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From: olcott <polcott333@gmail.com>
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Subject: =?UTF-8?Q?Re=3A_A_simulating_halt_decider_applied_to_the_The_Peter_?=
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Date: Sun, 2 Jun 2024 08:07:25 -0500
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On 6/2/2024 2:36 AM, Mikko wrote:
> On 2024-06-01 14:37:01 +0000, olcott said:
> 
>> On 6/1/2024 2:52 AM, Mikko wrote:
>>> On 2024-05-31 15:35:18 +0000, olcott said:
>>
>>>>
>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>
>>> Of those two lines one is false.
>>> As embedded_H is a copy of H both lines imply that H is not a halt 
>>> decider.
>>>
>>>> *Formalizing the Linz Proof structure*
>>>> ∃H  ∈ Turing_Machines
>>>> ∀x  ∈ Turing_Machine_Descriptions
>>>> ∀y  ∈ Finite_Strings
>>>> such that H(x,y) = Halts(x,y)
>>>
>>> As already noted, the above is not a part of a proof structure.
>>>
>>
>> Unless and until you provide reasoning to back that up it counts
>> as if you said nothing about it.
> 
> If there are no more questions about the details of the reasoning
> we may assume that the presiented reasoning is sufficieant.
> 

The above <is> the structure of his proof your empty assertion utterly 
bereft of any supporting (EAUBoaSR)) reasoning counts for zilch.

Linz claims that of every Turing Machine there are none that solve the 
halting problem.

∃!H  ∈ Turing_Machines  (What Richard was saying)
would say that there does not exist exactly one Turing Machine that
solves the halting problem thus fails if there are more than one.

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