This sentence is false! (The Liar Paradox)

Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/

Dan

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

It can be even strongly rejected, i.e. is refutable. I easily find
that the "Revenge Liar Paradox in Dan Christensens Trichotomy", is an
Antinomy, i.e. always false and hence provably the negation can be proved:

∀x(Fx ∧ (Tx ∧ Ux)),
∀x¬(Fx ∧ Tx),
∀x¬(Fx ∧ Ux),
∀x¬(Tx ∧ Ux) entails ∀x¬(Tx ↔ (¬Tx ∧ ¬Ux)).
https://www.umsu.de/trees/#~6x%28Fx~1Tx~1Ux%29,~6x%28~3%28Fx~1Tx%29%29,~6x%28~3%28Fx~1Ux%29%29,~6x%28~3%28Tx~1Ux%29%29|=~6x~3%28Tx~4~3Tx~1~3Ux%29

Should not be much of a problem to translate the above proof
into set theory, so that Dan Christensens t, f and m can be used.
For simplicity I did it with predicates.

Also in predicates I use U, for undefined, instead m.
You find for example U in Kleenes 3-valued logic. But
there is no stipulation that the underlying logic

is Kleenes 3-valued logic. Could be also something
else. Anything that satisfies Dan Christensens Trichotomy.

Dan Christensen schrieb am Donnerstag, 7. September 2023 um 19:50:24 UTC+2:
> This sentence is false! (The Liar Paradox)
>
> Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Or a more positive formulation, not using any negation, except
in "Dan Christensens Trichotomy", works also:

∀x(Fx ∧ (Tx ∧ Ux)),
∀x¬(Fx ∧ Tx),
∀x¬(Fx ∧ Ux),
∀x¬(Tx ∧ Ux)
entails ∀x¬(Tx ↔ (Fx ∧ (Fx ∨ Tx))).
https://www.umsu.de/trees/#~6x%28Fx~1Tx~1Ux%29,~6x%28~3%28Fx~1Tx%29%29,~6x%28~3%28Fx~1Ux%29%29,~6x%28~3%28Tx~1Ux%29%29|=~6x~3%28Tx~4Fx~1%28Fx~2Tx%29%29

There might be a few more variants, maybe some don't work?
I guess the different variants are different "Revenge Paradoxes",
semantically not 100% identical. How would we measure that?

Study non-classical logics, start with multi-valued logics?

LoL

Mild Shock schrieb am Montag, 11. September 2023 um 18:51:55 UTC+2:
> It can be even strongly rejected, i.e. is refutable. I easily find
> that the "Revenge Liar Paradox in Dan Christensens Trichotomy", is an
> Antinomy, i.e. always false and hence provably the negation can be proved:
>
> ∀x(Fx ∧ (Tx ∧ Ux)),
> ∀x¬(Fx ∧ Tx),
> ∀x¬(Fx ∧ Ux),
> ∀x¬(Tx ∧ Ux) entails ∀x¬(Tx ↔ (¬Tx ∧ ¬Ux)).
> https://www.umsu.de/trees/#~6x%28Fx~1Tx~1Ux%29,~6x%28~3%28Fx~1Tx%29%29,~6x%28~3%28Fx~1Ux%29%29,~6x%28~3%28Tx~1Ux%29%29|=~6x~3%28Tx~4~3Tx~1~3Ux%29
>
> Should not be much of a problem to translate the above proof
> into set theory, so that Dan Christensens t, f and m can be used.
> For simplicity I did it with predicates.
>
> Also in predicates I use U, for undefined, instead m.
> You find for example U in Kleenes 3-valued logic. But
> there is no stipulation that the underlying logic
>
> is Kleenes 3-valued logic. Could be also something
> else. Anything that satisfies Dan Christensens Trichotomy.
> Dan Christensen schrieb am Donnerstag, 7. September 2023 um 19:50:24 UTC+2:
> > This sentence is false! (The Liar Paradox)
> >
> > Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/
> >
> > 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, September 11, 2023 at 12:51:55 PM UTC-4, Mild Shock wrote:
> It can be even strongly rejected, i.e. is refutable. I easily find
> that the "Revenge Liar Paradox in Dan Christensens Trichotomy", is an
> Antinomy, i.e. always false and hence provably the negation can be proved:
>
> ∀x(Fx ∧ (Tx ∧ Ux)),
> ∀x¬(Fx ∧ Tx),
> ∀x¬(Fx ∧ Ux),
> ∀x¬(Tx ∧ Ux) entails ∀x¬(Tx ↔ (¬Tx ∧ ¬Ux)).
> https://www.umsu.de/trees/#~6x%28Fx~1Tx~1Ux%29,~6x%28~3%28Fx~1Tx%29%29,~6x%28~3%28Fx~1Ux%29%29,~6x%28~3%28Tx~1Ux%29%29|=~6x~3%28Tx~4~3Tx~1~3Ux%29
>
[snip]

How is this relevant to the Liar Paradox?

Dan

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

Thats the point of a revenge paradox, to create a new
contradiction. There are more options for revenge
paradoxes, which make use of the richer language that

you created via your "resolution". Here are some Revenge
Paradoxes you might choose from:

R : This sentences is not true and not indeterminate.
(Fy → Ty) ∧ ((Ty ∨ Uy) → Fy)

R' : This sentences is true iff it is false or indeterminate.
(Ty ↔ Fy ∨ Uy)

I guess my version (R) is the more complicated, the
version (R') from Julio Di Egidio is more elegant,
especially it its translation into a formula.

BTW: Both Revenge Paradoxes insert Uy at some
point, starting from the Liar Paradox:

L : This sentences is true iff it is false
(Ty ↔ Fy)

For Julio Di Egidio version its immediate that
(Ty ↔ Fy ∨ Uy) is Uy inserted in the right hand side
of the bi-conditional. For my version you have to

split the bi-conditional into two material implications:

L : This sentences is true iff it is false
(Fy → Ty) ∧(Ty → Fy)

And then you see that (Fy → Ty) ∧ ((Ty ∨ Uy) → Fy) is
Uy inserted in the left hand side of the second conjunction
component.

