Justifying the truth table for the IMPLIES-operator

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Dan Christensen

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Sep 21, 2022, 1:22:18 AMSep 21
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Here is the truth table for the IMPLIES-operator:

A B A=>B
T T T
T F F
F T T
F F T

Judging by frequent questions at MSE (at least once every day) on this topic, there seems to be much confusion about this one, simple table.

This table is often used to "define" the IMPLIES-operator in introductory textbooks. No real explanation is usually given, just a few examples. There is a good reason for this, namely that to truly understand why it works, you must understand some basic methods of proof, i.e. the following rules of inference in propositional logic:

1. Premise
2. Conclusion (intro =>, intro ~)
3. Join (intro &)
4. Split (elim &)
5. Detachment (elim =>)
6. Remove ~~

(For an excellent introduction to these rules of inference, you need only work your way through only the first 3 (of 13) examples in the DC Proof tutorial.)

The above truth table is really just a table of the following 4 theorems in propositional logic:

1. A & B => [A => B]

2. A & ~B => ~[A => B]

3. ~A & B => [A => B]

4. ~A & ~B => [A => B]

*****************************************************************

Thm 1: A & B => [A => B]

1. A & B
Premise

2. A
Premise

3. B
Split, 1

4. A => B
Conclusion, 2

5. A & B => [A => B]
Conclusion, 1

*****************************************************************

Thm 2: A & ~B => ~[A => B]

1. A & ~B
Premise

2. A => B
Premise

3. A
Split, 1

4. ~B
Split, 1

5. B
Detach, 2, 3

6. ~B & B
Join, 4, 5

7. ~[A => B]
Conclusion, 2

8. A & ~B => ~[A => B]
Conclusion, 1

*****************************************************************

Thm 3: ~A & B => [A => B]

1. ~A & B
Premise

2. A
Premise

3. B
Split, 1

4. A => B
Conclusion, 2

5. ~A & B => [A => B]
Conclusion, 1

*****************************************************************

Thm 4: ~A & ~B => [A => B]

1. ~A & ~B
Premise

2. A
Premise

3. ~B
Premise

4. ~A
Split, 1

5. A & ~A
Join, 2, 4

6. ~~B
Conclusion, 3

7. B
Rem DNeg, 6

8. A => B
Conclusion, 2

9. ~A & ~B => [A => B]
Conclusion, 1

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

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Sep 21, 2022, 3:04:48 AMSep 21
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If you know how material implications works, why do you write
nonsense like this here:

> > y=\=x => ~Bijection(id_y,x,x)
> Not true in general. What is your point, Jan Burse?

Do you have a pair y,x where the above statement
y=\=x => ~Bijection(id_y,x,x) is not true?

Hint: Must be a statement of the second row:

A B A=>B
T F F

Dan Christensen

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Sep 21, 2022, 9:56:33 AMSep 21
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> If you know how material implications works, why do you write
> nonsense like this here:
>
> > > y=\=x => ~Bijection(id_y,x,x)
> > Not true in general. What is your point, Jan Burse?
>
> Do you have a pair y,x where the above statement
> y=\=x => ~Bijection(id_y,x,x) is not true?
>

Suppose x = {0} and y = {0, 1}. Then x=/=y and for all z in x, id_y(z) =z, i.e. id_y is a bijection on set x.

Ross A. Finlayson

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Sep 21, 2022, 11:47:39 AMSep 21
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What's 0^0 again?
Message has been deleted

Fritz Feldhase

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Sep 21, 2022, 12:08:55 PMSep 21
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On Wednesday, September 21, 2022 at 5:47:39 PM UTC+2, Ross A. Finlayson wrote:

> What's 0^0 again?

1 .

