Re: True on the basis of meaning --- Good job Richard ! ---Socratic method

On 5/20/2024 9:24 PM, Richard Damon wrote:

On 5/20/24 9:54 PM, olcott wrote:

On 5/20/2024 7:57 PM, Richard Damon wrote:

On 5/20/24 2:59 PM, olcott wrote:

On 5/19/2024 6:30 PM, Richard Damon wrote:

On 5/19/24 4:12 PM, olcott wrote:

On 5/19/2024 12:17 PM, Richard Damon wrote:

On 5/19/24 9:41 AM, olcott wrote:

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True(L,x) is always a truth bearer.
when x is defined as True(L,x) then x is not a truth bearer.

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So, x being DEFINED to be a certain sentence doesn't make x to have the same meaning as the sentence itself?
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What does it mean to define a name to a given sentence, if not that such a name referes to exactly that sentence?
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p = ~True(L,p) // p is not a truth bearer because its refers to itself

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Then ~True(L,p) can't be a truth beared as they are the SAME STATEMENT, just using different "names".

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Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
p = ~True(L,p) Truthbearer(L,p) is false
q = ~True(L,p) Truthbearer(L,q) is true

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Irrelvent.
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If Truthbearer(L, p) is FALSE, and since p is just a NAME for the statement ~True(L, p), that means that True(L. p) is not a truth bearer and True has failed to be the required truth predicate.
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That is the same thing as saying that
True(English, "this sentence is not true") is false
proves that True(L,x) is not a truthbearer.

 Nope, why do you say that?
 What logic are you even TRYING to use to get there?
 I think you don't understand what defining a label to represent a statement means.
 

I did not said the above part exactly precisely to address
your objection.
p is defined as ~True(L,p)
LP is defined as "this sentence is not true" in English.
Thus True(L,p) ≡ True(English,LP) and
Thus True(L,~p) ≡ True(English,~LP)

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If you are defining your "=" symbol to be "is defined as" so the left side is now a name for the right side, you statement above just PROVES that your logic system is inconsistant as the same expression (with just different names) has contradicory values.
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You are just showing you utter lack of understanding of the fundamentals of Formal Logic.
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    ϕ(x) there is a sentence ψ such that S ⊢ ψ ↔ ϕ⟨ψ⟩.
The sentence ψ is of course not self-referential in a strict sense, but mathematically it behaves like one. https://plato.stanford.edu/entries/self-reference/#ConSemPar

 So? Can you show that it is NOT true? or is it just that you don't want it to be true, so you assume it isn't?
 

defined as is the way to go.

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No what it shows is that formal logic gets the wrong answer because
formal logic does not evaluate actual self-reference.

 No, you don't understand what you are talking about.
 

Formal logic NEVER EVER gets to
epistemological antinomies ARE NOT TRUTH BEARERS

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Just like (with context) YOU can be refered to a PO, Peter, Peter Olcott or Olcott, and all the reference get to the exact same entity, so any "name" for the express
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True(L,p)  is false
True(L,~p) is false
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So since True(L, p) is false, then ~True(L, p) is true.
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~True(True(L,p)) is true and is referring to the p that refers
to itself it is not referring to its own self.
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*ONE LEVEL OF INDIRECT REFERENCE MAKES ALL THE DIFFERENCE*

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Why add the indirection? p is the NAME of the statement, which means exactly the same thing as the statement itself.
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p = ~True(L,p)
does not mean that same thing as True(L, ~True(L,p))
The above ~True(L, p) has another ~True(L,p) embedded in p.
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Is the definition of an English word one level LESS of indirection than the word itself?
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This sentence is not true("This sentence is not true") is true.

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Right, that is a sentence about another sentence (that is part of itself)
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Likewise with ~True(L, ~True(L, p)) where p is defined as ~True(L, p)
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 So? Yes ~True(L, ~True(L, p)) IS a different sentence than ~True(L, p) even with p defined a ~True(L, p), BUT they are logically connected as the first follows as a consequence of the second and the definition of p.
 

p defined as ~True(L, p) isn't a sentence refering to ~True(L, p), it is assigning a name to the sentence to allow OTHER sentences to refer to it by name,
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Yet when p refers to its own name this creates infinite recursion.
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 So? What's wrong with that?

Sure any programs that get stuck in infinite loops are a feature that
everyone likes even when it means that payroll is two weeks late and
you missed your mortgage payment.

Note, it is recursion that doesn't HAVE to be followed. You seem to be stuck at counting the fingers level math, while trying to talk about trigonometry.
 

Any expression "standing for some kind of infinite structure."
CANNOT BE EVALUATED THUS CANNOT POSSIBLY BE A TRUTH BEARER
THUS <IS> A TYPE MISMATCH ERROR FOR EVERY SYSTEM OF BIVALENT LOGIC

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I don't think you understand what it means to define something.
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x := y means x is defined to be another name for y
https://en.wikipedia.org/wiki/List_of_logic_symbols
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LP := ~True(L, LP)
specifies ~True(~True(~True(~True(~True(...)))))

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Nope.
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When LP refers to its own name this creates infinite recursion.

