Bijectivity is Transitive, Formal Proof

[Dan Christensen]

As requested...

THEOREM: Bijectivity is Transitive

ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)

=> [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
& EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]

=> EXIST(h):Bijection(h,t,v)]]

PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)

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]

You could define:

u ~ v :<=> EXIST(f):[ f : u -> v, f bijective]

Now you only proved (but not using set-like f rather
did you use your class like f, i.e. FOL symbols, which
is ironic, since you constantly fight classes,

but in the present problem it probably doesn't matter
so much, since we will anyway only quantify over
the sets t, u, v, etc..):

/* transitivity */
t ~ u & u ~ v => t ~ v

How about these two, missing?

/* reflexivity */
t ~ t
/* symmetry */
t ~ u => u ~ t

And then the cardinality thingy?

ALL(t):[u ~ t <=> v ~ t] <=> u ~ v

[Mostowski Collapse]

Since this equality here, is an existential type:

u ~ v :<=> EXIST(f):[ f : u -> v, f bijective]

It anyway begs the question why you use your
nonsense function axiom. You can directly operate
with the graphs, no need to apply the function axiom,

to obtain results such as:

/* transitivity */
t ~ u & u ~ v => t ~ v

Your nonsense proof:
https://dcproof.com/BijectionsTransitive.htm (275 lines)

Would get much shorter without the detour
over your function axiom nonsense. I still have
the feeling you are a crank or an utter imbecil,

that prefers selling his moot nonsense, even if
it implies doing things utterly complicated. You
just sting to your self invented wonky math.

If you wouldn't sting to your self invented wonky
math, you could reduce the proof considerable, from
275 lines to much less. After all ordered pairs admit

composition, thats quite easy, (x,y) such that (x,z)
and (z,y). Extremly trivial, also an existential type. Using
your function axiom only to later be able to write f(g(x))

as you use to do, is totally senseless garbage.

[WM]

Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:

> THEOREM: Bijectivity is Transitive

> PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
>

There is no bijection between infinite sets because infinite sets contain dark numbers.

Poof for ℕ (5 lines):
If ℕ has no last element, then the intersection ∩{E(k) : k ∈ ℕ} cannot be empty. Up to every endsegment E(k) it is not empty, because the successor endsegment E(k+1) will further reduce it. The empty set however cannot be reduced. Since every natnumber of the endless sequence has a successor the intersection will never be empty. Therefore ∩{E(k) : k ∈ ℕ} = { } requires an empty endsegment and therefore dark numbers.

Regards, WM

[Fritz Feldhase]

Leidest Du an Brechdurchfall, Mückenheim?

[ Are you suffering from vomiting diarrhea, Mückenheim? ]

[Dan Christensen]

On Friday, July 1, 2022 at 6:08:11 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:
> > THEOREM: Bijectivity is Transitive
> > PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
> >
> There is no bijection between infinite sets because infinite sets contain dark numbers.
>

"Dark" numbers??? You have really gone off the deep end this time, Mucke!

There is a bijection between the set of natural numbers and the even numbers: For all x in N, f(x) = 2x. Both sets are infinite. Deal with it.

[Mostowski Collapse]

Cantor showed that:

N ~ N x N

Thats a classical result. Just try:

f : N x N -> N
https://mathworld.wolfram.com/PairingFunction.html

Because of this property that Dan Christensen did not yet prove:

/* Symmetry */
u ~ v => v ~ u

The existence of f, also implies the existence of some g : N -> N x N.

[Dan Christensen]

On Friday, July 1, 2022 at 5:05:00 AM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> Since this equality here, is an existential type:
> u ~ v :<=> EXIST(f):[ f : u -> v, f bijective]
> It anyway begs the question why you use your
> nonsense function axiom.

Jan Burse is pissed off that my standard definition (line 29) does not include his "dark" elements à la Muckenheim (WM). He wants to make logical inferences about functions outside of their domains of definition. Usually they are simply said to be undefined outside of their domains. Jan Burse still doesn't get it.

[snip]

> Would get much shorter without the detour
> over your function axiom nonsense.

Not really. Have you never had to prove the existence of a function? The function axiom establishes the standard requirements for the graph set of any function. That would be unavoidable even in your wonky system, Jan Burse.

[snip]

[Mostowski Collapse]

You can shorten your proofs considerably, when you don't
use your function axiom. The definition of ~ which was

at the center of much investigation by Cantor and others
100 years ago doesn't need any function axiom. You just

define it as follows:

u ~ v :<=> EXIST(f): [ f : u -> v , f bijective ]

Where f is a set-like functions aka a graph. You can directly
use graphs, didn't you know dumbo?

[Dan Christensen]

On Saturday, July 2, 2022 at 5:22:37 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Freitag, 1. Juli 2022 um 19:03:16 UTC+2:
> > On Friday, July 1, 2022 at 5:05:00 AM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> > > Since this equality here, is an existential type:
> > > u ~ v :<=> EXIST(f):[ f : u -> v, f bijective]
> > > It anyway begs the question why you use your
> > > nonsense function axiom.
> > Jan Burse is pissed off that my standard definition (line 29) does not include his "dark" elements à la Muckenheim (WM). He wants to make logical inferences about functions outside of their domains of definition. Usually they are simply said to be undefined outside of their domains. Jan Burse still doesn't get it.
> >
> > [snip]
> > > Would get much shorter without the detour
> > > over your function axiom nonsense.
> > Not really. Have you never had to prove the existence of a function? The function axiom establishes the standard requirements for the graph set of any function. That would be unavoidable even in your wonky system, Jan Burse.
> >

> You can shorten your proofs considerably, when you don't
> use your function axiom. The definition of ~ which was
>
> at the center of much investigation by Cantor and others
> 100 years ago doesn't need any function axiom. You just
>
> define it as follows:
> u ~ v :<=> EXIST(f): [ f : u -> v , f bijective ]

Now, you need to formally define "f: u -> v" and "bijective".

> Where f is a set-like functions aka a graph. You can directly

> use graphs...

Very awkward for most applications and prone to some wonky results, e.g. your "dark" elements. Better and much more convenient to have: ALL(a):ALL(b):[a in dom & b in cod => [f(a)=b <=> (a, b) in graph]]. Then f(x) is undefined/unspecified for x not in dom.

[WM]

Dan Christensen schrieb am Freitag, 1. Juli 2022 um 16:40:11 UTC+2:
> On Friday, July 1, 2022 at 6:08:11 AM UTC-4, WM wrote:

> > There is no bijection between infinite sets because infinite sets contain dark numbers.
> >
> "Dark" numbers??? You have really gone off the deep end this time, Mucke!
>
> There is a bijection between the set of natural numbers and the even numbers: For all x in N, f(x) = 2x. Both sets are infinite. Deal with it.

I did and showed dark numbers. Here is a proof (one of many):
For all definable numbers we have the intersection of their endsegments
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀

For all natural numbers, we have the empty intersection of their endsegments:
∩{E(k) : k ∈ ℕ} = { }

The endsegments can only shrink by on number per endsegment
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

What natural numbers reduce the intersection from ℵ₀ to zero? Can they be named?

Regards, WM

[Mostowski Collapse]

Thats not extremly difficult:

f : u -> v has the following definition:
- f relation: f ⊆ u x v,
- f total: ∀x e u ∃ y e v (x,y) e f
- f univalent: ∀x e u ∀ y,z e v ((x,y) e f & (x,z) e f => y = z)
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Total,_univalent_relation

f bijective has the following definition:
- f injective: ∀x,t e u ∀ y e v ((x,y) e f & (t,y) e f => x = t)
- f surjective: ∀y e v ∃ x e u (x,y) e f

Whats your prololoblem...

[Dan Christensen]

On Saturday, July 2, 2022 at 10:29:10 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Samstag, 2. Juli 2022 um 15:11:40 UTC+2:
> > Now, you need to formally define "f: u -> v" and "bijective".

> Thats not extremly difficult:
>
> f : u -> v has the following definition:
> - f relation: f ⊆ u x v,
> - f total: ∀x e u ∃ y e v (x,y) e f
> - f univalent: ∀x e u ∀ y,z e v ((x,y) e f & (x,z) e f => y = z)
> https://en.wikipedia.org/wiki/Function_%28mathematics%29#Total,_univalent_relation
>

These are the requirements for the graph of a function. See line 29. Missing from your "definition" is the connection between the graph set and the commonly used f(x)=y notation.

> f bijective has the following definition:

See line 3.

> - f injective: ∀x,t e u ∀ y e v ((x,y) e f & (t,y) e f => x = t)

See line 1 using the the prefix operator notation.

> - f surjective: ∀y e v ∃ x e u (x,y) e f
>

See line 2.

[Mostowski Collapse]

What is prefix operator notation?
Do you mean function application?
What you call "commonly used f(x)=y"

Well it doesn't exist for set-like functions
in the form of FOL function application.
Because a set-like function has:

f : u -> v has the following definition:
- f relation: f ⊆ u x v,

What is the value of f(x) when ~(x e u)?

A way out was already proposed by
Fritz, namely f(x) := U { y | ∃ y (x,y) e f }
you then have f(x) = {} for ~(x e u).

But how would the notation f(x) be relevant
if you want to prove these 4. Can you tell
is dumbo. As I said it gets easier without f(x):

/* Transitivity */
u ~ v & v ~ t => u ~ t

/* Symmetry */
u ~ v => v ~ u

/* Reflexivity */
v ~ v
/* Classes */
∀t(u ~ t <=> v ~ t) <=> u ~ v

Where do you see f(x)? You are not very
good at Occams Razor. You introduce concepts
that are not needed.

Thats also why you cannot prove:

~(2^1 ~ 2^0)

[Mostowski Collapse]

The connection is not needed to connect graph and function,
thats also an error that you are subject to. The graph is already
the function. See here wikipedia:

a function is a binary relation that is univalent and total.
https://en.wikipedia.org/wiki/Function_%28mathematics%29#Total,_univalent_relation

You don't need to connect graph and function, via
f(x) = y. f(x) = y is also defined on a graph. graph and
function are synonymous. Whats wrong with you?

But its funny, it seems wikipedia did an update. In the
past I was referencing wikipedia here on sci.logic, and
they didn't write "total" instead they wrote "serial",

and they didn't write "univalent" instead they wrote
"functional". Anyway, I don't mind. The message is the
same, a function is a relation on u x v, and

not the Dan-O-Matik nonsense.

[Mostowski Collapse]

Its rather an error of DC Proof that you cannot
invoke f(x) for a graph aka set-like function and
that you need a function axiom, and two

different symbols. Thats utter nonsense. f(x)
is of course when f is a graph and x e dom(f).
When ~(x e dom(f)) there is the option

of Fritz, defining it f(x)={}. A set theoretic
way to do it is to make the following definition:

/* Variant of the Peano Apostroph */
f(x) := U { y | (x,y) e f }

We would then not hava a function axiom, but
a function application axiom defining the
notion _(_) represented by Peano Apostroph _'_:

/* function application axiom */
ALL(f):ALL(x):ALL(y):ALL(z):[z e f'x <=> ∃ y [(x,y) e f & z e y]]

So what went wrong in DC Proof: 3 things
a) there is a function axiom which makes an
artificial distinction between graph and function,

introducing nonsense FOL symbols and quantifying
over them, which is anyway not possible in FOL,
b) there is no function application axiom.

c) last but not least many things don't need
function application anyway, they can be directly
proved from the graph, example equivalence

relation stuff. It gets much easier.

P.S.: The Peano Apostroph is also well behaved,
just like FOL symbols function application, you
can also use substitution and whatever.

