You can try yourself, you reject p | ~p, you cannot prove:

ALL(b)[b e s => b e t | b e f]
The counter example is b e m. Come on, you seem to have
reached a phase of denial in Scott and Jaffe’s model:

1. Denial
This is the first stage of Scott and Jaffe’s Change Model.
We enter this phase when we first learn about the change.
2. Resistance
3. Exploration
4. Commitment

So how many decades will it take, until you accept non-classical
logics? You are using them by yourself already here.

LMAO!

> > In logic, the law of excluded middle (or the principle of excluded middle)
> > states that for every proposition, either this proposition or its negation is true.
> > https://en.wikipedia.org/wiki/Law_of_excluded_middle

On Tuesday, July 25, 2023 at 9:08:17 PM UTC+3, Dan Christensen wrote:
> Some sentences are neither true nor false. We can say that their truth values are _indeterminate_ , not unlike the numerical value of 1/0. Some examples:
>
> - What time is it?
>
> - Wash the dishes.
>
> - This sentence is false.
>
> The latter will take some explaining: Suppose we have a set of sentences s. Define subsets t, f and m of set s such that
>
> t = the subset of "true" sentences
> f = the subset of "false" sentences
> m = the subset of sentences of indeterminate truth value
>
> Each element of s will be an element of precisely one of these subsets.
>
> Now, "This sentence is false" is problematic. It is assumed to be true if and only if it is false. Therefore, its truth value must be indeterminate.
>
> Formal proof: https://dcproof.com/LiarParadox2.htm (only 44 lines in DC Proof format)
>
> Your comments?
>
> Dan

What kind of mythematical hallucinations you are talking about?

0 was proven as no number, but you still want to divide one by zero, don't you?

There isn't any Paradox here but only mouthmatical endless hallucinations for sure

Bassam Karzeddin 🔊

On Wednesday, August 30, 2023 at 12:03:49 PM UTC-4, Mild Shock wrote:
> You can try yourself, you reject p | ~p, you cannot prove:
>
> ALL(b)[b e s => b e t | b e f]
>
[snip]

I assume, among other things that: ALL(a):[a in s => [a in t | a in f | a in m]

You seem to be denying the possibility that a sentence like "Wash your hangs" can be of indeterminate truth value (a in m). Try again. No, you do NOT need to postulate a third truth value.

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, August 30, 2023 at 1:04:36 PM UTC-4, bassam karzeddin wrote:
> On Tuesday, July 25, 2023 at 9:08:17 PM UTC+3, Dan Christensen wrote:
> > Some sentences are neither true nor false. We can say that their truth values are _indeterminate_ , not unlike the numerical value of 1/0. Some examples:
> >
> > - What time is it?
> >
> > - Wash the dishes.
> >
> > - This sentence is false.
> >
> > The latter will take some explaining: Suppose we have a set of sentences s. Define subsets t, f and m of set s such that
> >
> > t = the subset of "true" sentences
> > f = the subset of "false" sentences
> > m = the subset of sentences of indeterminate truth value
> >
> > Each element of s will be an element of precisely one of these subsets.
> >
> > Now, "This sentence is false" is problematic. It is assumed to be true if and only if it is false. Therefore, its truth value must be indeterminate.
> >
> > Formal proof: https://dcproof.com/LiarParadox2.htm (only 44 lines in DC Proof format)
> >
> > Your comments?
> >
> > Dan
> What kind of mythematical hallucinations you are talking about?
>
> 0 was proven as no number....

Oh, really? I guess we can stop right here.

