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From: olcott <polcott333@gmail.com>
Newsgroups: sci.logic
Subject: Re: The key undecidable instance that I know about --- Truth-bearers
 ONLY
Date: Sat, 15 Mar 2025 22:34:15 -0500
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On 3/15/2025 10:05 PM, dbush wrote:
> On 3/15/2025 10:52 PM, olcott wrote:
>> On 3/15/2025 9:41 PM, dbush wrote:
>>> On 3/15/2025 10:27 PM, olcott wrote:
>>>> On 3/15/2025 9:18 PM, dbush wrote:
>>>>> On 3/15/2025 9:47 PM, olcott wrote:
>>>>>> On 3/15/2025 8:27 PM, dbush wrote:
>>>>>>> On 3/15/2025 9:03 PM, olcott wrote:
>>>>>>>> On 3/15/2025 2:25 PM, dbush wrote:
>>>>>>>>> On 3/15/2025 3:16 PM, olcott wrote:
>>>>>>>>>> On 3/15/2025 2:05 PM, dbush wrote:
>>>>>>>>>>> On 3/15/2025 2:55 PM, olcott wrote:
>>>>>>>>>>>> On 3/15/2025 12:39 PM, dbush wrote:
>>>>>>>>>>>>> On 3/15/2025 1:08 PM, olcott wrote:
>>>>>>>>>>>>>> On 3/10/2025 9:49 PM, dbush wrote:
>>>>>>>>>>>>>>> On 3/10/2025 10:39 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 3/10/2025 9:21 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>> On 3/10/25 9:45 PM, olcott wrote:
>>>>>>>>>>>>>>>>>> On 3/10/2025 5:45 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>>> On 3/9/25 11:39 PM, olcott wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> LP := ~True(LP)  DOES SPECIFY INFINITE RECURSION.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> WHich is irrelevent, as that isn't the statement in 
>>>>>>>>>>>>>>>>>>> view, only what could be shown to be a meaning of the 
>>>>>>>>>>>>>>>>>>> actual statement.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> The Liar Paradox PROPERLY FORMALIZED <is> Infinitely 
>>>>>>>>>>>>>>>>>> recursive
>>>>>>>>>>>>>>>>>> thus semantically incorrect.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> But is irrelevent to your arguement.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> "It would then be possible to reconstruct the antinomy 
>>>>>>>>>>>>>>>>>> of the liar
>>>>>>>>>>>>>>>>>>   in the metalanguage, by forming in the language 
>>>>>>>>>>>>>>>>>> itself a sentence"
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Right, the "Liar" is in the METALANGUAGE, not the 
>>>>>>>>>>>>>>>>> LANGUAGE where the predicate is defined.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You are just showing you don't understand the concept 
>>>>>>>>>>>>>>>>> of Metalanguage.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Thus anchoring his whole proof in the Liar Paradox 
>>>>>>>>>>>>>>>>>> even if
>>>>>>>>>>>>>>>>>> you do not understand the term "metalanguage" well enough
>>>>>>>>>>>>>>>>>> to know this.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Yes, there is a connection to the liar's paradox, and 
>>>>>>>>>>>>>>>>> that is that he shows that the presumed existance of a 
>>>>>>>>>>>>>>>>> Truth Predicate forces the logic system to have to 
>>>>>>>>>>>>>>>>> resolve the liar's paradox.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> bool True(X)
>>>>>>>>>>>>>>>> {
>>>>>>>>>>>>>>>>    if (~unify_with_occurs_check(X))
>>>>>>>>>>>>>>>>      return false;
>>>>>>>>>>>>>>>>    else if (~Truth_Bearer(X))
>>>>>>>>>>>>>>>>     return false;
>>>>>>>>>>>>>>>>    else
>>>>>>>>>>>>>>>>     return IsTrue(X);
>>>>>>>>>>>>>>>> }
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> LP := ~True(LP)
>>>>>>>>>>>>>>>> True(LP) resolves to false.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> ~True(LP) resolves to true
>>>>>>>>>>>>>>> LP := ~True(LP) resolves to true
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Therefore the assumption that a correct True() predicate 
>>>>>>>>>>>>>>> exists is proven false.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> When you stupidly ignore Prolog and MTT that
>>>>>>>>>>>>>> both prove there is a cycle in the directed graph
>>>>>>>>>>>>>> of their evaluation sequence. If you have no idea
>>>>>>>>>>>>>> what "cycle", "directed graph" and "evaluation sequence"
>>>>>>>>>>>>>> means then this mistake is easy to make.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> That doesn't change the fact that 
>>>>>>>>>>>>
>>>>>>>>>>>> You have just proven you are clueless about these things
>>>>>>>>>>>> by your next statement.