[Dan Christensen]

On Saturday, July 2, 2022 at 1:57:05 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> Its rather an error of DC Proof that you cannot
> invoke f(x) for a graph aka set-like function and
> that you need a function axiom, and two
>
> different symbols.

It's no error, Jan Burse. Using different symbols for the graph and the function name avoids overloading the symbol for the function name. One is a set of ordered pairs, the other is an operator. It avoids confusion, maybe some wonky results. And, sorry, Jan Burse, but it also avoids your cherished "dark" elements. As in most math textbooks, if x is not in the domain of function f, then the definition of f is not applicable for x. f(x) is simply undefined or unspecified, as you would expect.

[Mostowski Collapse]

Well you doing it too much complicated. Thats not how
standard mathematics does it. You establish two worlds:

World1: graphs g, elements from the domain of discourse,
in particular they satisfy Set(g)
World2: functions f, not elements from the domain of discourse!
They are the usual FOL function symbols. They do not
satisfy Set(f), would lead to Russell Paradox

Now since you insist working with these two worlds,
you to it utterly complicated, your function axioms does
two things at once:

Dan O Matiks Function Axiom:
a) It says that for g there is some f
b) and it says what f(x) is

Now guess what. The standard mathematics works much
easier, it only does one thing:

Standard Mathematics Function Notation:
a) It says what g(x) is

[Mostowski Collapse]

FOL function symbols lead to Russell Paradox, if you
would try to put them into the usual domain of discourse,
since they lead to Universal Set. You can prove:

∀x∃y f(x) = y

This implies that dom(f) is the full domain of discourse.
So as a set of pairs, its graph { (x,f(x)) } is much too
large for an ordinary set.

A FOL function symbol is total on the domain of discourse.
But graphs are not total on the domain of discourse, a graph from:

g : u -> v

is only total on u, its domain, but not on the domain
of discourse. So your function axiom is utter nonsense,
the error is that it does two things:

Dan O Matiks Function Axiom:
a) It says that for g there is some f

b) and it says what f(x) is in relation to g

But usual mathematics needs only to do one thing:

Standard Mathematics Function Notation:
a) It says what g(x) is

Also the graph of g is g. You don't need to compute
{ (x,f(x)) }, the graph is directly { (x,y) | (x,y) e g } which is
g itself. Graph computation is only a school distinction,

since functions are sometimes given intensionally,
by some algorithm, mapping notation, etc.. so f refers
to this intensional thingy, and you want to express that

there is also an extensional side of it. You don't have
an intensional aspect in DC Proof, FOL function symbols
are also extensional, they are only too big for sets. So

I don't see any argument in favor of a second world,
if the second world were intensional this would lead to
some interesting new logic. But set-like functions

are anyway extensional objects. Thats all you need to do,
in ZFC you can do it as follows:

/* function application axiom */

∀f∀x∀y∀z(z e f'x <=> ∃ y ((x,y) e f & z e y))

Extremly short and elegant the Fritz approach. In DC
Proof, since it is close to ZFCU, i.e. ZFC with Urelements,
its more complicated. You cannot use Fritz approach.

Making a proposal for g(x) in ZFCU is left as an exercise.

[Dan Christensen]

On Saturday, July 2, 2022 at 4:19:58 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:

> Dan Christensen schrieb am Samstag, 2. Juli 2022 um 20:52:20 UTC+2:
> > On Saturday, July 2, 2022 at 1:57:05 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> > > Its rather an error of DC Proof that you cannot
> > > invoke f(x) for a graph aka set-like function and
> > > that you need a function axiom, and two
> > >
> > > different symbols.
> > It's no error, Jan Burse. Using different symbols for the graph and the function name avoids overloading the symbol for the function name. One is a set of ordered pairs, the other is an operator. It avoids confusion, maybe some wonky results. And, sorry, Jan Burse, but it also avoids your cherished "dark" elements. As in most math textbooks, if x is not in the domain of function f, then the definition of f is not applicable for x. f(x) is simply undefined or unspecified, as you would expect.

> Well you doing it too much complicated. Thats not how
> standard mathematics does it.

[snip]

For any function f, standard mathematics leaves f(x) undefined for x outside of its domain of definition. Its definition is simply not a applicable. This notion is not possible if functions are indistinguishable from their graph sets as you would have it. That's why we have, for any function f mapping set A to set B, its definition is of the form:

ALL(x):ALL(y):[x in A & y in B => [f(x)=y <=> P(x, y)]]

Such a definition is simply NOT APPLICABLE for x not in A. I know it pisses you off, Jan Burse, but it is a fact. Deal with it.

[Dan Christensen]

On Saturday, July 2, 2022 at 4:32:36 PM UTC-4, Mostowski Collapse wrote:
> FOL function symbols lead to Russell Paradox, if you
> would try to put them into the usual domain of discourse,
> since they lead to Universal Set. You can prove:
>
> ∀x∃y f(x) = y
>

[snip]

Not in DC Proof. Try it.

[Dan Christensen]

On Sunday, July 3, 2022 at 8:04:28 AM UTC-4, Dan Christensen wrote:
> On Saturday, July 2, 2022 at 4:19:58 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
>
> > Dan Christensen schrieb am Samstag, 2. Juli 2022 um 20:52:20 UTC+2:
> > > On Saturday, July 2, 2022 at 1:57:05 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> > > > Its rather an error of DC Proof that you cannot
> > > > invoke f(x) for a graph aka set-like function and
> > > > that you need a function axiom, and two
> > > >
> > > > different symbols.
> > > It's no error, Jan Burse. Using different symbols for the graph and the function name avoids overloading the symbol for the function name. One is a set of ordered pairs, the other is an operator. It avoids confusion, maybe some wonky results. And, sorry, Jan Burse, but it also avoids your cherished "dark" elements. As in most math textbooks, if x is not in the domain of function f, then the definition of f is not applicable for x. f(x) is simply undefined or unspecified, as you would expect.
> > Well you doing it too much complicated. Thats not how
> > standard mathematics does it.
> [snip]
>
> For any function f, standard mathematics leaves f(x) undefined for x outside of its domain of definition. Its definition is simply not a applicable. This notion is not possible if functions are indistinguishable from their graph sets as you would have it. That's why we have, for any function f mapping set A to set B, its definition is of the form:
>
> ALL(x):ALL(y):[x in A & y in B => [f(x)=y <=> P(x, y)]]
>

From which we can infer:

1. ALL(x):[x in A => f(x) in B]

2 ALL(x):[x in A => P(x, f(x))]

[Mostowski Collapse]

Still waiting for proofs in DC Proof of the
following theorems about equipolence:

/* Transitivity */
u ~ v & v ~ t => u ~ t
/* Symmetry */
u ~ v => v ~ u
/* Reflexivity */
v ~ v
/* Classes */
∀t(u ~ t <=> v ~ t) <=> u ~ v

from graphs alone. Whats the prololoblem?
No Pairing Axiom in DC Proof?

[Mostowski Collapse]

The following axioms from set theory are missing
in DC Proof. Namely Pairing Axiom and Replacement Axiom:

Axiom of pairing
https://en.wikipedia.org/wiki/Axiom_of_pairing

Axiom schema of replacement
https://en.wikipedia.org/wiki/Axiom_schema_of_replacement

The Pairing Axioms gives unrestrictedly an unordered
pair, namely we have the following so to speak:

ALL(a):ALL(b):EXIST(c):[c = {a,b}]

But the above cannot be used in DC Proof, since
the tweaking of forall in DC Proof, makes it hard
to apply the above. You cannot show from the above

for example that f(x) e {f(x)}. But in ordinary set
theory, the unordered pairing can be used to bootstrap
ordered pairing, again unrestricted:

ALL(a):ALL(b):EXIST(c):[c = <a,b>]

Again missing in DC Proof. The Replacement Axiom
is especially handy to prove the theorems about
equipolence. But one can also prove it without the

Replacement Axiom if the boostrapping of ordered
pairs is specified, and if the boostrapping allows
computing an upper bound for ordered pairs,

then the usual subset axiom aka Axiom schema of
specification can be used.

[Mostowski Collapse]

Works on my side:

1 f(x) ε singleton(f(x))
Axiom

2 f(x)=f(x)
Reflex, 1

3 EXIST(y):f(x)=y
E Gen, 2

Since we want to use the Pairing Axiom to create
the function spaces especially the set-like functions
aka graphs, we will have provable:

ALL(a):[a e {a}]

So that the axiom can be discharged.
Or do you say the Pairing Axiom is not available
in DC Proof. I asked this already,

its wasn't many days ago.

[Dan Christensen]

On Monday, July 4, 2022 at 8:13:43 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Sonntag, 3. Juli 2022 um 14:08:00 UTC+2:
> > On Saturday, July 2, 2022 at 4:32:36 PM UTC-4, Mostowski Collapse wrote:
> > > FOL function symbols lead to Russell Paradox, if you
> > > would try to put them into the usual domain of discourse,
> > > since they lead to Universal Set. You can prove:
> > >
> > > ∀x∃y f(x) = y
> > >
> > [snip]
> >
> > Not in DC Proof. Try it.

> Works on my side:
>
> 1 f(x) ε singleton(f(x))
> Axiom
> > 2 f(x)=f(x)
> Reflex, 1
>
> 3 EXIST(y):f(x)=y
> E Gen, 2
>

Doesn't work. What you have actually proven here:

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

NOT what was required.

> Since we want to use the Pairing Axiom to create
> the function spaces especially the set-like functions
> aka graphs, we will have provable:
>
> ALL(a):[a e {a}]
>

Not an axiom or theorem in DC Proof.

> So that the axiom can be discharged.

> Or do you say the Pairing Axiom is not available
> in DC Proof.

Correct, but you could introduce the axiom:

ALL(a):ALL(b):EXIST(c):[Set(c) & ALL(d):[d in c <=> d=a | d=b]]

There is no guarantee, however, that it won't lead to some wonky results or worse. And you would also have to introduce axioms about the existence of some set(s) or other object(s) since DC Proof does NOT assume a non-empty domain of discourse.

If you want to derive ∀x∃y f(x) = y in any formal system (like DC Proof), you cannot be adding new axioms to force it. Try again, without introducing any new axioms.

[Mostowski Collapse]

You only tell us that this here is not provable:

ALL(f):ALL(x):EXIST(y):f(x) = y

But what about this here:

EXIST(f):ALL(x):EXIST(y):f(x) = y

Obviously the Peano Unit, written as {x}, which is also
used in Zermelos Axiom from 1908, satisfies the above
second variant. If you take g(x) = {x}, you have in g

an existence witness for EXIST(f), since we have:

ALL(x):EXIST(y):{x}=y

Or do you deny the later. Its a consequence of the Pairing Axiom.

[Mostowski Collapse]

In ordinary FOL + ZFC you would make this definition:

ALL(x):ALL(z):[z e {x} <=> z = x]

I guess you can then show that {_} is a function using
ZFC, like extensionality and paring axiom, and that the
above is a conservative extension. Namely you should be

able to verify this criteria of conservative extension:

∀x1 … ∀xn ∃!yϕ(y, x1 , … , xn)
https://en.wikipedia.org/wiki/Extension_by_definitions#Definition_of_function_symbols

Can you prove in DC Poop the following:

∀x ∃!y ∀z (z e y <=> z = x)

Please show us....

[Mostowski Collapse]

What are you bragging about?
Are you now bat shit crazy again?

Dan Christensen schrieb am Montag, 4. Juli 2022 um 17:00:17 UTC+2:
> If you want to derive ∀x∃y f(x) = y in any formal system
> (like DC Proof), you cannot be adding new axioms to force it.