By two different definitions:

a.) ^ defined recursively on IN:

Let m e IN:

m^0 = 1
m^(n+1) = m^n * m (for all n e IN)

b.) "set-theoretically" on IN (where IN is defined due to von Neumann):

m^n = card({f : n --> m}) (for all n, m e IN)

Dan Christensen

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Sep 21, 2022, 1:28:57 PMSep 21
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On Wednesday, September 21, 2022 at 1:22:18 AM UTC-4, Dan Christensen wrote:
> Here is the truth table for the IMPLIES-operator:
>
> A B A=>B
> T T T
> T F F
> F T T
> F F T
>
> Judging by frequent questions at MSE (at least once every day) on this topic, there seems to be much confusion about this one, simple table.
>
> This table is often used to "define" the IMPLIES-operator in introductory textbooks. No real explanation is usually given, just a few examples. There is a good reason for this, namely that to truly understand why it works, you must understand some basic methods of proof, i.e. the following rules of inference in propositional logic:
>
> 1. Premise
> 2. Conclusion (intro =>, intro ~)
> 3. Join (intro &)
> 4. Split (elim &)
> 5. Detachment (elim =>)
> 6. Remove ~~
>
> (For an excellent introduction to these rules of inference, you need only work your way through only the first 3 (of 13) examples in the DC Proof tutorial.)
>
> The above truth table is really just a table of the following 4 theorems in propositional logic:
>
> 1. A & B => [A => B]
>
> 2. A & ~B => ~[A => B]
>
> 3. ~A & B => [A => B]
>
> 4. ~A & ~B => [A => B]
>

Started working on an update to the DC Proof tutorial to include these theorems as exercises with hints and full solutions. Should have thought of this years ago! The perfect proofs for the beginner--no more than 9 lines each and you don't even need that much logic. And IMHO quite profound when taken as a whole.

Dan Christensen

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Sep 21, 2022, 2:01:08 PMSep 21
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On Wednesday, September 21, 2022 at 11:47:39 AM UTC-4, Ross A. Finlayson wrote:

[snip]

> What's 0^0 again?

Huh? Oh, I see... You would rather change the subject. Can't blame you, really.

See: https://dcproof.wordpress.com/2013/10/09/oh-the-ambiguity-2/

Start a new thread if you want to rehash it after all these years. Could be fun.

Dan Christensen

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Sep 21, 2022, 7:55:31 PMSep 21
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On Wednesday, September 21, 2022 at 1:28:57 PM UTC-4, Dan Christensen wrote:
> On Wednesday, September 21, 2022 at 1:22:18 AM UTC-4, Dan Christensen wrote:
> > Here is the truth table for the IMPLIES-operator:
> >
> > A B A=>B
> > T T T
> > T F F
> > F T T
> > F F T
> >
> > Judging by frequent questions at MSE (at least once every day) on this topic, there seems to be much confusion about this one, simple table.
> >
> > This table is often used to "define" the IMPLIES-operator in introductory textbooks. No real explanation is usually given, just a few examples. There is a good reason for this, namely that to truly understand why it works, you must understand some basic methods of proof, i.e. the following rules of inference in propositional logic:
> >
> > 1. Premise
> > 2. Conclusion (intro =>, intro ~)
> > 3. Join (intro &)
> > 4. Split (elim &)
> > 5. Detachment (elim =>)
> > 6. Remove ~~
> >
> > (For an excellent introduction to these rules of inference, you need only work your way through only the first 3 (of 13) examples in the DC Proof tutorial.)
> >
> > The above truth table is really just a table of the following 4 theorems in propositional logic:
> >
> > 1. A & B => [A => B]
> >
> > 2. A & ~B => ~[A => B]
> >
> > 3. ~A & B => [A => B]
> >
> > 4. ~A & ~B => [A => B]
> >
> Started working on an update to the DC Proof tutorial to include these theorems as exercises with hints and full solutions. Should have thought of this years ago! The perfect proofs for the beginner--no more than 9 lines each and you don't even need that much logic. And IMHO quite profound when taken as a whole.

Now available at my homepage.

Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com <------- Here

Fritz Feldhase

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Sep 21, 2022, 8:00:20 PMSep 21
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In the context of set theory, we may define addition, multiplication and the power operation on IN quite naturally using set theoretic means:

Let n, m e IN. Then

n + m = card(n x {0} u m x {1}) ,
n * m = card(n x m) ,
n ^ m = card({f | f : m --> n}) .

n = 0 and/or m = 0 is no "special case" here.

Actually, we would "artificially" have to treat 0^0 as a special case to "avoid" the result 1. But there is absolutely no reason to do so (in this context).

Ross A. Finlayson

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Sep 21, 2022, 9:54:01 PMSep 21
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No you brickhead clod, that was changing _your_ subject,
just pointing out that your stipulations and
"showing your work" about fail given alternatives.

