On Monday, June 13, 2022 at 5:06:49 PM UTC-4, Mostowski Collapse wrote:

>
> ALL(a):[a e n => [Even(a) <=> EXISTS(b):[b e n & 2*b = a]]]
>

That's right. Here, evenness (even or not even) is defined ONLY for the natural numbers, as it should be. If you want to redefine it for everything in the universe, that's your problem, Jan Burse.

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:

> On Monday, June 13, 2022 at 5:06:49 PM UTC-4, Mostowski Collapse wrote:
>> ALL(a):[a e n => [Even(a) <=> EXISTS(b):[b e n & 2*b = a]]]
>
> That's right. Here, evenness (even or not even) is defined ONLY for the
> natural numbers, as it should be. If you want to redefine it for

not so. The war in "ukraine" goes like this:

"you give us the weapons, we give the deaths."

Nothing problematic, just the common definition of even number.
It is stated here word for word, having an AND aka conjunction.
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]]

Where do you have a verbose definition of your nonsense
that interchanges AND by an IF aka implication.
Can you find that on the internet?

Maybe stop calling things problematic that are in common use
and ask yourself whether what you were doing isnt
common use and rather exotic balderdash.

Dan Christensen schrieb am Dienstag, 14. Juni 2022 um 04:58:27 UTC+3:
> On Monday, June 13, 2022 at 5:06:49 PM UTC-4, Mostowski Collapse wrote:
>
> >
> > ALL(a):[a e n => [Even(a) <=> EXISTS(b):[b e n & 2*b = a]]]
> >
>
> That's right. Here, evenness (even or not even) is defined ONLY for the natural numbers, as it should be. If you want to redefine it for everything in the universe, that's your problem, Jan Burse.
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Same would apply to prime number, this would be a definition:

ALL(a):[Prime(a) <=> a e n & ~EXIST(b):[b e n & b≠1 & b≠a & b | a]]

Can you prove this here with DC poop?

ALL(a):[Prime(a) & a>3 => 12 | a^2-1]
https://mobile.twitter.com/pickover/status/1536796647128276994

Mostowski Collapse schrieb am Mittwoch, 15. Juni 2022 um 10:37:58 UTC+3:
> Nothing problematic, just the common definition of even number.
> It is stated here word for word, having an AND aka conjunction.
> 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]]
>
> Where do you have a verbose definition of your nonsense
> that interchanges AND by an IF aka implication.
> Can you find that on the internet?
>
> Maybe stop calling things problematic that are in common use
> and ask yourself whether what you were doing isnt
> common use and rather exotic balderdash.
> Dan Christensen schrieb am Dienstag, 14. Juni 2022 um 04:58:27 UTC+3:
> > On Monday, June 13, 2022 at 5:06:49 PM UTC-4, Mostowski Collapse wrote:
> >
> > >
> > > ALL(a):[a e n => [Even(a) <=> EXISTS(b):[b e n & 2*b = a]]]
> > >
> >
> > That's right. Here, evenness (even or not even) is defined ONLY for the natural numbers, as it should be. If you want to redefine it for everything in the universe, that's your problem, Jan Burse.
> >
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

Ok, lets make it more dramatic:

A mathematician emerges from a cave, hands you the piece of
paper below, and asks "Is this true?" What is your response?

If a is prime number > 3, then a^2 - 1 is a multiple of 12

Mostowski Collapse schrieb am Mittwoch, 15. Juni 2022 um 10:46:40 UTC+3:
> Same would apply to prime number, this would be a definition:
>
> ALL(a):[Prime(a) <=> a e n & ~EXIST(b):[b e n & b≠1 & b≠a & b | a]]
>
> Can you prove this here with DC poop?
>
> ALL(a):[Prime(a) & a>3 => 12 | a^2-1]
> https://mobile.twitter.com/pickover/status/1536796647128276994
> Mostowski Collapse schrieb am Mittwoch, 15. Juni 2022 um 10:37:58 UTC+3:
> > Nothing problematic, just the common definition of even number.
> > It is stated here word for word, having an AND aka conjunction.
> > 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]]
> >
> > Where do you have a verbose definition of your nonsense
> > that interchanges AND by an IF aka implication.
> > Can you find that on the internet?
> >
> > Maybe stop calling things problematic that are in common use
> > and ask yourself whether what you were doing isnt
> > common use and rather exotic balderdash.
> > Dan Christensen schrieb am Dienstag, 14. Juni 2022 um 04:58:27 UTC+3:
> > > On Monday, June 13, 2022 at 5:06:49 PM UTC-4, Mostowski Collapse wrote:
> > >
> > > >
> > > > ALL(a):[a e n => [Even(a) <=> EXISTS(b):[b e n & 2*b = a]]]
> > > >
> > >
> > > That's right. Here, evenness (even or not even) is defined ONLY for the natural numbers, as it should be. If you want to redefine it for everything in the universe, that's your problem, Jan Burse.
> > >
> > > 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, June 15, 2022 at 3:37:58 AM UTC-4, Mostowski Collapse wrote:
> Nothing problematic, just the common definition of even number.
> It is stated here word for word, having an AND aka conjunction.
> 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/
>