Its provable in FOL without adding new axioms:

∀x∃yf(x)=y is valid.
https://www.umsu.de/trees/#~6x~7yf%28x%29=y

Whats wrong with you?
Its a bug that its not anymore provable in DC Poop.

[Dan Christensen]

> You only tell us that this here is not provable:
>
> ALL(f):ALL(x):EXIST(y):f(x) = y
>

You have not proven this using DC Proof.

> But what about this here:
>
> EXIST(f):ALL(x):EXIST(y):f(x) = y
>

Also not proven using DC Proof.

> Obviously the Peano Unit, written as {x}, which is also
> used in Zermelos Axiom from 1908, satisfies the above
> second variant. If you take g(x) = {x}, you have in g
>
> an existence witness for EXIST(f), since we have:
>
> ALL(x):EXIST(y):{x}=y
>

Maybe you can prove this using ZFC, but you were making claims about DC Proof here. So much for those claims, eh, Jan Burse?

[Dan Christensen]

On Monday, July 4, 2022 at 6:04:04 PM UTC-4, Mostowski Collapse wrote:
> In ordinary FOL + ZFC you would make this definition:
>
> ALL(x):ALL(z):[z e {x} <=> z = x]
>
> I guess you can then show that {_} is a function using
> ZFC, like extensionality and paring axiom, and that the
> above is a conservative extension. Namely you should be
>
> able to verify this criteria of conservative extension:
>
> ∀x1 … ∀xn ∃!yϕ(y, x1 , … , xn)
> https://en.wikipedia.org/wiki/Extension_by_definitions#Definition_of_function_symbols
>

> Can you prove in DC Proof the following:

>
> ∀x ∃!y ∀z (z e y <=> z = x)
>

Not without introducing additional axioms.

It is possible, however, to prove:

ALL(a):[Set(a) => ALL(b):[b in a => EXIST(c):[Set(c) & b in c & ALL(d):[d in c <=> d=b]]]]

using the built-in Subset Axiom. Not what you wanted? Oh, well....

[Dan Christensen]

On Monday, July 4, 2022 at 6:06:29 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:

> Its provable in FOL without adding new axioms:
>
> ∀x∃yf(x)=y is valid.

I prefer to use the rules of logic implicitly used in most proofs in math textbooks. They are much more intuitive and simpler to use than the standard FOL that you use IMHO. Deal with it, Jan Burse.

[Mostowski Collapse]

What is simpler if you cannot prove.
You are talking nonsense as usual.

> Its provable in FOL without adding new axioms:
> ∀x∃yf(x)=y is valid.

> https://www.umsu.de/trees/#~6x~7yf%28x%29=y

Your inference rules are more complex,
they have more side conditions than FOL.
And why? Only because you cannot deal

with function spaces correctly. You even
don't have a pairing axiom. Its neither fish
nor chicken, neither set theory nor type theory,

maybe it has some type theory, because
f(x) requires often f(x) e a. A typical type theory
term forming rule is, related to modus ponens:

f : A -> B, x : A
-----------------------------
f(x) : B

But your nonsense has no term forming
rules. You can enter f(x), and it barks when
it doesn't find f(x) e a somewhere. Its utter

nonsense, your inference rules are a cludge
that comes much much too later. Normale type
systems check f(x) when f(x) is built. Your

DC Proof is utter nonsense.

[Mostowski Collapse]

Dan Christensen haluzinated:

> It is possible, however, to prove:

> ALL(a):[Set(a) => ALL(b):[b in a => EXIST(c):[Set(c) & b in c & ALL(d):[d in c <=> d=b]]]]

No, thats restricted singleton. Basically you
have something like b e a => ∃c c= {b}. But the
singleton exists unrestrictedly.

Can you get rid of b e a? The singleton derived
from pairing axiom in ZFC, by extending ZFC
conservatively by the function symbol {_}:

ALL(x):EXIST(y):{x}=y

Unrestricted, for every x there is such an y.
For every x, we do not need to know whether x is
some member of some set.

You can check yourself:

∀ A ∀ B ∃ C ∀ D [ D ∈ C ⟺ ( D = A ∨ D = B ) ]
https://en.wikipedia.org/wiki/Axiom_of_pairing

There is no nonsense of your quantifier restriction
that we see all over the place. Got it?

[Dan Christensen]

On Tuesday, July 5, 2022 at 9:03:10 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Dienstag, 5. Juli 2022 um 06:29:37 UTC+2:
> > On Monday, July 4, 2022 at 6:06:29 PM UTC-4, Mostowski Collapse (Jan Burse) wrote:
> >
> > > Its provable in FOL without adding new axioms:
> > >
> > > ∀x∃yf(x)=y is valid.
> > I prefer to use the rules of logic implicitly used in most proofs in math textbooks. They are much more intuitive and simpler to use than the standard FOL that you use IMHO. Deal with it, Jan Burse.

> What is simpler if you cannot prove.
> You are talking nonsense as usual.
> > Its provable in FOL without adding new axioms:
> > ∀x∃yf(x)=y is valid.

No domain or codomain is specified. It's quirkiness like this that may explain why so-called mathematical logic is not required course in university pure math program, e.g. at MIT. In the real world, definitions of functions are of the form:

ALL(a):ALL(b):[a in dom & b in cod => [f(a)=b <=> P(a,b)]]

Example from Terence Tao's "Analysis I," p. 49:

"Definition 3.3.1 (Functions). Let X, Y be sets, and let P(x, y) be a property pertaining to an object x ∈ X and an object y ∈ Y , such that for every x ∈ X, there is exactly one y ∈ Y for which P(x, y) is true (this is sometimes known as the vertical line test). Then we define the function f : X → Y defined by P on the domain X and range Y to be the object which, given any input x ∈ X, assigns an output f(x) ∈ Y, defined to be the unique object f(x) for which P(x, f(x)) is true. Thus, for any x ∈ X and y ∈ Y, y = f(x) ⇐⇒ P(x, y) is true."

> Your inference rules are more complex,
> they have more side conditions than FOL.

Nonsense.

> And why? Only because you cannot deal
> with function spaces correctly.

Wrong again, Jan Burse. For functions of 1 variable, the Function Space Axiom (on the Sets menu) gives us:

1. ALL(dom):ALL(cod):[Set(dom) & Set(cod)
=> EXIST(fsp):[Set(fsp) & ALL(f):[f in fsp <=> ALL(a1):[a1 in dom => f(a1) in cod]]]]
Fn Space

> You even
> don't have a pairing axiom.

In mathematical proofs, the Subset Axiom (on the Sets menu) will usually suffice. On the rare occasion that users may required a true pairing axiom, they can introduce the following axiom at the beginning of their proof using the Create Axiom rule (on the Logic menu):

1. ALL(a):ALL(b):EXIST(c):[Set(c) & ALL(d):[d in c <=> d=a | d=b]]
Axiom

[snip]

> But your nonsense has no term forming
> rules. You can enter f(x), and it barks when
> it doesn't find f(x) e a somewhere.

What is f(x) if it is not an element of set? It is probably a mistake.

[WM]

Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:
> As requested...
>
> THEOREM: Bijectivity is Transitive
>
> ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
>
> => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
> & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
>
> => EXIST(h):Bijection(h,t,v)]]
>
> PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)

Bijectivity does not exist between infinite sets. For instance consider Cantors's
"proof " of the bijection between ℕ and ℚ using the function

k = (m + n - 1)(m + n - 2)/2 + m

with the resulting sequence of fractions
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...

If we check it by bijecting first the natnumbers and the integers fractions in the matrix

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
...

putting the indices X in the first column
XOOO...
XOOO...
XOOO...
XOOO...
...

(the not yet indexed fractions are expressed by O's), then indexing the fractions according to Cantor's function

1/1, 1/2,

XXOO...
OOOO...
XOOO...
XOOO...
...

1/1, 1/2, 2/1,

XXOO...
XOOO...
OOOO...
XOOO...
...

1/1, 1/2, 2/1, 1/3,

XXXO...
XOOO...
OOOO...
OOOO...
...

should cover the whole infinite matrix with X's. But obviously that cannot happen.

Regards, WM

[Ross A. Finlayson]

On Friday, July 1, 2022 at 7:49:37 AM UTC-7, Mostowski Collapse wrote:
> Cantor showed that:
>
> N ~ N x N
>
> Thats a classical result. Just try:
>
> f : N x N -> N
> https://mathworld.wolfram.com/PairingFunction.html
>
> Because of this property that Dan Christensen did not yet prove:

>
> /* Symmetry */
> u ~ v => v ~ u
>

> The existence of f, also implies the existence of some g : N -> N x N.

> WM schrieb am Freitag, 1. Juli 2022 um 12:08:11 UTC+2:
> > Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:

> > > THEOREM: Bijectivity is Transitive

> > > PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
> > >

> > There is no bijection between infinite sets because infinite sets contain dark numbers.
> >
> > Poof for ℕ (5 lines):
> > If ℕ has no last element, then the intersection ∩{E(k) : k ∈ ℕ} cannot be empty. Up to every endsegment E(k) it is not empty, because the successor endsegment E(k+1) will further reduce it. The empty set however cannot be reduced. Since every natnumber of the endless sequence has a successor the intersection will never be empty. Therefore ∩{E(k) : k ∈ ℕ} = { } requires an empty endsegment and therefore dark numbers.
> >
> > Regards, WM

In set theory "infinite bijections" are transitive,
while in number theory and the concrete,
in set theory, they're organized.

[Dan Christensen]

On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:
> > As requested...
> >
> > THEOREM: Bijectivity is Transitive
> >
> > ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
> >
> > => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
> > & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
> >
> > => EXIST(h):Bijection(h,t,v)]]
> >
> > PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
>
> Bijectivity does not exist between infinite sets.

Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?

[WM]

Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:
> On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
> > Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:
> > > As requested...
> > >
> > > THEOREM: Bijectivity is Transitive
> > >
> > > ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
> > >
> > > => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
> > > & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
> > >
> > > => EXIST(h):Bijection(h,t,v)]]
> > >
> > > PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
> >
> > Bijectivity does not exist between infinite sets.
> Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?

I proved that Cantor's famous bijection fails. Otherwise it would be possible to change the positions of the X's in the matrix

XOOO...
XOOO...
XOOO...
XOOO...
...

such that all positions are occupied by X's. That is obviously impossible because for every step in Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... the number of places not occupied by X's remains absolutely constant. And I do not accept that all occupying happens "in the limit" after all finite steps.

Cantor's failure casts doubt on other attempts.

Regards, WM

[WM]

Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:

> On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:

> > Bijectivity does not exist between infinite sets.
> Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?

No. If there was a bijection possible, then in the matrix

XXXXXXXXXXXXXXXXXXXXXXX...
O O O O O O O O O O O O O...

we could change the positions of the X's such that they covered the whole matrix. Up to every finite step the number of not covered places remains constant.

Regards, WM

[Dan Christensen]

You had better check your work, Mucke. Something is seriously wrong with it if you cannot prove that f(x)=2x is a bijection mapping the natural numbers to the even numbers.

BTW, did you ever manage to prove that 1=/=2 in your system of natural numbers? If not, maybe that is at the root of your problem--your system just doesn't work.

[Fritz Feldhase]

On Saturday, July 2, 2022 at 4:17:44 PM UTC+2, WM wrote:

Slightly corrected your nonsense:

> For [any finite non-empty set of natural] numbers [M] we have[:]
> | ∩{E(k) : k ∈ M} | = ℵ₀.

> [On the other hand] we have:
> | ∩{E(k) : k ∈ ℕ} | = 0.

