Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.logic
Subject: Re: The key undecidable instance that I know about --- Truth-bearers ONLY
Date: Mon, 17 Mar 2025 11:08:21 +0200
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On 2025-03-15 17:08:33 +0000, olcott said:

> 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.

Prolog does not prove anything other than what you ask. I don't think
you can ask Prolog whether there is a cycle in LP after LP = not(true(LP)).

-- 
Mikko