The Liar Paradox: My latest blog posting

[Dan Christensen]

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

[Mild Shock]

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.

[Mild Shock]

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

[Dan Christensen]

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?

[Mild Shock]

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.

[Mild Shock]

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:

[Mild Shock]

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 ---------------------------

[mitchr...@gmail.com]

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

[Dan Christensen]

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.

[Mild Shock]

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]

[Mild Shock]

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

[Dan Christensen]

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.

[Mild Shock]

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.

[Mild Shock]

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]

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.

[Dan Christensen]

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)."

[Mild Shock]

There is also moron island for people like you.

[Mild Shock]

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

[Dan Christensen]

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

[bassam karzeddin]

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

[Dan Christensen]

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.

[Mild Shock]

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]

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.

[Mild Shock]

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.

[bassam karzeddin]

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

[Dan Christensen]

On Thursday, September 14, 2023 at 7:26:56 AM UTC-4, Mild Shock wrote:

> Dan Christensen schrieb am Donnerstag, 14. September 2023 um 12:00:02 UTC+2:
[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.

> No, I don't want to use T, F and U.
> I only want to use T,F.

[snip]

I see...

> ∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
> ∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.

[snip]

Hmmm... What's this??? Looks like a "U" !!! Four occurrences at that!

It sure is handy to have that extra predicate, isn't it, Mr. Collapse. (Hee, hee!)

>
> So you don't need y e m.

You don't like to make things look simple, do you, Mr. Collapse? Do you also insist on writing SSSSS0 instead of 5? As usual, you seem to be grasping at straws here.

[Dan Christensen]

On Thursday, September 14, 2023 at 8:56:36 AM UTC-4, bassam karzeddin wrote:
[snip]

> Belive me that whatever Trolls 🧌 ...

Speaking of trolls...

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.

[Mild Shock]

It shows that:

¬Ty ∧ ¬Fy

Is logically equal **over Trichotomy** to this:

Uy

Whats wrong with you?

Dan Christensen schrieb am Donnerstag, 14. September 2023 um 17:29:48 UTC+2:
> > ∀x(Fx ∨ (Tx ∨ Ux)), ∀x¬(Fx ∧ Tx),
> > ∀x¬(Fx ∧ Ux), ∀x¬(Tx ∧ Ux) entails (¬Ty ∧ ¬Fy) ↔ Uy.

[Mild Shock]

How many years will it take you until you have
simplified your Liar Paradox resolution?

Days, months, years, decades, centuries, etc..?
Now you can use this to reduce the problem from:
T F U

---------
0 0 1
0 1 0
1 0 0

To this problem, U is redundant:
T F

-----
0 0
0 1
1 0

Mild Shock schrieb:

[Mild Shock]

The Liar Problem gets way way simpler with T and F
only than when you use T, F and U.

This is the Liar Paradox with T and F only:

Trichtomoy: ∀x~(Tx & Fx)
Liar: Ty <-> Fy
Indeterminate: ~Ty & ~Fy

You can then prove:

Trichtomoy |- Liar -> Indeterminate

Its much much easier:

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

Dan Christensen schrieb:

> This sentence is false! (The Liar Paradox)
>
> Posted yesterday at WordPress: https://dcproof.wordpress.com/2023/09/06/this-sentence-is-false/
>

[Dan Christensen]

On Thursday, September 14, 2023 at 12:29:09 PM UTC-4, Mild Shock wrote:
> The Liar Problem gets way way simpler with T and F
> only than when you use T, F and U.
>

I prefer sets instead of predicates. Again, the advantage of using set theory is that you can actually prove the existence of a trichotomy 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. Yes, it must be frustrating as hell for you. (Hee, hee!)

> This is the Liar Paradox with T and F only:
>

> Trichotomoy: ∀x~(Tx & Fx)

Makes no sense. What are you quantifying over? You should at least name the domain of quantification.

Here is how I define the trichotomy rules (like the usual ones with 3 categories):

ALL(a):[a in s => [a in t | a in f | a in m]

& ~[a in t & a in f]
& ~[a in t & a in m]
& ~[a in f & a in m]]

Where we can interpret:

s = a set of sentences, not necessarily every sentence
t = the subset of true sentences in s
f = the subset of false sentence in s
m = the set of sentences of indeterminate truth value in s

Simple.

