >
> > You are forgetting: WAR OF AGRESSION = NAZI
> Idiot, wars ARE aggression, but the nazi is you.

Tell it to the judge Hr Goebbels.

[Mostowski Collapse]

It becomes evident that Dan-O-Matiks psychosis is
based on his age old confusion of "the" with "a".

He writes axioms like:

1 ALL(dom):ALL(cod):ALL(f):ALL(g):[Set(dom) & Set(cod) &
ALL(a):[a e dom => f(a) e cod] & ALL(a):[a e dom => g(a) e cod]
=> [f=g <=> ALL(a):[a e dom => f(a)=g(a)]]]
Axiom

But thinks about them in terms of:

1 ALL(f):ALL(g):THE(dom):THE(cod):[Set(dom) & Set(cod) &
ALL(a):[a e dom => f(a) e cod] & ALL(a):[a e dom => g(a) e cod]
=> [f=g <=> ALL(a):[a e dom => f(a)=g(a)]]]
Axiom

Just plain crazy....

THE: Is his mind reading quantifier.

[Mostowski Collapse]

So Dan Christensen still does not
believe that you cannot infer:

f : X -> X <=> ALL(a):[a in X => f(a) in X]

Easy counter example f = {(1,1),(2,2)} and X={1}, it
satisfies ALL(a):[a in X => f(a) in X], but it does not
satisfy f : X -> X, since dom(f)={1,2} =/= X.

Dan Christensen schrieb am Montag, 18. April 2022 um 19:26:50 UTC+2:
> > Hence "f: X --> X" means "f is a function with domain X and codomain X."
> You can also infer this from: ALL(a):[a in x => f(a) in x].
> Ultimately, it must come to this even with your "set theoretic" approach.

[Dan Christensen]

On Monday, April 18, 2022 at 4:32:28 PM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> So Dan Christensen still does not
> believe that you cannot infer:
>
> f : X -> X <=> ALL(a):[a in X => f(a) in X]
>
> Easy counter example f = {(1,1),(2,2)} and X={1}, it
> satisfies ALL(a):[a in X => f(a) in X], but it does not
> satisfy f : X -> X, since dom(f)={1,2} =/= X.
>

No "counter-example" here, Jan Burse. If X = {1}, then ALL(a):[a in X => f(a) in X] and f: X --> X mean the same thing.

Maybe you didn't know, but {1} =/= {1, 2}. We are not talking about your mystical "dark elements" here, Jan Burse.

[Fritz Feldhase]

On Tuesday, April 19, 2022 at 6:13:53 AM UTC+2, Dan Christensen wrote:
> On Monday, April 18, 2022 at 4:32:28 PM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> >

> > So Dan Christensen still does not believe that [one] cannot infer:

> >
> > f : X -> X <=> ALL(a):[a in X => f(a) in X]
> >
> > Easy counter example f = {(1,1),(2,2)} and X={1}, it
> > satisfies ALL(a):[a in X => f(a) in X], but it does not
> > satisfy f : X -> X, since dom(f)={1,2} =/= X.
> >

> No "counter-example" here, Jan Burse. If X = {1}, then <bla bla bla>

Holy shit. You are in the same category as WM, JG and AP now, Dan.

Hint: JB/MC defined a function f: {1, 2} -> {1, 2}.

Now Ax(x e {1} -> f(x) e {1}), but f: {1} --> {1} does not hold, since f is NOT a function from {1} to {1} (by definition).

[Mostowski Collapse]

The biggest problem, he denies his own U Spec inference rule:

Dan Christensen schrieb am Donnerstag, 21. April 2022 um 16:30:36 UTC+2:
> > the ad hoc axiom can prove g ~ h with g(-1/2)=1/2 & h(-1/2)=1/2,
> Another domain? Make up your mind, Jan Burse!

Thats just like Putin, no we didn't invade Ukraine. Balant denial.

