Re: This function proves that only the outermost HHH examines the execution trace

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