I know you "have to have it your way", but when it comes
to usual set-theoretic disambiguation, you're just a usual
shill and fool.

Showing that now your "implies is inference" also fails
your other failings about the maintenance of the truth-valued
in truth tables and no "false antecedents",
you false antecedent.

Just change the order of stipulations of your proofs they
change what they say. Don't you think inference should
be a little more reliable than assuming what you set out to prove?

No offense, it's a little "show your work" tool for some reasoning,
but it's weak and compounded by "false antecedents in a world
of truth tables", it's further "garbage-in garbage-out".

Then, really I do not care what you think "0^0" is,
just pointing out that you think you do,
but there's more than one answer, so,
one of yours must be wrong.


Dan Christensen

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Sep 21, 2022, 10:37:33 PMSep 21
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On Wednesday, September 21, 2022 at 9:54:01 PM UTC-4, Ross A. Finlayson wrote:
> On Wednesday, September 21, 2022 at 11:01:08 AM UTC-7, Dan Christensen wrote:
> > On Wednesday, September 21, 2022 at 11:47:39 AM UTC-4, Ross A. Finlayson wrote:
> >
> > [snip]
> >
> > > What's 0^0 again?
> >
> > Huh? Oh, I see... You would rather change the subject. Can't blame you, really.
> >
> > See: https://dcproof.wordpress.com/2013/10/09/oh-the-ambiguity-2/
> >
> > Start a new thread if you want to rehash it after all these years. Could be fun.

[snip]

So, you would rather not. Very wise.

> Don't you think inference should
> be a little more reliable than assuming what you set out to prove?
>

[snip]

Where do you imagine that I assume ~A & B => [A => B] ?

Here again is my proof:

1. ~A & B
Premise

2. A
Premise

3. B
Split, 1

4. A => B
Conclusion, 2

5. ~A & B => [A => B]
Conclusion, 1

Ross A. Finlayson

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Sep 22, 2022, 11:30:55 AMSep 22
to
You have no concept of time, the modal, or change.

For example, A as "A is premise 1", i.e. whether A or not A,
then "A is premise 2", it's an example of a generic premise
that according to its introduction, and what so follows,
doesn't admit false antecedents, at all.

What otherwise you have there as a template for "wrong".

Ross A. Finlayson

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Sep 22, 2022, 11:59:57 AMSep 22
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It's like implication is a bullet, and you are leaving bullets around in
all these guns that shoot backward.

Then you forget which ones you put where, and none of them are safe.

You got no reliable reverse "material" implication.

What you've written there is "not-A not-implies-B", not, "not-A not not-implies-not-B",
that a false antecedent is always "not implies".

Which is not necessarily "implies".

Leave "material" "implication" out of things, direct implication only. It's about
one of the stupidest kinds of notational laziness.

Don't you have any kind of object in your theory that changes?
Can't a _change_ even possibly be an object in your little theory?


Then it's like you say "whu all these are constants" then it's like
"you mean your variables, ...".


Mostowski Collapse

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Sep 22, 2022, 12:42:51 PMSep 22
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Nope, id_y isn't a bijection x -> x, it has missing

dom(id_y) = x

id_y is not member of this function space:

{ f : x -> x | f bijective }

Whats wrong with you?

Dan Christensen

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Sep 22, 2022, 12:47:29 PMSep 22
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On Thursday, September 22, 2022 at 11:30:55 AM UTC-4, Ross A. Finlayson wrote:
> On Wednesday, September 21, 2022 at 7:37:33 PM UTC-7, Dan Christensen wrote:
> > On Wednesday, September 21, 2022 at 9:54:01 PM UTC-4, Ross A. Finlayson wrote:
> > > On Wednesday, September 21, 2022 at 11:01:08 AM UTC-7, Dan Christensen wrote:
> > > > On Wednesday, September 21, 2022 at 11:47:39 AM UTC-4, Ross A. Finlayson wrote:
> > > >
> > > > [snip]
> > > >
> > > > > What's 0^0 again?
> > > >
> > > > Huh? Oh, I see... You would rather change the subject. Can't blame you, really.
> > > >
> > > > See: https://dcproof.wordpress.com/2013/10/09/oh-the-ambiguity-2/
> > > >
> > > > Start a new thread if you want to rehash it after all these years. Could be fun.
> > [snip]
> >
> > So, you would rather not. Very wise.
> > > Don't you think inference should
> > > be a little more reliable than assuming what you set out to prove?
> > >
> > [snip]
> >
> > Where do you imagine that I assume ~A & B => [A => B] ?
> >

No comment?