[Dan Christensen]

On Tuesday, April 19, 2022 at 12:13:53 AM UTC-4, Dan Christensen wrote:
> On Monday, April 18, 2022 at 4:32:28 PM UTC-4, Mostowski Collapse (aka Jan Burse) wrote:
> > So Dan Christensen still does not
> > believe that you cannot infer:
> >
> > f : X -> X <=> ALL(a):[a in X => f(a) in X]
> >

The arrow notation is not built into DC Proof. If I want to formally state that f is a function mapping the elements of set x to the elements of set y, I would write:

ALL(a):[a in x => f(a) in y]

I would only use f: x --> y as an informal abbreviation.

[Mostowski Collapse]

But ALL(a):[a in x => f(a) in y] and f : x -> y are not
logical equivalent. So your are not allowed to replace
the later by the former in all circumstances.

[Sam Kaloxylos]

Mostowski Collapse wrote:

> But ALL(a):[a in x => f(a) in y] and f : x -> y are not logical
> equivalent. So your are not allowed to replace
> the later by the former in all circumstances.

here's your nazis you are starving for. Beating children, disgusting sons
of the bitches. Strange your govt is not allying with Russia removing
these nazis wherever they are. There are shitholes countries, *knowingly*
arming the nazis with hard weaponry. The new *_bio_weapon_labs_* nazis.

Proof of Ukrainian Neo-Nazi Satanism and Underground Biolabs. Is Putin at
War With The NWO https://www.bitchute.com/video/Wj54U4w9JbeP/

[Sam Kaloxylos]

Dan Christensen wrote:

> The arrow notation is not built into DC Proof. If I want to formally
> state that f is a function mapping the elements of set x to the elements
> of set y, I would write: ALL(a):[a in x => f(a) in y]

you look like a nazi. Here are your nazis, you are starving for. Beating

[Mostowski Collapse]

Thank you micro penis, for displaying the morron you are,
this is quite some recurring laugh. BTW they found a dead
mouse in a Putler copter:

Marvelous Russian "Flying Tank"
https://www.youtube.com/watch?v=z7G8RKJBH1w

Don't know whether fake or real... LoL

[Mostowski Collapse]

LoL, **demilitarized** Ka-52 "Alligator"

[Sam Kaloxylos]

Mostowski Collapse wrote:

> this is quite some recurring laugh. BTW they found a dead mouse in a
> Putler copter:

You don't undrestand engilsh. You, *fake_money* shithole nazi countries,
turned anglo-saxon santa clauss "christianity" overnight, are to be
dismissed. And so, the russian army are treating humanly these satanists.

*_"those_nazi_should_be_shot,_and_that_would_be_merciful"_*.

*_our_govt_are_paying_for_that,_and_want_to_make_ukraine_big_israel_*

[Mostowski Collapse]

micro penis got also some function axiom wrong...

[Mostowski Collapse]

Are you a gator? No!
But DC Proof looks dead, so I come over and investigate.
But that makes you an investigator.

Why isn’t gator attacking you?!?!
https://www.youtube.com/watch?v=3nTbQlZsQv0

[Dong Vassilikos]

Mostowski Collapse wrote:

> Are you a gator? No!
> But DC Proof looks dead, so I come over and investigate.
> But that makes you an investigator.

You are an idiot. A stupid nazi. Proofs:

Die Wahrheit über den Schauspieler Selenski der Euch in den dritten
Weltkrieg führen will🤮 https://www.bitchute.com/video/qCCl5XnOCTP6/

[Mostowski Collapse]

Ding Dong Vaseline. Thats a good match for micro penis.

LoL

[Dong Vassilikos]

Also, you are stupid nazi "micro penises" obsessed pedophile.

[Mostowski Collapse]

What sits in the corner and cries all days, because
Selenski is not his homo partner? Its Luigi micro penis!

Maybe try Putler as your bed warmer ... Or Kyrill ...

[Mostowski Collapse]

Doesn't make any sense to prove existence of add/3 with 762 proof lines.
The problem is you have now Peano induction schema:

P(0) & ALL(x):[P(x+1) => P(x)]
------------------------------------------------
ALL(y):P(y)

You can prove it much shorter if you have such a schema. Thats
usually what they teach in University.

Dan Christensen schrieb am Samstag, 30. April 2022 um 22:39:44 UTC+2:
> This is how it is done: https://dcproof.com/ConstructAddFunction.htm

[Mostowski Collapse]

Corr.: Typo

> The problem is you have no Peano induction schema:
>
> P(0) & ALL(x):[P(x) => P(x+1)]
> ------------------------------------------------
> ALL(y):P(y)

Sorry, its already late...

[Mostowski Collapse]

Here is a much shorter proof, I playing around
with the maps to symbol |-> , just take it as a ternary
relation written as _,_ |-> _ , and alternative to your add.

Now dont define add as the smallest fixpoint, only
as a fixpoint should be enough. So do something
else than boring Dan Christensen:

x,y |-> z <=> (y=0 & z=x) | EXIST(u):EXIST(v):(y=s(u) & x,u |-> v & z=s(v))

Now prove:

Seriality: ALL(x):ALL(y):EXIST(z):x,y |-> z
Proof by induction on y.

Functionality: ALL(x):ALL(y):ALL(z):ALL(t):(x,y |-> z & x,y |-> t => z = t)
Proof by induction on y.

[Mostowski Collapse]

Now since we have proved:

ALL(x):ALL(y):EXISTUNIQUE(z):x,y |-> z

We are licensed to define, without harm:

add(x,y) = z <=> x,y |-> z

And some very easy proofs reveale:

add(x,0,x)
add(x,s(u))=s(add(x,u))

Universal quantifiers were omitted in some places,
to render the post more readable and shorter.

[Mostowski Collapse]

But we must point out that there are more problems.

Doesn't make any sense to prove existence of add/3 with 762 proof lines.

Dan Christensen schrieb am Samstag, 30. April 2022 um 22:39:44 UTC+2:
> This is how it is done: https://dcproof.com/ConstructAddFunction.htm

And where is a proof of EXIST(n):Peano(n) ? You don't have a powerful
set theory, like ZFC would be, that could prove that. In ZFC you
could do such a construction, by means of the Axiom of Infinity (AOI).

You are just a crank. You don't understand that mathematics works:

- definition
- lemma
- theorem

This building block is hardly used:

- axiom

Now you have axioms for Peano, and don't prove EXIST(n):Peano(n).

[Mostowski Collapse]

A variation of the definition discipline, found in many modern
theorem provers, reads as follows:

- meta /* provided by the tool */
- def_dom /* example use <n,0,s> */
- def_ind /* example use x,y |-> z */
- def_rec /* example use add(x,y) = z */
- lemma
- theorem

This building block is then hardly used:

- axiom

The meta theory assures that the domains from def_dom, inductive
predicate definitions from def_ind and recursive functions definitions
from def_rec are all well . The meta theory and logic framework assures

things like EXIST(n):Peano(n), similar like in ZFC via Axiom of
Infinity (AOI). You can check out Coq, Isabelle/HOL, etc.. how its
done. def_rec, def_ind and def_dom are then setup so that

they are more reasoning friendly, to better support lemma and
theorem construction. Ultimately we want to prove some theorems
and not only construct some domains, predicates and functions.

Although the construction of the basic things in itself could be also an end.

[Mostowski Collapse]

Dan Christensen is currently in some UNDEFINED
delirium. He says this is Wonky:

Dan Christensen schrieb am Dienstag, 10. Mai 2022 um 01:44:05 UTC+2:
> Wonky result:
> 11. ALL(a):[~a in n => ~Even(a)]
> Conclusion, 2
https://groups.google.com/g/sci.logic/c/gV50jEU6fvQ/m/jRRRllUEAgAJ

But every sane mathematician will agree:

~(a e {0,1,2,3,...}) => ~(a e {0,2,4,...})

LoL

[Dan Christensen]

On Tuesday, May 10, 2022 at 3:17:38 AM UTC-4, Mostowski Collapse wrote:
> Dan Christensen is currently in some UNDEFINED
> delirium. He says this is Wonky:
>
> Dan Christensen schrieb am Dienstag, 10. Mai 2022 um 01:44:05 UTC+2:
> > Wonky result:
> > 11. ALL(a):[~a in n => ~Even(a)]
> > Conclusion, 2

This is the wonky result of Jan Burse's wonky definition of Even( ). A cautionary tale that you really do need to restrict quantifiers -- to the set of natural numbers in this case.

