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 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 can work you 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

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

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 schrieb am Mittwoch, 21. September 2022 um 07:23:28 UTC+2:
> 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

On Wednesday, September 21, 2022 at 3:06:09 AM UTC-4, Mostowski Collapse wrote:
> 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.

Dan

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

On Wednesday, September 21, 2022 at 7:46:59 PM UTC+2, Dan Christensen wrote:
> On Wednesday, September 21, 2022 at 3:06:09 AM UTC-4, Mostowski Collapse wrote:
> > 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?

Of course, true in general.

> > 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.

"id_y is a bijection on set x" - huh?!

No, it isn't!

Hint: id_y is the identity function (defined) on y _by definition_. In other words, id_y is the function f: y --> y defined with f(z) = z for all z e y.

Again: id_y is NOT "a bijection from x onto x" (in symbols "id_y : x >->> x") but a bijection from y onto y: id_y: y >->> y.

Remember: Bijection(f, X, Y) :<-> Function(f, X, Y) & Bijective(f)

Another hint: id_y | x (i. e. the restriction of id_y to x) would be a bijection on x.

It seems that you are too dumb to understand basic math:

"[...] if f : R → R is the function f(x) := x^2, and g := f | [1, 2] is the restriction of f to the interval [1, 2],
then f and g both perform the operation of squaring, i.e., f(x) = x^2 and g(x) = x^2, but the two
functions f and g are not considered the same function, f =/= g, because they have different domains."
--Terrance Tao, "Analysis I, Third Edition", 2016

On Wednesday, September 21, 2022 at 2:25:55 PM UTC-4, Fritz Feldhase wrote:
> On Wednesday, September 21, 2022 at 7:46:59 PM UTC+2, Dan Christensen wrote:

[snip]

> > 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.
>
> "id_y is a bijection on set x" - huh?!
>
[snip]

In this case, we have:

ALL(a):[a in x => id_y(a) in x]
& ALL(a):ALL(b):[a in x & b in x => [id_y(a)=id_y(b) => a=b]]
& ALL(a):[a in x => EXIST(b):[b in x & id_y(b)=a]]

Make up your own definitions if you want, Fritz, but I think this is what most math textbook writers mean by function id_y being a bijection on the set x.

EOD

Dan

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

On Wednesday, September 21, 2022 at 9:29:58 PM UTC+2, Dan Christensen wrote:

> > > 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.
> >
> > "id_y is a bijection on set x" - huh?!
> >
> > No, it isn't!
> >
> > Hint: id_y is the identity function on y _by definition_. In other words, id_y is the function f: y --> y defined with f(z) = z for all z e y.
> >
> Make up your own definitions if you want, Fritz, but I think this is what most math textbook writers mean by function id_y being a bijection on the set x.

Look, you silly crank:

"Formally, if M is a set, the identity function f on M is defined to be a function with M as its domain and codomain, satisfying f(X) = X for all elements X in M.
[...] The identity function f on M is often denoted by id_M." (Wikipedia)

FUCK OF YOU SILLY ASSHOLE! YOU ARE WAY TOO DUMB FOR MATH.

Dan Christensen <Dan_Christensen@sympatico.ca> writes:

> On Wednesday, September 21, 2022 at 2:25:55 PM UTC-4, Fritz Feldhase wrote:
>> On Wednesday, September 21, 2022 at 7:46:59 PM UTC+2, Dan Christensen wrote:
>
> [snip]
>
>> > 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.
>>
>> "id_y is a bijection on set x" - huh?!
>>
> [snip]
>
> In this case, we have:
>
> ALL(a):[a in x => id_y(a) in x]
> & ALL(a):ALL(b):[a in x & b in x => [id_y(a)=id_y(b) => a=b]]
> & ALL(a):[a in x => EXIST(b):[b in x & id_y(b)=a]]
>
> Make up your own definitions if you want, Fritz, but I think this is
> what most math textbook writers mean by function id_y being a
> bijection on the set x.

Look at exercise 3.3.8 part (c) in your favourite math textbook and
you'll see that Tao disagrees (as does everyone else).

--
Ben.

On 9/21/2022 12:48 AM, 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 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:
>
With two inputs and one output, there are 16 possible truth tables. 8 of
them are the inverse of the other 8, with the output inverted (AND
becomes NAND). Of these, some are degenerate (always true, or ignore A
or B), three are the familiar AND OR XOR. I don't know if the remainder
have names like that.

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 schrieb am Mittwoch, 21. September 2022 um 19:46:59 UTC+2:
> On Wednesday, September 21, 2022 at 3:06:09 AM UTC-4, Mostowski Collapse wrote:
> > 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.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

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!

Mostowski Collapse schrieb am Donnerstag, 22. September 2022 um 18:43:53 UTC+2:
> 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 schrieb am Mittwoch, 21. September 2022 um 19:46:59 UTC+2:
> > On Wednesday, September 21, 2022 at 3:06:09 AM UTC-4, Mostowski Collapse wrote:
> > > 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.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

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

Dan

On Thursday, September 22, 2022 at 12:50:11 PM UTC-4, Mostowski Collapse wrote:
> 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!
> Mostowski Collapse schrieb am Donnerstag, 22. September 2022 um 18:43:53 UTC+2:
> > 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 schrieb am Mittwoch, 21. September 2022 um 19:46:59 UTC+2:
> > > On Wednesday, September 21, 2022 at 3:06:09 AM UTC-4, Mostowski Collapse wrote:
> > > > 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.
> > > Dan
> > >
> > > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > > Visit my Math Blog at http://www.dcproof.wordpress.com

Dan Christensen 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 schrieb am Donnerstag, 22. September 2022 um 20:38:50 UTC+2:
> See my reply just now to your identical posting at sci.logic.
>
> Dan
> On Thursday, September 22, 2022 at 12:50:11 PM UTC-4, Mostowski Collapse wrote:
> > 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!
> > Mostowski Collapse schrieb am Donnerstag, 22. September 2022 um 18:43:53 UTC+2:
> > > 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 schrieb am Mittwoch, 21. September 2022 um 19:46:59 UTC+2:
> > > > On Wednesday, September 21, 2022 at 3:06:09 AM UTC-4, Mostowski Collapse wrote:
> > > > > 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.
> > > > Dan
> > > >
> > > > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > > > Visit my Math Blog at http://www.dcproof.wordpress.com

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

Dan

On Thursday, September 22, 2022 at 4:06:56 PM UTC-4, Mostowski Collapse wrote:
> 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.
>
[snip]

<yawn!>

On 9/21/2022 6:22 PM, Michael Moroney wrote:

> With two inputs and one output,
> there are 16 possible truth tables.
> 8 of them are the inverse of the other 8,
> with the output inverted (AND becomes NAND).
> Of these, some are degenerate
> (always true, or ignore A or B),
> three are the familiar AND OR XOR.
> I don't know if the remainder have
> names like that.

Let's see what we can do.

0 0 1 1 | A
0 1 0 1 | B
---------
0 0 0 0 | ⊥
0 0 0 1 | A ∧ B
0 0 1 0 | A ↛ B U+219B rightwards arrow with stroke
0 0 1 1 | A
0 1 0 0 | A ↚ B U+219A leftwards arrow with stroke
0 1 0 1 | B
0 1 1 0 | A ⊻ B
0 1 1 1 | A ∨ B
1 0 0 0 | A ⊽ B
1 0 0 1 | A ↔ B U+2194 left right arrow
1 0 1 0 | ¬ B
1 0 1 1 | A ← B U+2192 leftwards arrow
1 1 0 0 | ¬ A
1 1 0 1 | A → B U+2192 rightwards arrow
1 1 1 0 | A ⊼ B
1 1 1 1 | ⊤

I admit I picked these out of a Unicode character table:
0 0 1 0 | A ↛ B U+219B rightwards arrow with stroke
0 1 0 0 | A ↚ B U+219A leftwards arrow with stroke
1 0 1 1 | A ← B U+2192 leftwards arrow

But I think they're reasonable extensions.

