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

On 2024-05-25 18:13:02 +0000, olcott said:

On 5/25/2024 3:01 AM, Mikko wrote:

On 2024-05-24 19:16:47 +0000, olcott said:
 

On 5/24/2024 3:18 AM, Mikko wrote:

On 2024-05-23 13:32:51 +0000, olcott said:
 

On 5/23/2024 3:09 AM, Mikko wrote:

On 2024-05-23 01:03:44 +0000, Richard Damon said:
 

On 5/22/24 7:55 PM, olcott wrote:

On 5/22/2024 6:01 PM, Richard Damon wrote:

On 5/22/24 3:52 PM, olcott wrote:

On 5/22/2024 11:58 AM, Mikko wrote:

On 2024-05-22 15:55:39 +0000, olcott said:
 

On 5/22/2024 2:57 AM, Mikko wrote:

On 2024-05-21 14:36:29 +0000, olcott said:
 

On 5/21/2024 3:05 AM, Mikko wrote:

On 2024-05-20 17:48:40 +0000, olcott said:
 

On 5/20/2024 2:55 AM, Mikko wrote:

On 2024-05-19 14:15:51 +0000, olcott said:
 

On 5/19/2024 9:03 AM, Mikko wrote:

On 2024-05-19 13:41:56 +0000, olcott said:
 

On 5/19/2024 6:55 AM, Richard Damon wrote:

On 5/18/24 11:47 PM, olcott wrote:

On 5/18/2024 6:04 PM, Richard Damon wrote:

On 5/18/24 6:47 PM, olcott wrote:

On 5/18/2024 5:22 PM, Richard Damon wrote:

On 5/18/24 4:00 PM, olcott wrote:

On 5/18/2024 2:57 PM, Richard Damon wrote:

On 5/18/24 3:46 PM, olcott wrote:

On 5/18/2024 12:38 PM, Richard Damon wrote:

On 5/18/24 1:26 PM, olcott wrote:

On 5/18/2024 11:56 AM, Richard Damon wrote:

On 5/18/24 12:48 PM, olcott wrote:

On 5/18/2024 9:32 AM, Richard Damon wrote:

On 5/18/24 10:15 AM, olcott wrote:

On 5/18/2024 7:43 AM, Richard Damon wrote:

No, your system contradicts itself.
 

 You have never shown this.
The most you have shown is a lack of understanding of the
Truth Teller Paradox.

 No, I have, but you don't understand the proof, it seems because you don't know what a "Truth Predicate" has been defined to be.
 

 My True(L,x) predicate is defined to return true or false for every
finite string x on the basis of the existence of a sequence of truth
preserving operations that derive x from

 And thus, When True(L, p) established a sequence of truth preserving operations eminationg from ~True(L, p) by returning false, it contradicts itself. The problem is that True, in making an answer of false, has asserted that such a sequence exists.
 

On 5/13/2024 9:31 PM, Richard Damon wrote:
 > On 5/13/24 10:03 PM, olcott wrote:
 >> On 5/13/2024 7:29 PM, Richard Damon wrote:
 >>>
 >>> Remember, p defined as ~True(L, p) ...
 >>
 >> Can a sequence of true preserving operations applied
 >> to expressions that are stipulated to be true derive p?
 > No, so True(L, p) is false
 >>
 >> Can a sequence of true preserving operations applied
 >> to expressions that are stipulated to be true derive ~p?
 >
 > No, so False(L, p) is false,
 >
 *To help you concentrate I repeated this*
The Liar Paradox and your formalized Liar Paradox both
contradict themselves that is why they must be screened
out as type mismatch error non-truth-bearers *BEFORE THAT OCCURS*

 And the Truth Predicate isn't allowed to "filter" out expressions.
 