So what? Trivial set-theoretic facts.

Hint: If E is a non-empty _finite_ set of endsegments, then ∩E = min(E), hence |∩E| = ℵ₀. If, on the other hand, E is an _infinite_ set of endsegments, then ∩E = { }, hence |∩E| = 0.

[WM]

Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 15:29:31 UTC+2:
> On Sunday, July 10, 2022 at 9:04:28 AM UTC-4, WM wrote:
> > Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:
> > > On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
> >
> > > > Bijectivity does not exist between infinite sets.
> > > Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?
> > No. If there was a bijection possible, then in the matrix
> >
> > XXXXXXXXXXXXXXXXXXXXXXX...
> > O O O O O O O O O O O O O...
> >
> > we could change the positions of the X's such that they covered the whole matrix. Up to every finite step the number of not covered places remains constant.
> >

> Something is seriously wrong with it if you cannot prove that f(x)=2x is a bijection mapping the natural numbers to the even numbers.

It is correct in potential infinity. It is wrong in actual infinity, independent of whether you can understand the above argument.

>
> BTW, did you ever manage to prove that 1=/=2 in your system of natural numbers?

There this is included in the definition.

Regards, WM

[WM]

Fritz Feldhase schrieb am Montag, 11. Juli 2022 um 08:39:03 UTC+2:
> On Saturday, July 2, 2022 at 4:17:44 PM UTC+2, WM wrote:

> > [On the other hand] we have:
> > | ∩{E(k) : k ∈ ℕ} | = 0.

(1) Every endsegment has all its elements in common with all its predecessors.
(2) All infinite endsegments have infinitely many elements in common with all their predecessors.
(3) A sequence of infinite endsegments consists only of infinite predecessors because every endsegment has an infinite successor.
(4) Therefore all infinite endsegments have an infinite set of elements in common.

Regards, WM

[Fritz Feldhase]

On Monday, July 11, 2022 at 3:06:34 PM UTC+2, WM wrote:

> (1) Every endsegment has all its elements in common with all its predecessors. [Hence:]
> (2) All endsegments have infinitely many elements in common with all their predecessors.
> (3) A sequence of endsegments consists only of predecessors [...]

Ok, so far.

> (4) Therefore

Hint: Your "therefore" almost always precedes a wrong claim.

> all infinite endsegments have an infinite set of elements in common.

See?!

1 + 2 = 3, therefore 1 = 2?

Look, dumbo, in math you have to PROVE your claims.

Never attended any math lesen in university?

[Dan Christensen]

On Monday, July 11, 2022 at 9:03:31 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 15:29:31 UTC+2:
> > On Sunday, July 10, 2022 at 9:04:28 AM UTC-4, WM wrote:
> > > Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:
> > > > On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
> > >
> > > > > Bijectivity does not exist between infinite sets.
> > > > Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?
> > > No. If there was a bijection possible, then in the matrix
> > >
> > > XXXXXXXXXXXXXXXXXXXXXXX...
> > > O O O O O O O O O O O O O...
> > >
> > > we could change the positions of the X's such that they covered the whole matrix. Up to every finite step the number of not covered places remains constant.
> > >
> > Something is seriously wrong with it if you cannot prove that f(x)=2x is a bijection mapping the natural numbers to the even numbers.

> It is correct in potential infinity. It is wrong in actual infinity

Not sure sure what YOU mean by infinite, but a set X is said to be infinite iff there exists an injective function f: X-->X that is not surjective (see Dedekind). By that definition, both N and 2N are infinite.

Maybe you should try to invent your own definition to make all your bizarre claims somehow "true." Just don't expect that anyone else will use it. I'm guessing it would be about as useless as your "axioms" for the natural numbers.

> >
> > BTW, did you ever manage to prove that 1=/=2 in your system of natural numbers?

> There this is included in the definition.
>

Not in the definition you originally posted here some time ago. Have you subsequently, added 1=/=2, 1=/=3, 1=/=4, ... 2=/=3, 2=/=4. ....? (HA, HA, HA!!!)

[Fritz Feldhase]

On Monday, July 11, 2022 at 5:08:23 PM UTC+2, Dan Christensen wrote:

> Maybe you should try to invent your own definition to make all your bizarre claims somehow "true." Just don't expect that anyone else will use it.

Reminds me to your "function equality axiom".

[WM]

Fritz Feldhase schrieb am Montag, 11. Juli 2022 um 16:42:38 UTC+2:
> On Monday, July 11, 2022 at 3:06:34 PM UTC+2, WM wrote:
>
> > (1) Every endsegment has all its elements in common with all its predecessors. [Hence:]
> > (2) All endsegments have infinitely many elements in common with all their predecessors.
> > (3) A sequence of endsegments consists only of predecessors [...]
>
> Ok, so far.
>

> > (4) Therefore all infinite endsegments have an infinite set of elements in common.

Not yet grasped?

All endsegments have infinitely many elements in common with all their predecessors. This also holds for all infinite successors. Therefore it holds for all endsegments.

If you can't understand, try to find a counterexample.

Regards, WM

[WM]

Dan Christensen schrieb am Montag, 11. Juli 2022 um 17:08:23 UTC+2:
> On Monday, July 11, 2022 at 9:03:31 AM UTC-4, WM wrote:
> > Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 15:29:31 UTC+2:
> > > On Sunday, July 10, 2022 at 9:04:28 AM UTC-4, WM wrote:
> > > > Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:
> > > > > On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
> > > >
> > > > > > Bijectivity does not exist between infinite sets.
> > > > > Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?
> > > > No. If there was a bijection possible, then in the matrix
> > > >
> > > > XXXXXXXXXXXXXXXXXXXXXXX...
> > > > O O O O O O O O O O O O O...
> > > >
> > > > we could change the positions of the X's such that they covered the whole matrix. Up to every finite step the number of not covered places remains constant.
> > > >
> > > Something is seriously wrong with it if you cannot prove that f(x)=2x is a bijection mapping the natural numbers to the even numbers.
>
> > It is correct in potential infinity. It is wrong in actual infinity
> Not sure sure what YOU mean by infinite, but a set X is said to be infinite iff there exists an injective function f: X-->X that is not surjective (see Dedekind). By that definition, both N and 2N are infinite.

Yes, that definition is wrong ( cannot be satisfied by infinite sets). See the above matrix. Only a fool can try to cover the whole matrix by simple shuffling the X's.

> Not in the definition you originally posted here some time ago. Have you subsequently, added 1=/=2,

Only a fool would think this is necessary.

Regards, WM

[Dan Christensen]

Perhaps you could write to Terence Tao and explain to him why his definition his wrong.

[Fritz Feldhase]

On Monday, July 11, 2022 at 8:37:33 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 11. Juli 2022 um 16:42:38 UTC+2:
> > On Monday, July 11, 2022 at 3:06:34 PM UTC+2, WM wrote:
> >
> > > (1) Every endsegment has all its elements in common with all its predecessors. [Hence:]
> > > (2) All endsegments have infinitely many elements in common with all their predecessors.
> > > (3) A sequence of endsegments consists only of predecessors [...]
> >
> > Ok, so far.
> >
> > (4) Therefore

Hint: Your "therefore" almost always precedes a wrong claim.

> > all infinite endsegments have an infinite set of elements in common.

See?!

1 + 2 = 3, therefore 1 = 2?

Look, dumbo, __in math you have to PROVE your claims__

Never attended any math lesen in university? Or just genuinely dumb like shit?

[Fritz Feldhase]

On Monday, July 11, 2022 at 9:52:34 PM UTC+2, Dan Christensen wrote:
> On Monday, July 11, 2022 at 1:01:12 PM UTC-4, Fritz Feldhase wrote:
> > On Monday, July 11, 2022 at 5:08:23 PM UTC+2, Dan Christensen wrote:
> > >
> > > Maybe you should try to invent your own definition to make all your bizarre claims somehow "true." Just don't expect that anyone else will use it.
> > >
> > Reminds me to your "function equality axiom".
> >
> Perhaps you could

You might ask Prof. Dr. Mückenheim for help. I'm sure he will support your approach.

[Dan Christensen]

On Monday, July 11, 2022 at 2:40:53 PM UTC-4, WM wrote:
> Dan Christensen schrieb am Montag, 11. Juli 2022 um 17:08:23 UTC+2:
> > On Monday, July 11, 2022 at 9:03:31 AM UTC-4, WM wrote:
> > > Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 15:29:31 UTC+2:
> > > > On Sunday, July 10, 2022 at 9:04:28 AM UTC-4, WM wrote:
> > > > > Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:
> > > > > > On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
> > > > >
> > > > > > > Bijectivity does not exist between infinite sets.
> > > > > > Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?
> > > > > No. If there was a bijection possible, then in the matrix
> > > > >
> > > > > XXXXXXXXXXXXXXXXXXXXXXX...
> > > > > O O O O O O O O O O O O O...
> > > > >
> > > > > we could change the positions of the X's such that they covered the whole matrix. Up to every finite step the number of not covered places remains constant.
> > > > >
> > > > Something is seriously wrong with it if you cannot prove that f(x)=2x is a bijection mapping the natural numbers to the even numbers.
> >
> > > It is correct in potential infinity. It is wrong in actual infinity
> > Not sure sure what YOU mean by infinite, but a set X is said to be infinite iff there exists an injective function f: X-->X that is not surjective (see Dedekind). By that definition, both N and 2N are infinite.
> Yes, that definition is wrong ( cannot be satisfied by infinite sets).

[snip]

Wrong again, Mucke. The successor function S: N-->N, is both injective and not surjective. Therefore, by the above definition, N is infinite.

If you are having trouble distinguishing finite and infinite sets, see my blog posting at

https://dcproof.wordpress.com/2014/09/17/infinity-the-story-so-far/

[sergi o]

On 7/10/2022 7:19 AM, WM wrote:
> Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:
>> On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
>>> Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:
>>>> As requested...
>>>>
>>>> THEOREM: Bijectivity is Transitive
>>>>
>>>> ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
>>>>
>>>> => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
>>>> & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
>>>>
>>>> => EXIST(h):Bijection(h,t,v)]]
>>>>
>>>> PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
>>>
>>> Bijectivity does not exist between infinite sets.
>> Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?
>
> I proved that Cantor's famous bijection fails.

Wrong, you held a swaparoofestival which is your spoof, not a proof.

<snip crap>

>
> Cantor's failure casts doubt on other attempts.
>
> Regards, WM
>

So WM sidestepped the question, so he can not agree.

[sergi o]

so, if it doesn't work at all, just define it that way, and all better then.

[Jim Burns]

On 7/11/2022 2:37 PM, WM wrote:
> Fritz Feldhase schrieb
> am Montag, 11. Juli 2022 um 16:42:38 UTC+2:
>> On Monday, July 11, 2022 at 3:06:34 PM UTC+2,
>> WM wrote:

>>> (1) Every endsegment has all its elements
>>> in common with all its predecessors. [Hence:]
>>> (2) All endsegments have infinitely many elements
>>> in common with all their predecessors.
>>> (3) A sequence of endsegments consists only of
>>> predecessors [...]
>>
>> Ok, so far.
>>
>>> (4) Therefore all infinite endsegments have
>>> an infinite set of elements in common.
>
> Not yet grasped?
>
> All endsegments have infinitely many elements in common
> with all their predecessors.
> This also holds for all infinite successors.

No,
the set of elements in common with all end-segment
predecessors is not
the set of elements in common with all end-segment
successors.


If elements exist which are _in_
the intersection of all end segments,
then end segments exist which are _not in_
the collection of all end segments.