Not a formal definition. You need something with "even if and only if."

Dan

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

When you proved this:

|B ^ A| = |B| ^ |A|

You could then prove this, for finite cardinalities,
does not hold for infinite cardinalities, since for
example | ω ^ 1 | = | ω ^ 2 | (Cantor Pairing):

|B| > 1, |A| < |C| => |B ^ A| < |B ^ C|

As a corollary you would have:

|2 ^ 1| ≠ |2 ^ 0|

Or was this already answered?

Dan Christensen schrieb am Mittwoch, 15. Juni 2022 um 17:53:53 UTC+2:
> On Wednesday, June 15, 2022 at 3:37:58 AM UTC-4, Mostowski Collapse wrote:
> > Nothing problematic, just the common definition of even number.
> > It is stated here word for word, having an AND aka conjunction.
> > 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/
> >
> Not a formal definition. You need something with "even if and only if."
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

On Monday, July 11, 2022 at 5:23:08 PM UTC-4, Mostowski Collapse wrote:
> When you proved this:
>
> |B ^ A| = |B| ^ |A|
>

[snip]

IIUC cardinal numbers require a form of class theory that is not supported by set theory alone, not even by the ZFC axioms. Maybe it is possible to devise some additional axioms for use in DC Proof to implement class theory, but it is not a project that interests me.

Dan

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

Its amazing how one error leads to another error.
Now Dan Christensen is as stubborn as he is
mentaly inhibited, he does not understand that

his insistence to the following:
1) extrinsic property
2) modelled as toplevel ALL

leads to inconsistent systems. So then lets make
a simple example where everybody can relate. Lets
say there are black and white horses, and the color

black is different from the color white, and there is
a horse ajax which has white color, and we want
to prove the following:

white_horse(ajax)

Based on a definition of white_horse?

Mostowski Collapse schrieb am Donnerstag, 25. November 2021 um 10:15:39 UTC+1:
> 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?

Here is how the troll Dan Christensen would do it:

ALL(color):ALL(x):[white_horse(x) <=> horse(x) & color = white]

And he would then proudly present a proof of:

white_horse(ajax)

Because alas the horse is white, so he can instantiate
the ALL quantifier for color with white, and gets white = white,
which is true, so that he gets white_horse(ajax).

Like 50% of his published proofs, especially the recent ones
where he struggles with function spaces are of this type.
So whats the problem? Well we can also prove this one:

~white_color(ajax)

Just use the same proof schema like before, but this time
instantiate color with black, which gives black = white,
which gives false. So whats the cure?

The cure is simple:
- use intrinsic properties
- have definite descriptions (yes its Russell) for these properties

Applied to the white horse problem, the correct way to
model what a white horse is, is then using this definition,
which does not anymore have an ALL color toplevel:

ALL(x):[white_horse(x) <=> horse(x) & color_of(x) = white]

Mostowski Collapse schrieb am Mittwoch, 13. Juli 2022 um 12:32:20 UTC+2:
> Its amazing how one error leads to another error.
> Now Dan Christensen is as stubborn as he is
> mentaly inhibited, he does not understand that
>
> his insistence to the following:
> 1) extrinsic property
> 2) modelled as toplevel ALL
>
> leads to inconsistent systems. So then lets make
> a simple example where everybody can relate. Lets
> say there are black and white horses, and the color
>
> black is different from the color white, and there is
> a horse ajax which has white color, and we want
> to prove the following:
>
> white_horse(ajax)
>
> Based on a definition of white_horse?
>
> Mostowski Collapse schrieb am Donnerstag, 25. November 2021 um 10:15:39 UTC+1:
> > 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?

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

Dan

On Wednesday, July 13, 2022 at 6:32:20 AM UTC-4, Mostowski Collapse wrote:
> Its amazing how one error leads to another error.