Dan Christensen schrieb am Dienstag, 12. September 2023 um 17:27:56 UTC+2:
> On Monday, September 11, 2023 at 12:51:55 PM UTC-4, Mild Shock wrote:
> > It can be even strongly rejected, i.e. is refutable. I easily find
> > that the "Revenge Liar Paradox in Dan Christensens Trichotomy", is an
> > Antinomy, i.e. always false and hence provably the negation can be proved:
> >
> > ∀x(Fx ∧ (Tx ∧ Ux)),
> > ∀x¬(Fx ∧ Tx),
> > ∀x¬(Fx ∧ Ux),
> > ∀x¬(Tx ∧ Ux) entails ∀x¬(Tx ↔ (¬Tx ∧ ¬Ux)).
> > https://www.umsu.de/trees/#~6x%28Fx~1Tx~1Ux%29,~6x%28~3%28Fx~1Tx%29%29,~6x%28~3%28Fx~1Ux%29%29,~6x%28~3%28Tx~1Ux%29%29|=~6x~3%28Tx~4~3Tx~1~3Ux%29
> >
> [snip]
>
> How is this relevant to the Liar Paradox?
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Try this one, this is funny:

R'' : This sentences is true iff (if it is true then it is false).
(Ty ↔ (Ty → Fy))

You can prove these two theorems:

=> ALL(b):[b ε s => [[b ε t <=> [b ε t => b ε f]] => b ε m]]]]

=> ALL(b):[b ε s => [[b ε t <=> [b ε t => b ε f]] => ~b ε m]]]]

I guess we need Quadrotomy, since Trichotomy is not enough...

LMAO!

Dan Christensen schrieb am Donnerstag, 7. September 2023 um 19:50:24 UTC+2:
> This sentence is false! (The Liar Paradox)
>
> Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Like in Dan Christensens famous Generalized Drinker
Paradox, we can prove, with Q an arbitrary formula:

60 ALL(s):ALL(t):ALL(f):ALL(m):[ALL(a):[a @ s => [a @ t | a @ f | a @ m]
& ~[a @ t & a @ f]
& ~[a @ t & a @ m]
& ~[a @ f & a @ m]]
=> ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]]
4 Conclusion, 1

Proof was done by first proving. I guess something went
exploding, I hope its not Dan Christensens brain!

42 ALL(a):[a @ s => ~[a @ t <=> a @ t => a @ f]]
Rem DNeg, 41

Maybe a short proof is possible?

----------------------------- begin proof ---------------------------

1 ALL(a):[a @ s => [a @ t | a @ f | a @ m]
& ~[a @ t & a @ f]
& ~[a @ t & a @ m]
& ~[a @ f & a @ m]]
Premise

2 ~ALL(a):[a @ s => ~[a @ t <=> a @ t => a @ f]]
Premise

3 ~~EXIST(a):~[a @ s => ~[a @ t <=> a @ t => a @ f]]
Quant, 2

4 EXIST(a):~[a @ s => ~[a @ t <=> a @ t => a @ f]]
Rem DNeg, 3

5 EXIST(a):~~[a @ s & ~~[a @ t <=> a @ t => a @ f]]
Imply-And, 4

6 EXIST(a):[a @ s & ~~[a @ t <=> a @ t => a @ f]]
Rem DNeg, 5

7 EXIST(a):[a @ s & [a @ t <=> a @ t => a @ f]]
Rem DNeg, 6

8 u @ s & [u @ t <=> u @ t => u @ f]
E Spec, 7

9 u @ s
Split, 8

10 u @ t <=> u @ t => u @ f
Split, 8

11 u @ s => [u @ t | u @ f | u @ m]
& ~[u @ t & u @ f]
& ~[u @ t & u @ m]
& ~[u @ f & u @ m]
U Spec, 1

12 [u @ t | u @ f | u @ m]
& ~[u @ t & u @ f]
& ~[u @ t & u @ m]
& ~[u @ f & u @ m]
Detach, 11, 9

