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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: Tue, 21 May 2024 21:46:50 -0400
Organization: i2pn2 (i2pn.org)
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On 5/21/24 11:00 AM, olcott wrote:
> On 5/21/2024 6:50 AM, Richard Damon wrote:
>> On 5/21/24 1:52 AM, olcott wrote:
>>> On 5/20/2024 10:37 PM, Richard Damon wrote:
>>>> 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?
>>>>
>>>
>>> p := ~True(L, p)  is false
>>> p := ~True(L, ~p) is false
>>>
>>> p is tossed out on its ass as a type mismatch error for every system
>>> of bivalent logic before it gets any chance to be evaluated in any
>>> other way.
>>
>> Not ALLOWED. p is DEFINED to be something, so it is that/.
>>
> 
> 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.
> 
> On 5/13/2024 7:29 PM, Richard Damon wrote:
>  > Remember, p defined as ~True(L, p) is BY DEFINITION a
>  > truth bearer, as True must return a Truth Value for
>  > all inputs, and ~ a truth valus is always the other
>  > truth value.
> 
> p defined as ~True(L, p) is rejected as a syntax error.

NOT ALLOWED.

So, your are just admitting that your logic system doesn't meet the 
requirements of Tarski, and thus your claims are just blatant lies.

> 
> https://www.swi-prolog.org/pldoc/man?predicate=unify_with_occurs_check/2
> or rejected as
> 
>    equal(X, X).
>    ?- equal(foo(Y), Y). ...
>    So Y ends up standing for some kind of infinite structure.
>    (Clocksin & Mellish 2003:254)
> 
> By
> 
> The SWI-Prolog implementation of unify_with_occurs_check/2 is cycle-safe 
> and only guards against creating cycles,
> https://www.swi-prolog.org/pldoc/man?predicate=unify_with_occurs_check/2
> 

And PROLOG is not the definition of what is allowed, so you just prove 
that you are too stupid to understand logic.