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

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.
Its state and rules specify and action that should be performed. Whether
it actually is performed does not affect the meaning.
The qestion to be answerd is whether the computation specified by the
input ever terminates if fully executed. Whether the computation is
eve fully executed does not alter the meaning of the question but may
help to find out whether any given answer is correct.
--
Mikko