13 u @ t | u @ f | u @ m
Split, 12

14 ~[u @ t & u @ f]
Split, 12

15 ~[u @ t & u @ m]
Split, 12

16 ~[u @ f & u @ m]
Split, 12

17 [u @ t => [u @ t => u @ f]] & [u @ t => u @ f => u @ t]
Iff-And, 10

18 u @ t => [u @ t => u @ f]
Split, 17

19 u @ t => u @ f => u @ t
Split, 17

20 u @ f
Premise

21 ~u @ t | u @ f
Arb Or, 20

22 ~~u @ t => u @ f
Imply-Or, 21

23 u @ t => u @ f
Rem DNeg, 22

24 u @ t
Detach, 19, 23

25 u @ t & u @ f
Join, 24, 20

26 ~[u @ t & u @ f] & [u @ t & u @ f]
Join, 14, 25

27 ~u @ f
4 Conclusion, 20

28 ~[u @ t => u @ f] => ~u @ t
Contra, 18

29 ~~[u @ t & ~u @ f] => ~u @ t
Imply-And, 28

30 u @ t & ~u @ f => ~u @ t
Rem DNeg, 29

31 u @ t
Premise

32 u @ t & ~u @ f
Join, 31, 27

33 ~u @ t
Detach, 30, 32

34 u @ t & ~u @ t
Join, 31, 33

35 ~u @ t
4 Conclusion, 31

36 ~u @ t | u @ f
Arb Or, 35

37 ~~u @ t => u @ f
Imply-Or, 36

38 u @ t => u @ f
Rem DNeg, 37

39 u @ t
Detach, 19, 38

40 ~u @ t & u @ t
Join, 35, 39

41 ~~ALL(a):[a @ s => ~[a @ t <=> a @ t => a @ f]]
4 Conclusion, 2

42 ALL(a):[a @ s => ~[a @ t <=> a @ t => a @ f]]
Rem DNeg, 41

43 ~ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]
Premise

44 ~~EXIST(a):~[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]
Quant, 43

45 EXIST(a):~[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]
Rem DNeg, 44

46 EXIST(a):~~[a @ s & ~[[a @ t <=> a @ t => a @ f] => Q]]
Imply-And, 45

47 EXIST(a):[a @ s & ~[[a @ t <=> a @ t => a @ f] => Q]]
Rem DNeg, 46

48 v @ s & ~[[v @ t <=> v @ t => v @ f] => Q]
E Spec, 47

49 v @ s
Split, 48

50 ~[[v @ t <=> v @ t => v @ f] => Q]
Split, 48

51 v @ s => ~[v @ t <=> v @ t => v @ f]
U Spec, 42

52 ~[v @ t <=> v @ t => v @ f]
Detach, 51, 49

53 ~~[[v @ t <=> v @ t => v @ f] & ~Q]
Imply-And, 50

54 [v @ t <=> v @ t => v @ f] & ~Q
Rem DNeg, 53

55 v @ t <=> v @ t => v @ f
Split, 54

56 ~Q
Split, 54

57 [v @ t <=> v @ t => v @ f]
& ~[v @ t <=> v @ t => v @ f]
Join, 55, 52

58 ~~ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]
4 Conclusion, 43

59 ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]
Rem DNeg, 58

60 ALL(s):ALL(t):ALL(f):ALL(m):[ALL(a):[a @ s => [a @ t | a @ f | a @ m]
& ~[a @ t & a @ f]
& ~[a @ t & a @ m]
& ~[a @ f & a @ m]]
=> ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]]
4 Conclusion, 1

----------------------------- end proof ---------------------------

Mild Shock schrieb am Dienstag, 12. September 2023 um 20:31:43 UTC+2:
> Try this one, this is funny:
>
> R'' : This sentences is true iff (if it is true then it is false).
> (Ty ↔ (Ty → Fy))
>
> You can prove these two theorems:
>
> => ALL(b):[b ε s => [[b ε t <=> [b ε t => b ε f]] => b ε m]]]]
>
> => ALL(b):[b ε s => [[b ε t <=> [b ε t => b ε f]] => ~b ε m]]]]
>
> I guess we need Quadrotomy, since Trichotomy is not enough...
>
> LMAO!
> Dan Christensen schrieb am Donnerstag, 7. September 2023 um 19:50:24 UTC+2:
> > This sentence is false! (The Liar Paradox)
> >
> > Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/
> >
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

On Thursday, September 7, 2023 at 10:50:24 AM UTC-7, Dan Christensen wrote:
> This sentence is false! (The Liar Paradox)
>
> Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/
>
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

The liar lies and Einstein said it was science.
He wrote it down on his last day.

On Tuesday, September 12, 2023 at 2:49:59 PM UTC-4, Mild Shock (aka Mr. Collapse) wrote:
> Like in Dan Christensen's famous Generalized Drinker
> Paradox, we can prove, with Q an arbitrary formula:
>

Similarly, we can prove 0=1 => Q. So what?

> 60 ALL(s):ALL(t):ALL(f):ALL(m):[ALL(a):[a @ s => [a @ t | a @ f | a @ m]
> & ~[a @ t & a @ f]
> & ~[a @ t & a @ m]
> & ~[a @ f & a @ m]]
> => ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]]
> Conclusion, 1
>

[snip]

How is that relevant? We still have my theorem:

ALL(s):ALL(t):ALL(f):ALL(m):[Set(s) & Set(t) & Set(f) & Set(m)

=> [ALL(a):[a in t => a in s] <---------------- 3 disjoint subsets of s
& ALL(a):[a in f => a in s]
& ALL(a):[a in m => a in s]

& ALL(a):[a in s => [a in t | a in f | a in m] <---------- Trichotomy rules assumed on these sets
& ~[a in t & a in f]
& ~[a in t & a in m]
& ~[a in f & a in m]]

=> ALL(b):[b in s => [[b in t <=> b in f] => b in m]]]] <-------- "This sentence is false" is of indeterminate truth value

Deal with it, Mr. Collapse. Just admit you were wrong.

Dan

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

But your Trichotomy is not enough. There are more sentences,
which are not in your s. For example this sentence here is not in your s:

R'' : This sentences is true iff (if it is true then it is false).
(Ty ↔ (Ty → Fy))

So I don't think you have resolved anything. The above Liar
sentences falls out of your framework, doesn't get resolved.

P.S.: Thats an easy corollary of:

42 ALL(a):[a @ s => ~[a @ t <=> a @ t => a @ f]]
Rem DNeg, 41

Just take the contraposition and you get:

ALL(a):[[a @ t <=> a @ t => a @ f] => ~a @ s]

Dan Christensen schrieb am Dienstag, 12. September 2023 um 22:55:41 UTC+2:
> On Tuesday, September 12, 2023 at 2:49:59 PM UTC-4, Mild Shock (aka Mr. Collapse) wrote:
> > Like in Dan Christensen's famous Generalized Drinker
> > Paradox, we can prove, with Q an arbitrary formula:
> >
> Similarly, we can prove 0=1 => Q. So what?
> > 60 ALL(s):ALL(t):ALL(f):ALL(m):[ALL(a):[a @ s => [a @ t | a @ f | a @ m]
> > & ~[a @ t & a @ f]
> > & ~[a @ t & a @ m]
> > & ~[a @ f & a @ m]]
> > => ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]]
> > Conclusion, 1
> >
>
> [snip]
>
> How is that relevant? We still have my theorem:
>
> ALL(s):ALL(t):ALL(f):ALL(m):[Set(s) & Set(t) & Set(f) & Set(m)
>
> => [ALL(a):[a in t => a in s] <---------------- 3 disjoint subsets of s
> & ALL(a):[a in f => a in s]
> & ALL(a):[a in m => a in s]
>
> & ALL(a):[a in s => [a in t | a in f | a in m] <---------- Trichotomy rules assumed on these sets
> & ~[a in t & a in f]
> & ~[a in t & a in m]
> & ~[a in f & a in m]]
>
> => ALL(b):[b in s => [[b in t <=> b in f] => b in m]]]] <-------- "This sentence is false" is of indeterminate truth value
>
> Deal with it, Mr. Collapse. Just admit you were wrong.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

There is a famous 4 Valued Logic, called Belnaps FOUR.
Maybe the Liar Paradox R'' gets resolved in this logic?

R'' : This sentences is true iff (if it is true then it is false).
(Ty ↔ (Ty → Fy))

The logic has been especially developed for contradictions:

"Nuel Belnap considered the challenge of question answering
by computer in 1975. Noting human fallibility, he was concerned
with the case where two contradictory facts were loaded into
memory, and then a query was made. "We all know about the
fecundity of contradictions in two-valued logic: contradictions are
never isolated, infecting as they do the whole system."[1] Belnap
proposed a four-valued logic as a means of containing contradiction."
https://en.wikipedia.org/wiki/Four-valued_logic

But your Trichotomy seems to be insuffienct, there is something
missing. Which isn't a surprise, since from syntactic consequence
we already know that there are FOUR positibilities:

1) not G |- A and not G |- ~A
2) G |- A and not G |- ~A
3) not G |- A and G |- ~A
4) G |- A and G |- ~A

Mild Shock schrieb am Mittwoch, 13. September 2023 um 01:19:02 UTC+2:
> But your Trichotomy is not enough. There are more sentences,
> which are not in your s. For example this sentence here is not in your s:
> R'' : This sentences is true iff (if it is true then it is false).
> (Ty ↔ (Ty → Fy))
> So I don't think you have resolved anything. The above Liar
> sentences falls out of your framework, doesn't get resolved.
>
> P.S.: Thats an easy corollary of:
>
> 42 ALL(a):[a @ s => ~[a @ t <=> a @ t => a @ f]]
> Rem DNeg, 41
>
> Just take the contraposition and you get:
>
> ALL(a):[[a @ t <=> a @ t => a @ f] => ~a @ s]
> Dan Christensen schrieb am Dienstag, 12. September 2023 um 22:55:41 UTC+2:
> > On Tuesday, September 12, 2023 at 2:49:59 PM UTC-4, Mild Shock (aka Mr. Collapse) wrote:
> > > Like in Dan Christensen's famous Generalized Drinker
> > > Paradox, we can prove, with Q an arbitrary formula:
> > >
> > Similarly, we can prove 0=1 => Q. So what?
> > > 60 ALL(s):ALL(t):ALL(f):ALL(m):[ALL(a):[a @ s => [a @ t | a @ f | a @ m]
> > > & ~[a @ t & a @ f]
> > > & ~[a @ t & a @ m]
> > > & ~[a @ f & a @ m]]
> > > => ALL(a):[a @ s => [[a @ t <=> a @ t => a @ f] => Q]]]
> > > Conclusion, 1
> > >
> >
> > [snip]
> >
> > How is that relevant? We still have my theorem:
> >
> > ALL(s):ALL(t):ALL(f):ALL(m):[Set(s) & Set(t) & Set(f) & Set(m)
> >
> > => [ALL(a):[a in t => a in s] <---------------- 3 disjoint subsets of s
> > & ALL(a):[a in f => a in s]
> > & ALL(a):[a in m => a in s]
> >
> > & ALL(a):[a in s => [a in t | a in f | a in m] <---------- Trichotomy rules assumed on these sets
> > & ~[a in t & a in f]
> > & ~[a in t & a in m]
> > & ~[a in f & a in m]]
> >
> > => ALL(b):[b in s => [[b in t <=> b in f] => b in m]]]] <-------- "This sentence is false" is of indeterminate truth value
> >
> > Deal with it, Mr. Collapse. Just admit you were wrong.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

On Tuesday, September 12, 2023 at 7:19:02 PM UTC-4, Mild Shock wrote:
> But your Trichotomy is not enough.

Essentially, there are (1) true sentences, (2) false sentences, (3) every other sentence. What more do you require?

> There are more sentences,
> which are not in your s. For example this sentence here is not in your s:
> R'' : This sentences is true iff (if it is true then it is false).
> (Ty ↔ (Ty → Fy))
> So I don't think you have resolved anything.

I have shown that "This sentence is false" is just one of many other sentences that have indeterminate truth values. In some sense, I think I may have demystified it.

>The above Liar
> sentences falls out of your framework, doesn't get resolved.
>
[snip]

Just as Russell found that not every formula corresponds to a set, not every string of characters corresponds to a sentence. Maybe you will will want to study such things, but I am talking here about sets of sentences. I am applying set theory to linguistics, leaving "sentence," "true" and "false" undefined.

Dan

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

Your Island of Liars Creta, is not the only Greek Island.
You are simply cheating your way out of only considering
one Greek Island. Trichotomy is not enough, since this here:
(Ty ↔ (Ty → Fy))

Is classical logic equivalent to:
Ty ∧ Fy

So there is a fourth possibility. But you have only 3
possibility, in your Trichotomy:
Ty Fy Uy
F F T
T F F
F T F

Where is Ty and Fy simultaneously true?

BTW: You could model your Trichotomy also without
Uy, where Uy is simply S \ T \ F:
Ty Fy
F F
T F
F T

Simply looking at Ty and Fy is enough for your Trichotomy.
But its evident that your Island of Liars Creta, is not the
only Greek Island. And your "resolution" is cheating.

Dan Christensen schrieb am Mittwoch, 13. September 2023 um 02:36:54 UTC+2:
> On Tuesday, September 12, 2023 at 7:19:02 PM UTC-4, Mild Shock wrote:
> > But your Trichotomy is not enough.
> Essentially, there are (1) true sentences, (2) false sentences, (3) every other sentence. What more do you require?
> > There are more sentences,
> > which are not in your s. For example this sentence here is not in your s:
> > R'' : This sentences is true iff (if it is true then it is false).
> > (Ty ↔ (Ty → Fy))
> > So I don't think you have resolved anything.
> I have shown that "This sentence is false" is just one of many other sentences that have indeterminate truth values. In some sense, I think I may have demystified it.
> >The above Liar
> > sentences falls out of your framework, doesn't get resolved.
> >
> [snip]
>
> Just as Russell found that not every formula corresponds to a set, not every string of characters corresponds to a sentence. Maybe you will will want to study such things, but I am talking here about sets of sentences. I am applying set theory to linguistics, leaving "sentence," "true" and "false" undefined.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Take the Island of Santorini, from Belnaps Logic FOUR.
Where you would have:
Ty Fy
F F
T F
F T
T T

All Paradoxes go away, not only the Liar (Ty ↔ Fy), but also
the Liar (Ty ↔ (Ty → Fy)). These are very typical but also very
different Liars. You managed to allow the first Liar,

but you failed to allow the second Liar. What are these Liars
logically, well its pretty simply your Liar can be satisfied by
rejecting LEM or rejecting LNC, my Liar can be only satisfied

by rejecting LNC, rejecting LEM doesn't work anymore in my example:

LEM: Law of Excluded Middle
https://en.wikipedia.org/wiki/Law_of_excluded_middle

LNC: Law of noncontradiction
https://en.wikipedia.org/wiki/Law_of_noncontradiction

Mild Shock schrieb am Mittwoch, 13. September 2023 um 08:37:04 UTC+2:
> Your Island of Liars Creta, is not the only Greek Island.
> You are simply cheating your way out of only considering
> one Greek Island. Trichotomy is not enough, since this here:
> (Ty ↔ (Ty → Fy))
>
> Is classical logic equivalent to:
> Ty ∧ Fy
>
> So there is a fourth possibility. But you have only 3
> possibility, in your Trichotomy:
> Ty Fy Uy
> F F T
> T F F
> F T F
>
> Where is Ty and Fy simultaneously true?
>
> BTW: You could model your Trichotomy also without
> Uy, where Uy is simply S \ T \ F:
> Ty Fy
> F F
> T F
> F T
>
> Simply looking at Ty and Fy is enough for your Trichotomy.
> But its evident that your Island of Liars Creta, is not the
> only Greek Island. And your "resolution" is cheating.
> Dan Christensen schrieb am Mittwoch, 13. September 2023 um 02:36:54 UTC+2:
> > On Tuesday, September 12, 2023 at 7:19:02 PM UTC-4, Mild Shock wrote:
> > > But your Trichotomy is not enough.
> > Essentially, there are (1) true sentences, (2) false sentences, (3) every other sentence. What more do you require?
> > > There are more sentences,
> > > which are not in your s. For example this sentence here is not in your s:
> > > R'' : This sentences is true iff (if it is true then it is false).
> > > (Ty ↔ (Ty → Fy))
> > > So I don't think you have resolved anything.
> > I have shown that "This sentence is false" is just one of many other sentences that have indeterminate truth values. In some sense, I think I may have demystified it.
> > >The above Liar
> > > sentences falls out of your framework, doesn't get resolved.
> > >
> > [snip]
> >
> > Just as Russell found that not every formula corresponds to a set, not every string of characters corresponds to a sentence. Maybe you will will want to study such things, but I am talking here about sets of sentences. I am applying set theory to linguistics, leaving "sentence," "true" and "false" undefined.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

Belnaps Logic FOUR rejects both:

Belnap notes that "paradoxes of implication" (A&~A)→B
and A→(B∨~B) are avoided in his 4-valued system.
https://en.wikipedia.org/wiki/Four-valued_logic

Remark: You could also consider a further Liar,
Ty ↔ (Fy → Ty)), this one would also work by rejecting
LEM but not by rejecting LNC, it is a little weaker than your

(Ty ↔ Fy), your (Ty ↔ Fy) would also work by rejecting LNC.

So your (Ty ↔ Fy) would also work in the Island of Hydra:
Ty Fy
T F
F T
T T

There are so many Islands in Greek, Creta, Santorini, Hydra,
Skiathos, Lesbos, Corfu, Rhodos, Mikonons, etc.. We have
only touched the surface of Non-Classical Logics.