Dan

On Wednesday, August 30, 2023 at 1:59:34 PM UTC-4, Dan Christensen wrote:
> On Wednesday, August 30, 2023 at 1:04:36 PM UTC-4, bassam karzeddin wrote:
> > On Tuesday, July 25, 2023 at 9:08:17 PM UTC+3, Dan Christensen wrote:
> > > Some sentences are neither true nor false. We can say that their truth values are _indeterminate_ , not unlike the numerical value of 1/0. Some examples:
> > >
> > > - What time is it?
> > >
> > > - Wash the dishes.
> > >
> > > - This sentence is false.
> > >
> > > The latter will take some explaining: Suppose we have a set of sentences s. Define subsets t, f and m of set s such that
> > >
> > > t = the subset of "true" sentences
> > > f = the subset of "false" sentences
> > > m = the subset of sentences of indeterminate truth value
> > >
> > > Each element of s will be an element of precisely one of these subsets.
> > >
> > > Now, "This sentence is false" is problematic. It is assumed to be true if and only if it is false. Therefore, its truth value must be indeterminate.
> > >
> > > Formal proof: https://dcproof.com/LiarParadox2.htm (only 44 lines in DC Proof format)
> > >
> > > Your comments?
> > >
> > > Dan
> > What kind of mythematical hallucinations you are talking about?
> >
> > 0 was proven as no number....
>
> Oh, really? I guess we can stop right here.
>

I nearly forgot...

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.

You rejected LEM. Which isn't ordinary logic.
According to your own account ordinary logic is
only classical logic. But you rejected LEM, since

LEM is not anymore provable:

ALL(b):[b e s => b e t | b e f]

Commands, like "Don't be stupid!", or questions, "Why
are you stupid?", are not affected, since they are not
statements. They anyway don't belong into the set s.

A statement is something that has a truth value.
But comands and questions don't have truth values.
A command c or a question q is not member of s,

it is anyway the case:

/* Commands are not Statements */
~c e s
/* Questions are not Statements */
~q e s

But since m is a partition of s, i.e. m ⊆ s, we can follow:

/* Commands are not "Dan O Matik Indeterminate" */
~c e m
/* Questions are not "Dan O Matik Indeterminate" */
~q e m

Thats quite easy to proof:

7 ~c @ m
4 Conclusion, 3

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

1 ~c @ s
Axiom

2 ALL(b):[b @ m => b @ s]
Axiom

3 c @ m
Premise

4 c @ m => c @ s
U Spec, 2

5 c @ s
Detach, 4, 3

6 ~c @ s & c @ s
Join, 1, 5

7 ~c @ m
4 Conclusion, 3

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

Dan Christensen schrieb am Mittwoch, 30. August 2023 um 19:56:37 UTC+2:
> On Wednesday, August 30, 2023 at 12:03:49 PM UTC-4, Mild Shock wrote:
> > You can try yourself, you reject p | ~p, you cannot prove:
> >
> > ALL(b)[b e s => b e t | b e f]
> >
> [snip]
>
> I assume, among other things that: ALL(a):[a in s => [a in t | a in f | a in m]
>
> You seem to be denying the possibility that a sentence like "Wash your hangs" can be of indeterminate truth value (a in m). Try again. No, you do NOT need to postulate a third truth value.
> 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, August 30, 2023 at 4:59:36 PM UTC-4, Mild Shock wrote:
> You rejected LEM. Which isn't ordinary logic.
> According to your own account ordinary logic is
> only classical logic. But you rejected LEM, since
>
> LEM is not anymore provable:
> ALL(b):[b e s => b e t | b e f]

[snip]

Where are you getting this nonsense? What is so hard to understand? Once again, in any set of natural language sentences, some may be true, some may be false and the rest would be of indeterminate truth value. This suggests that the natural trichotomy that I model in my proof. "This sentence is false" turns out to fall into the latter category.

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

Dan Christensen halcuinated:
> Recall that you were unable to demonstrate here that a trichotomy
> on any given set is somehow inconsistent with LEM.