 YOU ALREADY KNOW THAT IT DOESN'T
WE HAVE BEEN OVER THIS AGAIN AND AGAIN
THE FORMAL SYSTEM USES THE TRUE AND FALSE PREDICATE
TO FILTER OUT TYPE MISMATCH ERROR
 The first thing that the formal system does with any
arbitrary finite string input is see if it is a Truth-bearer:
Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))

 No, we can ask True(L, x) for any expression x and get an answer.
 

 The system is designed so you can ask this, yet non-truth-bearers
are rejected before True(L, x) is allowed to be called.
  

 Not allowed.
 

 My True(L,x) predicate is defined to return true or false for every
finite string x on the basis of the existence of a sequence of truth
preserving operations that derive x from
 A set of finite string semantic meanings that form an accurate
verbal model of the general knowledge of the actual world that
form a finite set of finite strings that are stipulated to have
the semantic value of Boolean true.
 *This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
*This is computable* Truthbearer(L,x) ≡ (True(L,x) ∨ True(L,~x))
 

 So, for a statement x to be false, it says that there must be a sequence of truth perserving operations that derive ~x from, right?
 

Yes we must build from mutual agreement, good.
 

So do you still say that for p defined in L as ~True(L, p) that your definition will say that True(L, p) will return false?
 

 It is the perfectly isomorphic to this:
True(English, "This sentence is not true")
 

  Nope, Because "This sentece is not true" can be a non-truth-bearer, but by its definition, True(L, x) can not.
 

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

 When x is defined as True(L,x) then x is what True(L,x) is,
in this case a truth bearer.

 

This is known as the Truth Teller Paradox

 Doesn't matter. But ir you say that "x is not a truth bearer" then,
by a truth preserving transformation, you imply that True(L,x) is

 True(English, "a cat is an animal) is true
LP := ~True(L, LP) expands to ~True(~True(~True(~True(...))))

 No, it doesn't. It is a syntax error to have the same symbol on
both sides ":=" so the expansion is not justified.

 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

 Your quote omitted important details. One is that the claim is not
true about every theory but is about first order arithmetic and its
extension. Another one is that ϕ(x) is that the claim is about
every formula ϕ(x).
 

 *The whole article is about self-reference*
The ONLY detail that I am referring to is that it is conventional to formalize self-reference incorrectly.
 *Richard and both fixed that*
 On 5/13/2024 9:31 PM, Richard Damon wrote:
 > On 5/13/24 10:03 PM, olcott wrote:
 >> On 5/13/2024 7:29 PM, Richard Damon wrote:
 >>>
 >>> Remember, p defined as ~True(L, p) ...
 x := y means x is defined to be another name for y

 Another name for the meaning of y. Therefore any pair of sentences that
are otherwise equal but one contains x where rhe other contains y is a pair
of equally true sentences. For example, if p defined as ~True(L, ⟨p⟩)

 I have no idea what you mean by the weird ⟨p⟩ quotes.
I AM ABSOLUTELY NOT TALKING ABOUT ANY FREAKING Gödel NUMBERS
I AM ABSOLUTELY NOT TALKING ABOUT ANY FREAKING Gödel NUMBERS
I AM ABSOLUTELY NOT TALKING ABOUT ANY FREAKING Gödel NUMBERS
 I AM TALKING ABOUT THE EXISTENCE OR NON-EXISTENCE OF
AN ACTUAL SEQUENCE OF TRUTH PRESERVING OPERATIONS FROM
EXPRESSIONS OF LANGUAGE KNOWN TO BE TRUE

 So, you aren't talking about Tarski's proof of the impossibility to define a Truth Predicate per his definition?
 

 

then Truthbearer(L,p) has the same truth value as Truthbearer(L,~True(L, ⟨p⟩)).
 

 When p defined as ~True(L, p)
Then ~True(L, p) is true, thus a truth-bearer.

 Which means that True(L, p) is false, so your True just erred in describing a true statement as false.
 Remeber, you just said that ~True(L, p) which has been given the name of p IS a truth-bearer.
 