[snip]

Dan Christensen wrote:

> See my recent reply to your identical posting at sci.logic Dan

you answer twice to identical posting, idiot.

> On Wednesday, July 13, 2022 at 6:32:20 AM UTC-4, Mostowski Collapse
> wrote:
>> Its amazing how one error leads to another error.

Maybe this will help you. They are in the same relation:

- The color of a horse (assuming a horse has a single color)
- The domain of a function

They are both definite descriptions. If you cannot
turn this into some function dom(f), in your proof
system DC Proof, then there is no use of

your proof system DC Proof, you will always produce
nonsense ad hoc axioms, that lead to an inconsistent
system, especially if you do your ALL nonsense.

Dan Christensen schrieb am Mittwoch, 13. Juli 2022 um 17:13:07 UTC+2:
> See my recent reply to your identical posting at sci.logic
>
> Dan
> On Wednesday, July 13, 2022 at 6:32:20 AM UTC-4, Mostowski Collapse wrote:
> > Its amazing how one error leads to another error.
> [snip]

And you are a fucking liar. I don't use different
domains for f and g. You can also provoke,
other people here did this also:

f=g & ~(f=g)

By using for x1,y1 for both f and g, and then
using x2,y2 for both f and g.

To difficult for you to understand? People
here already made an example of

such f,g,x1,y1,x2,y2. Thats extremly easy
to show that your Eg Function axiom

is nonsense.

Mostowski Collapse schrieb am Mittwoch, 13. Juli 2022 um 20:14:11 UTC+2:
> Maybe this will help you. They are in the same relation:
>
> - The color of a horse (assuming a horse has a single color)
> - The domain of a function
>
> They are both definite descriptions. If you cannot
> turn this into some function dom(f), in your proof
> system DC Proof, then there is no use of
>
> your proof system DC Proof, you will always produce
> nonsense ad hoc axioms, that lead to an inconsistent
> system, especially if you do your ALL nonsense.
> Dan Christensen schrieb am Mittwoch, 13. Juli 2022 um 17:13:07 UTC+2:
> > See my recent reply to your identical posting at sci.logic
> >
> > Dan
> > On Wednesday, July 13, 2022 at 6:32:20 AM UTC-4, Mostowski Collapse wrote:
> > > Its amazing how one error leads to another error.
> > [snip]

Its extremly trivial, use your idiotic axiom:

ALL(dom):ALL(cod):ALL(f1):ALL(f2):[Set(dom) & Set(cod)
& ALL(a):[a in dom => f1(a) in cod]
& ALL(a):[a in dom => f2(a) in cod]
=> [f1=f2 <=> ALL(a):[a in dom => f1(a)=f2(a)]]]

With these two functions:

f1(0)=0
f2(0)=1

Then use it first with dom={},cod={0,1} and you will get:

f1 = f2

Then use it with dom={0}, cod={0,1} and you will get:

~(f1 = f2)

You can also chase demons, and forbid empty domains,
we will still be able to construct f=g & ~(f=g) by using
another pair f,g.

Mostowski Collapse schrieb am Mittwoch, 13. Juli 2022 um 20:18:02 UTC+2:
> And you are a fucking liar. I don't use different
> domains for f and g. You can also provoke,
> other people here did this also:
>
> f=g & ~(f=g)
>
> By using for x1,y1 for both f and g, and then
> using x2,y2 for both f and g.
>
> To difficult for you to understand? People
> here already made an example of
>
> such f,g,x1,y1,x2,y2. Thats extremly easy
> to show that your Eg Function axiom
>
> is nonsense.
> Mostowski Collapse schrieb am Mittwoch, 13. Juli 2022 um 20:14:11 UTC+2:
> > Maybe this will help you. They are in the same relation:
> >
> > - The color of a horse (assuming a horse has a single color)
> > - The domain of a function
> >
> > They are both definite descriptions. If you cannot
> > turn this into some function dom(f), in your proof
> > system DC Proof, then there is no use of
> >
> > your proof system DC Proof, you will always produce
> > nonsense ad hoc axioms, that lead to an inconsistent
> > system, especially if you do your ALL nonsense.
> > Dan Christensen schrieb am Mittwoch, 13. Juli 2022 um 17:13:07 UTC+2:
> > > See my recent reply to your identical posting at sci.logic
> > >
> > > Dan
> > > On Wednesday, July 13, 2022 at 6:32:20 AM UTC-4, Mostowski Collapse wrote:
> > > > Its amazing how one error leads to another error.
> > > [snip]

środa, Mostowski Collapse napisał(a):

> Maybe this will help you. They are in the same relation:
> - The color of a horse (assuming a horse has a single color)
> - The domain of a function

that's impossible, idiot. A "single color" means a point on the chroma
scale, a frequency. Hence its width is zero. What you call a single color
are infact many, say infinity many, spreading from the single frequency
you think its alone. Your khakhole education is close to zero. What a
shithole of a country.