> Liar: Ty <-> Fy
> Indeterminate: ~Ty & ~Fy
>
> You can then prove:
>
> Trichtomoy |- Liar -> Indeterminate
>
> Its much much easier:
>
> ∀x¬(Fx ∧ Tx) entails (Ty ↔ Fy) → (¬Ty ∧ ¬Fy).

Makes no sense. Better is:

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

Where s, t, f and m are interpreted as above.

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

> https://www.umsu.de/trees/#~6x%28~3%28Fx~1Tx%29%29|=%28Ty~4Fy%29~5%28~3Ty~1~3Fy%29
>

Not a proof. Again, you seem to be grasping at straws here. Just admit that your version makes no sense.

[Mild Shock]

Your "resolution" is much way to complicated.
You are a lazy bastard. Only using f and t is
much simpler than using, f, t and m.

Move your ass, don't be like Julio Di Egidio

[Dan Christensen]

On Thursday, September 14, 2023 at 2:02:57 PM UTC-4, Mild Shock wrote:

> Dan Christensen schrieb am Donnerstag, 14. September 2023 um 19:22:35 UTC+2:
> > > Its much much easier:
> > > ∀x¬(Fx ∧ Tx) entails (Ty ↔ Fy) → (¬Ty ∧ ¬Fy).

That's not a trichotomy. It's not even a dichotomy!

If you want a trichotomy with predicates, you should have something like:

ALL(a):[T(a) => S(a)]
& ALL(a):[F(a) => S(a)]
& ALL(a):[M(a) => S(a)]

& ALL(a):[S(a) => [T(a) | F(a) | M(a)]
& ~[T(a) & F(a)]
& ~[T(a) & M(a)]
& ~[F(a) & M(a)]]

Where S, T, F and M can be interpreted as follows:

S(x) = x is a sentence
T(x) = x is true
F(x) = x is false
M(x) = x is of indeterminate truth value

A dichotomy would be:

ALL(a):[T(a) => S(a)]
& ALL(a):[F(a) => S(a)]

& ALL(a):[S(a) => [T(a) | F(a) ] <-------- You missed this in your DICHOTOMY, Mr. Collapse!
& ~[T(a) & F(a)]]

> > Makes no sense. Better is:
> > ALL(b):[b in s => [[b in t <=> b in f] => b in m]]

> Your "resolution" is much way too complicated.

[snip childish abuse]

Not really. It works. And yours is a pathetic joke, Mr. Collapse!

[Mild Shock]

Or instead of using Wolfgang Schwarz tree tool, you
can also look at truth tables, and see that it is the same:

Equivalence & Trichotomy II
Fs Ts Us (((¬Ts ∧ ¬Fs) ↔ Us) ∧ ¬(Ts ∧ Fs))
F F T T
F T F T
T F F T
https://web.stanford.edu/class/cs103/tools/truth-table-tool/

Trichotomy I
Fs Ts Us ((Ts ∨ (Fs ∨ Us)) ∧ (¬(Ts ∧ Fs) ∧ (¬(Ts ∧ Us) ∧ ¬(Fs ∧ Us))))
F F T T
F T F T
T F F T
https://web.stanford.edu/class/cs103/tools/truth-table-tool/

Q.E.D.

[Dan Christensen]

> Or instead of using Wolfgang Schwarz tree tool, you
> can also look at truth tables, and see that it is the same:
>
> Equivalence & Trichotomy II
> Fs Ts Us (((¬Ts ∧ ¬Fs) ↔ Us) ∧ ¬(Ts ∧ Fs))

What a mess!

