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From: Richard Damon <richard@damon-family.org>
Newsgroups: sci.logic,comp.theory
Subject: Re: True on the basis of meaning --- Good job Richard ! ---Socratic
 method
Date: Mon, 20 May 2024 23:37:54 -0400
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On 5/20/24 10:56 PM, olcott wrote:
> 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:
>>>>>>>>>
>>>>>>>>> True(L,x) is always a truth bearer.
>>>>>>>>> when x is defined as True(L,x) then x is not a truth bearer.
>>>>>>>>
>>>>>>>> So, x being DEFINED to be a certain sentence doesn't make x to 
>>>>>>>> have the same meaning as the sentence itself?
>>>>>>>>
>>>>>>>> What does it mean to define a name to a given sentence, if not 
>>>>>>>> that such a name referes to exactly that sentence?
>>>>>>>>
>>>>>>>
>>>>>>> p = ~True(L,p) // p is not a truth bearer because its refers to 
>>>>>>> itself
>>>>>>
>>>>>> Then ~True(L,p) can't be a truth beared as they are the SAME 
>>>>>> STATEMENT, just using different "names".
>>>>>
>>>>>
>>>>> 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
>>>>
>>>> Irrelvent.
>>>>
>>>> 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.
>>>>
>>>
>>> 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)

So, you admit that you did not answer the problem.

And that you think Strawmen and Red Herring are valid forms of logic.

How does p defined as ~True(L, p) NOT generate the shown contradiction 
when you begin by saying True(L, p) must not be true (and thus false) 
because p has not chain to truthbears?

You are just showing that you think it is ok for logical system to have 
contradictions in them.

> 
>>>
>>>> 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.
>>>>
>>>> You are just showing you utter lack of understanding of the 
>>>> fundamentals of Formal Logic.
>>>>
>>>
>>>     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.

Which mean?

And what does it have to do with the original statement?


Remember, if your goal is to just show that conventonal logic is just 
broken, you are going to need to make a much more convincing arguement 
to scrap it, unless you have a FULLY DEVELOPED alternative that does better.

Just remember, once you throw out the foundations, you need to start 
from a brand new foundation, and unless you have been lying about your 
prognossis, and sand-bagging about your logical abilities, your chance 
of actually proving somethiing like that is just about zero.

> 
>>>
>>> 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

Of course it does.

You just don't understand what you are reading.

In fact, Tarski points out the BECAUSE he can show that the existance of 
a Truth Primative forces an epistemological antinomy to have a truth 
value, that there can not be an existing Truth Primative.

YOU just don't understand logic,

> 
>>>
>>>
>>>>
>>>>>>
>>>>>> 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
>>>>>>
>>>>>>> True(L,p)  is false
>>>>>>> True(L,~p) is false
>>>>>>>
>>>>>>
>>>>>> So since True(L, p) is false, then ~True(L, p) is true.
>>>>>>
>>>>>>> ~True(True(L,p)) is true and is referring to the p that refers
>>>>>>> to itself it is not referring to its own self.
>>>>>>>
>>>>>>> *ONE LEVEL OF INDIRECT REFERENCE MAKES ALL THE DIFFERENCE*
>>>>>>
>>>>>> Why add the indirection? p is the NAME of the statement, which 
>>>>>> means exactly the same thing as the statement itself.
>>>>>>
>>>>>
>>>>> 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.
>>>>>
>>>>>> Is the definition of an English word one level LESS of indirection 
>>>>>> than the word itself?
>>>>>>
>>>>>
>>>>> This sentence is not true("This sentence is not true") is true.
>>>>
>>>> Right, that is a sentence about another sentence (that is part of 
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