 So? As I said, it doesn't HAVE to be fully expanded, as each level is doing a logical step of indirection
 

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It means that LP is defined to be the sentence ~True(L, LP)
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replacing the LP in the sentence with a copy of LP IS a level of indirection, so you can get the infinite expansion if you keep or derefencing the reference in the statement.
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"Definition by example" is worse than "Proof by example", at least proof by example can be correct if the assertion is that there exists, and not for all.
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A simpler isomorphism of the same thing is proof by analogy.
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Which isn't a valid proof in a formal system. You seem to think Formal System are a loosy goosy with proofs as Philosophy.
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True(English, "this sentence is not true") is false
Is 100% perfectly isomorphic to its formalized version
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LP is defined as ~True(L, LP)
True(L, LP) is false

 Nope. Because "this sentence" refers to the statement in quotes, not the logical statement using True.
 

The English is formalized as LP is defined as ~True(L, LP)
before it is analyzed.

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It is merely easier to see that "this sentence is not true"
cannot be true because that makes it false and
can't be false because that makes it true.

 And it is a different sentence.
 

No it is not.
The English is formalized as
LP is defined as ~True(L, LP) before it is analyzed.

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LP is defined as ~True(L, LP)
works this same yet yet it is not as intuitive.

 You are right that it causes problems, and the problem it causes is that it shows that the True Predicate can not exist.
 

Not at all.
It shows that no truth bearers must be rejected as
a type mismatch error for any system of bivalent logic.

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So we see that the above is a correct formalization
of the English and that gives us the cognitive leverage
of intuition.

 Nope, can't because the English sentence doesn't attach a "name" to the whole expression.
 

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A level of indirection:
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p: "This sentence is true", which is exactly the same as "p is true" since "this sentence" IS p
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p := True(L,p)
specifies True(True(True(True(True(...)))))

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Nope, it is equivelent to that, but doesn't SPECIFY that.
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LP := ~True(L, LP) means that every instance of LP
in the RHS is the same as the RHS.
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Clocksin & Mellish say this same thing.
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BEGIN:(Clocksin & Mellish 2003:254)
Finally, a note about how Prolog matching sometimes differs from the
unification used in Resolution. Most Prolog systems will allow you to
satisfy goals like:

 And how Prolog does it is irrelevent,
 

Not at all.
Prolog sees that LP is defined as ~True(LP) is nonsense
and rejects it.

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equal(X, X).
?- equal(foo(Y), Y).
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that is, they will allow you to match a term against an uninstantiated
subterm of itself. In this example, foo(Y) is matched against Y, which
appears within it. As a result, Y will stand for foo(Y), which is
foo(foo(Y)) (because of what Y stands for), which is foo(foo(foo(Y))),
and so on. So Y ends up standing for some kind of infinite structure.
END:(Clocksin & Mellish 2003:254)
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As I said above that is expanding levels of indirecction.
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*Prolog sees the same infinite recursion and rejects it*
?- TT = true(TT).
TT = true(TT).
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?- unify_with_occurs_check(TT, true(TT)).
false.
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Right, because prolog can't handle any levels of self referencing, and thus is not suitable for logic that can do that.
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Nothing can handle "some kind of infinite structure."

 Wrong. There are lots of logics that handle certain "infinte structures". After all, Mathematics is BASED on logic on infinite structures.
 

No expression that itself has an infinite structure can be
evaluated in finite time. that is what "infinite structure"
is defined to mean.

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You have been told this, but don't seem to understand it. My guess is you can't understand any logic more complicated than what Prolog handles, so don't realize how much it just doesn't handle.

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No the whole problem seems to be that you simply don't
bother to pay close enough attention the EXACTLY what I say.

 No, you don't use the words in the way they are properly defined, so of course people can't understand what you mean.
 We have to guess, and point out the errors that are clearly there.
 

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When I prove my point you simply ignore that I proved my point
and baselessly assume that I must be wrong. You will probably
completely "forget" my Clocksin & Mellish quote immediately after
you read it, or skip over it and assume that they are wrong.
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 Nope, you have yet to present an actual Formal proof.

A proof need not be formal.
A proof is any statement where its negation is unsatisfiable.

You seem to think that a Philosophical Arguement can substitute for a Formal Proof. YOu are just using the wrong tools that don't work in the system.
 Maybe if you actually tried to pay attention to what people say an not assume that your ideas, built on your assumptions of how things must work, have to be correct.
 

Try to "prove" that "2" really does stand for a number
without resorting to any definitions.
The definition itself is the complete proof, no steps required.

It seems you don't even have the tools to try to explain what you mean, but just like to throw out snipits quoted from places that you don;t really understand, but seem to say something sort of like what you are trying to say.
 All you have done is proved your ignorance.

Most of the best experts in the world are not sure that the Liar Paradox
is not a truth bearer. At least you know this much.
When we get to the formalized Liar Paradox this seems too difficult
for you, yet you are still doing better than most experts in the world.
You are even better at formalizing the Liar Paradox than most experts
in the field. They try to get away with this crap: LP ↔ ~True(LP).
You understand that this is the correct way: p defined as ~True(L, p).
So it is still: Good job Richard !
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer