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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: Mon, 29 Jul 2024 11:55:46 -0500
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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.

When a simulating partial halt decider is comprised of the notions of a:
(a) UTM
(b) decider
(c) Turing machine description

Then it is easy to see that any UTM that simulates less than
all of the steps of a TM Description has aborted this simulation.



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