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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Two dozen people were simply wrong --- Try to prove otherwise
Date: Mon, 3 Jun 2024 08:16:12 -0500
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On 6/3/2024 5:03 AM, Mikko wrote:
> On 2024-05-31 15:13:02 +0000, olcott said:
> 
>> On 5/31/2024 3:30 AM, Mikko wrote:
>>> On 2024-05-31 01:54:52 +0000, olcott said:
>>>
>>>> On 5/30/2024 8:37 PM, Richard Damon wrote:
>>>>> On 5/30/24 9:31 AM, olcott wrote:
>>>>>> On 5/30/2024 2:40 AM, Mikko wrote:
>>>>>>> On 2024-05-30 01:15:21 +0000, olcott said:
>>>>
>>>>>>>> x <is> a finite string Turing machine description that SPECIFIES 
>>>>>>>> behavior. The term: "representing" is inaccurate.
>>>>>>>
>>>>>>> No, x is a description of the Turing machine that specifies the 
>>>>>>> behaviour
>>>>>>> that H is required to report.
>>>>>>
>>>>>> That is what I said.
>>>>>
>>>>> Note, the string doesn't DIRECTLY specify behavior, but only 
>>>>> indirectly as a description/representation of the Turing Mach
>>>>>
>>>>
>>>> The string directly SPECIFIES behavior to a UTM or to
>>>> any TM based on a UTM.
>>>
>>> An UTM interpretes the string as a specification of behaviour
>>
>> YES, exactly !!!
>>
>>> and another Turing machine may interprete likewise. But in a
>>> different context the interpretation is different.
>>>
>>
>> When Ĥ is applied to ⟨Ĥ⟩
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>> When embedded_H <is> a UTM or <is> a halting computation based on a
>> UTM then the ⟨Ĥ⟩ ⟨Ĥ⟩ input to embedded_H SPECIFIES that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
>> simulated by embedded_H cannot possibly reach its own simulated final
>> state at ⟨Ĥ.qn⟩.
> 
> There is no requirement to follow the pecifications by embedded_H.


In other words embedded_H can simply play bingo and never halt and still 
correctly decide the halt status of its input?

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