ℕ = {definable} = ⋃ₙ{definer of n}

| Assume m ∈ ⋂ₙ{E(n) ⊆ ℕ}
|
| m ∈ ℕ
| ⟨0...i,i+1...m⟩ exists
|
| Define a split of ⟨0...i,i+1...m⟩ such that
| AFTER contains all j in ⋂ₙ{E(n) ⊆ ℕ}
| BEFORE contains all j not-in ⋂ₙ{E(n) ⊆ ℕ}
|
| Because ⟨0...i,i+1...m⟩ is a definer,
| some step i,i+1 exists such that
| i is last in BEFORE and
| i+1 is first in AFTER
|
| i+1 is in ⋂ₙ{E(n) ⊆ ℕ}
| i+1 is each end segment {E(n) ⊆ ℕ}
| Thus,
| i+1 is not in E(i+2)
| Thus,
| E(i+2) is not in {E(n) ⊆ ℕ}

Therefore,
if ⋂ₙ{E(n) ⊆ ℕ} is not empty,
then ⋂ₙ{E(n) ⊆ ℕ} does not have all end segments.

> Therefore it holds for all endsegments.
>
> If you can't understand, try to find a counterexample.

If elements exist which are _in_
the intersection of all end segments,
then end segments exist which are _not in_
the collection of all end segments.

[WM]

Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 00:14:13 UTC+2:

> The successor function S: N-->N, is both injective and not surjective. Therefore, by the above definition, N is infinite.

The matrix

XOOO...
XOOO...
XOOO...
XOOO...
...

is also infinite.
Nevertheless everybody believing that the X's by changing places with O'scan cover the whole matrix is a fOOl.

Rrgards, WM

[WM]

Jim Burns schrieb am Dienstag, 12. Juli 2022 um 07:22:35 UTC+2:
> On 7/11/2022 2:37 PM, WM wrote:

> > All endsegments have infinitely many elements in common
> > with all their predecessors.
> > This also holds for all infinite successors.
> No,
> the set of elements in common with all end-segment
> predecessors is not
> the set of elements in common with all end-segment
> successors.

But both sets are infinite sets, and the set in common with all infinite successors is a subset of the set in common with al predecessors. That is very simple to see, because all sets are infinite and no element is added that was missing in E(1) = ℕ.

>
> If elements exist which are _in_
> the intersection of all end segments,
> then end segments exist which are _not in_
> the collection of all end segments.

No element exists which is in the intersection of all endsegments, but infinitely many elements exist which are in the intersection of all infinite endsegments.

Your proof that for every n there exists an endsegment not containing holds only for all endsegments. It fails for infinitely many elements contained in _infinite_ endsegments. It would really require a sleight of hand if all natnumbers could be deleted by endsegments which leave infinitely many natnumbers undeleted.

> >
> > If you can't understand, try to find a counterexample.
> If elements exist which are _in_
> the intersection of all end segments,

No, such elements do not exist.

> then end segments exist which are _not in_
> the collection of all end segments.

There exist endsegments which are not in the collection of all _infinite_ endsegments. They reduce their infinite intersection to the empty set.

Regards, WM

[sergi o]

You are out of Math again, Your Swaparoofestivaladoodle is not in Math.

It is diversion.

[Dan Christensen]

On Tuesday, July 12, 2022 at 8:23:59 AM UTC-4, WM wrote:
> Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 00:14:13 UTC+2:
>
> > The successor function S: N-->N, is both injective and not surjective. Therefore, by the above definition, N is infinite.
> The matrix
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
> is also infinite.

No-one cares about your goofy matrices, Mucke. We are talking about finite and infinite sets and how to distinguish them. After all these years, you still don't seem to have a clue.

Dan

[WM]

Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 14:51:13 UTC+2:
> On Tuesday, July 12, 2022 at 8:23:59 AM UTC-4, WM wrote:
> > Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 00:14:13 UTC+2:
> >
> > > The successor function S: N-->N, is both injective and not surjective. Therefore, by the above definition, N is infinite.
> > The matrix
> > XOOO...
> > XOOO...
> > XOOO...
> > XOOO...
> > ...
> > is also infinite.
> No-one cares about your goofy matrices,

Then have a good day.

Regards, WM

[Ross A. Finlayson]

How about, "a third of the integers are multiples of three".

[Dan Christensen]

On Friday, July 1, 2022 at 12:48:59 AM UTC-4, Dan Christensen wrote:
> As requested...
>
> THEOREM: Bijectivity is Transitive
>
> ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
>
> => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
> & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
>
> => EXIST(h):Bijection(h,t,v)]]
>

> PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
>

Using a similar approach, it is easy to prove the reflexivity and symmetry of bijections.

[Jim Burns]

On 7/12/2022 8:32 AM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 12. Juli 2022 um 07:22:35 UTC+2:
>> On 7/11/2022 2:37 PM, WM wrote:

>>> All endsegments have infinitely many elements in common
>>> with all their predecessors.
>>> This also holds for all infinite successors.
>>
>> No,
>> the set of elements in common with all end-segment
>> predecessors is not
>> the set of elements in common with all end-segment
>> successors.
>
> But both sets are infinite sets,

Also,
if elements exist which are in the intersection of
all _infinite_ end segments,
then _infinite_ end segments exist which are not in

the collection of all _infinite_ end segments.

Thus, no elements exist in the second set.

> and the set in common with all infinite successors
> is a subset of the set in common with al predecessors.
> That is very simple to see, because
> all sets are infinite and
> no element is added that was missing in E(1) = ℕ.

Yes,
no element ∉ ℕ is added to get {}
{} isn't infinite.

>> If elements exist which are _in_
>> the intersection of all end segments,
>> then end segments exist which are _not in_
>> the collection of all end segments.
>
> No element exists which is in the intersection of
> all endsegments,
> but infinitely many elements exist which are in
> the intersection of all infinite endsegments.

No.
If elements exist which are in the intersection of
all _infinite_ end segments,
then _infinite_ end segments exist which are not in

the collection of all _infinite_ end segments.

Thus, no elements exist in the intersection of
all _infinite_ end segments.

> Your proof that for every n

> there exists an endsegment not containing [n]

> holds only for all endsegments.
> It fails for infinitely many elements
> contained in _infinite_ endsegments.

I address your concern:

If elements exist which are in the intersection of
all _infinite_ end segments,
then _infinite_ end segments exist which are not in
the collection of all _infinite_ end segments.

ℕ = {definable} = ⋃ₙ{definer of n}

IE(n) = { j ∈ ℕ : j >= n }

n ∈ IE(n)
IE(n) ≠ {}

if
j ∈ IE(n)
then
j ∈ ℕ
j ∈ ⟨0...i,i+1...j⟩
j+1 ∈ ⟨0...i,i+1...j+1⟩
j+1 ∈ ℕ
j < j+1
j+1 ∈ IE(n)

IE(n) is an infinite end segment of ℕ

{IE(n) ⊆ ℕ} is the collection of all
infinite end segments of ℕ

⋂ₙ{IE(n) ⊆ ℕ} is the intersection of all
infinite end segments of ℕ

| Assume m ∈ ⋂ₙ{IE(n) ⊆ ℕ}

|
| m ∈ ℕ
| ⟨0...i,i+1...m⟩ exists
|
| Define a split of ⟨0...i,i+1...m⟩ such that

| AFTER contains all j in ⋂ₙ{IE(n) ⊆ ℕ}
| BEFORE contains all j not-in ⋂ₙ{IE(n) ⊆ ℕ}

|
| Because ⟨0...i,i+1...m⟩ is a definer,
| some step i,i+1 exists such that
| i is last in BEFORE and
| i+1 is first in AFTER
|

| i+1 is in ⋂ₙ{IE(n) ⊆ ℕ}
| i+1 is in each infinite end segment in {IE(n) ⊆ ℕ}
| However,
| i+1 is not in IE(i+2)
| Thus,
| IE(i+2) is not in {IE(n) ⊆ ℕ}

Therefore,
if ⋂ₙ{IE(n) ⊆ ℕ} is not empty,
then {IE(n) ⊆ ℕ} does not hold all
infinite end segments.

However,
{IE(n) ⊆ ℕ} holds all infinite end segments.
Contradiction.
Thus, ⋂ₙ{IE(n) ⊆ ℕ} is empty.

> It would really require a sleight of hand
> if all natnumbers could be deleted by endsegments
> which leave infinitely many natnumbers undeleted.

I have an idea of what we do which you might call
"sleight of hand". However, we are not concealing,
we are revealing infinitely-many.

We are finite. And yet we can know something is
true of each one of infinitely-many. How can this be?

Part of what we know comes from knowing what we are
talking about. We _know_ a right triangle is
a triangle with a right angle. We don't need to
know which one it is of infinitely-many in order to
know that.

Part of what we know (the newer, revealing part)
comes from NOT advancing to newer claims UNLESS
that advance is perfectly secure.

| What I'm not saying here:
| I'm not saying the older claims must be perfectly
| secure, only the advance itself must be.
|
| The envelop in my hand contains a geometric
| figure. The figure may or may not be a right
| triangle. If it's not, we would not be justified
| if we applied the Pythagorean theorem. If it is,
| we know that the theorem is correct in this case.
|
| I don't know if the envelop contains a right
| triangle. However, I know a fact about infinitely-
| -many right triangles. That's not everything,
| but it's a long way from nothing.

[Jim Burns]

On 7/12/2022 11:03 AM, Ross A. Finlayson wrote:

> How about,
> "a third of the integers are multiples of three".

When it's put that way, there is a risk of mistaking
how many integers or how many multiples of three for
reallyreallyreallyreallyreallyreally large numbers.

It is also true that a third of the integers are
as many as (are able to match) all the integers.
And this is not a paradox, it is being not
a reallyreallyreallyreallyreallyreally large number.

[Ross A. Finlayson]

These days at least it's called 'density" again,
where for example historically it's also "frequency".

One notion is just building an arithmetic of objects
that instead of integers or indivisible, have thirds,
then to add to all the theorems that implement
models of arithmetic, this one that divides by thirds
first then the rest of ratio.

Still this would be for adding all the theorems of
models of arithmetic together, then for a proper
reasoning that all the paths of inference about
them are valid.

[Fritz Feldhase]

On Tuesday, July 12, 2022 at 3:24:21 PM UTC+2, WM wrote:
> Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 14:51:13 UTC+2:
> > On Tuesday, July 12, 2022 at 8:23:59 AM UTC-4, WM wrote:
> > > Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 00:14:13 UTC+2:
> > > >

> > > > The successor function S: N --> N, is both injective and not surjective. Therefore, by the above definition, N is infinite.

Indeed!

> > > The matrix <bla>

> > >
> > No-one cares about your goofy matrices,
> >
> Then have a good day.

It's still true.

Btw. (in the context of set theory) the identity function on IN is just the set of ordered pairs {(n, n) : n e IN}. (The axioms of ZFC ensure its existence.)
This function is a bijection from IN onto IN, and IN is an infinite set. Hence your claim

| "Bijectivity does not exist between infinite sets." [WM]

ist just NONSENSE, wie üblich.

[Mostowski Collapse]

Its always easy to prove some from garbage.
If you have nowhere an example of a pair a,b
with a proof of:

~(a ~ b) /* in works a and b are not in bijection */

Then nobody knows whether your relation is
not trivial. Whether you have not defined a relation
which anyway has the following for arbitrary a,b

a ~ b

A relation that is full as above, also satisfies
all laws of an equivalence relation, since it
has only one equivalence class. a full

relation satisfies:

a ~ a
a~ b => b ~ a
a ~ b & b ~ c => a ~ c

So its mandatory in equivalence class theories to
also give example where elements are not equivalent,
to show that the relation is not full.