Mild Shock schrieb am Mittwoch, 13. September 2023 um 08:49:16 UTC+2:
> Take the Island of Santorini, from Belnaps Logic FOUR.
> Where you would have:
> Ty Fy
> F F
> T F
> F T
> T T
>
> All Paradoxes go away, not only the Liar (Ty ↔ Fy), but also
> the Liar (Ty ↔ (Ty → Fy)). These are very typical but also very
> different Liars. You managed to allow the first Liar,
>
> but you failed to allow the second Liar. What are these Liars
> logically, well its pretty simply your Liar can be satisfied by
> rejecting LEM or rejecting LNC, my Liar can be only satisfied
>
> by rejecting LNC, rejecting LEM doesn't work anymore in my example:
>
> LEM: Law of Excluded Middle
> https://en.wikipedia.org/wiki/Law_of_excluded_middle
>
> LNC: Law of noncontradiction
> https://en.wikipedia.org/wiki/Law_of_noncontradiction
> Mild Shock schrieb am Mittwoch, 13. September 2023 um 08:37:04 UTC+2:
> > Your Island of Liars Creta, is not the only Greek Island.
> > You are simply cheating your way out of only considering
> > one Greek Island. Trichotomy is not enough, since this here:
> > (Ty ↔ (Ty → Fy))
> >
> > Is classical logic equivalent to:
> > Ty ∧ Fy
> >
> > So there is a fourth possibility. But you have only 3
> > possibility, in your Trichotomy:
> > Ty Fy Uy
> > F F T
> > T F F
> > F T F
> >
> > Where is Ty and Fy simultaneously true?
> >
> > BTW: You could model your Trichotomy also without
> > Uy, where Uy is simply S \ T \ F:
> > Ty Fy
> > F F
> > T F
> > F T
> >
> > Simply looking at Ty and Fy is enough for your Trichotomy.
> > But its evident that your Island of Liars Creta, is not the
> > only Greek Island. And your "resolution" is cheating.
> > Dan Christensen schrieb am Mittwoch, 13. September 2023 um 02:36:54 UTC+2:
> > > On Tuesday, September 12, 2023 at 7:19:02 PM UTC-4, Mild Shock wrote:
> > > > But your Trichotomy is not enough.
> > > Essentially, there are (1) true sentences, (2) false sentences, (3) every other sentence. What more do you require?
> > > > There are more sentences,
> > > > which are not in your s. For example this sentence here is not in your s:
> > > > R'' : This sentences is true iff (if it is true then it is false).
> > > > (Ty ↔ (Ty → Fy))
> > > > So I don't think you have resolved anything.
> > > I have shown that "This sentence is false" is just one of many other sentences that have indeterminate truth values. In some sense, I think I may have demystified it.
> > > >The above Liar
> > > > sentences falls out of your framework, doesn't get resolved.
> > > >
> > > [snip]
> > >
> > > Just as Russell found that not every formula corresponds to a set, not every string of characters corresponds to a sentence. Maybe you will will want to study such things, but I am talking here about sets of sentences. I am applying set theory to linguistics, leaving "sentence," "true" and "false" undefined.
> > > 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 13, 2023 at 2:37:04 AM UTC-4, Mild Shock wrote:
> Your Island of Liars Creta, is not the only Greek Island.

See my recent posting here, "Resolution of Epimenides' Paradox (in DC Proof)."

Dan

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

There is also moron island for people like you.

Dan Christensen schrieb am Mittwoch, 13. September 2023 um 17:38:16 UTC+2:
> On Wednesday, September 13, 2023 at 2:37:04 AM UTC-4, Mild Shock wrote:
> > Your Island of Liars Creta, is not the only Greek Island.
> See my recent posting here, "Resolution of Epimenides' Paradox (in DC Proof)."
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Your proof is much much too complicated. Its total
overkill for a very trivial matter. You use a way too
complicated trichotomy I as follows:
T F M
----------------
0 0 1
0 1 0
1 0 0

You can also use this trichotomy II as follows,
the column "M" is redundant:
T F
-------
0 0
0 1
1 0

Trichtomoty II has a very simple formalization:
∀x~(Tx & Fx)

You can then establish this theorem, were
indeterminate is literally ~Ty & ~Fy:
(Ty <-> Fy) -> ~Ty & ~Fy

∀x¬(Tx ∧ Fx) entails (Ty ↔ Fy) → (¬Ty ∧ ¬Fy)
https://www.umsu.de/trees/#~6x~3%28Tx~1Fx%29|=%28Ty~4Fy%29~5%28~3Ty~1~3Fy%29

Much shorter Liar Resolution!

LoL

Mild Shock schrieb am Mittwoch, 13. September 2023 um 18:52:18 UTC+2:
> There is also moron island for people like you.
> Dan Christensen schrieb am Mittwoch, 13. September 2023 um 17:38:16 UTC+2:
> > On Wednesday, September 13, 2023 at 2:37:04 AM UTC-4, Mild Shock wrote:
> > > Your Island of Liars Creta, is not the only Greek Island.
> > See my recent posting here, "Resolution of Epimenides' Paradox (in DC Proof)."
> > 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 13, 2023 at 8:12:15 PM UTC-4, Mild Shock wrote:
> Your proof is much much too complicated. Its total
> overkill for a very trivial matter. You use a way too
> complicated trichotomy I as follows:
> T F M
> ----------------
> 0 0 1
> 0 1 0
> 1 0 0
>
> You can also use this trichotomy II as follows,
> the column "M" is redundant:
> T F
> -------
> 0 0
> 0 1
> 1 0
>
> Trichtomoty II has a very simple formalization:
> ∀x~(Tx & Fx)
>
> You can then establish this theorem, were
> indeterminate is literally ~Ty & ~Fy

[snip]

A dead end. Now what? From this we can infer that sentence y, like "What time is it?" and "Wash your hands", is a sentence of indeterminate truth value. Deal with it.

See https://dcproof.com/LiarParadox2.htm

Dan

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

On Thursday, September 14, 2023 at 4:18:13 AM UTC+3, Dan Christensen wrote:
> On Wednesday, September 13, 2023 at 8:12:15 PM UTC-4, Mild Shock wrote:
> > Your proof is much much too complicated. Its total
> > overkill for a very trivial matter. You use a way too
> > complicated trichotomy I as follows:
> > T F M
> > ----------------
> > 0 0 1
> > 0 1 0
> > 1 0 0
> >
> > You can also use this trichotomy II as follows,
> > the column "M" is redundant:
> > T F
> > -------
> > 0 0
> > 0 1
> > 1 0
> >
> > Trichtomoty II has a very simple formalization:
> > ∀x~(Tx & Fx)
> >
> > You can then establish this theorem, were
> > indeterminate is literally ~Ty & ~Fy
> [snip]
>
> A dead end. Now what? From this we can infer that sentence y, like "What time is it?" and "Wash your hands", is a sentence of indeterminate truth value. Deal with it.
>
> See https://dcproof.com/LiarParadox2.htm
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

Belive me that intermediate theorem is quite false & no theorem at all since it is solely based upon false misleading conclusions & never any true rigorous proof

Also believe me that both of you were well-known on sci.math as being stubborn deniers & Trolls 🧌, I.e not cranks yet, since cranks at least have points 😉!