>>>>>>>>>>>>
>>>>>>>>>>>>> that ~True(LP) evaluates to true.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> When
>>>>>>>>>>>> LP  := ~True(LP)  True_Bearer(LP) == FALSE
>>>>>>>>>>>
>>>>>>>>>>> And by the above function, because True_Bearer(LP) == FALSE:
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Which means that LP cannot possibly be either TRUE or FALSE
>>>>>>>>>> and instead must be rejected as invalid input to a True(X)
>>>>>>>>>> predicate.
>>>>>>>>>
>>>>>>>>> False.  The True() predicate must return "true" for true 
>>>>>>>>> statements and false for *all other statements*.
>>>>>>>>>
>>>>>>>>> The fact that the True() you've defined *does* accept non-truth 
>>>>>>>>> bearers and returns false for them shows you know this, but are 
>>>>>>>>> being deliberately deceptive.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> True(LP) == FALSE, then
>>>>>>>>>>> ~True(LP) == TRUE, so
>>>>>>>>>>> LP == TRUE
>>>>>>>>>>>
>>>>>>>>>>> Contradiction.  Therefore the assumption that a correct 
>>>>>>>>>>> True() predicate exists is proven false
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Likewise Truth_Bearer("ksdnf34589jknsdf34r87&%78^%78") == FALSE
>>>>>>>>>> <sarcasm>
>>>>>>>>>> and on that basis we know that True(X) predicates cannot
>>>>>>>>>> exist because True(X) predicates must correctly determine
>>>>>>>>>> whether random gibberish is true or false.
>>>>>>>>>> </sarcasm>
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> True(X) predicates must correctly determine
>>>>>>>>> whether random gibberish is true or *not true*.  And because 
>>>>>>>>> random gibberish is not true, 
>>>>>>>>> True("ksdnf34589jknsdf34r87&%78^%78") must return false
>>>>>>>>>
>>>>>>>>
>>>>>>>> That is fine, and makes Tarski wrong.
>>>>>>>>
>>>>>>>
>>>>>>> Nope.  Tarski uses a proof by contradiction.  You know, that type 
>>>>>>> of proof you still don't understand 50 years after learning it.
>>>>>>>
>>>>>>> He starts by assuming that a True() predicate exists in a system 
>>>>>>> that can express the full properties of natural numbers.
>>>>>>>
>>>>>>> He then shows that it's possible to create in the system that can 
>>>>>>> be shown in a meta system to effectively mean:
>>>>>>>
>>>>>>> LP  := ~True(LP)
>>>>>>>
>>>>>>> Given that True(LP) == false, we then have ~True(LP) == true.  
>>>>>>> And since ~True(LP) is the same as LP, that gives us LP == true.
>>>>>>>
>>>>>>> Contradiction. 
>>>>>>
>>>>>> True(LP) == FALSE 
>>>>>
>>>>> And ~True(LP) == TRUE
>>>>> Therefore LP == TRUE
>>>>>
>>>>> Contradiction.
>>>>>
>>>>> Therefore the assumption that a True() predicate exists is proven 
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