> > Here again is my proof:
> > 1. ~A & B
> > Premise
> >
> > 2. A
> > Premise
> >
> > 3. B
> > Split, 1
> >
> > 4. A => B
> > Conclusion, 2
> >
> > 5. ~A & B => [A => B]
> > Conclusion, 1

> You have no concept of time, the modal, or change.
>

I am looking the state of "the world" at some instant in time (usually the present). A snapshot. It may sound limiting, but it turns out to be extremely useful in the real world and in mathematics.

The statement, "If it is raining then it is cloudy" means only that, at present, it is not both raining and not cloudy. It does NOT mean that rain causes cloudiness. Or that, historically, whenever it was raining it was also cloudy.

> For example, A as "A is premise 1", i.e. whether A or not A,
> then "A is premise 2", it's an example of a generic premise
> that according to its introduction, and what so follows,
> doesn't admit false antecedents, at all.
>

A premise need not be consistent with previous ones if that is what you are getting at. A premise could even be a contradiction.

EXAMPLE

1 A & ~A
Premise

2 ~[A & ~A]

Mostowski Collapse

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Sep 22, 2022, 12:49:00 PMSep 22
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Usually in set theory this here should hold for product function spaces:

Definition: 4.2.6 The Exponentiation Axiom (abbreviated Exp) postulates
that for sets a; b the class of all functions from a to b forms a set:
∀a∀b∃c ∀f(f e c <-> (f : a -> b))
https://www1.maths.leeds.ac.uk/~rathjen/book.pdf

You can even strengthen it to ∃!c. Now it seems in
DC Spoiled, the notion of product function space is
shape shifting, and so are function spaces that are

subset of the product function space, like this here:

{ f : x -> x | f bijective }

If id_y is element of the above function space, this
would imply that x -> x and y -> y are not disjoint,
for x =\= y. But usually they are disjoint, even in the

set theoretic approach, and also in the Bourbaki approach.

LMAO!

Dan Christensen

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Sep 22, 2022, 1:13:28 PMSep 22
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On the contrary. If A => B, then we can infer ~B => ~A (the rule of contrapositive).

> What you've written there is "not-A not-implies-B", not, "not-A not not-implies-not-B",
> that a false antecedent is always "not implies".
>

Guessing at your meaning here: ~[A => B] <=> A & ~B

> Which is not necessarily "implies".
>
> Leave "material" "implication" out of things, direct implication only. It's about
> one of the stupidest kinds of notational laziness.
>

If you accept the "first principles" (rules of inference) that I list above, material implication is inevitable. Here they are again:

1. Premise
2. Conclusion (intro =>, intro ~)
3. Join (intro &)
4. Split (elim &)
5. Detachment (elim =>)
6. Remove ~~

Which do you propose we limit or eliminate altogether?

> Don't you have any kind of object in your theory that changes?

[snip]

You could interpret some independent variable t as time with different values of t corresponding to different points in space if that's what you mean? But that's getting into the realm of science.

Dan Christensen

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Sep 22, 2022, 1:29:14 PMSep 22
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On Thursday, September 22, 2022 at 12:49:00 PM UTC-4, Mostowski Collapse wrote:

[snip]

> Usually in set theory this here should hold for product function spaces:
>
> Definition: 4.2.6 The Exponentiation Axiom (abbreviated Exp) postulates
> that for sets a; b the class of all functions from a to b forms a set:
> ∀a∀b∃c ∀f(f e c <-> (f : a -> b))
> https://www1.maths.leeds.ac.uk/~rathjen/book.pdf
>

Not unlike the DC Proof Function Space axiom:

ALL(dom):ALL(cod):[Set(dom) & Set(cod)
=> EXIST(fsp):[Set(fsp) & ALL(f):[f in fsp <=> Function(f,dom,cod) & ALL(a1):[a1 in dom => f(a1) in cod]]]]

> You can even strengthen it to ∃!c. Now it seems in
> DC Proof, the notion of product function space is
> shape shifting, and so are function spaces that are
> subset of the product function space, like this here:
> { f : x -> x | f bijective }
> If id_y is element of the above function space, this
> would imply that x -> x and y -> y are not disjoint,
> for x =\= y.