Whats wrong with you? I demonstrated that already:

∃xTx, ∃xFx, ∃xMx,
∀x(Tx ∨ (Fx ∨ Mx)),
∀x¬(Tx ∧ Fx), ∀x¬(Tx ∧ Mx), ∀x¬(Fx ∧ Mx)
does not entail ∀x(Tx ∨ Fx).
https://www.umsu.de/trees/#~7xTx,~7xFx,~7xMx,~6x(Tx~2Fx~2Mx),~6x(~3(Tx~1Fx)),~6x(~3(Tx~1Mx)),~6x(~3(Fx~1Mx))|=~6x(Tx~2Fx)

Or without the ∃xTx, ∃xFx, ∃xMx:

∀x(Tx ∨ (Fx ∨ Mx)),
∀x¬(Tx ∧ Fx), ∀x¬(Tx ∧ Mx), ∀x¬(Fx ∧ Mx)
does not entail ∀x(Tx ∨ Fx).
https://www.umsu.de/trees/#~6x(Tx~2Fx~2Mx),~6x(~3(Tx~1Fx)),~6x(~3(Tx~1Mx)),~6x(~3(Fx~1Mx))|=~6x(Tx~2Fx)

Trichotomy rejects LEM. It does not fully embrace
i.e. accept LEM, LEM doesn't entail.

Dan Christensen schrieb am Donnerstag, 31. August 2023 um 04:18:19 UTC+2:
> On Wednesday, August 30, 2023 at 4:59:36 PM UTC-4, Mild Shock wrote:
> > You rejected LEM. Which isn't ordinary logic.
> > According to your own account ordinary logic is
> > only classical logic. But you rejected LEM, since
> >
> > LEM is not anymore provable:
> > ALL(b):[b e s => b e t | b e f]
> [snip]
>
> Where are you getting this nonsense? What is so hard to understand? Once again, in any set of natural language sentences, some may be true, some may be false and the rest would be of indeterminate truth value. This suggests that the natural trichotomy that I model in my proof. "This sentence is false" turns out to fall into the latter category.
>
> 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

Does your Liar sentence have a name?

BTW: I nowhere deined that there might be sentences like
"Wash your hands" or "This sentence is false" and that
somebody might proclaim them as indeterminate.

For example some guy by the name Dan Christensen did that here:
https://dcproof.com/LiarParadox2.htm

So where would I deny that. Can you show me a post
where I denied that somebody might proclaim their indeterminacy.
The question is only, is it necessary.

If you would only use your brain a split second, you would
see that your theorem:

=> ALL(b):[b e s => [[b e t <=> b e f] => b e m]]]]
https://dcproof.com/LiarParadox2.htm

Can be strengthed. Does your Liar sentence have a name?
For example b0 ? Lets say b0 is the name of the Liar sentences.
Then the above theorem can be strengthed to:

=> ALL(b):[b e s' => [[b e t <=> b e f] => b e m]]]]
https://dcproof.com/LiarParadox2.htm

Where s' = s ∩ {b0}

Yes or no? What does this say about the relevance of sentences b=\=b0 ?
Namely sentences b different from the Liar Paradox b0 ? Are they
relevant, you can form a theorem using s' without them.

LMAO!

You had more luck in the Epimenides Paradox, using a name x for
the statement. Your Liar Paradox is a bad imitation. But here not only
didn't you give a name, we also don't know whether

its a sentence. If you further assume that Liar sentence b0 is element
of s, which you nowhere assume so far, then you can strengthen
the theorem to:

=> [[b0 e t <=> b0 e f] => b0 e m]]]]
https://dcproof.com/LiarParadox2.htm

Meaning any sentences b different from the Liar sentence itself
b0, i.e. b =\= b0, fall out of the equation, are ultimately
completely irrelevant.

Proof:
Thats not so difficult to see. If s' = s ∩ {b0} and if
b0 ∈ s, then we know that s' = {b0}. But then the
theorem of the form:

=> ALL(b):[b e s' => [[b e t <=> b e f] => b e m]]]]
https://dcproof.com/LiarParadox2.htm

Is logically equivalent to the theorem:

=> [[b0 e t <=> b0 e f] => b0 e m]]]]
https://dcproof.com/LiarParadox2.htm

Because for s' = {b0}, we have that b e s' is logical
equivalent to b = b0, and then we can eliminate b.
Q.E.D.

