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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2
Date: Fri, 26 Jul 2024 11:45:43 +0300
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On 2024-07-24 13:33:55 +0000, olcott said:

> On 7/24/2024 3:57 AM, Mikko wrote:
>> On 2024-07-23 13:31:35 +0000, olcott said:
>> 
>>> On 7/23/2024 1:32 AM, 0 wrote:
>>>> On 2024-07-22 13:46:21 +0000, olcott said:
>>>> 
>>>>> On 7/22/2024 2:57 AM, Mikko wrote:
>>>>>> On 2024-07-21 13:34:40 +0000, olcott said:
>>>>>> 
>>>>>>> On 7/21/2024 4:34 AM, Mikko wrote:
>>>>>>>> On 2024-07-20 13:11:03 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> On 7/20/2024 3:21 AM, Mikko wrote:
>>>>>>>>>> On 2024-07-19 14:08:24 +0000, olcott said:
>>>>>>>>>> 
>>>>>>>>>>> When we use your incorrect reasoning we would conclude
>>>>>>>>>>> that Infinite_Loop() is not an infinite loop because it
>>>>>>>>>>> only repeats until aborted and is aborted.
>>>>>>>>>> 
>>>>>>>>>> You and your HHH can reason or at least conclude correctly about
>>>>>>>>>> Infinite_Loop but not about DDD. Possibly because it prefers to
>>>>>>>>>> say "no", which is correct about Infinte_loop but not about DDD.
>>>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> *Because this is true I don't understand how you are not simply lying*
>>>>>>>>> int main
>>>>>>>>> {
>>>>>>>>>    DDD();
>>>>>>>>> }
>>>>>>>>> 
>>>>>>>>> Calls HHH(DDD) that must abort the emulation of its input
>>>>>>>>> or {HHH, emulated DDD and executed DDD} never stop running.
>>>>>>>> 
>>>>>>>> You are the lying one.
>>>>>>>> 
>>>>>>>> If HHH(DDD) abrots its simulation and returns true it is correct as a
>>>>>>>> halt decider for DDD really halts.
>>>>>>>> 
>>>>>>> 
>>>>>>> (b) We know that a decider is not allowed to report on the behavior
>>>>>>> computation that itself is contained within.
>>>>>> 
>>>>>> No, we don't. There is no such prohibition.
>>>>>> 
>>>>> 
>>>>> Turing machines never take actual Turing machines as inputs.
>>>>> They only take finite strings as inputs and an actual executing
>>>>> Turing machine is not itself a finite string.
>>>> 
>>>> The definition of a Turing machine does not say that a Turing machine
>>>> is not a finite string. It is an abstract mathematical object without
>>>> a specification of its exact nature. It could be a set or a finite
>>>> string. Its exact nature is not relevant to the theory of computation,
>>>> which only cares about certain properties of Turing machines.
>>>> 
>>>>> Therefore It is not allowed to report on its own behavior.
>>>> 
>>>> Anyway, that does not follow. The theory of Turing machines does not
>>>> prohibit anything.
>>>> 
>>>>> Another different TM can take the TM description of this
>>>>> machine and thus accurately report on its actual behavior.
>>>> 
>>>> If a Turing machine can take a description of a TM as its input
>>>> or as a part of its input it can also take its own description.
>>>> Every Turing machine can be given its own description as input
>>>> but a Turing machine may interprete it as something else.
>>>> 
>>> In this case we have two x86utm machines that are identical
>>> except that DDD calls HHH and DDD does not call HHH1.
>> 
>> That DDD calls HHH and DDD does not call HHH1 is not a difference
>> between two unnamed turing machines.
>> 
> 
> The same thing happens at the Peter Linz Turing Machine level
> I will provide that more difficult example if and only if you
> prove that you understand this one.

However, Peter Linz does not call taht same thing a difference.

-- 
Mikko