If, however, x is a proper subset of y, we can say that id_y (the identity function on set y) is also a bijection on set x .

Mostowski Collapse

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Sep 22, 2022, 4:06:24 PMSep 22
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Dan Christensen halucinated on Thursday, 22. September 2022:
> If, however, x is a proper subset of y, we can say that id_y (the identity function on set y) is also a bijection on set x .

No you cannot say that. People don't say that.

What does exist in mathematics, is the operation restriction:

f | A = { (x,y) e f | x e A }

And you can for example say:

id_X = id_Y | X

for X ⊆ Y. And subsequetly you can also say id_Y | X is a X -> X bijection.

This is all explained in the very first chapters here:

Basic Set Theory (Dover Books on Mathematics)
Azriel Levy - Revised Edition (13. August 2002)
https://www.amazon.com/dp/0486420795

Or read wikipedia:

In mathematics, the restriction of a function f is a new function, denoted f | A
https://en.wikipedia.org/wiki/Restriction_%28mathematics%29


Dan Christensen

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Sep 22, 2022, 5:09:31 PMSep 22
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On Thursday, September 22, 2022 at 4:06:24 PM UTC-4, Mostowski Collapse wrote:

> > If, however, x is a proper subset of y, we can say that id_y (the identity function on set y) is also a bijection on set x .

> No you cannot say that. People don't say that.
>
> What does exist in mathematics, is the operation restriction:
>
> f | A = { (x,y) e f | x e A }
>
> And you can for example say:
>
> id_X = id_Y | X
>
> for X ⊆ Y. And subsequetly you can also say id_Y | X is a X -> X bijection.
>

In other words, like I said, "id_y is a bijection on set x."

Thanks for clearing that up, Jan Burse. (Hee, hee!)

Mostowski Collapse

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Sep 22, 2022, 5:32:17 PMSep 22
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Nope, id_y | x is a bijection on x.
id_y isn't a bijection on x.

Whats wrong with you?

Dan Christensen schrieb:

Dan Christensen

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Sep 22, 2022, 6:56:36 PMSep 22
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On Thursday, September 22, 2022 at 5:32:17 PM UTC-4, Mostowski Collapse wrote:
> Nope, id_y | x is a bijection on x.
> id_y isn't a bijection on x.
>

It is. Deal with it, Jan Burse.

Mostowski Collapse

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Sep 22, 2022, 8:12:35 PMSep 22
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Not in math text books. Did your mama
notice that you are brain damaged?

The word "is a" translates to membership,
so the two sentences translate to:

id_y | x e { f : x -> x | f bijective }
id_y e { f : x -> x | f bijective }

For x ⊆ y & x =\= y only the first membership
is true, the second membership is false.

Dan Christensen schrieb:

Dan Christensen

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Sep 22, 2022, 10:30:51 PMSep 22
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On Thursday, September 22, 2022 at 8:12:35 PM UTC-4, Mostowski Collapse wrote:


> For x ⊆ y & x =\= y only the first membership
> is true, the second membership is false.
>
> Dan Christensen schrieb:
> > On Thursday, September 22, 2022 at 5:32:17 PM UTC-4, Mostowski Collapse wrote:
> >> Nope, id_y | x is a bijection on x.
> >> id_y isn't a bijection on x.
> >>
> >
> > It is. Deal with it, Jan Burse.
> >

> Not in math text books.

[snip childish abuse]

>
> The word "is a" translates to membership,
> so the two sentences translate to:
>
> id_y | x e { f : x -> x | f bijective }
> id_y e { f : x -> x | f bijective }
>

I don't know why you are bringing function spaces into this other than to perhaps muddy the waters.

Here is a the usual definition/abbreviation for the bijectivity predicate:

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

It's use is optional. It is used only to improve readability. Instead of writing the abbreviation Bijective(f,a,b), you could write it out in full as:

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

From the definitions of x, y and id_y above, we have Bijective(id_y,x,x). Deal with it, Jan Burse. Just admit you were wrong.

Fritz Feldhase

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Sep 22, 2022, 11:27:42 PMSep 22
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On Friday, September 23, 2022 at 4:30:51 AM UTC+2, Dan Christensen wrote:

> Here is a

nonsensical

> definition/abbreviation for the bijectivity predicate:
>
> ALL(f):ALL(a):ALL(b):[Bijective(f,a,b) <=>
> [ALL(c):ALL(d):[c in a & d in a => [f(c)=f(d) => c=d]] & ALL(c):[c in b => EXIST(d):[d in a & f(d)=c]]]

Yeah, it's nonsense.

Dan Christensen

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Sep 22, 2022, 11:50:47 PMSep 22
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On Thursday, September 22, 2022 at 11:27:42 PM UTC-4, Fritz Feldhase wrote:
> On Friday, September 23, 2022 at 4:30:51 AM UTC+2, Dan Christensen wrote:
>
> > Here is definition/abbreviation for the bijectivity predicate:
> >
> > ALL(f):ALL(a):ALL(b):[Bijective(f,a,b) <=>
> > [ALL(c):ALL(d):[c in a & d in a => [f(c)=f(d) => c=d]] & ALL(c):[c in b => EXIST(d):[d in a & f(d)=c]]]

> Yeah, it's nonsense.

Oh, really? How do YOU define Bijective(f,a,b), Fritz?

Ross A. Finlayson

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Oct 6, 2022, 9:49:48 PMOct 6
to
For-any a in A, there exists unique b in B such that F(a) = b.

And, for any b in B, there exists unique a in A such that F^-1(b) = a.

There exists F.

Here the unique part of course means for-any a, exists b, not exists
c =/= b, s.t. F(a) = c, and vice versa.

These days often set theorists are lazy and let Cantor/Schroeder/Bernstein thm.
work out there "exists" such an F, while of course it must result the above.

Ross A. Finlayson

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Oct 6, 2022, 9:52:44 PMOct 6
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Yeah you brought a great game from home, but nobody's playing it,
and it's pretty clear the rules are absolutely arbitrary.

Ross A. Finlayson

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Oct 6, 2022, 10:56:08 PMOct 6
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Ah, that's function theory, where all functions have generalized inverses in terms of their value.

You did say "function", right?

You got DesCartes, and for DesCartes, it was the time of geometry defining function.
So, the Cartesian function, it's the space of all the single value mappings from domain
to codomain. (Domain, image, domain, range, domain, codomain, image.)

This is where in geometry the only "space" is supports here is a line in geometry. According
to, "Euclid", the line is what exists and there are points on it. So, that's "not" a Cartesian function.

(I.e. functions like it without otherwise the multivalent are "discontinuous" functions.)

Which among them are series, ....



Ross A. Finlayson

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Oct 6, 2022, 11:25:02 PMOct 6
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Maybe it'd be easier if you built a world where "my material implications with these
extra conditions will never reverse on me when taken the wrong way".

Also called "science".


Yeah, why don't you put "science" first in all your theorems.

The "science": part is only for the "non-logical",
the "logical" is already "logical".

Of course there's syllogism, where words are abstract non-logically.

Mostly "fulfills question words with respect to is or has".

Ross A. Finlayson

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Oct 6, 2022, 11:31:29 PMOct 6
to
Hey no worries, Dan, just you know, in your face.

Dan Christensen

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Oct 6, 2022, 11:59:38 PMOct 6
to
Huh???

>
> Yeah, why don't you put "science" first in all your theorems.
>
[snip]

Can you give even a single example of an actual error in science, engineering or any other human endeavour arising strictly from the application of faulty rules of logic that you imagine? What were the consequences? How many lives were lost or ruined, etc.

Ross A. Finlayson

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Oct 7, 2022, 12:05:40 AMOct 7
to
If you ever forget that I'd feel sorry for you.

It's about the shortest way to say it, ....



(... insert Wiki ....)


Dan Christensen

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Oct 7, 2022, 12:18:15 AMOct 7
to
On Friday, October 7, 2022 at 12:05:40 AM UTC-4, Ross A. Finlayson wrote:

> >
> > Can you give even a single example of an actual error in science, engineering or any other human endeavour arising strictly from the application of faulty rules of logic that you imagine? What were the consequences? How many lives were lost or ruined, etc.

> If you ever forget that I'd feel sorry for you.
>
> It's about the shortest way to say it, ....
>
>
>
> (... insert Wiki ....)

