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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: sci.logic
Subject: =?utf-8?Q?Re:_ZFC_solution_to_incorrect_questions:_reject_them_--G=C3=B6del--?=
Date: Mon, 18 Mar 2024 11:57:23 +0200
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On 2024-03-17 17:17:42 +0000, olcott said:

> On 3/17/2024 10:44 AM, Mikko wrote:
>> On 2024-03-15 14:28:49 +0000, olcott said:
>> 
>>> On 3/15/2024 6:07 AM, Mikko wrote:
>>>> On 2024-03-13 14:25:10 +0000, olcott said:
>>>> 
>>>>> On 3/13/2024 5:03 AM, Mikko wrote:
>>>>>> On 2024-03-12 20:38:34 +0000, olcott said:
>>>>>> 
>>>>>>> On 3/12/2024 3:31 PM, immibis wrote:
>>>>>>>> On 12/03/24 20:02, olcott wrote:
>>>>>>>>> On 3/12/2024 1:31 PM, immibis wrote:
>>>>>>>>>> On 12/03/24 19:12, olcott wrote:
>>>>>>>>>>> ∀ H ∈ Turing_Machine_Deciders
>>>>>>>>>>> ∃ TMD ∈ Turing_Machine_Descriptions  |
>>>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>>>> 
>>>>>>>>>>> There is some input TMD to every H such that
>>>>>>>>>>> Predicted_Behavior(H, TMD) != Actual_Behavior(TMD)
>>>>>>>>>> 
>>>>>>>>>> And it can be a different TMD to each H.
>>>>>>>>>> 
>>>>>>>>>>> When we disallow decider/input pairs that are incorrect
>>>>>>>>>>> questions where both YES and NO are the wrong answer
>>>>>>>>>> 
>>>>>>>>>> Once we understand that either YES or NO is the right answer, the whole 
>>>>>>>>>> rebuttal is tossed out as invalid and incorrect.
>>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
>>>>>>>>> BOTH YES AND NO ARE THE WRONG ANSWER FOR EVERY Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>>>> 
>>>>>>>> 
>>>>>>>> Once we understand that either YES or NO is the right answer, the whole 
>>>>>>>> rebuttal is tossed out as invalid and incorrect.
>>>>>>>> 
>>>>>>>>>>> Does the barber that shaves everyone that does not shave
>>>>>>>>>>> themselves shave himself? is rejected as an incorrect question.
>>>>>>>>>> 
>>>>>>>>>> The barber does not exist.
>>>>>>>>> 
>>>>>>>>> Russell's paradox did not allow this answer within Naive set theory.
>>>>>>>> 
>>>>>>>> Naive set theory says that for every predicate P, the set {x | P(x)} 
>>>>>>>> exists. This axiom was a mistake. This axiom is not in ZFC.
>>>>>>>> 
>>>>>>>> In Turing machines, for every non-empty finite set of alphabet symbols 
>>>>>>>> Γ, every b∈Γ, every Σ⊆Γ, every non-empty finite set of states Q, every 
>>>>>>>> q0∈Q, every F⊆Q, and every δ:(Q∖F)×Γ↛Q×Γ×{L,R}, ⟨Q,Γ,b,Σ,δ,q0,F⟩ is a 
>>>>>>>> Turing machine. Do you think this is a mistake? Would you remove this 
>>>>>>>> axiom from your version of Turing machines?
>>>>>>>> 
>>>>>>>> (Following the definition used on Wikipedia: 
>>>>>>>> https://en.wikipedia.org/wiki/Turing_machine#Formal_definition)
>>>>>>>> 
>>>>>>>>>> The following is true statement:
>>>>>>>>>> 
>>>>>>>>>> ∀ Barber ∈ People. ¬(∀ Person ∈ People. Shaves(Barber, Person) ⇔ 
>>>>>>>>>> ¬Shaves(Person, Person))
>>>>>>>>>> 
>>>>>>>>>> The following is a true statement:
>>>>>>>>>> 
>>>>>>>>>> ¬∃ Barber ∈ People. (∀ Person ∈ People. Shaves(Barber, Person) ⇔ 
>>>>>>>>>> ¬Shaves(Person, Person))
>>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> That might be correct I did not check it over and over
>>>>>>>>> again and again to make sure.
>>>>>>>>> 
>>>>>>>>> The same reasoning seems to rebut Gödel Incompleteness:
>>>>>>>>> ...We are therefore confronted with a proposition which
>>>>>>>>> asserts its own unprovability. 15 ... (Gödel 1931:43-44)
>>>>>>>>> ¬∃G ∈ F | G := ~(F ⊢ G)
>>>>>>>>> 
>>>>>>>>> Any G in F that asserts its own unprovability in F is
>>>>>>>>> asserting that there is no sequence of inference steps
>>>>>>>>> in F that prove that they themselves do not exist in F.
>>>>>>>> 
>>>>>>>> The barber does not exist and the proposition does not exist.
>>>>>>>> 
>>>>>>> 
>>>>>>> When we do this exact same thing that ZFC did for self-referential
>>>>>>> sets then Gödel's self-referential expressions that assert their
>>>>>>> own unprovability in F also cease to exist.
>>>>>> 
>>>>>> Although Russel's set cannot be costructed in in ZFC Gödel's set can,
>>>>>> thus proving that ZFC is incomplete and ZFC augmented with additional
>>>>>> axioms is either incomplete or inconsistent.
>>>>>> 
>>>>> 
>>>>> That is not how it works at all. Russell's paradox pointed out
>>>>> incoherence in the notion of a set. ZFC fixed that.
>>>>> 
>>>>> The inability to show the a self-contradictory sentence is true or false
>>>>> is merely the inability to do the logically impossible and places no
>>>>> actual limit on anyone or anything.
>>>> 
>>>> If a theory is complete there is a simple computable method to find out
>>>> whether a particular sentence is a theorem or not. That method does not
>>>> work with incomplete theories, and in many cases, including ZF and ZFC,
>>>> no method works.
>>>> 
>>> 
>>> To say that anything or anyone is in anyway limited or incomplete
>>> because they lack the ability to do the logically impossible is
>>> incorrect.
>>> 
>>> Human knowledge is not incomplete on the basis of the lack of the 
>>> ability to prove the Liar Paradox is true or false.
>>> "This sentence is not true." is not true, yet neither true nor false.
>>> 
>>> The Liar Paradox is not truth bearer thus has no truth value.
>>> Tarski concluded that True(L,x) cannot be defined because it
>>> gets stumped on the Liar Paradox.
>> 
>> Every sentence in the language of ZFC either is or is not provable.
>> Anything else is a logical impossibility.
>> 
>> There is no complete method to determine whether a sentence in the
>> language of ZFC is provable. Existence of such method is a logical
>> impossibility.
>> 
> 
> ...We are therefore confronted with a proposition which
> asserts its own unprovability. 15 ...(Gödel 1931:43-44)
> 
> ?- G = not(provable(F, G)).
> G = not(provable(F, G)).
> 
> ?- unify_with_occurs_check(G, not(provable(F, G))).
> false.
> 
> Prolog correctly detects a cycle in the evaluation graph of
> the above expression.
> 
> *Is unprovable in F because of its pathological self-reference*
> Not because F is in any way incomplete.

Per definitium F is incomplete if there is in the language
of F a sentence that is neither a theorem nor the negation
or any theorem. ZFC is that way incomplete.

A theory is called unsovable if there is no complete method
to determine whether a particular sentence is a theorem.
In this sense ZFC is unsovable.

-- 
Mikko