>
> But every sane mathematician will agree:
>
> ~(a e {0,1,2,3,...}) => ~(a e {0,2,4,...})

A problem potentially arises when you try to define a predicate Even(x) (means: x is even) and ~Even(x) means x is odd. As you would expect, if you are are talking only about the natural numbers should probably restrict quantifiers to the natural numbers. But not Jan Burse. He wants to be able to make inferences about the evenness of every object in the universe -- his "dark elements."

Here is the standard way to define Even on the natural numbers:

ALL(a):[a in N => [Even(a) <=> EXIST(b):[b in N & a=2*b]]]

Poor Jan Burse gets in backwards, doesn't explicitly restrict the first quantifier and gets into trouble. He writes:

ALL(a):[Even(a) <=> a in N & EXIST(b):[b in N & a=2*b]]

It may look OK to his untrained eye, but it does lead to some major wonkiness with ALL(a):[~a in N => ~Even(a)]. Does ~Even(x) mean x is odd? Not necessarily, according to Jan Burse.

This confusion can be easily avoided by starting with a proper definition, one in which each quantifier is explicitly restricted to the set of natural numbers. Jan Burse would rather die.

[Mostowski Collapse]

You are crazy, this definition:

ALL(a):[Even(a) <=> a in N & EXIST(b):[b in N & a=2*b]]

You can even write it as:
ALL(a):[Even(a) <=> EXIST(b):[b in N & a=2*b]]

Gives nothing else than Even = {0, 2, 4, …}. What should it
give else, so that ALL(a):[~a in n => ~Even(a)] would not be valid?

[Mostowski Collapse]

Or do you see some b e N, so that this expression:

a = 2*b

gives something else than what is listed Even = {0, 2, 4, …}.

BTW: We can now establish a third form of defining Even and Odd,
also mentioned in Euclid, and the most useful for math:

Case 3:

ALL(a):[Even(a) <=> EXIST(b):[b in N & a=2*b]]

ALL(a):[Odd(a) <=> EXIST(b):[b in N & a=2*b+1]]

[Mostowski Collapse]

Case 3 is equivalent to Case 2.
Case 3 is not equivalent to Case 1.

Since Euclid mentioned Case 2 and Case 3, I guess Case 1 is a no go.
But the main argument against Case 1 is that it is a non-well defined
definition, whereas Case 2 and Case 3 are both well defined.

https://en.m.wikipedia.org/wiki/Well-defined_expression

[Colt Hiyama]

Dan Christensen wrote:

>> > 11. ALL(a):[~a in n => ~Even(a)]
>> > Conclusion, 2
>
> This is the wonky result of Jan Burse's wonky definition of Even( ). A
> cautionary tale that you really do need to restrict quantifiers -- to the
> set of natural numbers in this case.

you two nazis sure??

Ukraine’s ‘senseless PR action’ at Snake Island led to heavy losses – Russia
Kiev lost 30 drones, 14 aircraft and 3 ships in a ‘senseless PR action’ to retake island, the Russian military says
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Ukraine turns off Europe-bound gas
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https://www.rt.com/russia/555265-gazprom-gas-ukraine-europe/

Millions of UK homes face no heat this winter, power chief warns
https://www.rt.com/news/555258-uk-energy-heating-cost/

British TV censored Zelensky Nazi pic – Russian diplomat
Sky News slot ended after Dmitry Polyansky showed an image of a Nazi symbol posted by Ukraine’s leader
https://www.rt.com/russia/555225-russian-diplomat-sky-news/

UN Has 'Credible Information' on Ukrainian Troops' Torture of Russian PoWs
https://sputniknews.com/20220510/un-has-credible-information-on-ukrainian-troops-torture-of-russian-pows-1095406731.html
"We have received credible information of torture, ill-treatment and incommunicado detention by [the] Ukrainian Armed Forces of prisoners of war belonging to the Russian Armed Forces and affiliated armed groups", Bogner said in a press briefing on Tuesday.