> F F T T
> F T F T
> T F F T
> https://web.stanford.edu/class/cs103/tools/truth-table-tool/
>
> Trichotomy I

> ... ((Ts ∨ (Fs ∨ Us)) ∧ (¬(Ts ∧ Fs) ∧ (¬(Ts ∧ Us) ∧ ¬(Fs ∧ Us))))

[snip]

If you don't want to use set theory, you must still use quantifiers, something like:

ALL(a):[T(a) => S(a)]
& ALL(a):[F(a) => S(a)]
& ALL(a):[M(a) => S(a)]

& ALL(a):[S(a) => [T(a) | F(a) | M(a)]
& ~[T(a) & F(a)]
& ~[T(a) & M(a)]
& ~[F(a) & M(a)]]

Recall that the Liar sentence (x) is characterized by the fact that it is a true sentence if and only if it is a false sentence:

S(x) & [T(x) <=> ~ F(x)]

What can you infer from this and the trichotomy rule? I was able to infer that M(x).

https://dcproof.com/LiarParadox2.htm

You are nowhere near finished. Let's see a complete proof from you.

[Dan Christensen]

Maybe this will help:

(∀x(Fx ∨ (Tx ∨ Ux)) ∧ (∀x¬(Fx ∧ Tx) ∧ (∀x¬(Fx ∧ Ux) ∧ ∀x¬(Tx ∧ Ux)))) → ∀x((Tx ↔ Fx) → Ux) is valid.

https://www.umsu.de/trees/#~6x(Fx~2Tx~2Ux)~1~6x(~3(Fx~1Tx))~1~6x(~3(Fx~1Ux))~1~6x(~3(Tx~1Ux))~5~6x((Tx~4Fx)~5Ux)

[Mild Shock]

I posted what you entered yourself into Wolfgang Scharz tree
tool already a dozen times. Do you want to get into
the guiness book of records being a slow thinker?

Its not the same like the simplified Liar:

Trichotomy II: ∀x¬(Fx ∧ Tx) [The formula has 3 different solutions]
Liar: (Ty ↔ Fy) [The formula has anyway no U]
Indeterminate II: (¬Ty ∧ ¬Fy) [The formula selects 1 of the 3 different solutions]

We can then show, using a trichotomy II that works
without using U, and using an indeterminate II that
works without using U, a much more simplified "resolution":

Trichotomy II |- Liar -> Indeterminate II

Its much much easier:

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

[Mild Shock]

You can use truth table generator, to verify:

Trichotomy II [The formula has 3 different solutions]
Fy Ty ¬(Fy ∧ Ty)
F F T
F T T
T F T
https://web.stanford.edu/class/cs103/tools/truth-table-tool/

Indeterminate II [The formula selects 1 of the 3 different solutions]

Fy Ty (¬Ty ∧ ¬Fy)

F F T
https://web.stanford.edu/class/cs103/tools/truth-table-tool/

But maybe you want to bullshit yourself into the guiness
book of records as the person with the most truth tables,
that he didn't understand?

Mild Shock schrieb:

[Dan Christensen]

On Friday, September 15, 2023 at 2:30:26 AM UTC-4, Mild Shock (aka Mr. Collapse) wrote:

> Dan Christensen schrieb am Freitag, 15. September 2023 um 01:59:31 UTC+2:
> > Maybe this will help:
> > (∀x(Fx ∨ (Tx ∨ Ux)) ∧ (∀x¬(Fx ∧ Tx) ∧ (∀x¬(Fx ∧ Ux) ∧ ∀x¬(Tx ∧ Ux)))) → ∀x((Tx ↔ Fx) → Ux) is valid.
> > https://www.umsu.de/trees/#~6x(Fx~2Tx~2Ux)~1~6x(~3(Fx~1Tx))~1~6x(~3(Fx~1Ux))~1~6x(~3(Tx~1Ux))~5~6x((Tx~4Fx)~5Ux)

> I posted what you entered yourself into Wolfgang Scharz tree
> tool already a dozen times.

Not the same. Look at your previous postings.

> Its not the same like the simplified Liar:
>
> Trichotomy II: ∀x¬(Fx ∧ Tx)

This only rules out sentences being both true and false. That is not a trichotomy.

> [The formula has 3 different solutions]
> Liar: (Ty ↔ Fy) [The formula has anyway no U]

That is why you are unable resolve the paradox. You need a trichotomy. You need to take into account sentences of indeterminate truth values, e.g. questions and instructions. It turns out that the Liar is just such a sentence. Deal with it, Mr. Collapse.