Also you were wasting your lives aimlessly for sure

BKK

On Wednesday, September 13, 2023 at 9:38:20 PM UTC-4, bassam karzeddin wrote:
> On Thursday, September 14, 2023 at 4:18:13 AM UTC+3, Dan Christensen wrote:
> > On Wednesday, September 13, 2023 at 8:12:15 PM UTC-4, Mild Shock wrote:
> > > Your proof is much much too complicated. Its total
> > > overkill for a very trivial matter. You use a way too
> > > complicated trichotomy I as follows:
> > > T F M
> > > ----------------
> > > 0 0 1
> > > 0 1 0
> > > 1 0 0
> > >
> > > You can also use this trichotomy II as follows,
> > > the column "M" is redundant:
> > > T F
> > > -------
> > > 0 0
> > > 0 1
> > > 1 0
> > >
> > > Trichtomoty II has a very simple formalization:
> > > ∀x~(Tx & Fx)
> > >
> > > You can then establish this theorem, were
> > > indeterminate is literally ~Ty & ~Fy
> > [snip]
> >
> > A dead end. Now what? From this we can infer that sentence y, like "What time is it?" and "Wash your hands", is a sentence of indeterminate truth value. Deal with it.
> >
> > See https://dcproof.com/LiarParadox2.htm
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com
> Belive me that intermediate theorem is quite false & no theorem at all since it is solely based upon false misleading conclusions & never any true rigorous proof
>
> Also believe me that both of you were well-known on sci.math as being stubborn deniers & Trolls 🧌, I.e not cranks yet, since cranks at least have points 😉!
>

From Psycho Troll BKK who also wrote here:

“Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now, for sure.”
-- BKK, Dec. 6, 2017

"The Devils deeds that are strictly and basically sourced from mathematicians like humans, FOR SURE!"
-- BKK, June 11, 2020

“You know certainly that I'm the man, and more specially the KING who is going to upside down most of your current false mathematics for all future generations.”
-- BKK, Nov. 22, 2018

“Despite thousands of years of continuous juggling and false definitions of what is truly the real number, they [us carbon-based lifeforms?] truly don't want to understand it as was discovered strictly by the *KING* [BKK Himself!]”
-- BKK, Nov. 28, 2019

“I don't believe even in one being a number”
-- BKK, Dec. 31, 2019

Math failure, BKK, doesn't believe in negative numbers, zero, one or numbers like pi and root 2. He doesn't even believe in 40 degree angles or circles. Simple speed-distance-time problems seem to be impossible for him. Really!

Needless to say his own goofy little system is getting nowhere and never will. As such he is insanely jealous of wildly successful mainstream mathematics. He seems to believe these super-intelligent artificial beings of his will somehow be enlisting his aid to "reform" mathematics worldwide when they take over the planet in the near future. He is truly delusional.

Dan

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

It proofs exactly the same you dumbo from moron island.
~Ty & ~Fy and My are the same. Or with set notation:

~y e t & ~y e f

and:

y e m

Are the same. This follows from your trichotomy I,
you imbecil. Especially from:

[a e t | a e f | a e m]
& ~[a ε t & a ε f]
& ~[a ε t & a ε m]
& ~[a ε f & a ε m]
https://www.dcproof.com/LiarParadoxResolution.htm

Thats quite easy:

∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.
https://www.umsu.de/trees/#~6x%28Fx~2Tx~2Ux%29,~6x~3%28Fx~1Tx%29,~6x~3%28Fx~1Ux%29,~6x~3%28Tx~1Ux%29|=%28~3Ty~1~3Fy~4Uy%29

So you don't need y e m.

Dan Christensen schrieb am Donnerstag, 14. September 2023 um 03:18:13 UTC+2:
> On Wednesday, September 13, 2023 at 8:12:15 PM UTC-4, Mild Shock wrote:
> > Your proof is much much too complicated. Its total
> > overkill for a very trivial matter. You use a way too
> > complicated trichotomy I as follows:
> > T F M
> > ----------------
> > 0 0 1
> > 0 1 0
> > 1 0 0
> >
> > You can also use this trichotomy II as follows,
> > the column "M" is redundant:
> > T F
> > -------
> > 0 0
> > 0 1
> > 1 0
> >
> > Trichtomoty II has a very simple formalization:
> > ∀x~(Tx & Fx)
> >
> > You can then establish this theorem, were
> > indeterminate is literally ~Ty & ~Fy
> [snip]
>
> A dead end. Now what? From this we can infer that sentence y, like "What time is it?" and "Wash your hands", is a sentence of indeterminate truth value. Deal with it.
>
> See https://dcproof.com/LiarParadox2.htm
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

On Thursday, September 14, 2023 at 3:30:56 AM UTC-4, Mild Shock wrote:
snip childish abuse]

> ~Ty & ~Fy and My are the same. Or with set notation:
>
> ~y e t & ~y e f
>
> and:
>
> y e m
>
> Are the same. This follows from your trichotomy I,

[snip childish abuse]

>
> [a e t | a e f | a e m]
> & ~[a ε t & a ε f]
> & ~[a ε t & a ε m]
> & ~[a ε f & a ε m]
> https://www.dcproof.com/LiarParadoxResolution.htm
>
> Thats quite easy:
> ∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
> ∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.
>
https://www.umsu.de/trees/#~6x%28Fx~2Tx~2Ux%29,~6x~3%28Fx~1Tx%29,~6x~3%28Fx~1Ux%29,~6x~3%28Tx~1Ux%29|=%28~3Ty~1~3Fy~4Uy%29
>
> So you don't need y e m.

