Here, I introduce the term _vacuously false_. First, I review what I mean by vacuously true both in words and in formal predicate logic.

*** VACUOUSLY TRUE ***

In Words: If there are no P's , then it is _vacuously TRUE_ that every P is a Q.

We could also say that it also vacuously true that every P is NOT a Q.

Symbolically: ~EXIST(a):P(a) => ALL(a):[P(a) => Q(a)]

Or equivalently: ~EXIST(a):P(a) => ALL(a):[P(a) => ~Q(a)]

"(¬∃aPa → ∀a(Pa → Qa)) ↔ (¬∃aPa → ∀a(Pa → ¬Qa)) is valid."
https://www.umsu.de/trees/#(~3~7aPa~0~5~6a(Pa~5Qa))~4(~3~7aPa~0~5~6a(Pa~5~3Qa))

PROOF (using a form of natural deduction in my DC Proof system)

1 ~EXIST(a):P(a)
Premise

2 P(y)
Premise

3 ~Q(y)
Premise

4 EXIST(a):P(a)
E Gen, 2

5 EXIST(a):P(a) & ~EXIST(a):P(a)
Join, 4, 1

6 ~~Q(y)
Conclusion, 3

7 Q(y)
Rem DNeg, 6

8 ALL(a):[P(a) => Q(a)]
Conclusion, 2

9 ~EXIST(a):P(a) => ALL(a):[P(a) => Q(a)]
Conclusion, 1

*** VACUOUSLY FALSE (my term) ***

In Words: If there are no P's, then it is _vacuously FALSE_ that some P is a Q.

We could also say that it is vacuously false that some P is NOT a Q.

Symbolically: ~EXIST(a):P(a) => ~EXIST(a):[P(a) & Q(a)]

Or equivalently, ~EXIST(a):P(a) => ~EXIST(a):[P(a) & ~Q(a)]

"(¬∃aPa → ¬∃a(Pa ∧ Qa)) ↔ (¬∃aPa → ¬∃a(Pa ∧ ¬Qa)) is valid."
https://www.umsu.de/trees/#(~3~7aPa~0~5~3~7a(Pa~0~1~0Qa))~4(~3~7aPa~0~5~3~7a(Pa~0~1~0~3Qa))

PROOF

1 ~EXIST(a):P(a)
Premise

2 P(x) & Q(x)
Premise

3 P(x)
Split, 2

4 EXIST(a):P(a)
E Gen, 3

5 EXIST(a):P(a) & ~EXIST(a):P(a)
Join, 4, 1

6 ~EXIST(a):[P(a) & Q(a)]
Conclusion, 2

7 ~EXIST(a):P(a) => ~EXIST(a):[P(a) & Q(a)]
Conclusion, 1

Your comments?

Dan

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

Unplugging that worthless Canadian oaf of science and logic, for the creep only knows how to stalk,-- never think a clear thought on any science.

Re: Proof of Kepler Packing//Jan Burse-Alzheimer faggot//ETH Zurich, Dietmar Salamon, Martin Schweizer, Mete Soner, looking at it
by Dan Christensen Jul 9, 2017, 11:34:05 PM

Unplugging Dan Christensen and entire Univ Western Ontario uwo from sci.math as a demented racist defaming poster

John Woods, Mario Bunge, Ted Honderich, Brendan Myers, Joachim Lambek, Carrie Jenkins, Katalin Bimbo, Doug Walton.

Re: 1.1Dr. John Baez is a failed mathematician-physicist with his proton of 938MeV when it is 840MeV, electron= muon //his ellipse is a conic when it never was// as phony in math and physics as kibo Parry Moroney's ellipse and Christensen 10 OR 4 by Dan Christensen Sep 22, 2019, 9:54:06 AM



Re: A newsgroup like sci.math is a pile of shit when you have paid stalkers like Kibo Parry M. or Dan Christensen lording over sci.math as if he owns the place-- stalking and attacking posters 7-24-365. This is why I now post a roadmap to AP's newsgr
by Alan Mackenzie Jun 29, 2021, 2:36:04 PM

Univ Western Ontario math dept
Janusz Adamus, Tatyana  Barron,   Dan Christensen, Graham Denham, Ajneet Dhillon, Matthias  Franz, John Jardine, Massoud Khalkhali, Nicole Lemire, Jan Mináč, Victoria Olds, Martin Pinsonnault, Lex Renner, David Riley, Rasul Shafikov, Gordon Sinnamon

Chancellor Linda Hasenfratz
President Alan Shepard
Amit Chakma (chem engr)

Univ. Western Ontario physics dept
Pauline Barmby, Shantanu Basu, Peter Brown, Alex Buchel, Jan Cami, Margret Campbell-Brown, Blaine Chronik, Robert Cockcroft, John R. de Bruyn, Colin Denniston, Giovanni Fanchini, Sarah Gallagher, Lyudmila Goncharova, Wayne Hocking, Martin Houde, Jeffrey L. Hutter, Carol Jones, Stan Metchev, Silvia Mittler, Els Peeters, Robert Sica, Aaron Sigut, Peter Simpson, Mahi Singh, Paul Wiegert, Eugene Wong, Martin Zinke-Allmang

Re: 1.2Analbuttfuckmanure Justin Trudeau,Sophie Gregoire, says Dan Christensen or Kibo//Stephen Lecce, Jan Mináč, Victoria Olds, Martin Pinsonnault, Lex Renner, David Riley, Rasul Shafikov, Gordon Sinnamon
by Hoofington P. McSnort Aug 26, 2020, 7:01:28 PM

Re: Bill Blair, David Vigneault please help us put Dan Christensen into a Canadian Asylum or psychiatric treatment where he belongs
By Donald Trump Mar 30, 2020, 11:30:25 PM

Re: 2- Dan Christensen on failed math MIT Gilbert Strang, with his scatterbrained Calculus books, no geometry proof of Fundamental Theorem of Calculus, because Gilbert never knew
by Chris M. Thomasson 6:09 PM On 6/25/2021

Re: Dan Christensen says close all math, physics, chemistry dept. in ALL Canadian Colleges with their fake electron (muon is real electron), fake geometry proof of calculus (Old Math sums up 0 width rectangles), fake Lewis 8 Structure (dissociation...
by Giant Radioactive Easter Bunnie Jan 21, 2021, 5:38:37 PM

Re: Stewart failed Calculus also, and his book should be removed
by Dan Christensen
Jun 24, 2018, 10:52:49 PM

Re: Markus Klyver greatest math fool of Sweden-- thinks an ellipse is a conic section
by
Dan Christensen Aug 2, 2017, 11:52:21 PM

On Monday, October 21, 2019 at 1:29:49 PM UTC-5, Dan Christensen wrote:
>
> Are you ready, kids??? Bend over, er...
>
> Dan

Re: 81,045-Student victims of Rose M. Patten Univ Toronto from stalker Dan Christensen teaching 10 OR 2 = 12 with AND as subtraction, never a geometry proof of Fundamental Theorem of Calculus Univ Toronto, physics, Gordon F.. West, Michael B. Walker
by Frank Cassa 12Apr2021 7:00 AM 

