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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: D correctly simulated by H cannot possibly halt --- templates and
 infinite sets
Date: Wed, 29 May 2024 22:25:03 -0400
Organization: i2pn2 (i2pn.org)
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On 5/29/24 9:55 PM, olcott wrote:
> On 5/29/2024 8:25 PM, Richard Damon wrote:
>> On 5/29/24 9:12 PM, olcott wrote:
>>> On 5/29/2024 8:02 PM, Richard Damon wrote:
>>>> On 5/29/24 8:53 PM, olcott wrote:
>>>>> On 5/29/2024 7:47 PM, Richard Damon wrote:
>>>>>> On 5/29/24 8:21 PM, olcott wrote:
>>>>>>> On 5/29/2024 7:09 PM, Richard Damon wrote:
>>>>>>>> On 5/29/24 8:01 PM, olcott wrote:
>>>>>>>>> On 5/29/2024 6:47 PM, Richard Damon wrote:
>>>>>>>>>>> *Formalizing the Linz Proof structure*
>>>>>>>>>>> ∃H  ∈ Turing_Machines
>>>>>>>>>>> ∀x  ∈ Turing_Machines_Descriptions
>>>>>>>>>>> ∀y  ∈ Finite_Strings
>>>>>>>>>>> such that H(x,y) = Halts(x,y)
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> And since NO H, can get right the H^ built to contradict IT, 
>>>>>>>>>> that claim is proven false.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> YOU KEEP TRYING TO GET AWAY WITH CHANGING THE SUBJECT
>>>>>>>>> THE ABOVE FORMALIZATION IS CORRECT
>>>>>>>>>
>>>>>>>>
>>>>>>>> How?
>>>>>>>>
>>>>>>>
>>>>>>> The above is the question that Linz asks and the he gets
>>>>>>> an answer of no, no such H exists.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> So, you now agree with Linz. Good.
>>>>>>
>>>>>
>>>>> I said that Linz says that. The point is that the Linz
>>>>> template examines an infinite set of Turing Machine / input
>>>>> pairs the same way my H/D template references an infinite set
>>>>> of C function / input pairs.
>>>>>
>>>>
>>>> The difference is, In Linz's formulation, each machine is 
>>>> INDIVIDUALLY EVALUTED with its inputs, 
>>>
>>>
>>> *No that is never the case*
>>
>>
>> Of course it is.
>>
>>
>>>
>>> When Ĥ is applied to ⟨Ĥ⟩
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> The entire category of every decider/input pair is examined ALL AT ONCE.
>>> No one is dumb enough to look at each element of an infinite set
>>> one at a time because they know this takes literally forever.
>>>
>>
>> Why do you say that?
>>
>> How do you run ALL the machines at once?
>>
> 
> When the category is examined all at once then there is no need
> to look at each individual element.

So, which one or ones gave the correct answer for their input?

Since for EVERY H^ (H^) based on an H that goes to its state qn when it 
is  applied to (H^) (H^), that H^ (H^) will also go to qn and halt, and 
thus that H was wrong.

> 
>> Maybe you can think of all of them running INDIVIDUALLY in parrallel, 
>> but each machine does what that machine does with the input that THAT 
>> machine was given.
>>
>> You just don't understand what you are talking about.
>>
> 
> Existential quantification always looks at all the elements
> of an infinite set.
> 

Ok, and for EVERY H(H^,H^) that goes to qn creates an H^ (H^) that 
halts, and thus ALL are wrong, so there doesn't exist ANY that are right.

Actually "Existential" qualifications look for just at least one 
example, they don't need to find all of them. The issue is that this 
problem has both an existential and universal qualifications. Since each 
H needs to handle ALL inputs, that includes the "pathological" input H^ 
that makes it wrong and thus no H to meet the requirements exist.

You are talking Turing machine and the actual Halting problem so none of 
this H can't simulate its input to a final state, the ACTUAL QUESTION in 
you description said "Halts" which is SPECIFICAL about the actual 
behavior of the actual specific machine given to it.

So, you are CATEGORIALLY incorrect.