> Indeterminate II: (¬Ty ∧ ¬Fy) [The formula selects 1 of the 3 different solutions]
>
> We can then show, using a trichotomy II that works
> without using U,

Maybe you didn't know, but in a TRI-chotomy, every element of a given set falls into one of THREE disjoint categories.

> and using an indeterminate II that
> works without using U, a much more simplified "resolution":
>

> Trichtomoy II |- Liar -> Indeterminate II

>
> Its much much easier:
>
> ∀x¬(Fx ∧ Tx) entails ∀x((Tx ↔ Fx) → (¬Tx ∧ ¬Fx)).

This only restates the paradox. No resolution here. Try again, Mr. Collapse.

[Dan Christensen]

Apologies. Actually, this it the basis for a useful lemma.

Using set theoretic notation in DC Proof, you can prove (24 lines):

ALL(s):ALL(t):ALL(f):[ALL(a):[a in s => ~[a in t & a in f]]

=> ALL(x):[x in s => [[x in t <=> x in f] => ~x in t & ~x in f]]]

PROOF OF LEMMA

Suppose...

1. ALL(a):[a in s => ~[a in t & a in f]]
Premise

Suppose...

2. x in s
Premise

3. x in s => ~[x in t & x in f]
U Spec, 1

4. ~[x in t & x in f]
Detach, 3, 2

5. ~~[x in t => ~x in f]
Imply-And, 4

6. x in t => ~x in f
Rem DNeg, 5

Suppose...

7. x in t <=> x in f
Premise

8. [x in t => x in f] & [x in f => x in t]
Iff-And, 7

9. x in t => x in f
Split, 8

10. x in f => x in t
Split, 8

Suppose to the contrary...

11. x in t
Premise

12. x in f
Detach, 9, 11

13. ~x in f
Detach, 6, 11

14. x in f & ~x in f
Join, 12, 13

15. ~x in t
Conclusion, 11

Suppose to the contrary...

16. x in f
Premise

17. x in t
Detach, 10, 16

18. ~x in f
Detach, 6, 17

19. x in f & ~x in f
Join, 16, 18

20. ~x in f
Conclusion, 16

21. ~x in t & ~x in f
Join, 15, 20

22. [x in t <=> x in f] => ~x in t & ~x in f
Conclusion, 7

23. ALL(x):[x in s => [[x in t <=> x in f] => ~x in t & ~x in f]]
Conclusion, 2

As required:

24. ALL(s):ALL(t):ALL(f):[ALL(a):[a in s => ~[a in t & a in f]]
=> ALL(x):[x in s => [[x in t <=> x in f] => ~x in t & ~x in f]]]
Conclusion, 1

This lemma can be applied to obtain, as in my proof:

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

=> [ALL(a):[a in t => a in 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]

& ~[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]]]]

[Dan Christensen]

Unfortunately, this approach requires significantly more lines of proof. 44 lines without lemma. 38+24 = 62 lines with lemma.

[Jeffrey Rubard]

On Thursday, September 14, 2023 at 11:53:59 PM UTC-7, Mild Shock wrote:
> You can use truth table generator, to verify:
>
> Trichotomy II [The formula has 3 different solutions]
> Fy Ty ¬(Fy ∧ Ty)
> F F T
> F T T
> T F T
> https://web.stanford.edu/class/cs103/tools/truth-table-tool/
>
> Indeterminate II [The formula selects 1 of the 3 different solutions]
>
> Fy Ty (¬Ty ∧ ¬Fy)
> F F T
> https://web.stanford.edu/class/cs103/tools/truth-table-tool/
> But maybe you want to bullshit yourself into the guiness
> book of records as the person with the most truth tables,
> that he didn't understand?

"What, like 'Ryoki Inoue' is the world's most prolific book author?"

[Mild Shock]

But the ¬(A ∧ B) Trichotomy is not the only resolution.