I imagine the OP question is some version of
"Why not those, too?"

I think it is a matter of taste, not logic.

In English, anyway,
text flows left-to-right.
There is an implicit time-order on the page.

There is also an implicit time-order _in proving_
(though all truths proved are always true).
We know A
We know A → B
We know B

If there's something more complicated involved,
I don't know what it would be.

Nope, id_y | x is a bijection on x.
id_y isn't a bijection on x.

Whats wrong with you?

Dan Christensen schrieb:
> See my reply just now to your identical posting at sci.logic
>
> Dan
>
>
> On Thursday, September 22, 2022 at 4:06:56 PM UTC-4, Mostowski Collapse wrote:
>> 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.
>>
> [snip]
>
> <yawn!>
>

Can DC Spoiled prove that this function space, where B = {0,1}:

F = { f : B x B -> B }

Has cardinality 16? I doubt so.

Jim Burns schrieb am Donnerstag, 22. September 2022 um 23:24:15 UTC+2:
> On 9/21/2022 6:22 PM, Michael Moroney wrote:
>
> > With two inputs and one output,
> > there are 16 possible truth tables.
> > 8 of them are the inverse of the other 8,
> > with the output inverted (AND becomes NAND).
> > Of these, some are degenerate
> > (always true, or ignore A or B),
> > three are the familiar AND OR XOR.
> > I don't know if the remainder have
> > names like that.
> Let's see what we can do.
>
> 0 0 1 1 | A
> 0 1 0 1 | B
> ---------
> 0 0 0 0 | ⊥
> 0 0 0 1 | A ∧ B
> 0 0 1 0 | A ↛ B U+219B rightwards arrow with stroke
> 0 0 1 1 | A
> 0 1 0 0 | A ↚ B U+219A leftwards arrow with stroke
> 0 1 0 1 | B
> 0 1 1 0 | A ⊻ B
> 0 1 1 1 | A ∨ B
> 1 0 0 0 | A ⊽ B
> 1 0 0 1 | A ↔ B U+2194 left right arrow
> 1 0 1 0 | ¬ B
> 1 0 1 1 | A ← B U+2192 leftwards arrow
> 1 1 0 0 | ¬ A
> 1 1 0 1 | A → B U+2192 rightwards arrow
> 1 1 1 0 | A ⊼ B
> 1 1 1 1 | ⊤
>
> I admit I picked these out of a Unicode character table:
> 0 0 1 0 | A ↛ B U+219B rightwards arrow with stroke
> 0 1 0 0 | A ↚ B U+219A leftwards arrow with stroke
> 1 0 1 1 | A ← B U+2192 leftwards arrow
>
> But I think they're reasonable extensions.
>
>
> I imagine the OP question is some version of
> "Why not those, too?"
>
> I think it is a matter of taste, not logic.
>
> In English, anyway,
> text flows left-to-right.
> There is an implicit time-order on the page.
>
> There is also an implicit time-order _in proving_
> (though all truths proved are always true).
> We know A
> We know A → B
> We know B
>
> If there's something more complicated involved,
> I don't know what it would be.

