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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: 195 page execution trace of DDD correctly simulated by HH0
Date: Wed, 26 Jun 2024 10:01:45 +0200
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Op 25.jun.2024 om 21:30 schreef olcott:
> On 6/25/2024 2:17 PM, Fred. Zwarts wrote:
>>
>> It might be true, but it is irrelevant, because the simulated H0 is 
>> aborted prematurely. The simulating H0 aborts after two cycles, 
> 
> *I am not even talking about a simulating halt decider yet dumbo*

Neither am I. Why do you mention a simulating halt decider? (Who is the 
dumbo?)

> If you can't begin to comprehend x86 emulators then our conversation
> is dead right here.

Fortunately, I am very well able to do so.
But it seems that you have to learn a few basic facts about simulation.

> 
> For every x86 emulator Ho that can possibly exist
> at machine address 0000217a...
> 
> _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]
> 
> The call from DDD to H0(DDD) when DDD is correctly emulated
> by H0 cannot possibly return.

So, you repeat your claim without showing any error in my reasoning.
Therefore, I repeat again:

It might be true hat H0 cannot return, but it is irrelevant, because the 
simulated H0 is aborted prematurely. The simulating H0 aborts after two 
cycles, when the simulated H0 has one cycle to go before it would return.
So, the only reason that the simulated H0 does not return is that it is 
aborted prematurely. A correct simulation would show that one cycle 
later it would return.
It seems that it is over your head that a prematurely aborted simulation 
is not a correct simulation.
There are a few reasons why a simulation can be incorrect:
1) Some instructions are incorrectly simulated.
2) Instructions are simulated out of order.
3) Some instructions are not simulated at all.
Your problem is 3).
You only prove that your H0 is unable to simulate itself correctly, 
because it aborts too soon and fails to simulate the end of itself.