You can define diagonals relative to X in Y as,
where Δ is the symmetric difference:
diag(X,Y) := { x e Y | ∀y(y e X => x Δ y = {}) }

You can use it to find Liar Paradoxs. For example
the Liar Paradox that Dan Christensen is using, is:
A <-> B

it is relatively easy to find: Use 0 to encode A,
and 1 to encode B, then X = {{0}, {1}} is classical
logic, and Y = {{}, {0}, {1}, {0, 1}} is Belnap FOUR,

diag(X,Y) = {{}, {0,1}}

which is the same as A <-> B. The diagonal is
automatically an Antinomy in X, but satisfiable
in Y. Anything between X and Y, excluding X and

including Y does also satisfy it, and in Dan
Christensens terminology consists a "resolution":

1) { {}, {0}, {1}}: 3-valued Logic with bottom, the solution proposed
by Dan Christensen, is a resolution
2) { {0}, {1}, {0,1}: 3-valued Logic with top, sometimes called
Logic of Paradox, would be also a resolution
3) { {}, {0}, {1}, {0,1}: 4-valued Logic with bottom and top, called
Belnap FOUR, would be also a resolution

Have Fun!

[Mild Shock]

What is the Logic of Paradox? It was described here:

The logic of paradox
by G Priest · 1979 · Cited by 1308
https://link.springer.com/article/10.1007/BF00258428

Google shows a huge number of citations. But Dan
Christensen doesn't read books, neither does Julio Di Egidio.
Or maybe he reads something but doesn't tell us?

LoL

Want to know more about this currious logic? It is 3-valued
doesn't have T, F and U. Instead it has T, F and P, where P
means simultaneously true and false. More info:

Analytic Proof Systems for Priest’s Logic Of
Paradox LP M. Ultlog - July 23, 2022
https://www.logic.at/multlog/paradox.pdf

[Dan Christensen]

On Friday, September 15, 2023 at 5:03:52 PM UTC-4, Mild Shock wrote:
> But the ¬(A ∧ B) Trichotomy is not the only resolution.
>

[snip]

I will stick to my proof at https://dcproof.com/LiarParadox2.htm Yours make no sense on its own, but thanks for your input. It has been helpful even if I ignored it in the end.

[Chris M. Thomasson]

On 9/15/2023 4:51 PM, Dan Christensen wrote:
> On Friday, September 15, 2023 at 5:03:52 PM UTC-4, Mild Shock wrote:
>> But the ¬(A ∧ B) Trichotomy is not the only resolution.
>>
> [snip]
>
> I will stick to my proof at https://dcproof.com/LiarParadox2.htm Yours make no sense on its own, but thanks for your input. It has been helpful even if I ignored it in the end.

Helpful, but ignored?

[Dan Christensen]

On Friday, September 15, 2023 at 10:07:42 PM UTC-4, Chris M. Thomasson wrote:
> On 9/15/2023 4:51 PM, Dan Christensen wrote:
> > On Friday, September 15, 2023 at 5:03:52 PM UTC-4, Mild Shock wrote:
> >> But the ¬(A ∧ B) Trichotomy is not the only resolution.
> >>
> > [snip]
> >
> > I will stick to my proof at https://dcproof.com/LiarParadox2.htm Yours make no sense on its own, but thanks for your input. It has been helpful even if I ignored it in the end.

> Helpful, but ignored.

I honestly considered JB's recommendation, but ultimately rejected it.

[Chris M. Thomasson]

On 9/15/2023 8:54 PM, Dan Christensen wrote:
> On Friday, September 15, 2023 at 10:07:42 PM UTC-4, Chris M. Thomasson wrote:
>> On 9/15/2023 4:51 PM, Dan Christensen wrote:
>>> On Friday, September 15, 2023 at 5:03:52 PM UTC-4, Mild Shock wrote:
>>>> But the ¬(A ∧ B) Trichotomy is not the only resolution.
>>>>
>>> [snip]
>>>
>>> I will stick to my proof at https://dcproof.com/LiarParadox2.htm Yours make no sense on its own, but thanks for your input. It has been helpful even if I ignored it in the end.
>> Helpful, but ignored.
>
> I honestly considered JB's recommendation, but ultimately rejected it.

So, its true that the recommendation made you considered it, that is
true. However, its false that you accepted it? Fair enough? This is fun
to model via probabilities.

[Mild Shock]

Corr.: Made a little Typo here, should say =\= {} instead = {}.

Better call it the anti set, i.e. we can define:
anti(X,Y) := { x e Y | ∀y(y e X => x Δ y =\= {}) }
Now if we have extensionality available, we find:
anti(X,Y) = { x e Y | ∀y(y e X => x =\= y) }
And by contraposition:
anti(X,Y) = { x e Y | ∀y(x = y => ~y e X) }
And by substitution:
anti(X,Y) = { x e Y | ~x e X }

Hence the anti set is just the set difference:
anti(X, Y) = Y \ X.

Diagonal then referes to a popular construction of a set,
namely this construction:
diag(X) := { x e X | ~(x e x) }

It can be used for Russells Paradox, in the form of Dan Christensens
Generalized Drinker paradox, we can use the construction to show
that anti(X,V) is non-empty, since it is provable:
diag(X) e anti(X,V)

Or it can be used for Cantors Theorem, to show:
diag(X) e anti(X,P(X))

Have Fun!

[Mild Shock]

In Belnap FOUR, the Liar Sentences is not diag(X).
In Belnap FOUR we have X = { {}, {0}, {1}, {0,1} },
the diag is way too set theoretic, it jumps out of

X cannot be used to analyse multi-valued logic.
diag(X) gives the set X itself. Namely we find
quite trivially that the diagonal is:
diag(X) = { {}, {0}, {1}, {0,1} } = X

And X is of course element of the universal class
V, so that we have X e V, and hence X e anti(X,V).
And X is also element of the power set, i.e.

X e P(X), and hence X e anti(X,P(X)). But for multi-valued
logic, we need to define another type of diagonal.
Or we go with the set anti(X,Y) directly.

A propositional formula identifies a set of propositional
variable assignments, the satisfaction set. So looking
for a formula with satisfaction set anti(X,Y) works fine.

Every formula L, where sat(L) has this property:
sat(L) =\= {} and sat(L) ⊆ anti(X,Y) and .

We can show two theorems:
1) L is an antinomy in X
2) L is resolved in Y

Proof: 1) sat(L) ⊆ anti(X,Y) is the same as sat(L) ⊆ Y \ X.
Now assume that there were a propositional variable assignment
x that would make L true, and that were in X. Then we would
have x e sat(L) and x e Y \ X. But by definition ~(x e Y \ X),
and hence it should be also ~(x e sat(L)), hence our assumption is false.
2) sat(L) implies that it is resolved. sat(L) ⊆ anti(X,Y) implies
that it is resolved in Y, since sat(L) ⊆ Y \ X also implies sat(L) ⊆ Y.

Q.E.D.

[Mild Shock]

Corr.: Typo in the proof:

have x e sat(L) and x e Y \ X. But by definition ~(x e Y \ X),

Should read:
have x e sat(L) and x e X. But by definition ~(x e Y \ X),

[Edison Lappo]

Mild Shock wrote:

> Corr.: Typo in the proof:
> have x e sat(L) and x e Y \ X. But by definition ~(x e Y \ X),
>
> Should read:
> have x e sat(L) and x e X. But by definition ~(x e Y \ X),

how should I read that, when 𝘆𝗼𝘂 𝘄𝗿𝗶𝘁𝗲 𝘀𝗼𝗺𝗲𝘁𝗵𝗶𝗻𝗴 𝗲𝗹𝘀𝗲. Which one shall I read
then?? Your country 𝗺𝗮𝗸𝗲𝘀 𝗳𝘂𝗻𝗻𝘆 𝗺𝗼𝗻𝗲𝘆 𝗼𝘂𝘁 𝗼𝗳 𝘁𝗵𝗲 𝘁𝗵𝗶𝗻 𝗮𝗶𝗿, on a computer,
𝘄𝗶𝘁𝗵𝗼𝘂𝘁 𝘄𝗼𝗿𝗸. Which is illegal. It is stealing 𝗳𝗿𝗼𝗺 𝗺𝗮𝗱𝗲 𝗽𝗼𝗼𝗿 𝗽𝗲𝗼𝗽𝗹𝗲, going
hungry in bed. You disgusting thief. It is written in The Bible that 𝘆𝗼𝘂
𝘀𝘁𝗲𝗮𝗹 𝗴𝗼𝗼𝗱𝘀 𝗮𝗻𝗱 𝘄𝗲𝗮𝗹𝘁𝗵, with the fake money you make. You disgusting
satanists.