Mild Shock schrieb am Montag, 4. September 2023 um 01:06:00 UTC+2:
> Does your Liar sentence have a name?
>
> BTW: I nowhere deined that there might be sentences like
> "Wash your hands" or "This sentence is false" and that
> somebody might proclaim them as indeterminate.
>
> For example some guy by the name Dan Christensen did that here:
> https://dcproof.com/LiarParadox2.htm
>
> So where would I deny that. Can you show me a post
> where I denied that somebody might proclaim their indeterminacy.
> The question is only, is it necessary.
>
> If you would only use your brain a split second, you would
> see that your theorem:
> => ALL(b):[b e s => [[b e t <=> b e f] => b e m]]]]
> https://dcproof.com/LiarParadox2.htm
> Can be strengthed. Does your Liar sentence have a name?
> For example b0 ? Lets say b0 is the name of the Liar sentences.
> Then the above theorem can be strengthed to:
>
> => ALL(b):[b e s' => [[b e t <=> b e f] => b e m]]]]
> https://dcproof.com/LiarParadox2.htm
>
> Where s' = s ∩ {b0}
>
> Yes or no? What does this say about the relevance of sentences b=\=b0 ?
> Namely sentences b different from the Liar Paradox b0 ? Are they
> relevant, you can form a theorem using s' without them.
>
> LMAO!

On top of your bullshit, you even don't understand why LEM
is rejected. Despite the fact that this here is not provable:

EXIST(b):[b e s]

LoL

Mild Shock schrieb am Montag, 4. September 2023 um 01:15:21 UTC+2:
> You had more luck in the Epimenides Paradox, using a name x for
> the statement. Your Liar Paradox is a bad imitation. But here not only
> didn't you give a name, we also don't know whether
>
> its a sentence. If you further assume that Liar sentence b0 is element
> of s, which you nowhere assume so far, then you can strengthen
> the theorem to:
>
> => [[b0 e t <=> b0 e f] => b0 e m]]]]
> https://dcproof.com/LiarParadox2.htm
>
> Meaning any sentences b different from the Liar sentence itself
> b0, i.e. b =\= b0, fall out of the equation, are ultimately
> completely irrelevant.
>
> Proof:
> Thats not so difficult to see. If s' = s ∩ {b0} and if
> b0 ∈ s, then we know that s' = {b0}. But then the
> theorem of the form:
> => ALL(b):[b e s' => [[b e t <=> b e f] => b e m]]]]
> https://dcproof.com/LiarParadox2.htm
> Is logically equivalent to the theorem:
>
> => [[b0 e t <=> b0 e f] => b0 e m]]]]
> https://dcproof.com/LiarParadox2.htm
>
> Because for s' = {b0}, we have that b e s' is logical
> equivalent to b = b0, and then we can eliminate b.
> Q.E.D.
> Mild Shock schrieb am Montag, 4. September 2023 um 01:06:00 UTC+2:
> > Does your Liar sentence have a name?
> >
> > BTW: I nowhere deined that there might be sentences like
> > "Wash your hands" or "This sentence is false" and that
> > somebody might proclaim them as indeterminate.
> >
> > For example some guy by the name Dan Christensen did that here:
> > https://dcproof.com/LiarParadox2.htm
> >
> > So where would I deny that. Can you show me a post
> > where I denied that somebody might proclaim their indeterminacy.
> > The question is only, is it necessary.
> >
> > If you would only use your brain a split second, you would
> > see that your theorem:
> > => ALL(b):[b e s => [[b e t <=> b e f] => b e m]]]]
> > https://dcproof.com/LiarParadox2.htm
> > Can be strengthed. Does your Liar sentence have a name?
> > For example b0 ? Lets say b0 is the name of the Liar sentences.
> > Then the above theorem can be strengthed to:
> >
> > => ALL(b):[b e s' => [[b e t <=> b e f] => b e m]]]]
> > https://dcproof.com/LiarParadox2.htm
> >
> > Where s' = s ∩ {b0}
> >
> > Yes or no? What does this say about the relevance of sentences b=\=b0 ?
> > Namely sentences b different from the Liar Paradox b0 ? Are they
> > relevant, you can form a theorem using s' without them.
> >
> > LMAO!