In set theory its very easy, using set exponentiation:

| { f : B x B -> B } | = | B ^ (B x B) |

And since B and B x B are finite:

| B ^ (B x B) | = | B | ^ | (B x B) |

= | B | ^( | B | * | B | )

Now since | B | = 2 we have:

= 2 ^ (2 * 2) = 2 ^ 4 = 16

Mostowski Collapse schrieb am Donnerstag, 22. September 2022 um 23:55:14 UTC+2:
> Can DC Spoiled prove that this function space, where B = {0,1}:
>
> F = { f : B x B -> B }
>
> Has cardinality 16? I doubt so.
> Jim Burns schrieb am Donnerstag, 22. September 2022 um 23:24:15 UTC+2:
> > On 9/21/2022 6:22 PM, Michael Moroney wrote:
> >
> > > With two inputs and one output,
> > > there are 16 possible truth tables.
> > > 8 of them are the inverse of the other 8,
> > > with the output inverted (AND becomes NAND).
> > > Of these, some are degenerate
> > > (always true, or ignore A or B),
> > > three are the familiar AND OR XOR.
> > > I don't know if the remainder have
> > > names like that.
> > Let's see what we can do.
> >
> > 0 0 1 1 | A
> > 0 1 0 1 | B
> > ---------
> > 0 0 0 0 | ⊥
> > 0 0 0 1 | A ∧ B
> > 0 0 1 0 | A ↛ B U+219B rightwards arrow with stroke
> > 0 0 1 1 | A
> > 0 1 0 0 | A ↚ B U+219A leftwards arrow with stroke
> > 0 1 0 1 | B
> > 0 1 1 0 | A ⊻ B
> > 0 1 1 1 | A ∨ B
> > 1 0 0 0 | A ⊽ B
> > 1 0 0 1 | A ↔ B U+2194 left right arrow
> > 1 0 1 0 | ¬ B
> > 1 0 1 1 | A ← B U+2192 leftwards arrow
> > 1 1 0 0 | ¬ A
> > 1 1 0 1 | A → B U+2192 rightwards arrow
> > 1 1 1 0 | A ⊼ B
> > 1 1 1 1 | ⊤
> >
> > I admit I picked these out of a Unicode character table:
> > 0 0 1 0 | A ↛ B U+219B rightwards arrow with stroke
> > 0 1 0 0 | A ↚ B U+219A leftwards arrow with stroke
> > 1 0 1 1 | A ← B U+2192 leftwards arrow
> >
> > But I think they're reasonable extensions.
> >
> >
> > I imagine the OP question is some version of
> > "Why not those, too?"
> >
> > I think it is a matter of taste, not logic.
> >
> > In English, anyway,
> > text flows left-to-right.
> > There is an implicit time-order on the page.
> >
> > There is also an implicit time-order _in proving_
> > (though all truths proved are always true).
> > We know A
> > We know A → B
> > We know B
> >
> > If there's something more complicated involved,
> > I don't know what it would be.

On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski Collapse wrote:
> Can DC Spoiled prove that this function space, where B = {0,1}:
>
> F = { f : B x B -> B }
>
> Has cardinality 16?

I haven't done any work on the cardinalities of sets, finite or otherwise. Like ZFC, DC Proof has no axioms of cardinality.

Dan

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

Dan Christensen wrote:

> On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski Collapse
> wrote:
>> Can DC Spoiled prove that this function space, where B = {0,1}:
>> F = { f : B x B -> B } Has cardinality 16?
>
> I haven't done any work on the cardinalities of sets, finite or
> otherwise. Like ZFC, DC Proof has no axioms of cardinality.

you two are gays.

Dan Christensen wrote:

> On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski Collapse
> wrote:
>> Can DC Spoiled prove that this function space, where B = {0,1}:
>> F = { f : B x B -> B } Has cardinality 16?
>
> I haven't done any work on the cardinalities of sets, finite or
> otherwise. Like ZFC, DC Proof has no axioms of cardinality.

lol, Swiss hospitals advised to cut power use
https://www.rt.com/business/563275-swiss-hospitals-power-rationing/

khazar *fake_money* shithole completely fucked up.

“Hospitals are exempt from electricity and gas quotas but they can also
consider how they can reduce their electricity consumption,” Schwark told
the Swiss newspaper Neue Zürcher Zeitung on Wednesday.