On Wednesday, July 13, 2022 at 2:25:27 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> Its extremly trivial, use your idiotic axiom:
>
> ALL(dom):ALL(cod):ALL(f1):ALL(f2):[Set(dom) & Set(cod)
> & ALL(a):[a in dom => f1(a) in cod]
> & ALL(a):[a in dom => f2(a) in cod]
> => [f1=f2 <=> ALL(a):[a in dom => f1(a)=f2(a)]]]
>
> With these two functions:
>
> f1(0)=0
> f2(0)=1
>
> Then use it first with dom={},cod={0,1} and you will get:
>
> f1 = f2
>
> Then use it with dom={0}, cod={0,1} and you will get:
>
> ~(f1 = f2)
>

Pay attention, Jan Burse! When you construct a function with a different domain, you must give it a another name, i.e. not f1 or f2 in this case since they have already been used. (See https://en.wikipedia.org/wiki/Existential_instantiation)

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:

>> Then use it with dom={0}, cod={0,1} and you will get:
>>
>> ~(f1 = f2)
>
> Pay attention, Jan Burse! When you construct a function with a different
> domain, you must give it a another name, i.e. not f1 or f2 in this case
> since they have already been used. (See

it's called "mapping", not construct, you puerile deplorable behavioural.
That's what you do with function, map. You guys are mathematically
irrelevant.

On Thursday, July 14, 2022 at 6:49:45 AM UTC-4, Jere De angelis wrote:
> Dan Christensen wrote:
>
> >> Then use it with dom={0}, cod={0,1} and you will get:
> >>
> >> ~(f1 = f2)
> >
> > Pay attention, Jan Burse! When you construct a function with a different
> > domain, you must give it a another name, i.e. not f1 or f2 in this case
> > since they have already been used. (See

> it's called "mapping", not construct, you puerile deplorable behavioural.
> That's what you do with function, map. You guys are mathematically
> irrelevant.

Is that the best you can do, to quibble about terminology? What a loser.

Dan Christensen wrote:

>> it's called "mapping", not construct, you puerile deplorable
>> behavioural.
>> That's what you do with function, map. You guys are mathematically
>> irrelevant.
>
> Is that the best you can do, to quibble about terminology? What a loser.

you ladymen from america don't know how traitors looks like. Here we go.

Ukrainians riding in an American Humvee in Donbass
https://www.bitchute.com/video/C0KIw4tM0CuC/

On Thursday, July 14, 2022 at 5:43:44 PM UTC-4, Emil Bulgarelli wrote:
> Dan Christensen wrote:
>
> >> it's called "mapping", not construct, you puerile deplorable
> >> behavioural.
> >> That's what you do with function, map. You guys are mathematically
> >> irrelevant.
> >
> > Is that the best you can do, to quibble about terminology? What a loser.
> you ladymen from america don't know how traitors looks like. Here we go.
>
> Ukrainians riding in an American Humvee in Donbass

Did you find a good war crimes lawyer yet? You will need one, Nazi boy!

Dan Christensen wrote:

>> >> it's called "mapping", not construct, you puerile deplorable
>> >> behavioural.
>> >> That's what you do with function, map. You guys are mathematically
>> >> irrelevant.
>> >
>> > Is that the best you can do, to quibble about terminology? What a
>> > loser.
>> you ladymen from america don't know how traitors looks like. Here we
>> go.
>>
>> Ukrainians riding in an American Humvee in Donbass
>
> Did you find a good war crimes lawyer yet? You will need one, Nazi boy!

yet another _nazi_nato_ shithole country involved. Let it begin.