For example in Peano its a routine exercise to shoe:

1 ≠ 0

Can you prove it. Now I demanded for a proof of:

~(2^0 ~ 2^1)

But still no response from DC Proof? Why? Because
various people of already pointed out that the nonsense
of DC Proof with an extrinsic parameter leads to

full relation, and therefore to a useless definition.

[Mostowski Collapse]

If Dan Christensen is lucky, his bull shit
leads only to a full relation. but prople
have already pointed out that it leads to:

a ~ b & ~(a ~ b)

by producing multiple derivations with different
doms. Logic doesnt care what Dan Christensen things
the correct dom is, for some extrinsic parameter.

The problem is that the definition of a ~ b in Dan
Christensens world is totally botched. Basically
the ordinary definition is:

a ~ b :<=> EXIST(f):[ …. f … a … b ….]

which should be bivalent, either at least one f exists
or not. But Dan Christensen managed to produce this nonsene.
Just analyze his nonsense, maybe switch sides of an

implication, and you find that he did:

ALL(dom):[ a ~ b <=> EXIST(f):[ … dom …. f … a … b ….]]

Now we can derive a ~ b and ~(a ~ b).

P.S.: The simplest schematic abstraction of his nonsense, his
unforgivable blooper because of some type 1 definitions aberration
is this „definition“ in QBF, ask Pehou:

ALL(Q):[P <=> Q]

Is basically „quodlibet“ without the „ex falso“. You are basically
directly jumping into the open sword, commiting logical
harakiri to your theory. rendering it useless.

[Ross A. Finlayson]

On Saturday, July 9, 2022 at 5:11:59 PM UTC-7, Ross A. Finlayson wrote:
> On Friday, July 1, 2022 at 7:49:37 AM UTC-7, Mostowski Collapse wrote:
> > Cantor showed that:
> >
> > N ~ N x N
> >
> > Thats a classical result. Just try:
> >
> > f : N x N -> N
> > https://mathworld.wolfram.com/PairingFunction.html
> >
> > Because of this property that Dan Christensen did not yet prove:
> >
> > /* Symmetry */
> > u ~ v => v ~ u
> >
> > The existence of f, also implies the existence of some g : N -> N x N.
> > WM schrieb am Freitag, 1. Juli 2022 um 12:08:11 UTC+2:
> > > Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:
> > > > THEOREM: Bijectivity is Transitive

> > > > PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
> > > >

> > > There is no bijection between infinite sets because infinite sets contain dark numbers.
> > >
> > > Poof for ℕ (5 lines):
> > > If ℕ has no last element, then the intersection ∩{E(k) : k ∈ ℕ} cannot be empty. Up to every endsegment E(k) it is not empty, because the successor endsegment E(k+1) will further reduce it. The empty set however cannot be reduced. Since every natnumber of the endless sequence has a successor the intersection will never be empty. Therefore ∩{E(k) : k ∈ ℕ} = { } requires an empty endsegment and therefore dark numbers.
> > >
> > > Regards, WM
>
> In set theory "infinite bijections" are transitive,
> while in number theory and the concrete,
> in set theory, they're organized.

Basically "set theory is not necessarily the proper domain of numbers,
for number theory", that only the relation of numbers in terms of sets,
and usually a chain of ordinals, is any kind of "set" that's not an "inconsistent
multiplicity" or "not casted out of set-class distinction logically".

Here because "any two infinite sets have a bijection", is that
"all the rest of 1-1 relation defined is lattice what builds them".

I.e. it's constructively weak to not prove "and all the examples
that could possibly be".

I.e., then it's a "bijection" between any two sets, that according
to numbers, is however functions _always exists between and
even witin, the function_, that hte relation is satisfied only numerically
and quantitatively.

Also none of the fallacies of the infinite can be used in argument in the finite.

[Mostowski Collapse]

Did just a herpes blister blast? You are also
inclined to write nonsense day in day out:

> Here because "any two infinite sets have a bijection", is that

Thats not true. The set if natural numbers and the set of
functions from natural numbers to natural numbers,

are not in bijection. Could be also a nice exercise for a pair
a, b where we have the following:

~(a ~ b)

Basically we have:

~(N ~ N^N)

Proof hint: Show N < 2^N by cantors theorem A < P(A).
This basically shows that the equivalence relation is also
not full among the infinities,

there is not a single infinity, there are many.

[Mostowski Collapse]

I omit the cardinality bars | | when doing proofs.
To complete the proofs you even dont have to
know whether its 2^N < N^N or 2^N =< N^N.

Since we already have N < 2^N. BTW 2^N =< N^N
is not so difficult to prove. In all said so B^A
denotes the set theoretic exponentiation,

which is defined as the function space:

B^A = { f | f graph, f : A -> B }

It is not the quack from DC proof.

[Ross A. Finlayson]

"countably infinite, ....".

What I saw here of a proof didn't already bring in
full-blown set theory.

I don't know if you know about Skolem and Levy
and Lowenheim when you mention Mostowski where
Levy already figured, for generic extension and collapse,
if for example "between these two infinite sets there are
some models of them that biject, ..., they're sets, ...".
(It's not valid to claim that extensionality though.)

Here though it's simply that I meant "countable",
and all this for number theory relations, is constructible.

Then, really Burse, I don't know why you're the expert in herpes,
I simply don't have any symptoms and never really have, so
please to not be jinxing people with cold sores.

What's all the ordinals right on through?

Anyways number theory is much more structured.

Anyways the "Peano's successors and bijection,
toy numbers different and numbers same", it's
a rather strong fact that integers are regular,
more ways than one.

Anyways sorry about your pimples, Burse, I'd commiserate
but really can't identify.

[Dan Christensen]

On Tuesday, July 12, 2022 at 5:31:25 PM UTC-4, Mostowski Collapse wrote:
> If Dan Christensen is lucky, his bull shit
> leads only to a full relation. but prople
> have already pointed out that it leads to:
>
> a ~ b & ~(a ~ b)
>
> by producing multiple derivations with different
> doms.

You apparently didn't know that we can only determine whether a pair of functions is equal or not if and ONLY if they have the SAME domain and the SAME codomain. When will you learn, Jan Burse? No wonder you kept getting contradictions!

[snip]

> The problem is that the definition of a ~ b in Dan
> Christensens world is totally botched. Basically
> the ordinary definition is:
>
> a ~ b :<=> EXIST(f):[ …. f … a … b ….]
>

As you can see in my proof, I use these standard definitions:

ALL(f):ALL(a):ALL(b):[Set(a) & Set(b) & ALL(c):[c in a => f(c) in b]
=> [Injection(f,a,b) <=> ALL(c):ALL(d):[c in a & d in a => [f(c)=f(d) => c=d]]]]

ALL(f):ALL(a):ALL(b):[Set(a) & Set(b) & ALL(c):[c in a => f(c) in b]
=> [Surjection(f,a,b) <=> ALL(c):[c in b => EXIST(d):[d in a & f(d)=c]]]]

ALL(f):ALL(a):ALL(b):[Set(a) & Set(b) & ALL(c):[c in a => f(c) in b]
=> [Bijection(f,a,b) <=> Injection(f,a,b) & Surjection(f,a,b)]]

Read 'em and weep, Jan Burse!

[Dan Christensen]

On Tuesday, July 12, 2022 at 5:09:12 PM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Dienstag, 12. Juli 2022 um 17:22:43 UTC+2:
> > On Friday, July 1, 2022 at 12:48:59 AM UTC-4, Dan Christensen wrote:
> > > As requested...
> > >
> > > THEOREM: Bijectivity is Transitive
> > >
> > > ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
> > >
> > > => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
> > > & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
> > >
> > > => EXIST(h):Bijection(h,t,v)]]
> > >
> > > PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
> > >
> > Using a similar approach, it is easy to prove the reflexivity and symmetry of bijections.

> Its always easy to prove some from garbage.
> If you have nowhere an example of a pair a,b
> with a proof of:
>
> ~(a ~ b) /* in works a and b are not in bijection */
>

I was able to establish the reflexivity, symmetry and transitivity of the bijection relation. And, sadly for you, you were unable to in any meaningful way to fault the standard definitions I used. Must be frustrating as hell for you, Jan Burse.

[snip]

[Mostowski Collapse]

Well the "if they have the SAME domain and the SAME codomain"
works if the domain and codomain are intrinsic, hence unique.
But lets see what your nonsense says:

/* Fn Equality */
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)]]]

Its still extrinsic. So I can choose domain and codomain
x,y and the functions are equal. And then I can choose domain
and codomain x2,y2 and the functions are not equal.

This was also already demonstrated by people here,
that we can then derive from your nonsense ad-hoc axiom
the following utter nonsense for certain pairs f,g:

f = g & ~(f = g)

This was demonstrated like a 1000 times here.

[Mostowski Collapse]

You cannot make definitions of the form:

ALL(p):ALL(x):[p(x) <=> q(x,w)]

They mostlikely go wrong. A definition has only
ALL quantification for the arguments of the
definiendum:

ALL(x1):...:ALL(xn):[p(x1,..,xn) <=> A(x1,..,xn)]

Other ALL quantifications for parameters inside
A lead to the "quodlibet" paradox, the "quodlibet"
paradox can be summarized as an axiom of

the form. You only need ALL instantiation to
derive an inconsistency from it. You don't need
ex-falso quodlibet inference rule:

ALL(Q):[P <=> Q]

Make two ALL instantiations for Q:

ALL instantiation with Q = TRUE
[P <=> TRUE]
ALL instantiation with Q = FALSE
[P <=> FALSE]

Join them, and you get:

P & ~P

Woa! You are a Genius, just like Putin.

[WM]

Jim Burns schrieb am Dienstag, 12. Juli 2022 um 18:28:36 UTC+2:
> On 7/12/2022 8:32 AM, WM wrote:
> > Jim Burns schrieb
> > am Dienstag, 12. Juli 2022 um 07:22:35 UTC+2:
> >> On 7/11/2022 2:37 PM, WM wrote:
>
> >>> All endsegments have infinitely many elements in common
> >>> with all their predecessors.
> >>> This also holds for all infinite successors.
> >>
> >> No,
> >> the set of elements in common with all end-segment
> >> predecessors is not
> >> the set of elements in common with all end-segment
> >> successors.
> >
> > But both sets are infinite sets,
> Also,

I see, no counter argument.

> if elements exist which are in the intersection of
> all _infinite_ end segments,
> then _infinite_ end segments exist which are not in
> the collection of all _infinite_ end segments.

No, but endsegments exist, finite and infinite, which are not in the collection of defined endsegments accessible to you and, by you, are erroneously believed to be all endegments.

> Thus, no elements exist in the second set.

The decreasing sequence of endegments cannot have an empty intersection as long as all endsegments are infinite. Inclusion monotony cannot be killed by any argument.

> > That is very simple to see, because
> > all sets are infinite and
> > no element is added that was missing in E(1) = ℕ.
> Yes,
> no element ∉ ℕ is added to get {}
> {} isn't infinite.

Silly phrase.

> > No element exists which is in the intersection of
> > all endsegments,
> > but infinitely many elements exist which are in
> > the intersection of all infinite endsegments.
> No.

Of course. All endsegments satisfying
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
can be collected in the set {E(k) : k ∈ ℕ_def} and will satisfy
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀
by inclusion monotony.

> If elements exist which are in the intersection of
> all _infinite_ end segments,

That is fact with no doubt. Your conclusion is true because you don't know all infinite endsegments.

