Re: DDD correctly emulated by H0

Op 22.jun.2024 om 20:47 schreef olcott:

On 6/22/2024 1:39 PM, Fred. Zwarts wrote:

Op 21.jun.2024 om 15:21 schreef olcott:

On 6/21/2024 2:16 AM, Mikko wrote:

On 2024-06-20 15:04:35 +0000, olcott said:
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On 6/20/2024 9:42 AM, Mikko wrote:

On 2024-06-20 05:15:37 +0000, olcott said:
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On 6/20/2024 12:04 AM, Mikko wrote:

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Sitll inclear whether you know what "termination analyzer" means.

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I really don't care what you believe.
It is not about belief.
It is about correct reasoning.

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No, it is not. It is about language maintenance. If you cannot present
your reasoning in Common Language it does not matter whether your
reasoning is correct.

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I cannot possibly present my reasoning in a convincing way
to people that have already made up their mind and closed it
thus fail to trace through each step of this reasoning looking
for an error and finding none.

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If you can't convince the reviewers of a journal that your article is
well thought and well written you cannot get it published in a
respected journal.
>

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The trick is to get people that say I am wrong
to point out the exact mistake. When they really
try to do this they find no mistake and all of
their rebbutal was pure bluster with no actual basis.
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It seems you do not even try to answer questions to show errors in the reasoning of your opponents, in order to protect yourself against finding no errors in their rebuttal.

 I quit bothering to point out the strawman deception fake
rebuttal and instead I just ignore the whole reply. If you
change the subject away from DDD correctly emulated by H0
your reply will be ignored.
 When we stipulate that the only measure of a correct emulation
is the semantics of the x86 programming language then we see
that when DDD is correctly emulated by H0 that its call to
H0(DDD) cannot possibly return.
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 When we define H1 as identical to H0 except that DDD does not
call H1 then we see that when DDD is correctly emulated by H1
that its call to H0(DDD) does return. This is the same behavior
as the directly executed DDD().
 

Probably. Which shows that the simulation of H0 by H0 is incorrect. H1 shows that a correct simulation of H0 halts, but H0 is unable to simulate itself up to its final state. It aborts one cycle before the simulation would end, as is shown by H1.