 *You are just not paying close enough attention again*
 When p defined as ~True(L, p)
 True(L,p)  is false
 True(L,~p) is false
~True(L,~p) is true
 x := y means x is defined to be another name for y
https://en.wikipedia.org/wiki/List_of_logic_symbols

 Right, so since p is DEFINED to be ~True(L, p), which since True(L, p) is false, must be true, that means that you are claiming that
T(L, <a statement that has been shown to be true>) is false.
 Thus your True predicat is just broken.
 

 You ignored the part where Mikko agreed that
 p defined as ~True(L, p)
is a syntax error:

 So, what it the "Syntax Error"?
 Are we not allowed to negate an expression
 Or are we not allowed to assign an expression to a name.
 Note, "Syntax Error", by its definition doesn't look at Semantics,
 

 On 5/21/2024 3:05 AM, Mikko wrote:

On 2024-05-20 17:48:40 +0000, olcott said:

True(English, "a cat is an animal) is true
LP := ~True(L, LP) expands to ~True(~True(~True(~True(...))))

 No, it doesn't. It is a syntax error to have the same symbol on
both sides ":=" so the expansion is not justified.

 But it isn't.

 By the usual rules a definition of a symbol in terms of itself is not
an acceptable definition.
 

 One can either reject it as a syntax error or let it go ahead
and infinitely expand and reject it as a semantic error.

 It is a syntax error by the usual rules. If you want to use a different
syntax then you should specify one, preferably using a different symbol
instead of ":=". It is OK to extend the syntax but one should avoid any
conflict with the usual conventions. Also, if you change the syntax
rules you should not call it a "definition".
 

 LP := ~True(L, LP) is required to refer to itself on both sides
that is what actual self-reference means.
 *THE ACTUAL Stanford ARTICLE ON SELF-REFERENCE SAYS*
*THAT THEY MAKE SURE TO ENCODE IS INCORRECTLY*
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
 It is the standard convention throughout the literature to encode
self-reference incorrectly. When the standard convention is to do
these things incorrectly then the standard convention must be
superseded and replaced.
 That is one of the reasons what correctly analyzing these this is
so difficult.
 If you are correct that this is incorrect syntax
LP := ~True(L, LP)
that is yet another reason to reject the Liar Paradox
(and every other self-reference paradox) as ill-formed.
 

Or one can reject is as a self-contradictory epistemological antinomy
having no truth value thus a type mismatch error for any formal
system of bivalent logic.

 If that can be formulated as a syntax rule. Being an epistemological
antinomy is semantics as is being true or false but type mismach can
be handled as syntax error if the syntax rules have a type system.
 

 The formalized Liar Paradox
LP := ~True(L, LP) <is> an epistemological antinomy because assuming
that it is true makes it false and assuming that it is false makes it
true.
 That you want to also call it a syntax error seems reasonable to me.
 If it is not rejected as a syntax error then it does recursively
expand ~True(~True(~True(~True(~True(...))))) as Clocksin & Mellish
point out.
 BEGIN:(Clocksin & Mellish 2003:254)
   equal(X, X).
   ?- equal(foo(Y), Y).
    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)
 

Most of the greatest experts in the field are not even sure
that there is anything wrong with it.

 Nothing is inherently wrong in an uninterpreted formal system.
Something may be unsuitable for some purpose but still useful
for another purpose.
 

 You already said that this is a syntax error:
LP := ~True(L, LP)
please at least be consistent with yourself.

 I don't. Because of a syntax error "LP := ~True(L, LP)" is not an
expression in the formal system and not in contradiction that there
is nothing wrong in the formal system.
 

 This is where Tarski says that his proof is anchored in the Liar Paradox
https://liarparadox.org/Tarski_247_248.pdf

Nothing to that page contradicts anything I have said above.

When you look at my new thread (and completely understand what it says)
You will see when we correctly formalize Tarski's clumsy formalization
of the Liar Paradox

You don't formalize it correctly with a string that is not in the
language of the formnal system. A syntax error excludes all meaning
and in prticular the meaning that Tarksi's expressions have.
--
Mikko