> > Your proof that for every n
> > there exists an endsegment not containing [n]
> > holds only for all endsegments.
> > It fails for infinitely many elements
> > contained in _infinite_ endsegments.

> I address your concern:
>
> If elements exist which are in the intersection of
> all _infinite_ end segments,
> then _infinite_ end segments exist which are not in
> the collection of all _infinite_ end segments.

That is not addressing my concern but repeating your mistaken view.

Obviously you cannot even understand that you did not address my concern. Don't be sorry, most are as unable as you.

Regards, WM

[WM]

Jim Burns schrieb am Dienstag, 12. Juli 2022 um 18:51:03 UTC+2:

> It is also true that a third of the integers are
> as many as (are able to match) all the integers.
> And this is not a paradox, it is

simply a big mistake.
If you were right, then you could find a way to shuffle all X in

XOOO...
XOOO...
XOOO...
XOOO...
...

such that the whole matrix was covered by X'x. Or you could shuffle all integer fractions in
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
...
such that the whole matrix was covered by integer fractions.

Don't "prove" your belief, but show that you can do what I said or show why it is not modelling Cantor's bijection.

Regards, WM

[WM]

Fritz Feldhase schrieb am Dienstag, 12. Juli 2022 um 20:21:23 UTC+2:
> On Tuesday, July 12, 2022 at 3:24:21 PM UTC+2, WM wrote:

> > > No-one cares about your goofy matrices,
> > >
> > Then have a good day.
> It's still true.

It is true for most matheologians, but not for honest mathematicians. They are concerned with this argument: If infinite bijections existed, then you could find a way to shuffle all integer fractions in

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
...
such that the whole matrix was covered by integer fractions.

Don't try to "prove" your belief, but show that you can do what I said or show why it is not modelling Cantor's bijection.

> Btw. (in the context of set theory) the identity function on IN is just the set of ordered pairs {(n, n) : n e IN}. (The axioms of ZFC ensure its existence.)

You will see that this axiom is as silly as the axiom of a second even prime number - after you will have grasped the above argument.

> This function is a bijection from IN onto IN, and IN is an infinite set.

Potentially infinite.

Regards, WM

[Dan Christensen]

On Wednesday, July 13, 2022 at 2:55:49 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Mittwoch, 13. Juli 2022 um 04:29:05 UTC+2:
> > On Tuesday, July 12, 2022 at 5:31:25 PM UTC-4, Mostowski Collapse wrote:
> > > If Dan Christensen is lucky, his bull shit
> > > leads only to a full relation. but prople
> > > have already pointed out that it leads to:
> > >
> > > a ~ b & ~(a ~ b)
> > >
> > > by producing multiple derivations with different
> > > doms.
> > You apparently didn't know that we can only determine whether a pair of functions is equal or not if and ONLY if they have the SAME domain and the SAME codomain. When will you learn, Jan Burse? No wonder you kept getting contradictions!
> >
> > [snip]
> > > The problem is that the definition of a ~ b in Dan
> > > Christensens world is totally botched. Basically
> > > the ordinary definition is:
> > >
> > > a ~ b :<=> EXIST(f):[ …. f … a … b ….]
> > >
> > As you can see in my proof, I use these standard definitions:
> >
> > ALL(f):ALL(a):ALL(b):[Set(a) & Set(b) & ALL(c):[c in a => f(c) in b]
> > => [Injection(f,a,b) <=> ALL(c):ALL(d):[c in a & d in a => [f(c)=f(d) => c=d]]]]
> >
> > ALL(f):ALL(a):ALL(b):[Set(a) & Set(b) & ALL(c):[c in a => f(c) in b]
> > => [Surjection(f,a,b) <=> ALL(c):[c in b => EXIST(d):[d in a & f(d)=c]]]]
> >
> > ALL(f):ALL(a):ALL(b):[Set(a) & Set(b) & ALL(c):[c in a => f(c) in b]
> > => [Bijection(f,a,b) <=> Injection(f,a,b) & Surjection(f,a,b)]]
> >
> > Read 'em and weep, Jan Burse!

> Well the "if they have the SAME domain and the SAME codomain"
> works if the domain and codomain are intrinsic, hence unique.
> But lets see what your nonsense says:
>
> /* Fn Equality */
> 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)]]]
>
> Its still extrinsic. So I can choose domain and codomain
> x,y and the functions are equal. And then I can choose domain
> and codomain x2,y2 and the functions are not equal.
>
> This was also already demonstrated by people here,
> that we can then derive from your nonsense ad-hoc axiom
> the following utter nonsense for certain pairs f,g:
>
> f = g & ~(f = g)
>

You are grasping at straws here. Again, we can only determine whether a pair of functions is equal or not if and ONLY if they have the SAME domain and the SAME codomain. When will you learn, Jan Burse? No wonder you kept getting contradictions! They can be avoided by using standard formal definitions to construct functions and determine their equality.

[sergi o]

On 7/13/2022 6:27 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 12. Juli 2022 um 20:21:23 UTC+2:
>> On Tuesday, July 12, 2022 at 3:24:21 PM UTC+2, WM wrote:
>
>>>> No-one cares about your goofy matrices,
>>>>
>>> Then have a good day.
>> It's still true.
>

<snip delusional crap>

>
>> Btw. (in the context of set theory) the identity function on IN is just the set of ordered pairs {(n, n) : n e IN}. (The axioms of ZFC ensure its existence.)
>

>

>> This function is a bijection from IN onto IN, and IN is an infinite set.
>

>

> Regards, WM

no math at all.

[sergi o]

On 7/13/2022 6:15 AM, CLOWN SHOES WM wrote:
> Jim Burns schrieb am Dienstag, 12. Juli 2022 um 18:28:36 UTC+2:
>> On 7/12/2022 8:32 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 12. Juli 2022 um 07:22:35 UTC+2:
>>>> On 7/11/2022 2:37 PM, WM wrote:
>>
>>>>> All endsegments have infinitely many elements in common
>>>>> with all their predecessors.
>>>>> This also holds for all infinite successors.
>>>>
>>>> No,
>>>> the set of elements in common with all end-segment
>>>> predecessors is not
>>>> the set of elements in common with all end-segment
>>>> successors.
>>>
>>> But both sets are infinite sets,
>> Also,
>
> I see, no counter argument.
>
>> if elements exist which are in the intersection of
>> all _infinite_ end segments,
>> then _infinite_ end segments exist which are not in
>> the collection of all _infinite_ end segments.
>

> No, but endsegments exist, finite CLOWN SHOES and infinite, which are not in the collection of CLOWN SHOES endsegments accessible to you and, by you, are erroneously believed to be all endegments.

>
>> Thus, no elements exist in the second set.
>

> The decreasing sequence of endegments cannot have an empty intersection as long as all endsegments are infinite. Inclusion monotony CLOWN SHOES cannot be killed by any argument.

>
>>> That is very simple to see, because
>>> all sets are infinite and
>>> no element is added that was missing in E(1) = ℕ.
>> Yes,
>> no element ∉ ℕ is added to get {}
>> {} isn't infinite.
>
> Silly phrase.
>
>>> No element exists which is in the intersection of
>>> all endsegments,
>>> but infinitely many elements exist which are in
>>> the intersection of all infinite endsegments.
>> No.
>
> Of course. All endsegments satisfying
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

> can be CLOWN SHOES in the set {E(k) : k ∈ ℕ_def} and will satisfy

> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀

> by CLOWN SHOES inclusion monotony.

>
>> If elements exist which are in the intersection of
>> all _infinite_ end segments,
>

> That is fact with no CLOWN SHOES. Your conclusion is true because you don't know all infinite CLOWN SHOES.

>
>>> Your proof that for every n
>>> there exists an endsegment not containing [n]
>>> holds only for all endsegments.
>>> It fails for infinitely many elements
>>> contained in _infinite_ endsegments.
>
>> I address your concern:
>>
>> If elements exist which are in the intersection of
>> all _infinite_ end segments,
>> then _infinite_ end segments exist which are not in
>> the collection of all _infinite_ end segments.
>

> That is not addressing my CLOWN SHOES but repeating your mistaken view.
>
> Obviously you cannot even understand that you did not address my CLOWN SHOES. Don't be sorry, most CLOWN SHOES are as unable as you.
>
> Regards, CLOWN SHOES WM

[Mostowski Collapse]

You are a funcking liar. I wrote x,y choose for doman
and codomain, of course for BOTH functions f and g
you idiot. If you instantiate your axiom, it is

set for BOTH functions. And then use x2,y2 for domain
adn codomain, of course for BOTH functions f and g
yoi idiot. If you instantiate your axiom its is

again set for BOTH functions.

Whats wrong with you? This counter example was
already given by a couple of people here, weeks ago,
but you didn't understand it, or what?

[Mostowski Collapse]

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.

[Jim Burns]

ℕ = {definable} = ⋃ₙ{definer of n}

A definer ⟨0...i,i+1...n⟩ exists for definable n.
Definable n is _accessible_ ==
For each split on the way to n,
a step i,i+1 exists which crosses that split.

For an end segment E(n) of ℕ
E(n) is _accessible_ ==
An end-definer ⟨ℕ...E',E'\\...E(n)⟩ exists such that
for each split on the way to E(n),
a step E',E'\\ exists which crosses that split.

Notation: E'\\ = E'\{min(E')}
E(2) = E(1)\\
E(3) = E(2)\\
...


Consider all the end segments of ℕ

ℕ = {definable} = ⋃ₙ{definer of n}

An end segment is closed-upward.
An end segment is not empty.

Let 𝒰(ℕ) be the collection of closed-upward
subsets of ℕ
𝒰(ℕ) = { E ∈ 𝒫(ℕ) : ∀i ∈ E, i < j -> j ∈ E }

The set ENDS of all end segments is 𝒰(ℕ) --
except for {}
ENDS = 𝒰(ℕ)\{ {} }

| Assume E is an end segment of ℕ
| E ∈ ENDS
|
| E is a non-empty subset of ℕ
| k ∈ E and k ∈ ⟨0...i,i+1...k⟩
|
| Split ⟨0...i,i+1...k⟩ between
| BEFORE not in E and AFTER in E
| Let k₁ begin AFTER
| E = E(k₁) = { j ∈ ℕ : k₁ =< j }
|
| Consider
| { E(j) : j ∈ ⟨0...i,i+1...k₁⟩ }
| For each split on the way to E(k₁),
| a step E',E'\\ exists which crosses that split.
| E' = E(i) and E'\\ = E(i+1)
|
| { E(j) : j ∈ ⟨0...i,i+1...k₁⟩ } is
| an end-defined of E(k₁) = E
| E is accessible.

Therefore,
if E is an end segment of ℕ
then an end-definer of E exists
and E is accessible.

> No, but endsegments exist, finite and infinite,
> which are not in the collection of defined endsegments
> accessible to you and, by you, are erroneously believed
> to be all endegments.

No,
if E is an end segment of ℕ
then an end-definer of E exists
and E is accessible.

ℕ = {definable} = ⋃ₙ{definer of n}

Also,
if n is in ℕ
then { j ∈ ℕ : n < j } is
an infinite end segment which n is not in.

>> Thus, no elements exist in the second set.
>
> The decreasing sequence of endegments cannot have
> an empty intersection as long as
> all endsegments are infinite.

Each end segment is in the sequence of end segments.

Each end segment equals the intersection of
all end segments _up to_ that end segment.

Each end segment _does not equal_ the intersection of
all end segments _up to_ that end segment
_and the next end segment in the sequence_

Therefore,
each end segment _does not equal_ the intersection of
all end segments.

Therefore,
the intersection of all end segments is not
an end segment.

Empty intersection and infinite end segments
is not a contradiction.

> Inclusion monotony cannot be killed by any argument.

I don't doubt that that is literally true, for you.

| It is useless to attempt to reason a man out of
| a thing he was never reasoned into.
|
-- Jonathan Swift

[Fritz Feldhase]

"Cranks characteristically dismiss all evidence or arguments which contradict their own unconventional beliefs, making any rational debate a futile task and rendering them impervious to facts, evidence, and rational inference." (Wikipedia)

[sergi o]

I am labeling this "WM BOT Response #1" (matrix swaparoofestival)

[sergi o]

On 7/11/2022 1:40 PM, WM wrote:
> Dan Christensen schrieb am Montag, 11. Juli 2022 um 17:08:23 UTC+2:
>> On Monday, July 11, 2022 at 9:03:31 AM UTC-4, WM wrote:
>>> Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 15:29:31 UTC+2:
>>>> On Sunday, July 10, 2022 at 9:04:28 AM UTC-4, WM wrote:
>>>>> Dan Christensen schrieb am Sonntag, 10. Juli 2022 um 04:45:09 UTC+2:

>>>>>> On Saturday, July 9, 2022 at 8:59:01 AM UTC-4, WM wrote:
>>>>>
>>>>>>> Bijectivity does not exist between infinite sets.

>>>>>> Can you not agree that there is a bijection mapping the natural numbers to the set of even numbers, and that they are both infinite sets?
>>>>> No. If there was a bijection possible, then in the matrix
>>>>>
>>>>> XXXXXXXXXXXXXXXXXXXXXXX...
>>>>> O O O O O O O O O O O O O...
>>>>>
>>>>> we could change the positions of the X's such that they covered the whole matrix. Up to every finite step the number of not covered places remains constant.
>>>>>
>>>> Something is seriously wrong with it if you cannot prove that f(x)=2x is a bijection mapping the natural numbers to the even numbers.
>>
>>> It is correct in potential infinity. It is wrong in actual infinity
>> Not sure sure what YOU mean by infinite, but a set X is said to be infinite iff there exists an injective function f: X-->X that is not surjective (see Dedekind). By that definition, both N and 2N are infinite.
>
> Yes, that definition is wrong ( cannot be satisfied by infinite sets). See the above matrix. Only a fool can try to cover the whole matrix by simple shuffling the X's.
>
>> Not in the definition you originally posted here some time ago. Have you subsequently, added 1=/=2,
>
> Only a fool would think this is necessary.
>
> Regards, WM


I am labeling this "WM BOT Response #1a" (matrix swaparoofestival)

[sergi o]

On 7/9/2022 7:58 AM, WM wrote:
> Dan Christensen schrieb am Freitag, 1. Juli 2022 um 06:48:59 UTC+2:

>> As requested...
>>
>> THEOREM: Bijectivity is Transitive
>>
>> ALL(t):ALL(u):ALL(v):[Set(t) & Set(u) & Set(v)
>>
>> => [EXIST(f):[ALL(a):[a in t => f(a) in u] & Bijection(f,t,u)]
>> & EXIST(f):[ALL(a):[a in u => f(a) in v] & Bijection(f,u,v)]
>>
>> => EXIST(h):Bijection(h,t,v)]]
>>
>> PROOF: https://dcproof.com/BijectionsTransitive.htm (275 lines)
>

> Bijectivity does not exist between infinite sets. For instance consider Cantors's
> "proof " of the bijection between ℕ and ℚ using the function
>
> k = (m + n - 1)(m + n - 2)/2 + m
>
> with the resulting sequence of fractions
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
>
> If we check it by bijecting first the natnumbers and the integers fractions in the matrix

>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
>

> putting the indices X in the first column

> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>

> (the not yet indexed fractions are expressed by O's), then indexing the fractions according to Cantor's function
>
> 1/1, 1/2,
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> ...
>
> 1/1, 1/2, 2/1,
>
> XXOO...
> XOOO...
> OOOO...
> XOOO...
> ...
>
> 1/1, 1/2, 2/1, 1/3,
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> ...
>
> should cover the whole infinite matrix with X's. But obviously that cannot happen.
>
> Regards, WM
>
>
>


I am labeling this "WM BOT Response #1" (matrix swaparoofestival)

[WM]

Jim Burns schrieb am Mittwoch, 13. Juli 2022 um 20:51:02 UTC+2:

> Therefore,
> each end segment _does not equal_ the intersection of
> all end segments.
>
> Therefore,
> the intersection of all end segments is not
> an end segment.

Each endsegment equals the intersection up to that endsegment.
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k).
For infinite endsegments we have
∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

>
> Empty intersection and infinite end segments
> is not a contradiction.

You can repeat this falsehood as often as you like. It is and remains wrongt. Only real fools will accept it.

If this were right for all endsegments:

∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

then the intersection of all these endsegments would not be less.

By the way there cannot be ℵ₀ endsegments containing ℵ₀ natnumbers each, because there cannot exist two consecutive ℵ₀-infinite sets in ℕ.
Are you even unable to understand this?

> > Inclusion monotony cannot be killed by any argument.
> I don't doubt that that is literally true, for you.

It is the basis of mathematics. Everybody not brain-damaged by set theory will understand. I never met a student who didn't.

Regards, WM

[Jim Burns]

On 7/13/2022 7:21 AM, WM wrote:

> Or you could shuffle all integer fractions in
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
> such that the whole matrix was covered by
> integer fractions.

Each natural n has a unique prime factorization.

Each rational (not fraction) p/q has a unique
prime factorization, by negating exponents to
prime factors of the denominator.
(Common divisors cancel.)

For each positive exponent in the unique prime
factorization of n,
there is a positive/negative exponent in the
unique prime factorization of a rational p/q,
and,
for each positive/negative exponent in the
unique prime factorization of a rational p/q,
there is a positive exponent in the unique prime
factorization of n.

...like this:
1 2 3 4 5 6 ...
1 -1 2 -2 3 -3 ...

Unique-prime-exponents of naturals can match
unique-prime-exponents of rationals.

...
97^1 ~~ 97^1
2^1 7^2 ~~ 2^1 7^-1
3^2 11^1 ~~ 3^-1 11^1
2^2 5^2 ~~ 2^-1 5^-1
101^1 ~~ 101^1
...

Therefore,
naturals can match
rationals.
...
97 = 97^1 ~~ 97^1 = 97/1
98 = 2^1 7^2 ~~ 2^1 7^-1 = 2/7
99 = 3^2 11^1 ~~ 3^-1 11^1 = 11/3
100 = 2^2 5^2 ~~ 2^-1 5^-1 = 1/10
101 = 101^1 ~~ 101^1 = 101/1
...

[WM]

Jim Burns schrieb am Mittwoch, 13. Juli 2022 um 23:13:53 UTC+2:
> On 7/13/2022 7:21 AM, WM wrote:
>
> > Or you could shuffle all integer fractions in
> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, 3/2, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> > such that the whole matrix was covered by
> > integer fractions.
> Each natural n has a unique prime factorization.

Your "proofs" are irrelevant.
Please answer the question: Can the integer fractions cover all places of the matrix? An affirmative answer would prove your brain damage.

Regards, WM

[Fritz Feldhase]

On Wednesday, July 13, 2022 at 11:16:46 PM UTC+2, WM wrote:
>
> Your "proofs" are irrelevant.

Well, Mückenheim, we ALL know that you actually mean:

"Proofs are irrelevant."

Yes, that's certainly true in your wahnsystem.

> Please answer the question: Can the integer fractions cover all places of the matrix? An affirmative answer would prove your brain damage.

Sounds like a rather serious and fair question. Intellectual integrity is one of your main strength!

[Fritz Feldhase]

On Wednesday, July 13, 2022 at 11:13:32 PM UTC+2, WM wrote:
>
> "Inclusion monotony" [...] is the basis of mathematics. Everybody not brain-damaged by set theory will understand.

Errrr... ok.

Good to know.

[Jim Burns]

On 7/13/2022 5:16 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 13. Juli 2022 um 23:13:53 UTC+2:
>> On 7/13/2022 7:21 AM, WM wrote:

>>> Or you could shuffle all integer fractions in
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>> such that the whole matrix was covered by
>>> integer fractions.
>>
>> Each natural n has a unique prime factorization.

And each rational has a unique prime factorization.

> Your "proofs" are irrelevant.
> Please answer the question:
> Can the integer fractions cover all places of
> the matrix?
> An affirmative answer would prove your brain damage.

[Jim Burns]

On 7/13/2022 5:13 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 13. Juli 2022 um 20:51:02 UTC+2:

>> Therefore,
>> each end segment _does not equal_ the intersection of
>> all end segments.
>>
>> Therefore,
>> the intersection of all end segments is not
>> an end segment.
>
> Each endsegment equals the intersection up to
> that endsegment.
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k).

Each end segment _does not equal_ the intersection
up to _and one beyond_ that end segment.
∀k ∈ ℕ: ⋂{ E(1),E(2),...,E(k),E(k+1) } ≠ E(k).

Each end segment _does not equal_ the intersection
of all end segments.
∀k ∈ ℕ: ⋂{ E(1),E(2),E(3),... } ≠ E(k).

> For infinite endsegments we have
> ∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
>>
>> Empty intersection and infinite end segments
>> is not a contradiction.
>
> You can repeat this falsehood as often as you like.
> It is and remains wrongt.
> Only real fools will accept it.

What is the first end segment which equals
the intersection up to _and one beyond_
that end segment?

Assume ⋂{ E(1),E(2),...,E(k),E(k+1) } = E(k).
what is the first k for which this is true?

[Dan Christensen]

On Wednesday, July 13, 2022 at 2:23:42 PM UTC-4, Mostowski Collapse 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)

[Jim Burns]

On 7/13/2022 10:30 PM, Dan Christensen wrote:
> On Wednesday, July 13, 2022 at 2:23:42 PM UTC-4,
> Mostowski Collapse 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.

Jan Burse points out that you didn't write
_your axiom_ that way.

How do you ensure that,
wherever the domain of f1 is referred to,
it's the same?
Name a _function_ dom(f)
a uniquely-valued relation,
from functions to domains (sets)

If you want to say f is a function, try
Function(f)

It's a property f1 may or may mot have.
We can say f1 does have it this way
Function(f1)

"f1 has a property" doesn't say much.
but We can universally quantify 'Function(f)'
in order to say what it means to be a function.

For all functions, there is a unique domain.
It is a set.
For all functions, there is a unique codomain.
It is a set.

Say
∀f, Function(f) -> ∃!d, Set(d) ∧ d = dom(f)
∀f, Function(f) -> ∃!d, Set(d) ∧ d = cod(f)

*NOW* in order to refer to a function with
a different domain, you must give it a different
name. *NOW* dom(f1) = {0} and dom(f1) = {0,1}
is a contradiction. Your axiom doesn't do that.

Say what a domain and codomain are _for functions_
∀f, Function(f) ->
∀x, x ∈ dom(f) -> ∃!y, y ∈ cod(f) ∧ y = f(x)

Say what equality is _for functions_
∀f1 ∀f2, Function(f1) ∧ Function(f2) ->
f1 = f2 <->
dom(f1) = dom(f2) ∧
cod(f1) = cod(f2) ∧
∀x, x ∈ dom(f1) -> f1(x) = f2(x)

'dom(f1) = dom(f2)' uses equality _for sets_
which is defined elsewhere, no doubt.

In my opinion, it's better to present these
claims as separate statements. Among the reasons,
that way, you don't need to write a bunch of
claims about f1 and then the same claims about f2.
'Function(f1) ∧ Function(f2)' covers it.