The website contains a proof that the Yablo Paradox
is paradoxical as an assumption:

Yablo |- ⊥
"We can show that the Yablo sequence entails a contradiction
in a similar manner via the following informal argument:"
https://iep.utm.edu/yablo-pa/

Similar like we can prove:

Liar |- ⊥
"The Yablo sequence of sentences seem **paradoxical** in the
same sense as the more well-known Liar Paradox:"
https://iep.utm.edu/yablo-pa/

Nobody uses the phrase "indeterminate", everybody calls these
paradoxes **paradoxical**. A corollary of the above two proofs,
in classical logic, is that both are Antinomies:

|- ~Yablo

|- ~Liar

Dan Christensen schrieb am Dienstag, 25. Juli 2023 um 20:08:17 UTC+2:
> Some sentences are neither true nor false. We can say that their truth values are _indeterminate_ , not unlike the numerical value of 1/0. Some examples:
>
> - What time is it?
>
> - Wash the dishes.
>
> - This sentence is false.
>
> The latter will take some explaining: Suppose we have a set of sentences s. Define subsets t, f and m of set s such that
>
> t = the subset of "true" sentences
> f = the subset of "false" sentences
> m = the subset of sentences of indeterminate truth value
>
> Each element of s will be an element of precisely one of these subsets.
>
> Now, "This sentence is false" is problematic. It is assumed to be true if and only if it is false. Therefore, its truth value must be indeterminate.
>
> Formal proof: https://dcproof.com/LiarParadox2.htm (only 44 lines in DC Proof format)
>
> Your comments?
>
> Dan

Dan Christensen isn't the smartest tool in the shed.
How many weeks will it take until he understands
that this here:

~y e f & ~y e t

and this here:

y e m

Are logically equivalent over Trichotomy?

On Thursday, September 14, 2023 at 2:30:20 PM UTC+3, Mild Shock wrote:
> Dan Christensen isn't the smartest tool in the shed.
> How many weeks will it take until he understands
> that this here:
>
> ~y e f & ~y e t
>
> and this here:
>
> y e m
>
> Are logically equivalent over Trichotomy?

See the Trolls 🧌 endless fight & hallucinations about nothing in logic or philosophy 🤔!

BKK

On Thursday, September 14, 2023 at 8:47:21 AM UTC-4, bassam karzeddin wrote:
[snip]

> See the Trolls 🧌 endless fight & hallucinations about nothing in logic or philosophy 🤔!
>

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.

Dan

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

I am not talking about the Revenge Paradox you moron.
I am demonstrating how to reduce the problem from this:
T F U
---------
0 0 1
0 1 0
1 0 0

To this problem:
T F
-----
0 0
0 1
1 0

Obviously the column U is redundant, it can be computed as ~Ty & ~Fy,
which even the stanford truth table tool knows:
A B (¬A ∧ ¬B)
F F T
F T F
T F F
https://web.stanford.edu/class/cs103/tools/truth-table-tool/

But morons like Dan Christensen and Julio Di Egidio can even
not read truth tables, neither they understand what negation ~
and conjunction & does. Whats wrong with you guys,

do you have zero brain cells?

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(~3(Fx~1Tx))|=(Ty~4Fy)~5(~3Ty~1~3Fy)

Mild Shock schrieb am Donnerstag, 14. September 2023 um 18:20:19 UTC+2:
> I am not talking about the Revenge Paradox you moron.
> I am demonstrating how to reduce the problem from this:
> T F U
> ---------
> 0 0 1
> 0 1 0
> 1 0 0
>
> To this problem:
> T F
> -----
> 0 0
> 0 1
> 1 0
>
> Obviously the column U is redundant, it can be computed as ~Ty & ~Fy,
> which even the stanford truth table tool knows:
> A B (¬A ∧ ¬B)
> F F T
> F T F
> T F F
> https://web.stanford.edu/class/cs103/tools/truth-table-tool/
>
> But morons like Dan Christensen and Julio Di Egidio can even
> not read truth tables, neither they understand what negation ~
> and conjunction & does. Whats wrong with you guys,
>
> do you have zero brain cells?

On Thursday, September 14, 2023 at 12:48:02 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.

Dan

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