[snip]

For your trichotomy, you want to use predicates T, F and U instead sets t, f and m respectively. The advantage of using set theory is that you can actually prove the existence of a trichotomies on any given set. See the Trichotomy Lemma at https://www.dcproof.com/LiarParadoxLemma.htm. It requires the Subset Axiom (lines 2 and 21). Also, set notation is more widely used and understood than predicate notation. Must be frustrating as hell for you, Mr. Collapse.

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 don't want to use T, F and U.
I only want to use T,F.

Translated to sets, I don't want to use t,f and m.
I only want to use t,f.

Whats wrong with you?

I explained that already:

~y e t & ~y e f

and

y e m

are logically equivalent for trichtomoy. Should be easy
to produce a proof in DC Poop. Here is a proof in
Wolfgang Schwarz tree tool:

∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.
https://www.umsu.de/trees/#~6x%28Fx~2Tx~2Ux%29,~6x~3%28Fx~1Tx%29,~6x~3%28Fx~1Ux%29,~6x~3%28Tx~1Ux%29|=%28~3Ty~1~3Fy~4Uy%29

So you don't need y e m.

Dan Christensen schrieb am Donnerstag, 14. September 2023 um 12:04:17 UTC+2:
> On Thursday, September 14, 2023 at 3:30:56 AM UTC-4, Mild Shock wrote:
> snip childish abuse]
> > ~Ty & ~Fy and My are the same. Or with set notation:
> >
> > ~y e t & ~y e f
> >
> > and:
> >
> > y e m
> >
> > Are the same. This follows from your trichotomy I,
> [snip childish abuse]
> >
> > [a e t | a e f | a e m]
> > & ~[a ε t & a ε f]
> > & ~[a ε t & a ε m]
> > & ~[a ε f & a ε m]
> > https://www.dcproof.com/LiarParadoxResolution.htm
> >
> > Thats quite easy:
> > ∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
> > ∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.
> >
> https://www.umsu.de/trees/#~6x%28Fx~2Tx~2Ux%29,~6x~3%28Fx~1Tx%29,~6x~3%28Fx~1Ux%29,~6x~3%28Tx~1Ux%29|=%28~3Ty~1~3Fy~4Uy%29
> >
> > So you don't need y e m.
> [snip]
>
> For your trichotomy, you want to use predicates T, F and U instead sets t, f and m respectively. The advantage of using set theory is that you can actually prove the existence of a trichotomies on any given set. See the Trichotomy Lemma at https://www.dcproof.com/LiarParadoxLemma.htm. It requires the Subset Axiom (lines 2 and 21). Also, set notation is more widely used and understood than predicate notation. Must be frustrating as hell for you, Mr. Collapse.
> Dan
>
> Download my DC Proof 2.0 freeware at http://www.dcproof.com
> Visit my Math Blog at http://www.dcproof.wordpress.com

On Thursday, September 14, 2023 at 2:26:56 PM UTC+3, Mild Shock wrote:
> No, I don't want to use T, F and U.
> I only want to use T,F.
>
> Translated to sets, I don't want to use t,f and m.
> I only want to use t,f.
>
> Whats wrong with you?
>
> I explained that already:
> ~y e t & ~y e f
>
> and
>
> y e m
> are logically equivalent for trichtomoy. Should be easy
> to produce a proof in DC Poop. Here is a proof in
> Wolfgang Schwarz tree tool:
> ∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
> ∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.
> https://www.umsu.de/trees/#~6x%28Fx~2Tx~2Ux%29,~6x~3%28Fx~1Tx%29,~6x~3%28Fx~1Ux%29,~6x~3%28Tx~1Ux%29|=%28~3Ty~1~3Fy~4Uy%29
>
> So you don't need y e m.
> Dan Christensen schrieb am Donnerstag, 14. September 2023 um 12:04:17 UTC+2:
> > On Thursday, September 14, 2023 at 3:30:56 AM UTC-4, Mild Shock wrote:
> > snip childish abuse]
> > > ~Ty & ~Fy and My are the same. Or with set notation:
> > >
> > > ~y e t & ~y e f
> > >
> > > and:
> > >
> > > y e m
> > >
> > > Are the same. This follows from your trichotomy I,
> > [snip childish abuse]
> > >
> > > [a e t | a e f | a e m]
> > > & ~[a ε t & a ε f]
> > > & ~[a ε t & a ε m]
> > > & ~[a ε f & a ε m]
> > > https://www.dcproof.com/LiarParadoxResolution.htm
> > >
> > > Thats quite easy:
> > > ∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
> > > ∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.
> > >
> > https://www.umsu.de/trees/#~6x%28Fx~2Tx~2Ux%29,~6x~3%28Fx~1Tx%29,~6x~3%28Fx~1Ux%29,~6x~3%28Tx~1Ux%29|=%28~3Ty~1~3Fy~4Uy%29
> > >
> > > So you don't need y e m.
> > [snip]
> >
> > For your trichotomy, you want to use predicates T, F and U instead sets t, f and m respectively. The advantage of using set theory is that you can actually prove the existence of a trichotomies on any given set. See the Trichotomy Lemma at https://www.dcproof.com/LiarParadoxLemma.htm. It requires the Subset Axiom (lines 2 and 21). Also, set notation is more widely used and understood than predicate notation. Must be frustrating as hell for you, Mr. Collapse.
> > Dan
> >
> > Download my DC Proof 2.0 freeware at http://www.dcproof.com
> > Visit my Math Blog at http://www.dcproof.wordpress.com

Belive me that whatever Trolls 🧌 who are a stupid belivevers of two successive integers are ultimately become suddenly equals can't comprehend anything true in logic, no matter if they use tons of symbolic chicken notations to impress others who are completely clueless as the common mainstream academic sheeples

You & him have a lot to learn, but alas, the time isn't much available to both of you due to your ages!

BKK