WATCH Ukrainian Neo-Nazis March to Celebrate Anniversary of UPA
https://sputniknews.com/20220510/watch-ukrainian-neo-nazis-march-to-celebrate-anniversary-of-upa-1095393516.html
https://videon.img.ria.ru/Out/Flv/20220509/2022_05_09_DONBASSGENOCIDE9051_kf5tkl44.fba.mp4

[Dan Christensen]

On Tuesday, May 10, 2022 at 4:38:58 PM UTC-4, Colt Hiyama wrote:
> Dan Christensen wrote:
>
> >> > 11. ALL(a):[~a in n => ~Even(a)]
> >> > Conclusion, 2
> >
> > This is the wonky result of Jan Burse's wonky definition of Even( ). A
> > cautionary tale that you really do need to restrict quantifiers -- to the
> > set of natural numbers in this case.
> you two nazis sure??
>

Remember:

WAR OF AGREESION = NAZI

That's you, Nazi boy! Their denials didn't help the original Nazis. Yours won't help you. Also remember to keep that length of rope handy for when the Nazi hunters track you down.

[Cecil Nitta]

Dan Christensen wrote:

>> > the set of natural numbers in this case.
>> you two nazis sure??
>
> Remember: WAR OF AGREESION = NAZI

this inbreed nazi still committing double negation. _Aggression_ already
is included the term _war_, you know nothing idiot.

[Mostowski Collapse]

Use your EvenNextOdd to prove:

f(1)=2
f(n)=f(n-1)+1 for n>1 & Even(n)
f(n)=f(n-2)+2 for n>1 & Odd(n)
=>
f(2017)=2018

[Mostowski Collapse]

Source:

MiniF2F: a cross-system benchmark for formal Olympiad-level mathematics,
Zheng, Kunhao and Han, Jesse Michael and Polu, Stanislas - 2021
https://github.com/openai/miniF2F

[Dan Christensen]

On Wednesday, May 11, 2022 at 4:03:39 PM UTC-4, Cecil Nitta wrote:
> Dan Christensen wrote:
>
> >> > the set of natural numbers in this case.
> >> you two nazis sure??
> >
> > Remember: WAR OF AGREESION = NAZI
> this inbreed nazi still committing double negation.

Tell it to the judges, Nazi boy.

See the Nuremberg judgement. Or has your boss has censored that, too? Must make him nervous.

[Mostowski Collapse]

The problem is from here:
https://artofproblemsolving.com/wiki/index.php/2017_AMC_12A_Problems#Problem_7

I guess its even not provable so easy from your Peano
Axioms. Course of value induction would be more handy.
But maybe you can map it to ordinary mathematical induction.

By proving Q(a) <=> ALL(b):[b < a => P(b)], via ordinary mathematical
induction. One gets from the ordinary mathematical induction
schema which reads:

Q(0) ALL(c):[Q(c) => Q(c+1)]
----------------------------------------------------
ALL(d):Q(d)

The following new schema. For observe that Q(0) is
vacusly true, since there is no b < 0. So we have to
only look what Q(c) => Q(c+1) says. But Q(c) says

ALL(b):[b < c => P(b)] and Q(c+1) says this here
ALL(b):[b < c+1 => P(b)]. The later is equivalent
to ALL(b)[b < c => P(b)] and P(c). But the ALL(b):[b < c=> P(b)]

in the conclusion Q(c+1) is redundant. Also a further
consideration tells that ALL(d):Q(d) implies ALL(e):P(e).
So basically we have a new schema:

ALL(c):[ALL(b)[b < c => P(b)] => P(c)]
-----------------------------------------------------------
ALL(e):P(e)

Course of value induction. This could help
in proving that we have:

ALL(a):[a e n => [a > 0 => f(a)=a+1]]

[Dock Ichimonji]

Dan Christensen wrote:

> On Wednesday, May 11, 2022 at 4:03:39 PM UTC-4, Cecil Nitta wrote:
>> Dan Christensen wrote:
>>
>> >> > the set of natural numbers in this case.
>> >> you two nazis sure??
>> >
>> > Remember: WAR OF AGREESION = NAZI
>> this inbreed nazi still committing double negation.

British mercenaries in Ukraine face death penalty
https://www.rt.com/news/556221-mercenaries-ukraine-death-penalty/
Viktor Gavrilov, the judiciary spokesman, revealed that an investigation
into the activities of “a group of foreign mercenaries” who allegedly
“took part in the preparation and implementation of hostilities against
the DPR,” had been completed and a criminal case had been “fully formed.”