𝗭𝗲𝗹𝗲𝗻𝘀𝗸𝘆_𝗽𝗼𝗹𝗶𝘁𝗶𝗰𝗮𝗹_𝗿𝗶𝘃𝗮𝗹_𝗰𝗵𝗮𝗿𝗴𝗲𝗱_𝘄𝗶𝘁𝗵_𝘁𝗿𝗲𝗮𝘀𝗼𝗻
https://r%74.com/ru%73%73ia/583051-ukraine-zelensky-shufrych-treason/

they have military bases around the world, dictating "democracy", then
accuse the President of China of being "dictator".

𝗚𝗲𝗿𝗺𝗮𝗻_𝗙𝗠_𝗹𝗮𝗯𝗲𝗹𝘀_𝗫𝗶_‘𝗮_𝗱𝗶𝗰𝘁𝗮𝘁𝗼𝗿’
The West must help Ukraine win its conflict against Russia to deter people
like the Chinese leader, 𝗔𝗻𝗻𝗮𝗹-ena Baerbock has claimed
https://r%74.com/news/583050-germany-baerbock-xi-dictator/

That's an insult coming from a real dictator and a nazi. Germans are
genetically genocidal and that's a fact, not an insult. They've proven it,
not once, not twice, but 7-8 times and are now proving in Bosnia with what
their nazi dictator is doing to Republic of Srpska.

She complains of Dictator at the same time she Dictates ...
western bureaucrat ignorance never ceases to amaze

Seems like the braincells that had been left also jumped ship.

Incredible shame that damn fool (Gorbachev) returned East Germany after
the Nazis lost it in a fair fight.

Typical German bimbo repeating the dumb Americans warmongers talking
points.

These 𝘄𝗲𝘀𝘁𝗲𝗿𝗻_𝗿𝗮𝗰𝗶𝘀𝘁_𝗰𝗿𝗲𝗮𝘁𝘂𝗿𝗲𝘀 think they can do or talk whatever they like
from their filthy mouth

The US & EU label other leaders they cannot control as dictators. Just
more demonization campaign against China.

Dumb FOOL cannot recognize that the REAL Dictator is in the US, 𝘄𝗶𝘁𝗵 𝗺𝗶𝗹𝗶𝘁𝗮𝗿𝘆
𝗯𝗮𝘀𝗲𝘀 𝗶𝗻 𝗵𝗲𝗿 𝗼𝘄𝗻 𝗰𝗼𝘂𝗻𝘁𝗿𝘆, dictating 𝘄𝗵𝗮𝘁 𝘁𝗵𝗶𝘀 𝘂𝗴𝗹𝘆 𝗯𝗶𝘁𝗰𝗵 𝗵𝗮𝘀 𝘁𝗼 𝘀𝗮𝘆. How much more
STUPID can you get. What a disgrace to the German Population

[Dan Christensen]

On Saturday, September 16, 2023 at 7:43:47 AM UTC-4, Edison Lappo wrote:
> Mild Shock wrote:
>
> > Corr.: Typo in the proof:
> > have x e sat(L) and x e Y \ X. But by definition ~(x e Y \ X),
> >
> > Should read:
> > have x e sat(L) and x e X. But by definition ~(x e Y \ X),

> how should I read that, when 𝘆𝗼𝘂 𝘄𝗿𝗶𝘁𝗲 𝘀𝗼𝗺𝗲𝘁𝗵𝗶𝗻𝗴 𝗲𝗹𝘀𝗲. Which one shall I read
> then?? Your country 𝗺𝗮𝗸𝗲𝘀 𝗳𝘂𝗻𝗻𝘆 𝗺𝗼𝗻𝗲𝘆 𝗼𝘂𝘁 𝗼𝗳 𝘁𝗵𝗲 𝘁𝗵𝗶𝗻 𝗮𝗶𝗿, on a computer,

[snip]

Somehow you have avoided being conscripted, Nazi Boy? How is that possible? When are you going to volunteer to die for Putin's political career?