MQ-9 Reaper UCAV of the US Air Force crashed near the Romanian airbase
Campia Turzii https://www.bitchute.com/video/M6QAYdkiNBwC/

A proof is here:

146 f1=f2 & ~f1=f2
Join, 65, 145
http://www.rubycap.ch/dcproof16_inconsistency.html

But question is, is your new proposal of an equaivalence
relation really an equality? From an equality we expect:

1) f = f
2) f = g => g = f
3) f = g & g = h => f = h
4) f = g & A(f) => A(g)

Number 4) is called substitution of equals for equals.
Or you can also call 4) by the following phrase:

The indiscernibility of identicals
https://en.wikipedia.org/wiki/Identity_of_indiscernibles#Identity_and_indiscernibility

Does your new equivalence relation satisfy 4)?
(BTW, if you have 4) then 2) and 3) are redundant)

Dan Christensen schrieb am Donnerstag, 14. Juli 2022 um 04:36:15 UTC+2:
> On Wednesday, July 13, 2022 at 2:25:27 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> > Its extremly trivial, use your idiotic axiom:
> >
> > ALL(dom):ALL(cod):ALL(f1):ALL(f2):[Set(dom) & Set(cod)
> > & ALL(a):[a in dom => f1(a) in cod]
> > & ALL(a):[a in dom => f2(a) in cod]
> > => [f1=f2 <=> ALL(a):[a in dom => f1(a)=f2(a)]]]
> >
> > With these two functions:
> >
> > f1(0)=0
> > f2(0)=1
> >
> > Then use it first with dom={},cod={0,1} and you will get:
> >
> > f1 = f2
> >
> > Then use it with dom={0}, cod={0,1} and you will get:
> >
> > ~(f1 = f2)
> >
> Pay attention, Jan Burse! When you construct a function with a different domain, you must give it a another name, i.e. not f1 or f2 in this case since they have already been used. (See https://en.wikipedia.org/wiki/Existential_instantiation)
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Against Dan O Matik nonsense claim "This definition says NOTHING
about the equality of pairs of functions with different domains.".
Wrong, an equality that satisfies 4), has inqueality:

Just take 4):

f = g & A(f) => A(g)

Now use contraposition:

A(f) => A(g) & f =\= g

Now use contraposition again:

A(f), ~A(g) => f =\= g

Now we have turned 4) into a way of deriving inequality.
Is Dan O Matiks new equivalence relation powerful enough
to derive this inequality. Or is it just another turd?

What would be an example of such an A(_) that
would spawn an inequality of two functions? (I have
a few suggestion, but I want to see whether Dan-O-Matik

understands the problem, since so far he doesn't)

Mostowski Collapse schrieb am Donnerstag, 28. Juli 2022 um 08:56:54 UTC+2:
> A proof is here:
>
> 146 f1=f2 & ~f1=f2
> Join, 65, 145
> http://www.rubycap.ch/dcproof16_inconsistency.html
>
> But question is, is your new proposal of an equaivalence
> relation really an equality? From an equality we expect:
>
> 1) f = f
> 2) f = g => g = f
> 3) f = g & g = h => f = h
> 4) f = g & A(f) => A(g)
>
> Number 4) is called substitution of equals for equals.
> Or you can also call 4) by the following phrase:
>
> The indiscernibility of identicals
> https://en.wikipedia.org/wiki/Identity_of_indiscernibles#Identity_and_indiscernibility
>
> Does your new equivalence relation satisfy 4)?
> (BTW, if you have 4) then 2) and 3) are redundant)
> Dan Christensen schrieb am Donnerstag, 14. Juli 2022 um 04:36:15 UTC+2:
> > On Wednesday, July 13, 2022 at 2:25:27 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> > > Its extremly trivial, use your idiotic axiom:
> > >
> > > ALL(dom):ALL(cod):ALL(f1):ALL(f2):[Set(dom) & Set(cod)
> > > & ALL(a):[a in dom => f1(a) in cod]
> > > & ALL(a):[a in dom => f2(a) in cod]
> > > => [f1=f2 <=> ALL(a):[a in dom => f1(a)=f2(a)]]]
> > >
> > > With these two functions:
> > >
> > > f1(0)=0
> > > f2(0)=1
> > >
> > > Then use it first with dom={},cod={0,1} and you will get:
> > >
> > > f1 = f2
> > >
> > > Then use it with dom={0}, cod={0,1} and you will get:
> > >
> > > ~(f1 = f2)
> > >
> > Pay attention, Jan Burse! When you construct a function with a different domain, you must give it a another name, i.e. not f1 or f2 in this case since they have already been used. (See https://en.wikipedia.org/wiki/Existential_instantiation)
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

See my reply just now to your identical posting at sci.logic

Dan

On Thursday, July 28, 2022 at 3:18:49 AM UTC-4, Mostowski Collapse wrote:
> Against Dan O Matik nonsense claim "This definition says NOTHING
> about the equality of pairs of functions with different domains.".
> Wrong, an equality that satisfies 4), has inqueality:
>
[snip]