Three suspected mercenaries from Britain and Morocco, who joined the
Ukrainian military and were later captured by the forces of the Donetsk
People’s Republic (DPR), could be facing the death penalty there, the
republic’s General Prosecutor’s Office said on Friday.

[Bodie Nakahara]

Mostowski Collapse wrote:


> The problem is from here:

amazing, you guys. An armed police of 5000 in the fucking *fake_money*
shithole country of switzarland. To "protect" the oligarchs "partners" to
the retard dementia hit claus schwab (anal), all "climate warming"
oriented, traveling in private jets.

here's another nazi oligarch, trying to escape justice.

Ex-president attempts to leave Ukraine
https://www.rt.com/russia/556224-ukraine-poroshenko-poland-border/

Former president of Ukraine Petro Poroshenko tried to cross the border
into Poland on Friday, Ukrainian media reported citing the state customs
service sources. Volodymyr Zelensky’s predecessor is currently facing
treason charges in Kiev.

[Dan Christensen]

On Tuesday, May 10, 2022 at 12:40:43 PM UTC-4, Mostowski Collapse wrote:
> You are crazy, this definition:
> ALL(a):[Even(a) <=> a in N & EXIST(b):[b in N & a=2*b]]
> You can even write it as:
> ALL(a):[Even(a) <=> EXIST(b):[b in N & a=2*b]]
>

You can play around with this wonky definition and get all the wonky results you want. Not interested, Jan Burse.

Dan

[Mostowski Collapse]

You should be able to prove it from Presburger arithmetic
only, thats quite easy, I sketched it already here:
https://groups.google.com/g/sci.logic/c/UABVyIeAlxc/m/O_kQbK6PAQAJ

Presburger arithmetic is Peano induction
plus addition. See here:

x + 0 = x
x + (y + 1) = (x + y) + 1
https://en.wikipedia.org/wiki/Presburger_arithmetic

No need to introduce multiplication which is not necessarely
multiplication on the natural numbers. Just use a+a in place of 2*a.

On the German Wikipedia of Presburger arithmetic,
they make Even as an example for Presburger arithmetic:

„ x ist gerade“ durch G ( x ) :⇔ ∃ z : z + z = x
https://de.wikipedia.org/wiki/Presburger-Arithmetik

[Dan Christensen]

See my reply yesterday to your identical posting elsewhere.

Dan

On Saturday, May 28, 2022 at 5:52:32 AM UTC-4, Mostowski Collapse wrote:
> You should be able to prove it from Presburger arithmetic
> only, thats quite easy, I sketched it already here:
> https://groups.google.com/g/sci.logic/c/UABVyIeAlxc/m/O_kQbK6PAQAJ
>
> Presburger arithmetic is Peano induction

> plus addition. See here: ...

[snip]

[Mostowski Collapse]

How would you prove:

ALL(a):[a e n => Even(a) v Even(3*a+1)]
https://www.tptp.org/cgi-bin/SeeTPTP?Category=Problems&Domain=ARI&File=ARI598=1.p

Is your EvenNextOdd.html strong enough?

[Mostowski Collapse]

You only want to glorify your dislexia. But I see an AND
here and not an IF, from the website cuemath.com:

„The even natural numbers are the numbers that are
even, exactly divisible by 2, and belong to the set N. So
the set of even natural numbers is {2,4,6,8,...}.“
https://www.cuemath.com/numbers/natural-numbers/

So this is wrong: Dan O Matiks EvenNextOdd
ALL(a):[a e n => [Even(a) <=> EXISTS(b):[b e n & 2*b = a]]]

And this is correct: Common Sense, gives {2,4,6,8,...}.
ALL(a):[Even(a) <=> a e n & EXISTS(b):[b e n & 2*b = a]]

[Levon Tsuda]

Mostowski Collapse wrote:

> You only want to glorify your dislexia. But I see an AND here and not an
> IF, from the website cuemath.com:

absolutely, this nazi oriented individual, named Dan Christensen, is
deleting the evidences and the proofs proving him wrong.