So, not a single example. A tempest in a teacup, then. Whew!

Dan

Ross A. Finlayson

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Oct 7, 2022, 12:19:23 AMOct 7
to
You did say function, right?

Dan Christensen

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Oct 7, 2022, 12:49:04 AMOct 7
to
> You did say function, right?

Get to the point, Ross.

Dan

Ross A. Finlayson

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Oct 7, 2022, 2:41:55 AMOct 7
to
2 +2 = 4, Ken.

Mostowski Collapse

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Oct 7, 2022, 7:42:46 PMOct 7
to
Maybe should launch a basic class in logic that
explains domain of discourse. Wonky man has
absolute no clue what it means for the Drinker Paradox

to deal with this issue:

"This proof illustrates several properties of classical
predicate logic which do not always agree with
ordinary language.
Non-Empty Domain"
https://paradox.fandom.com/wiki/Drinker_paradox

He still thinks some Russel Paradox and juggling with
sets is behind the Drinker Paradox. A complete
hopleless case wonky man aka Dan Christensen,

once a crank always a crank.

LMAO!

Mostowski Collapse

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Oct 7, 2022, 7:55:02 PMOct 7
to
Thats a recuring problem with wonky man, if you tell
him, that you work in the natural numbers, he starts
fiddling with this here:

x e N

Introducing another level, sets of natural numbers.
He is just plain crazy and doesn't understand elementary
wordings like these:

- Drinker Paradox: The people in a pub ...
- Liar Paradox: The people on an island ...
- Number Theory: The natural numbers ...
-

If mathematicians or logicians really want to quantify
over other stuff, they say so initially, like they mention sets
or functions, or they say they go higher order and not

first order. But if somebody say "The Natural Numbers",
then its not x e N, then its the domain of discourse is the
natural numbers. And you cannot have a set which is

the unversal set. If you introduce a first class object N,
and if it is a set, it cannot be the domain of discourse.
You can prove that with DC Spoiled itself, that introducing

N destroys the idea of working with natural numbers:

/* Previous Result http://www.dcproof.com/UniversalSet.htm */
1 ALL(s):[Set(s) => EXIST(a):~a ε s]
Axiom

2 Set(nat)
Axiom

3 Set(nat) => EXIST(a):~a ε nat
U Spec, 1

4 EXIST(a):~a ε nat
Detach, 3, 2

Oopsi, what is this thing outside of N?

LoL

Dan Christensen

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Oct 7, 2022, 8:24:50 PMOct 7
to
On Friday, October 7, 2022 at 7:42:46 PM UTC-4, Mostowski Collapse wrote:
> Maybe should launch a basic class in logic that
> explains domain of discourse.

Be sure to invite most mathematicians. There is no such thing in mathematics. For the most part, that it is in the domain of philosophers. Must be frustrating as hell for you,

> [snip childish abuse] has
> absolute no clue what it means for the Drinker Paradox
>

Jan Burse is still in denial. He should know better by now that, from ordinary set theory and logic, we know that for any set S and proposition Q (be it true or false), we have:

EXIST(x):[x in S => Q]

See my proof at the above link.

If S = the set of all drinkers, and Q is the proposition that all people in a given pub are drinkers, then we have the a set-theoretic version of the Drinker's Paradox. Deal with it, Jan Burse. Just admit your were wrong.

Mostowski Collapse

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Oct 8, 2022, 4:57:04 PMOct 8
to
The problem is you don't know what FOL is. Your
DC Spoiled is not FOL. And you cannot prove this
FOL theorem, namely the Drinker Paradox:

/* Not Provable in DC Spoiled, but Provable in FOL */
∃x(D(x) => ∀yD(y))

FOL is a modern development between 1900 - 1950.
In this periods a couple of publications appeared which
started classifying logics more clearly. One booklet

was Hilbert & Ackermann, Grundzüge der Theoretischen
Logik, which defined more clearly what FOL is. But you
find it also here in Gödels lecture, the simplest logics are:

1.1 Propositional logic
1.2 Predicate logic
Logic Lectures: Gödel's Basic Logic Course at Notre Dame
https://arxiv.org/abs/1705.02601

Neither of them can quatify function symbols as
DC Spoiled does. His lecture has a critical tone, and
you find sections like "Failure of traditional logic". First

thing Gödel does is this here:

1.2.5 Theorems and derived rules of the system for predicate logic
[74] ∀xPhi(x) ⊃ ∃xPhi(x) Syllogism

Which doesn't work in empty domain.