Re: 7,744-Student victims of Linda Hasenfratz Univ Western Ontario from stalker Dan Christensen teaching 10 OR 2 = 12 with AND as subtraction, never a geometry proof of Fundamental Theorem of Calculus Chancellor Linda Hasenfratz President Alan Shepard
11:53 AM 10Apr2021
by Wayne Decarlo

Re: 102,852-Student victims of Dominic Barton, Univ Waterloo from stalker Dan Christensen teaching 10 OR 2 = 12 with AND as subtraction, never a geometry proof of Fundamental Theorem of Calculus Dominic Barton, President Feridun Hamdullahpur physics
by konyberg Apr 15, 2021, 3:09:41 PM 

Re: 176,232-Student Victims of Michael Meighen McGill Univ by Dan Christensen teaching 10 OR 2 = 12 with AND as subtraction, never a geometry proof of Fundamental Theorem of Calculus... 0.5MeV electron when in truth it is the muon as the real electron
by Dan Christensen Jul 2, 2021, 9:47:42 AM

Re: 135,568 Student victims Queen's Univ. James Leech, Arthur B. McDonald by Dan Christensen teaching 10 OR 2 = 12 with AND as subtraction, never a geometry proof of Fundamental Theorem of Calculus-- his mindless electron =0.5MeV when real electron of
May 10, 2021
by Professor Wordsmith

Re: 135,566 Student victims Queen's Univ. James Leech, Arthur B. McDonald by Dan Christensen teaching 10 OR 2 = 12 with AND as subtraction, never a geometry proof of Fundamental Theorem of Calculus-- his mindless electron =0.5MeV when real electron o
May 10, 2021
by Michael Moroney

Re: Proof of Kepler Packing//Jan Burse-Alzheimer faggot//ETH Zurich, Dietmar Salamon, Martin Schweizer, Mete Soner, looking at it
by Dan Christensen Jul 9, 2017, 11:34:05 PM

Re: *Fire the entire Univ Western Ontario math dept/ still teaching that the contradictory sine graph as sinusoid when it is really semicircle
by Dan Christensen Nov 21, 2017,

Re: Shit for brains in Logic Morons of logic George Boole, William Jevons, Bertrand Russell, Kurt Godel, Rudolf Carnap, Jan Burse& Dan Christensen as moronette cadets of Boole, Ludwig Wittgenstein, Willard Quine, Alfred North Whitehead, Irving Copi
by zelos...@gmail.com Mar 25, 2021, 12:55:15 AM 

Re: Spamming fuckdog Dan Christensen says fire Stanford's John Lipa, Alexander Fetter, William Little, Douglas Osheroff, David Ritson, H. Alan Schwettman, John Turneaure, Robert Wagoner, Stanley Wojcicki, Mason Yearian rather than let them brainwash our
by Mutt Buncher

Re: 2-fuckdog boston's Kibo Parry Moroney subhuman *stalkers* Dan Christensen, Betsy DeVos, NSF Dr. Panchanathan, barry shein's world std paying how much to stalk AP??
Earl Biggs

Re: Subhuman Kibo Parry Moroney says freedom of speech means freedom to stalk, liar, cheat, steal, kill// Dan Christensen, Uncle Al, NSF Dr. Panchanathan, Dept Educ Betsy DeVos, John Baez, Port563 Gilbert Strang, Barry Shein
Earl Biggs

                              ..
            .- " `-.   ,..-'''  ```.....'`-..
           ,      . `.'            '        `.
         .'   .' `    `           '   `..     ;
         .   ;  .'                     . `.    ;
         ;   . '                       `.  .   '
          . '                            ` `.  |
        . '.                                  '
       .          0              0            ' `.
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     ;                                            `
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    '.      `    ``..,                   .'   :'    '
     ;       `        ;'...          ..-''    '     '  I am Christensen,I am such a stupid insane imp of math and logic that I thought a vertex has a derivative, that distinct means not-distinct, and that 2 OR 1 = 3 with AND as subtraction. And I love spam reading of vvgra and tomato hello, and I am a failure of academics and so I spend most days making out hate-lists of people who actually succeed in science for my pea sized brain is envious and jealous of those who succeed in science yet I failed in science.
      `       `        ;  ````'''""'  ;      '    '
       `       `        ;            ;      '    '
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         /  `       `.                  '    '        .
        /     `       ..            ..'    .'"""""...'
       /   .`   `       ``.........-'     .'` .....'''
      / .'' ;     `                    .'   `
  ...'.'    ;    .' `                .'      `
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           ; .'   |      `. . . . ' .  \       `
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                  |    '     `      '     `.    `
                  `   '       `     ;       `.  |
                  `.'          `    ;         `-'
                                `...'

STUDENTS BEWARE: Don't be a victim of AP's fake math and science

On Friday, October 22, 2021 at 4:20:59 PM UTC-4, Archimedes Plutonium (AP) wrote:
> Unplugging that worthless...

AP is a malicious internet troll who wants only to mislead and confuse you. He may not be all there, but his fake math and science can only be meant to promote failure in schools. One can only guess at his motives.

In AP's OWN WORDS here:

“Primes do not exist, because the set they were borne from has no division.”
--June 29, 2020

“The last and largest finite number is 10^604.”
--June 3, 2015

“0 appears to be the last and largest finite number”
--June 9, 2015

“0/0 must be equal to 1.”
-- June 9, 2015

“0 is an infinite irrational number.”
--June 28, 2015

“No negative numbers exist.”
--December 22, 2018

“Rationals are not numbers.”
--May 18, 2019

According to AP's “chess board math,” an equilateral triangle is a right-triangle.
--December 11, 2019

Which could explain...

“The value of sin(45 degrees) = 1.”
--May 31, 2019

AP deliberately and repeatedly presented the truth table for OR as the truth table for AND:

“New Logic
AND
T & T = T
T & F = T
F & T = T
F & F = F”
--November 9, 2019

AP seeks aid of Russian agents to promote failure in schools:

"Please--Asking for help from Russia-- russian robots-- to create a new, true mathematics [sic]. What I like for the robots to do, is list every day, about 4 Colleges ( of the West) math dept, and ask why that math department is teaching false and fake math, and if unable to change to the correct true math, well, simply fire that math department until they can find professors who recognize truth in math from fakery...."
--November 9, 2017

And if that wasn't weird enough...

“The totality, everything that there is [the universe], is only 1 atom of plutonium [Pu]. There is nothing outside or beyond this one atom of plutonium.”
--April 4, 1994

“The Universe itself is one gigantic big atom.”
--November 14, 2019

AP's sinister Atom God Cult of Failure???

