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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: This function proves that only the outermost HHH examines the
 execution trace
Date: Tue, 30 Jul 2024 18:27:24 -0500
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On 7/30/2024 2:15 AM, Mikko wrote:
> On 2024-07-29 16:55:46 +0000, olcott said:
> 
>> On 7/28/2024 4:10 AM, Mikko wrote:
>>> On 2024-07-27 18:14:52 +0000, Alan Mackenzie said:
>>>
>>>> olcott <polcott333@gmail.com> wrote:
>>>
>>>>> Stopping running is not the same as halting.
>>>>> DDD emulated by HHH stops running when its emulation has been aborted.
>>>>> This is not the same as reaching its ret instruction and terminating
>>>>> normally (AKA halting).
>>>
>>>> I think you're wrong, here.  All your C programs are a stand in for
>>>> turing machines.  A turing machine is either running or halted.  
>>>> There is
>>>> no third state "aborted".  An aborted C program certainly doesn't
>>>> correspond with a running turing machine - so it must be a halted 
>>>> turing
>>>> machine.
>>>
>>>> So aborted programs are halted programs.  If you disagree, perhaps you
>>>> could point out where in my arguments above I'm wrong.
>>>
>>> May I disagree? An "aborted" Turing machine is a runnung Turing machine.
>>
>> A Turing machine has no notion of being aborted.
> 
> That's correct. But you have used the word anyway.
> 

It seems that either you have ADD or are becoming a liar.
When we combine these conventional notions
(a) UTM
(b) halt decider
(c) Turing Machine description

into a simulating halt decider that bases it halt status
decision on the behavior of its simulated TMD
The the SHD can abort the simulation of its TMD.

The only reason that I suspect that you have become a
liar or have ADD is that I already completely explained
all this before and you act like I never said it.
before

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