LoL

Dan Christensen

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Oct 8, 2022, 6:27:42 PMOct 8
to
See my reply just now to your identical posting at sci.math

Dan


On Saturday, October 8, 2022 at 4:57:04 PM UTC-4, Mostowski Collapse wrote:
> The problem is you don't know what FOL is. Your
> DC Spoiled is not FOL. And you cannot prove this
> FOL theorem, namely the Drinker Paradox:
>

[snip]

Fritz Feldhase

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Oct 9, 2022, 12:22:53 AMOct 9
to
On Saturday, October 8, 2022 at 2:24:50 AM UTC+2, Dan Christensen wrote:

> from ordinary set theory and logic, we know that for any set S and proposition Q (be it true or false), we have:
>
> EXIST(x):[x in S => Q]

So what?A rather trivial observation.

Hint: In ZF(C) ~(S in S) for any set S due to the axiom of foundation.

Hence

S in S -> Q

for any proposition Q (since "ex falso quodlibet").

Hence

Ex(x in S -> Q).

No "paradox" at all. No need to refer to Russell's Paradox.

Has nothing to do with the Drinker Paradox.

Mostowski Collapse

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Oct 9, 2022, 5:28:45 AMOct 9
to
Dan Christensen is the uber-crank. He not only beats John Gabriel,
he is on par with "John Gabriel: look I reinvented calculus, here is
new calculus", in that he has "Dan Christensen: look I reinvented

logic, here is DC Proof". He also beats Archimedes Plutonium,
in that he has more Alzheimer than Archimedes Plutonium. You
can talk to Dan Christensen and he doesn't understand acronyms:

FOL = first order logic
FOL+SET = first order logic plus his Subset axiom schema

And he doesn't know facts like:

Drinker Paradox is a FOL Paradox
Drinker Paradox is not a FOL+SET Paradox
Russell Paradox is not a FOL Paradox
Russell Paradox is a FOL+SET Paradox

Please wonky many, show as the Drinker Paradox in FOL
only? What is FOL only? It is not FOL+SET!

Mostowski Collapse

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Oct 9, 2022, 5:53:35 AMOct 9
to
Why is the Russell Paradox viewed as an application of the
Diagonal Argument? He he, Alzheimer wonky man, so you don't
remember? Your brains totally blank, eaten by Corona Virus?

Because one can prove the Russell Paradox from a particular
instance of the Subset axiom schema:

∀k∃u(Su ∧ ∀a(Eau ↔ (Eak ∧ ¬Eaa))) → ∀s(Ss → ∃a¬Eas) is valid.
https://www.umsu.de/trees/#~6k~7u%28Su~1~6a%28Eau~4%28Eak~1~3Eaa%29%29%29~5~6s%28Ss~5~7a~3Eas%29

And this instance with ~Eaa corresponds to the construction
of an Anti-diagonal, as used in the Diagonal Argument. The
Regularity Axiom doesn't come into play, it is even not built-in

in DC Proof. See also here, similar Diagonal Argument used in Cantors Proof:

https://en.wikipedia.org/wiki/Cantor's_diagonal_argument#Uncountable_set

The Drinker Paradox doesn't need this argument. It
even doesn't need SET, and is also not supposed to use SET.
You can much more easily prove the Drinker Paradox

in FOL, by proof by cases:

"The proof begins by recognising it is true that either
everyone in the pub is drinking (in this particular round of drinks),
or at least one person in the pub isn't drinking."
https://paradox.fandom.com/wiki/Drinker_paradox

Dan Christensen

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Oct 9, 2022, 1:19:29 PMOct 9
to
See my reply earlier today to your identical posting at sci.math

Dan

On Sunday, October 9, 2022 at 5:28:45 AM UTC-4, Mostowski Collapse wrote:
> Dan Christensen is the uber-crank. He not only beats John Gabriel,
> he is on par with "John Gabriel: look I reinvented calculus, here is
> new calculus", in that he has "Dan Christensen: look I reinvented
>
>[snip]

Mostowski Collapse

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Oct 9, 2022, 4:31:54 PMOct 9