He added that even in a critical industry such as pharmaceuticals, cost-
saving measures could be implemented without significant cuts to
production.

“Currently, almost every industry requires exceptions for itself. But this
is not the right approach. Rather, each industry should think about what
it can do in a critical situation,” Schwark said, adding: “This could be,
for example, reducing hours of work. If this will cut electricity
consumption by another 10-20%, we will postpone the introduction of
quotas.”

On Friday, September 23, 2022 at 2:19:29 PM UTC-4, Jeiker Carboni wrote:
> Dan Christensen wrote:
>
> > On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski Collapse
> > wrote:
> >> Can DC Spoiled prove that this function space, where B = {0,1}:
> >> F = { f : B x B -> B } Has cardinality 16?
> >
> > I haven't done any work on the cardinalities of sets, finite or
> > otherwise. Like ZFC, DC Proof has no axioms of cardinality.
> lol, Swiss hospitals advised to cut power use

Unlike your fellow countrymen, the Swiss are not fleeing their country in droves. You would be wise to join them, Nazi boy. Your boss's days are numbered. No doubt you will charged with war crimes for your work here, like the original Nazi propagandists.

Dan Christensen wrote:

> On Friday, September 23, 2022 at 2:19:29 PM UTC-4, Jeiker Carboni wrote:
>> Dan Christensen wrote:
>>
>> > On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski
>> > Collapse wrote:
>> >> Can DC Spoiled prove that this function space, where B = {0,1}:
>> >> F = { f : B x B -> B } Has cardinality 16?
>> >
>> > I haven't done any work on the cardinalities of sets, finite or
>> > otherwise. Like ZFC, DC Proof has no axioms of cardinality.
>> lol, Swiss hospitals advised to cut power use
>
> Unlike your fellow countrymen, the Swiss are not fleeing their country
> in

imbecile, nobody is fleeing anything. That's khakhole propaganda. Lol,
which only "works" on morons till they shut down the electric current,
lol. You nazi khakhole wankers are ready to die without khakhole
propaganda. Bwahahahahaaaa. Go get your damn vaccine!

Dan Christensen wrote:

> On Friday, September 23, 2022 at 2:19:29 PM UTC-4, Jeiker Carboni wrote:
>> Dan Christensen wrote:
>> > On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski
>> > Collapse wrote:
>> > I haven't done any work on the cardinalities of sets, finite or
>> > otherwise. Like ZFC, DC Proof has no axioms of cardinality.
>> lol, Swiss hospitals advised to cut power use
>
> Unlike your fellow countrymen, the Swiss are not fleeing their country
> in

idiots. They made you, you two khakhole nazis, to take vaccines on
*pure_propaganda*. Face it, you are idiots. Not my fault. LOL.

Dan Christensen wrote:

> On Friday, September 23, 2022 at 2:19:29 PM UTC-4, Jeiker Carboni wrote:
>> Dan Christensen wrote:
>> > On Thursday, September 22, 2022 at 5:55:14 PM UTC-4, Mostowski
>> > Collapse wrote:
>> > I haven't done any work on the cardinalities of sets, finite or
>> > otherwise. Like ZFC, DC Proof has no axioms of cardinality.
>> lol, Swiss hospitals advised to cut power use
>
> Unlike your fellow countrymen, the Swiss are not fleeing their country

lol, this khazar goy, gay actor on cocaine, named "president" zelenske, is
shocked Israel in NOT sending free weapons to the *nazis*. LOL.

Zelensky ‘shocked’ at Israel’s lack of support
https://www.rt.com/russia/563422-zelensky-shocked-israel-weapons/

"I am shocked,” Zelensky told a group of reporters from numerous French
media outlets. “I don't understand. Israel has provided us with nothing,
nothing, zero,"

this khakhole gay actor imbecile, forgets the production of weapons of war
are in his attributes. You suck big dicks going to war without weapons,
YOU made, or you PAID for.