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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: D correctly simulated by H cannot possibly halt --- templates and infinite sets
Date: Mon, 3 Jun 2024 13:36:05 +0300
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On 2024-06-02 14:04:14 +0000, olcott said:

> On 6/2/2024 3:24 AM, Mikko wrote:
>> On 2024-06-01 14:52:54 +0000, olcott said:
>> 
>>> On 6/1/2024 3:20 AM, Mikko wrote:
>>>> On 2024-05-31 15:44:22 +0000, olcott said:
>>>> 
>>>>> On 5/31/2024 8:10 AM, Mikko wrote:
>>>>>> On 2024-05-28 16:16:48 +0000, olcott said:
>>>>>> 
>>>>>>> typedef int (*ptr)();  // ptr is pointer to int function in C
>>>>>>> 00       int H(ptr p, ptr i);
>>>>>>> 01       int D(ptr p)
>>>>>>> 02       {
>>>>>>> 03         int Halt_Status = H(p, p);
>>>>>>> 04         if (Halt_Status)
>>>>>>> 05           HERE: goto HERE;
>>>>>>> 06         return Halt_Status;
>>>>>>> 07       }
>>>>>>> 08
>>>>>>> 09       int main()
>>>>>>> 10       {
>>>>>>> 11         H(D,D);
>>>>>>> 12         return 0;
>>>>>>> 13       }
>>>>>>> 
>>>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>> 
>>>>>>> *Formalizing the Linz Proof structure*
>>>>>>> ∃H  ∈ Turing_Machines
>>>>>>> ∀x  ∈ Turing_Machines_Descriptions
>>>>>>> ∀y  ∈ Finite_Strings
>>>>>>> such that H(x,y) = Halts(x,x)
>>>>>>> 
>>>>>>> *Here is the same thing applied to H/D pairs*
>>>>>>> ∃H ∈ C_Functions
>>>>>>> ∀D ∈ x86_Machine_Code_of_C_Functions
>>>>>>> such that H(D,D) = Halts(D,D)
>>>>>>> 
>>>>>>> In both cases infinite sets are examined to see
>>>>>>> if any H exists with the required properties.
>>>>>> 
>>>>>> That says nothing about correct simulation. It says
>>>>>> something abuout some D but not whether it is correctly
>>>>>> simulated. Also nothing is said about templates or
>>>>>> infinite sets. At the end is claimed that some
>>>>>> infinite sets are examined but not who examined, nor
>>>>>> how, nor what was found in the alleged examination.
>>>>>> 
>>>>> 
>>>>> *Formalizing the Linz Proof structure*
>>>>> ∃H  ∈ Turing_Machines
>>>>> ∀x  ∈ Turing_Machines_Descriptions
>>>>> ∀y  ∈ Finite_Strings
>>>>> such that H(x,y) = Halts(x,x)
>>>> 
>>>> The above is the counter hypothesis for the proof. Proof structore
>>>> is that a contradiction is derived from the counter hypthesis.
>>>> 
>>>>> The above disavows Richard's claim based on a misinterpretation of
>>>>> Linz that the Linz proof is about a single specific Turing machine.
>>>> 
>>>> Your ∃H declares H as a new symbol for a specific Turing machine.
>>>> Therefore everything that follows refers to that specific Turing machine.
>>>> There may be others that could be discussed the same way but they aren't.
>>>> 
>>> 
>>> ∃H  ∈ Turing_Machines
>>> There exists at least one H
>>> from the infinite set of all Turing_Machines
>>> 
>>> ∃!H  ∈ Turing_Machines
>>> There exists a single unique H
>>> from the infinite set of all Turing_Machines
>> 
>> The symbol ∃! is non-standard and should be defined when used.
>> 
>> And there is no point to present some formulas without saying
>> anything about them.
>> 
>>>>>     The domain of this problem is to be taken as the set of
>>>>>     all Turing machines and all w; that is, we are looking
>>>>>     for a single Turing machine that, given the description
>>>>>     of an arbitrary M and w, will predict whether or not the
>>>>>     computation of M applied to w will halt.
>>>> 
>>>> Note the words "a single Turing machine".
>>> 
>>> I know that he said that yet he meant this
>>> ∃H ∈ Turing_Machines *and didn't mean this* ∃!H ∈ Turing_Machines
>>> or he would be contradicting every other HP proof.
>> 
>> He didn't mean either. Your false claim is merely an attempted
>> preparation of strawman deception.
>> 
>>>>> Linz <IS NOT> looking for a single machine that gets the wrong answer.
>>>>> Linz is looking for at least one Turing Machine that gets the right
>>>>> answer: ∃H ∈ Turing_Machines
>>>> 
>>>> Not at least one but exactly one. The Halting Problem asks for one
>>>> or a proof that there is none.
>>> 
>>> In other words when there are two machines that solve the halting
>>> problem then the halting problem IS NOT SOLVED?
>> 
>> If there are two then one of them is one and the other is irrelevant.
> 
> ∃!H  ∈ Turing_Machines
> ∀x  ∈ Turing_Machines_Descriptions
> ∀y  ∈ Finite_Strings
> such that H(x,y) = Halts(x,x)

The last line does not make sense. There is no point to have an
y on the left side but not on the right side. And there is no
point in a funciton that does not use its second argument.

Besides, there is no point to present a formula without
saying something about it.

> https://en.wikipedia.org/wiki/Uniqueness_quantification

The page discusses uniqueness quantification and presents two
symbols for it and has pointers to some sources but does not
claim aboöut either symbol that it be stndard.

> He was saying that Linz was saying that Linz was looking for exactly 
> one unique Turing machine that solved the halting problem.

In that context "he" does not denote so the sentence is meaningless.

>> When Linz says "we are looking for a single Turing machine that ..."
>> he implies that when he (or someone) finds one he is not going to
>> look for another one. When he finally proves that it is not possible
>> to find one it is obvious that it is not possible to find more.
>> 
>> And if you need to ask that you don't need to as why you look stupid.
>> 
> 
> Unless we attain 100% perfectly identical encoding and decoding of
> meanings the truth of what I am saying with slip through the cracks of
> ambiguity.

You are still far from 100% clarity and your meanings are obscure
by ambiguities and wrong encodings.

> ∃H  ∈ Turing_Machines
> ∀x  ∈ Turing_Machines_Descriptions
> ∀y  ∈ Finite_Strings
> such that H(x,y) = Halts(x,x)

Perhaps there should be an y on the right side, too?

> Does there exist at least one H that solves the halting problem?

That is as clear an unambiguous as a question needs be.
The answer is that there is not.

-- 
Mikko