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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 20:12:44 -0500
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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*

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.




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