“Since God-Pu is marching on.
Glory! Glory! Atom Plutonium!
Its truth is marching on.
It has sounded forth the trumpet that shall never call retreat;
It is sifting out the hearts of people before its judgment seat;
Oh, be swift, my soul, to answer it; be jubilant, my feet!
Our God-Pu is marching on.”
--December 15, 2018 (Note: Pu is the atomic symbol for plutonium)

Dan

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

On Friday, October 22, 2021 at 3:52:40 PM UTC-4, Dan Christensen wrote:
> Here, I introduce the term _vacuously false_. First, I review what I mean by vacuously true both in words and in formal predicate logic.
>
> *** VACUOUSLY TRUE ***
>
> In Words: If there are no P's , then it is _vacuously TRUE_ that every P is a Q.
>
> We could also say that it also vacuously true that every P is NOT a Q.
>
> Symbolically: ~EXIST(a):P(a) => ALL(a):[P(a) => Q(a)]
>
> Or equivalently: ~EXIST(a):P(a) => ALL(a):[P(a) => ~Q(a)]
>
> "(¬∃aPa → ∀a(Pa → Qa)) ↔ (¬∃aPa → ∀a(Pa → ¬Qa)) is valid."
> https://www.umsu.de/trees/#(~3~7aPa~0~5~6a(Pa~5Qa))~4(~3~7aPa~0~5~6a(Pa~5~3Qa))
>
> PROOF (using a form of natural deduction in my DC Proof system)
>
> 1 ~EXIST(a):P(a)
> Premise
>
> 2 P(y)
> Premise
>
> 3 ~Q(y)
> Premise
>
> 4 EXIST(a):P(a)
> E Gen, 2
>
> 5 EXIST(a):P(a) & ~EXIST(a):P(a)
> Join, 4, 1
>
> 6 ~~Q(y)
> Conclusion, 3
>
> 7 Q(y)
> Rem DNeg, 6
>
> 8 ALL(a):[P(a) => Q(a)]
> Conclusion, 2
>
> 9 ~EXIST(a):P(a) => ALL(a):[P(a) => Q(a)]
> Conclusion, 1
>
> *** VACUOUSLY FALSE (my term) ***
>
> In Words: If there are no P's, then it is _vacuously FALSE_ that some P is a Q.
>
> We could also say that it is vacuously false that some P is NOT a Q.
>
> Symbolically: ~EXIST(a):P(a) => ~EXIST(a):[P(a) & Q(a)]
>
> Or equivalently, ~EXIST(a):P(a) => ~EXIST(a):[P(a) & ~Q(a)]
>
> "(¬∃aPa → ¬∃a(Pa ∧ Qa)) ↔ (¬∃aPa → ¬∃a(Pa ∧ ¬Qa)) is valid."
> https://www.umsu.de/trees/#(~3~7aPa~0~5~3~7a(Pa~0~1~0Qa))~4(~3~7aPa~0~5~3~7a(Pa~0~1~0~3Qa))
>
> PROOF
>
> 1 ~EXIST(a):P(a)
> Premise
>
> 2 P(x) & Q(x)
> Premise
>
> 3 P(x)
> Split, 2
>
> 4 EXIST(a):P(a)
> E Gen, 3
>
> 5 EXIST(a):P(a) & ~EXIST(a):P(a)
> Join, 4, 1
>
> 6 ~EXIST(a):[P(a) & Q(a)]
> Conclusion, 2
>
> 7 ~EXIST(a):P(a) => ~EXIST(a):[P(a) & Q(a)]
> Conclusion, 1
>

Note that if it is vacuously true that every P is a Q, then it is vacuously false that some P is NOT a Q.

Likewise, if it is vacuously false that some P is a Q, then it is vacuously true that every P is NOT a Q.

> Your comments?
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Suppose there are no P's. Then we also have:

It is vacuously TRUE that:

1) EVERY P is a Q
2) EVERY P is NOT a Q

It is vacuously FALSE that:

1) SOME P is a Q
2) SOME P is NOT a Q

Symbolically, we have:

~EXIST(a):P(a)

=> ALL(a):[P(a) => Q(a)]
& ALL(a):[P(a) => ~Q(a)]

& ~EXIST(a):[P(a) & Q(a)]
& ~EXIST(a):[P(a) & ~Q(a)]

Dan

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

Why do you think Russell would use it? If you
translate "the X is Y" to this here, i.e. "Some X is Y":

EXISTS(a):[X(a) & Y(a)]

You don't have Non-Contradiction anymore. You cannot
prove this here anymore for all cases of "the X is Y":

~("the X is Y" & "the X is ~Y")

On the other hand if you for example use EXIST(a):[ALL(b):[X(b) <=>
b=a] & Y(a)], then you can prove Non-Contradiction.

So your translation doesn't have anything to do with Russell.
Especially his requirement of Non-Contradiction in

all cases of "the X is Y" is violated.

Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> Suppose there are no P's. Then we also have:
>
> It is vacuously TRUE that:
>
> 1) EVERY P is a Q
> 2) EVERY P is NOT a Q
>
> It is vacuously FALSE that:
>
> 1) SOME P is a Q
> 2) SOME P is NOT a Q
>
> Symbolically, we have:
> ~EXIST(a):P(a)
>
> => ALL(a):[P(a) => Q(a)]
> & ALL(a):[P(a) => ~Q(a)]
>
> & ~EXIST(a):[P(a) & Q(a)]
> & ~EXIST(a):[P(a) & ~Q(a)]
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

On Saturday, October 23, 2021 at 7:20:07 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> > Suppose there are no P's. Then we also have:
> >
> > It is vacuously TRUE that:
> >
> > 1) EVERY P is a Q
> > 2) EVERY P is NOT a Q
> >
> > It is vacuously FALSE that:
> >
> > 1) SOME P is a Q
> > 2) SOME P is NOT a Q
> >
> > Symbolically, we have:
> > ~EXIST(a):P(a)
> >
> > => ALL(a):[P(a) => Q(a)]
> > & ALL(a):[P(a) => ~Q(a)]
> >
> > & ~EXIST(a):[P(a) & Q(a)]
> > & ~EXIST(a):[P(a) & ~Q(a)]

> Why do you think Russell would use it? If you
> translate "the X is Y" to this here, i.e. "Some X is Y":
>
> EXISTS(a):[X(a) & Y(a)]
>
> You don't have Non-Contradiction anymore.

Again, to use Russell's example, if there does not exist someone who is a present king of France, then there also does not exist someone who is both a present king of France AND bald. There would also not exist someone who is both a present king of France AND NOT bald. What makes you think there is contradiction here?

Dan

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

Well Russell has not only the King of France example.
You suggest to translate the King of France example to:

EXISTS(a):[X(a) & Y(a)]

But you could also translate it to this here:

False

You can also prove, in even much fewer lines:

~EXISTS(a):[X(a)] => ~False

Whats the point of translating it not to False? Like for
example to your EXISTS(a):[X(a) & Y(a)], if you are not

interested also in when EXISTS(a):[X(a)] ?

Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 19:54:49 UTC+2:
> On Saturday, October 23, 2021 at 7:20:07 AM UTC-4, Mostowski Collapse wrote:
>
> > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> > > Suppose there are no P's. Then we also have:
> > >
> > > It is vacuously TRUE that:
> > >
> > > 1) EVERY P is a Q
> > > 2) EVERY P is NOT a Q
> > >
> > > It is vacuously FALSE that:
> > >
> > > 1) SOME P is a Q
> > > 2) SOME P is NOT a Q
> > >
> > > Symbolically, we have:
> > > ~EXIST(a):P(a)
> > >
> > > => ALL(a):[P(a) => Q(a)]
> > > & ALL(a):[P(a) => ~Q(a)]
> > >
> > > & ~EXIST(a):[P(a) & Q(a)]
> > > & ~EXIST(a):[P(a) & ~Q(a)]
> > Why do you think Russell would use it? If you
> > translate "the X is Y" to this here, i.e. "Some X is Y":
> >
> > EXISTS(a):[X(a) & Y(a)]
> >
> > You don't have Non-Contradiction anymore.
> Again, to use Russell's example, if there does not exist someone who is a present king of France, then there also does not exist someone who is both a present king of France AND bald. There would also not exist someone who is both a present king of France AND NOT bald. What makes you think there is contradiction here?
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

I mean you extrapolate a formula, which works on much
more "data points" than the King is Bald, from a single
"data point" like the King is Bald,

and the insist on this formula. Thats quite a moronic
thing to do. This is like solving boundary value problem
by taking into account only a small boundary, or

by considering a problem that isn't solvable from the
boundary. I really wonder whats going on in your head
Dan-O-Matik? It seems you are loosing your marbles.

Russelll puts more boundary and value constraints,
by also postulating for example Non-Contraditctory
for all possible "data points", and he can therefore

convince his readers. Your formula that you pulled
out of your ass, and that matches only one "data point"
is probably the worst machine learning example

ever done on this planet.

LoL

Mostowski Collapse schrieb am Samstag, 23. Oktober 2021 um 21:32:07 UTC+2:
> Well Russell has not only the King of France example.
> You suggest to translate the King of France example to:
>
> EXISTS(a):[X(a) & Y(a)]
>
> But you could also translate it to this here:
>
> False
>
> You can also prove, in even much fewer lines:
>
> ~EXISTS(a):[X(a)] => ~False
>
> Whats the point of translating it not to False? Like for
> example to your EXISTS(a):[X(a) & Y(a)], if you are not
>
> interested also in when EXISTS(a):[X(a)] ?
> Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 19:54:49 UTC+2:
> > On Saturday, October 23, 2021 at 7:20:07 AM UTC-4, Mostowski Collapse wrote:
> >
> > > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> > > > Suppose there are no P's. Then we also have:
> > > >
> > > > It is vacuously TRUE that:
> > > >
> > > > 1) EVERY P is a Q
> > > > 2) EVERY P is NOT a Q
> > > >
> > > > It is vacuously FALSE that:
> > > >
> > > > 1) SOME P is a Q
> > > > 2) SOME P is NOT a Q
> > > >
> > > > Symbolically, we have:
> > > > ~EXIST(a):P(a)
> > > >
> > > > => ALL(a):[P(a) => Q(a)]
> > > > & ALL(a):[P(a) => ~Q(a)]
> > > >
> > > > & ~EXIST(a):[P(a) & Q(a)]
> > > > & ~EXIST(a):[P(a) & ~Q(a)]
> > > Why do you think Russell would use it? If you
> > > translate "the X is Y" to this here, i.e. "Some X is Y":
> > >
> > > EXISTS(a):[X(a) & Y(a)]
> > >
> > > You don't have Non-Contradiction anymore.
> > Again, to use Russell's example, if there does not exist someone who is a present king of France, then there also does not exist someone who is both a present king of France AND bald. There would also not exist someone who is both a present king of France AND NOT bald. What makes you think there is contradiction here?
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

On Saturday, October 23, 2021 at 3:51:27 PM UTC-4, Mostowski Collapse wrote:

> > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 19:54:49 UTC+2:
> > > On Saturday, October 23, 2021 at 7:20:07 AM UTC-4, Mostowski Collapse wrote:
> > >
> > > > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> > > > > Suppose there are no P's. Then we also have:
> > > > >
> > > > > It is vacuously TRUE that:
> > > > >
> > > > > 1) EVERY P is a Q
> > > > > 2) EVERY P is NOT a Q
> > > > >
> > > > > It is vacuously FALSE that:
> > > > >
> > > > > 1) SOME P is a Q
> > > > > 2) SOME P is NOT a Q
> > > > >
> > > > > Symbolically, we have:
> > > > > ~EXIST(a):P(a)
> > > > >
> > > > > => ALL(a):[P(a) => Q(a)]
> > > > > & ALL(a):[P(a) => ~Q(a)]
> > > > >
> > > > > & ~EXIST(a):[P(a) & Q(a)]
> > > > > & ~EXIST(a):[P(a) & ~Q(a)]

> > > > Why do you think Russell would use it? If you
> > > > translate "the X is Y" to this here, i.e. "Some X is Y":
> > > >
> > > > EXISTS(a):[X(a) & Y(a)]
> > > >
> > > > You don't have Non-Contradiction anymore.

> > > Again, to use Russell's example, if there does not exist someone who is a present king of France, then there also does not exist someone who is both a present king of France AND bald. There would also not exist someone who is both a present king of France AND NOT bald. What makes you think there is contradiction here?

You haven't answered this question.

> Mostowski Collapse schrieb am Samstag, 23. Oktober 2021 um 21:32:07 UTC+2:
> > Well Russell has not only the King of France example.
> > You suggest to translate the King of France example to:
> >
> > EXISTS(a):[X(a) & Y(a)]
> >

It works.

> > But you could also translate it to this here:
> >
> > False
> >

You are making no sense here.

> > You can also prove, in even much fewer lines:
> >
> > ~EXISTS(a):[X(a)] => ~False
> >

You start the proof with a premise of the form ~EXISTS(a):[X(a)]. Then you introduce another predicate Y, and derive ~EXIST(a):[X(a) & Y(a)]. You could also as easily derive ~EXIST(a):[X(a) & ~Y(a)] (with a negation of the Y(a) term). There is no inconsistency here. That is main point of the exercise AFAICT.

I don't pretend this is an important result. It's the sort of exercise you might find in introductory textbooks. Inserting a term for uniqueness simply adds an annoying distraction. Dealing with it provides no new insight.

Dan

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

You can derive a lot from ~EXISTS(a):[X(a)],

for example:

~EXISTS(a):[X(a)] => ~EXISTS(a):[X(a) & a proved the Goldbach Conjecture & B(a)]

Its not related in any way to Russells "the X is Y".

Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 23:42:12 UTC+2:
> On Saturday, October 23, 2021 at 3:51:27 PM UTC-4, Mostowski Collapse wrote:
>
> > > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 19:54:49 UTC+2:
> > > > On Saturday, October 23, 2021 at 7:20:07 AM UTC-4, Mostowski Collapse wrote:
> > > >
> > > > > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> > > > > > Suppose there are no P's. Then we also have:
> > > > > >
> > > > > > It is vacuously TRUE that:
> > > > > >
> > > > > > 1) EVERY P is a Q
> > > > > > 2) EVERY P is NOT a Q
> > > > > >
> > > > > > It is vacuously FALSE that:
> > > > > >
> > > > > > 1) SOME P is a Q
> > > > > > 2) SOME P is NOT a Q
> > > > > >
> > > > > > Symbolically, we have:
> > > > > > ~EXIST(a):P(a)
> > > > > >
> > > > > > => ALL(a):[P(a) => Q(a)]
> > > > > > & ALL(a):[P(a) => ~Q(a)]
> > > > > >
> > > > > > & ~EXIST(a):[P(a) & Q(a)]
> > > > > > & ~EXIST(a):[P(a) & ~Q(a)]
>
> > > > > Why do you think Russell would use it? If you
> > > > > translate "the X is Y" to this here, i.e. "Some X is Y":
> > > > >
> > > > > EXISTS(a):[X(a) & Y(a)]
> > > > >
> > > > > You don't have Non-Contradiction anymore.
>
> > > > Again, to use Russell's example, if there does not exist someone who is a present king of France, then there also does not exist someone who is both a present king of France AND bald. There would also not exist someone who is both a present king of France AND NOT bald. What makes you think there is contradiction here?
> You haven't answered this question.
> > Mostowski Collapse schrieb am Samstag, 23. Oktober 2021 um 21:32:07 UTC+2:
> > > Well Russell has not only the King of France example.
> > > You suggest to translate the King of France example to:
> > >
> > > EXISTS(a):[X(a) & Y(a)]
> > >
> It works.
> > > But you could also translate it to this here:
> > >
> > > False
> > >
> You are making no sense here.
> > > You can also prove, in even much fewer lines:
> > >
> > > ~EXISTS(a):[X(a)] => ~False
> > >
> You start the proof with a premise of the form ~EXISTS(a):[X(a)]. Then you introduce another predicate Y, and derive ~EXIST(a):[X(a) & Y(a)]. You could also as easily derive ~EXIST(a):[X(a) & ~Y(a)] (with a negation of the Y(a) term). There is no inconsistency here. That is main point of the exercise AFAICT.
>
> I don't pretend this is an important result. It's the sort of exercise you might find in introductory textbooks. Inserting a term for uniqueness simply adds an annoying distraction. Dealing with it provides no new insight.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

You only hit a "data point" where Modern "Some X is Y"
and Russells "The X is Y" agree. But otherwise its irrelevant.

Maybe you have some mental issue with Russells "The X is Y".
Like you cannot relate to the uniqueness quantifier?

Maybe this T-Shirt can cheer you up?

Interactive theorem proving as a beach sport
https://twitter.com/AngelikiKoutso1/status/1442931588006768643

Mostowski Collapse schrieb am Sonntag, 24. Oktober 2021 um 00:15:00 UTC+2:
> You can derive a lot from ~EXISTS(a):[X(a)],
>
> for example:
>
> ~EXISTS(a):[X(a)] => ~EXISTS(a):[X(a) & a proved the Goldbach Conjecture & B(a)]
>
> Its not related in any way to Russells "the X is Y".
> Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 23:42:12 UTC+2:
> > On Saturday, October 23, 2021 at 3:51:27 PM UTC-4, Mostowski Collapse wrote:
> >
> > > > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 19:54:49 UTC+2:
> > > > > On Saturday, October 23, 2021 at 7:20:07 AM UTC-4, Mostowski Collapse wrote:
> > > > >
> > > > > > Dan Christensen schrieb am Samstag, 23. Oktober 2021 um 06:08:10 UTC+2:
> > > > > > > Suppose there are no P's. Then we also have:
> > > > > > >
> > > > > > > It is vacuously TRUE that:
> > > > > > >
> > > > > > > 1) EVERY P is a Q
> > > > > > > 2) EVERY P is NOT a Q
> > > > > > >
> > > > > > > It is vacuously FALSE that:
> > > > > > >
> > > > > > > 1) SOME P is a Q
> > > > > > > 2) SOME P is NOT a Q
> > > > > > >
> > > > > > > Symbolically, we have:
> > > > > > > ~EXIST(a):P(a)
> > > > > > >
> > > > > > > => ALL(a):[P(a) => Q(a)]
> > > > > > > & ALL(a):[P(a) => ~Q(a)]
> > > > > > >
> > > > > > > & ~EXIST(a):[P(a) & Q(a)]
> > > > > > > & ~EXIST(a):[P(a) & ~Q(a)]
> >
> > > > > > Why do you think Russell would use it? If you
> > > > > > translate "the X is Y" to this here, i.e. "Some X is Y":
> > > > > >
> > > > > > EXISTS(a):[X(a) & Y(a)]
> > > > > >
> > > > > > You don't have Non-Contradiction anymore.
> >
> > > > > Again, to use Russell's example, if there does not exist someone who is a present king of France, then there also does not exist someone who is both a present king of France AND bald. There would also not exist someone who is both a present king of France AND NOT bald. What makes you think there is contradiction here?
> > You haven't answered this question.
> > > Mostowski Collapse schrieb am Samstag, 23. Oktober 2021 um 21:32:07 UTC+2:
> > > > Well Russell has not only the King of France example.
> > > > You suggest to translate the King of France example to:
> > > >
> > > > EXISTS(a):[X(a) & Y(a)]
> > > >
> > It works.
> > > > But you could also translate it to this here:
> > > >
> > > > False
> > > >
> > You are making no sense here.
> > > > You can also prove, in even much fewer lines:
> > > >
> > > > ~EXISTS(a):[X(a)] => ~False
> > > >
> > You start the proof with a premise of the form ~EXISTS(a):[X(a)]. Then you introduce another predicate Y, and derive ~EXIST(a):[X(a) & Y(a)]. You could also as easily derive ~EXIST(a):[X(a) & ~Y(a)] (with a negation of the Y(a) term). There is no inconsistency here. That is main point of the exercise AFAICT.
> >
> > I don't pretend this is an important result. It's the sort of exercise you might find in introductory textbooks. Inserting a term for uniqueness simply adds an annoying distraction. Dealing with it provides no new insight..
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

On Saturday, October 23, 2021 at 6:19:08 PM UTC-4, Mostowski Collapse wrote:
> You only hit a "data point" where Modern "Some X is Y"
> and Russells "The X is Y" agree. But otherwise its irrelevant.
>
> Maybe you have some mental issue with Russells "The X is Y".
> Like you cannot relate to the uniqueness quantifier?
>

What theorem, expressed in FOL (i.e. no words), do you think Russell proved in this case?

Dan

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

Come one, its quite easy! He proved:

~("the X is Y" & "the X is ~Y")

If you translate "the X is Y" as EXIST(a):[ALL(b):
[X(b) <=> b=a] & Y(a)], then he proved:

~(EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)] &
EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)])

But you can also prove it with other translations of
"the X is Y", like for example what MSE shows.

Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 02:38:43 UTC+2:
> On Saturday, October 23, 2021 at 6:19:08 PM UTC-4, Mostowski Collapse wrote:
> > You only hit a "data point" where Modern "Some X is Y"
> > and Russells "The X is Y" agree. But otherwise its irrelevant.
> >
> > Maybe you have some mental issue with Russells "The X is Y".
> > Like you cannot relate to the uniqueness quantifier?
> >
> What theorem, expressed in FOL (i.e. no words), do you think Russell proved in this case?
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Russsells "the X is Y":

¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid.
https://www.umsu.de/trees

Dams "Some X is Y"

¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
https://www.umsu.de/trees

Mostowski Collapse schrieb am Sonntag, 24. Oktober 2021 um 11:16:02 UTC+2:
> Come one, its quite easy! He proved:
> ~("the X is Y" & "the X is ~Y")
> If you translate "the X is Y" as EXIST(a):[ALL(b):
> [X(b) <=> b=a] & Y(a)], then he proved:
>
> ~(EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)] &
> EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)])
>
> But you can also prove it with other translations of
> "the X is Y", like for example what MSE shows.
> Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 02:38:43 UTC+2:
> > On Saturday, October 23, 2021 at 6:19:08 PM UTC-4, Mostowski Collapse wrote:
> > > You only hit a "data point" where Modern "Some X is Y"
> > > and Russells "The X is Y" agree. But otherwise its irrelevant.
> > >
> > > Maybe you have some mental issue with Russells "The X is Y".
> > > Like you cannot relate to the uniqueness quantifier?
> > >
> > What theorem, expressed in FOL (i.e. no words), do you think Russell proved in this case?
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

On Sunday, October 24, 2021 at 5:16:02 AM UTC-4, Mostowski Collapse wrote:
> Come one, its quite easy! He proved:
> ~("the X is Y" & "the X is ~Y")
> If you translate "the X is Y" as EXIST(a):[ALL(b):
> [X(b) <=> b=a] & Y(a)], then he proved:
>
> ~(EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)] &
> EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)])
>

That's it??? If so, I wonder why Russell would bother to mention it. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. Using DC Proof, we have:

1 EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)]
& EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)]
Premise

2 EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)]
Split, 1

3 EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)]
Split, 1

4 ALL(b):[X(b) <=> b=t] & Y(t)
E Spec, 2

5 ALL(b):[X(b) <=> b=t]
Split, 4

6 Y(t)
Split, 4

11 X(t)
Detach, 9, 10

12 ALL(b):[X(b) <=> b=u] & ~Y(u)
E Spec, 3

13 ALL(b):[X(b) <=> b=u]
Split, 12

14 ~Y(u)
Split, 12

15 X(t) <=> t=u
U Spec, 13

16 [X(t) => t=u] & [t=u => X(t)]
Iff-And, 15

17 X(t) => t=u
Split, 16

18 t=u
Detach, 17, 11

19 ~Y(t)
Substitute, 18, 14

20 Y(t) & ~Y(t)
Join, 6, 19

21 ~[EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)]
& EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)]]
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

what is the inverse of Vacuously TRUE ?

Not Vacuously TRUE ? Totally True ? or Vacuously FALSE ?

On 10/22/2021 2:52 PM, Dan Christensen wrote:
> Here, I introduce the term _vacuously false_. First, I review what I mean by vacuously true both in words and in formal predicate logic.
>
> *** VACUOUSLY TRUE ***
>
> In Words: If there are no P's , then it is _vacuously TRUE_ that every P is a Q.
>
> We could also say that it also vacuously true that every P is NOT a Q.
>
> Symbolically: ~EXIST(a):P(a) => ALL(a):[P(a) => Q(a)]
>
> Or equivalently: ~EXIST(a):P(a) => ALL(a):[P(a) => ~Q(a)]
>
> "(¬∃aPa → ∀a(Pa → Qa)) ↔ (¬∃aPa → ∀a(Pa → ¬Qa)) is valid."
> https://www.umsu.de/trees/#(~3~7aPa~0~5~6a(Pa~5Qa))~4(~3~7aPa~0~5~6a(Pa~5~3Qa))
>
> PROOF (using a form of natural deduction in my DC Proof system)
>
> 1 ~EXIST(a):P(a)
> Premise
>
> 2 P(y)
> Premise
>
> 3 ~Q(y)
> Premise
>
> 4 EXIST(a):P(a)
> E Gen, 2
>
> 5 EXIST(a):P(a) & ~EXIST(a):P(a)
> Join, 4, 1
>
> 6 ~~Q(y)
> Conclusion, 3
>
> 7 Q(y)
> Rem DNeg, 6
>
> 8 ALL(a):[P(a) => Q(a)]
> Conclusion, 2
>
> 9 ~EXIST(a):P(a) => ALL(a):[P(a) => Q(a)]
> Conclusion, 1
>
> *** VACUOUSLY FALSE (my term) ***
>
> In Words: If there are no P's, then it is _vacuously FALSE_ that some P is a Q.
>
> We could also say that it is vacuously false that some P is NOT a Q.
>
> Symbolically: ~EXIST(a):P(a) => ~EXIST(a):[P(a) & Q(a)]
>
> Or equivalently, ~EXIST(a):P(a) => ~EXIST(a):[P(a) & ~Q(a)]
>
> "(¬∃aPa → ¬∃a(Pa ∧ Qa)) ↔ (¬∃aPa → ¬∃a(Pa ∧ ¬Qa)) is valid."
> https://www.umsu.de/trees/#(~3~7aPa~0~5~3~7a(Pa~0~1~0Qa))~4(~3~7aPa~0~5~3~7a(Pa~0~1~0~3Qa))
>
> PROOF
>
> 1 ~EXIST(a):P(a)
> Premise
>
> 2 P(x) & Q(x)
> Premise
>
> 3 P(x)
> Split, 2
>
> 4 EXIST(a):P(a)
> E Gen, 3
>
> 5 EXIST(a):P(a) & ~EXIST(a):P(a)
> Join, 4, 1
>
> 6 ~EXIST(a):[P(a) & Q(a)]
> Conclusion, 2
>
> 7 ~EXIST(a):P(a) => ~EXIST(a):[P(a) & Q(a)]
> Conclusion, 1
>
> Your comments?
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com
>

On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> Russsells "the X is Y":
>
> ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid.
> https://www.umsu.de/trees
>

I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???

> Dan's "Some X is Y"
>
> ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> https://www.umsu.de/trees

You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)

¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))

We also have:

¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)

I hope this helps.

Dan

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

On Sunday, October 24, 2021 at 12:04:36 PM UTC-4, Serg io wrote:
> what is the inverse of Vacuously TRUE ?
>

If ~EXIST(a):X(a), then ALL(a):[X(a) => Y(a)] is said be vacuously TRUE, and EXIST(a):[X(a) & Y(a)] is said to be vacuously FALSE.

I'm not sure that NOT vacuously true is a meaningful or useful notion.

Dan

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

No I didn't forget anything. You asked what is the FOL formulation
of the Russell requirement. The Russell requirement wants a
translation of "the X is Y" which satisfies the folllowing

~("the X is Y" & "the X is ~Y")

for any X and Y. If you translate "the X is Y" to EXIST(a):(X(a) & Y(a))
the Russell requirement is not anymore satisfied for any X and Y.
Even this online prover knows that fact

> Dan's "Some X is Y"
>
> ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> https://www.umsu.de/trees

If you prove on the other hand:

¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
https://www.umsu.de/trees

Then you don't deal with arbitrary X and Y anymore, and pick
some X and Y, some where the you have nevertheless Russells
requirement. But Russells requirement is not a synthetic

requirement which is supposed to be satisified sometimes.
He doesn't say there "are some X and Y" such that ~("the X is Y"
& "the X is ~Y"). He has it that he wants ~("the X is Y"

& "the X is ~Y") for arbirary X and Y. So that it becomes an
analytic truth, which holds always, irrespective of the situation.
Russell doesn't make the King is Bald example to

illustrate his analytic truth, he makes it only to illustrate
that "the X is Y" is false for the King is Bald example.

Got it?

Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 18:07:14 UTC+2:
> On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> > Russsells "the X is Y":
> >
> > ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid..
> > https://www.umsu.de/trees
> >
> I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???
>
> > Dan's "Some X is Y"
> >
> > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > https://www.umsu.de/trees
> You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)
>
> ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))
>
> We also have:
>
> ¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
> https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)
>
> I hope this helps.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

The material implication acts as a situation filter.
And weakens the theorem to those situations that
pass the material implication filter.

If you have a theorem:

.... X ... Y ... (B)

The meaning is for all X and Y. If on the other
hand if you prove a different theorem:

..... X ... Y ... (A) => ... X ... Y ... (B)

Then you prove (B) only for those X and Y that
satisfy (A). You don't prove the full scope of (B)
anymore. Predicate symbols/variables are universally

quantified in FOL theorems. One might thing this
leaves not much room to restrict predicate symbols/
variables in a theorem. But material implication

exactly does that. It can focus on a subset of possibly
X and Y configurations, a subset that can be characterized
via a first order formula itself.

Russells requirement is without such a situation filter.

Mostowski Collapse schrieb am Montag, 25. Oktober 2021 um 15:23:22 UTC+2:
> No I didn't forget anything. You asked what is the FOL formulation
> of the Russell requirement. The Russell requirement wants a
> translation of "the X is Y" which satisfies the folllowing
> ~("the X is Y" & "the X is ~Y")
> for any X and Y. If you translate "the X is Y" to EXIST(a):(X(a) & Y(a))
> the Russell requirement is not anymore satisfied for any X and Y.
> Even this online prover knows that fact
> > Dan's "Some X is Y"
> >
> > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > https://www.umsu.de/trees
> If you prove on the other hand:
> ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> https://www.umsu.de/trees
> Then you don't deal with arbitrary X and Y anymore, and pick
> some X and Y, some where the you have nevertheless Russells
> requirement. But Russells requirement is not a synthetic
>
> requirement which is supposed to be satisified sometimes.
> He doesn't say there "are some X and Y" such that ~("the X is Y"
> & "the X is ~Y"). He has it that he wants ~("the X is Y"
>
> & "the X is ~Y") for arbirary X and Y. So that it becomes an
> analytic truth, which holds always, irrespective of the situation.
> Russell doesn't make the King is Bald example to
>
> illustrate his analytic truth, he makes it only to illustrate
> that "the X is Y" is false for the King is Bald example.
>
> Got it?
> Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 18:07:14 UTC+2:
> > On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> > > Russsells "the X is Y":
> > >
> > > ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid..
> > > https://www.umsu.de/trees
> > >
> > I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???
> >
> > > Dan's "Some X is Y"
> > >
> > > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > > https://www.umsu.de/trees
> > You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)
> >
> > ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))
> >
> > We also have:
> >
> > ¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)
> >
> > I hope this helps.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

One study in mathematical logic reveals a further
relationship between logic, situation filtering and
material implication. The meta logical theorem is

the so called "Deduction theorem". It sheds some
light on situation filter via the material implication.
Namely the "Deduction theorem" reads:

if G, A |- B then G |- A => B

The other direction is Modus Ponens:

if G |- A => B then G, A |- B

In certain logics the "Deduction theorem works also
unrestricted, i.e. A and B don't need to be closed
formulas. If you put together the "Deduction Theorem"

and "Modus Ponens" you get:

G, A |- B <=> G |- A => B

So basically material implication acts like supposition
of premisses. Proving A => B is the same as assuming
the formula A and then proving B.

So when Slow-O-Matik proves:

¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
https://www.umsu.de/trees

This is the same as proving:

¬∃aXa |- ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya))

So Dan-O-Matik make the supposition aka assumption
that ¬∃aXa. But concerning his law of non-contradiction
Russell isn't interessted in any supposition aka assumption,

he wants a translation of "the X is Y" that works unconditionally,
and satisfies ~("the X is Y" & "the X is ~Y") unconditionally,
without the need for some supposition aka assumption.

Mostowski Collapse schrieb am Montag, 25. Oktober 2021 um 15:37:46 UTC+2:
> The material implication acts as a situation filter.
> And weakens the theorem to those situations that
> pass the material implication filter.
>
> If you have a theorem:
>
> ... X ... Y ... (B)
>
> The meaning is for all X and Y. If on the other
> hand if you prove a different theorem:
>
> .... X ... Y ... (A) => ... X ... Y ... (B)
>
> Then you prove (B) only for those X and Y that
> satisfy (A). You don't prove the full scope of (B)
> anymore. Predicate symbols/variables are universally
>
> quantified in FOL theorems. One might thing this
> leaves not much room to restrict predicate symbols/
> variables in a theorem. But material implication
>
> exactly does that. It can focus on a subset of possibly
> X and Y configurations, a subset that can be characterized
> via a first order formula itself.
>
> Russells requirement is without such a situation filter.
> Mostowski Collapse schrieb am Montag, 25. Oktober 2021 um 15:23:22 UTC+2:
> > No I didn't forget anything. You asked what is the FOL formulation
> > of the Russell requirement. The Russell requirement wants a
> > translation of "the X is Y" which satisfies the folllowing
> > ~("the X is Y" & "the X is ~Y")
> > for any X and Y. If you translate "the X is Y" to EXIST(a):(X(a) & Y(a))
> > the Russell requirement is not anymore satisfied for any X and Y.
> > Even this online prover knows that fact
> > > Dan's "Some X is Y"
> > >
> > > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > > https://www.umsu.de/trees
> > If you prove on the other hand:
> > ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees
> > Then you don't deal with arbitrary X and Y anymore, and pick
> > some X and Y, some where the you have nevertheless Russells
> > requirement. But Russells requirement is not a synthetic
> >
> > requirement which is supposed to be satisified sometimes.
> > He doesn't say there "are some X and Y" such that ~("the X is Y"
> > & "the X is ~Y"). He has it that he wants ~("the X is Y"
> >
> > & "the X is ~Y") for arbirary X and Y. So that it becomes an
> > analytic truth, which holds always, irrespective of the situation.
> > Russell doesn't make the King is Bald example to
> >
> > illustrate his analytic truth, he makes it only to illustrate
> > that "the X is Y" is false for the King is Bald example.
> >
> > Got it?
> > Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 18:07:14 UTC+2:
> > > On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> > > > Russsells "the X is Y":
> > > >
> > > > ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid.
> > > > https://www.umsu.de/trees
> > > >
> > > I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???
> > >
> > > > Dan's "Some X is Y"
> > > >
> > > > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > > > https://www.umsu.de/trees
> > > You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)
> > >
> > > ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> > > https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))
> > >
> > > We also have:
> > >
> > > ¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
> > > https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)
> > >
> > > I hope this helps.
> > > Dan
> > >
> > > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > > Visit my Math Blog at http://www.dcproof.wordpress.com

On Monday, October 25, 2021 at 9:23:22 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 18:07:14 UTC+2:
> > On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> > > Russsells "the X is Y":
> > >
> > > ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid..
> > > https://www.umsu.de/trees
> > >
> > I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???
> >
> > > Dan's "Some X is Y"
> > >
> > > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > > https://www.umsu.de/trees
> > You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)
> >
> > ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))
> >
> > We also have:
> >
> > ¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)
> >
> > I hope this helps.

> No I didn't forget anything. You asked what is the FOL formulation
> of the Russell requirement. The Russell requirement wants a
> translation of "the X is Y" which satisfies the folllowing
> ~("the X is Y" & "the X is ~Y")

You mean, ~[EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)] & EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)]], a trivial result as it turns out (see my proof above). Again, if you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y.

If, on the other hand, as in Russell king-of-France example, you have ~EXIST(a):K(a), then you can have both ~EXIST(a):[K(a) & B(a))] and ~EXIST(a):[K(a) & B(a))]. Or equivalently, ALL(a):[K(a) => B(a)] and ALL(a):[K(a) => ~B(a)]. This is not a contradiction since the antecedent K(a) is assumed to be false in both cases. See https://en.wikipedia.org/wiki/Vacuous_truth

Dan

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

On Monday, October 25, 2021 at 9:23:22 AM UTC-4, Mostowski Collapse wrote:

> Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 18:07:14 UTC+2:
> > On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> > > Russsells "the X is Y":
> > >
> > > ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid..
> > > https://www.umsu.de/trees
> > >
> > I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???
> >
> > > Dan's "Some X is Y"
> > >
> > > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > > https://www.umsu.de/trees
> > You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)
> >
> > ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))
> >
> > We also have:
> >
> > ¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
> > https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)
> >
> > I hope this helps.

> No I didn't forget anything. You asked what is the FOL formulation
> of the Russell requirement. The Russell requirement wants a
> translation of "the X is Y" which satisfies the folllowing
> ~("the X is Y" & "the X is ~Y")

You mean,

~[EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)] & EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)]]

A trivial result as it turns out (see my proof above). Again, if you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y.

If, on the other hand, as in Russell's king-of-France example, you have ~EXIST(a):K(a), then you can have both ~EXIST(a):[K(a) & ~B(a))] and ~EXIST(a):[K(a) & B(a))]. Or equivalently, ALL(a):[K(a) => B(a)] and ALL(a):[K(a) => ~B(a)]. This is not a contradiction since the antecedent K(a) is assumed to be false in both cases. See https://en.wikipedia.org/wiki/Vacuous_truth

Dan

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

Dan-O-Matik halucinated:
> If, on the other hand, as in Russell's king-of-France example, you
have ~EXIST(a):K(a), then you can have both ~EXIST(a):[K(a) & ~B(a))]
and ~EXIST(a):[K(a) & B(a))].

It has nothing to do with Russell, since Russell doesn't
translate "the X is Y" into EXIST(a):(X(a) & Y(a)).

Dan Christensen schrieb am Montag, 25. Oktober 2021 um 16:21:23 UTC+2:
> On Monday, October 25, 2021 at 9:23:22 AM UTC-4, Mostowski Collapse wrote:
>
> > Dan Christensen schrieb am Sonntag, 24. Oktober 2021 um 18:07:14 UTC+2:
> > > On Sunday, October 24, 2021 at 5:20:45 AM UTC-4, Mostowski Collapse wrote:
> > > > Russsells "the X is Y":
> > > >
> > > > ¬(∃a(∀b(Xb ↔ b=a) ∧ Ya) ∧ ∃a(∀b(Xb ↔ b=a) ∧ ¬Ya)) is valid.
> > > > https://www.umsu.de/trees
> > > >
> > > I posted my formal proof here just now. If you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y. So what???
> > >
> > > > Dan's "Some X is Y"
> > > >
> > > > ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is invalid.
> > > > https://www.umsu.de/trees
> > > You forgot to mention that there is no X in my examples. (As in Russell's examples where there is no present king of France.)
> > >
> > > ¬∃aXa → ¬(∃a(Xa ∧ Ya) ∧ ∃a(Xa ∧ ¬Ya)) is valid.
> > > https://www.umsu.de/trees/#~3~7aXa~0~5~3(~7a(Xa~0~1~0Ya)~0~1~0~7a(Xa~0~1~0~3Ya))
> > >
> > > We also have:
> > >
> > > ¬∃aXa → (¬∃a(Xa ∧ Ya) ∧ ¬∃a(Xa ∧ ¬Ya)) is valid.
> > > https://www.umsu.de/trees/#~3~7aXa~0~5~3~7a(Xa~0~1~0Ya)~0~1~0~3~7a(Xa~0~1~0~3Ya)
> > >
> > > I hope this helps.
> > No I didn't forget anything. You asked what is the FOL formulation
> > of the Russell requirement. The Russell requirement wants a
> > translation of "the X is Y" which satisfies the folllowing
> > ~("the X is Y" & "the X is ~Y")
> You mean,
>
> ~[EXIST(a):[ALL(b):[X(b) <=> b=a] & Y(a)] & EXIST(a):[ALL(b):[X(b) <=> b=a] & ~Y(a)]]
>
> A trivial result as it turns out (see my proof above). Again, if you have precisely one X, then, TRIVIALLY, it cannot be both a Y and not a Y.
>
> If, on the other hand, as in Russell's king-of-France example, you have ~EXIST(a):K(a), then you can have both ~EXIST(a):[K(a) & ~B(a))] and ~EXIST(a):[K(a) & B(a))]. Or equivalently, ALL(a):[K(a) => B(a)] and ALL(a):[K(a) => ~B(a)]. This is not a contradiction since the antecedent K(a) is assumed to be false in both cases. See https://en.wikipedia.org/wiki/Vacuous_truth
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com