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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: D correctly simulated by H cannot possibly halt --- templates and
 infinite sets
Date: Wed, 29 May 2024 20:55:44 -0500
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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.

> 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.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer