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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 21:02:32 -0400
Organization: i2pn2 (i2pn.org)
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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, and a specific input is shown to exist for 
each one that makes it wrong.

YOU try o merge them all together to claim that since none of them reach 
a final state in its partial simulation (for those that answer) that 
means that all the inputs represented non-halting machines,

THAT is just invalid logic.

You just don't understand how to do logic with qualifiers.