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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: 195 page execution trace of DDD correctly simulated by HH0
Date: Wed, 26 Jun 2024 08:30:16 -0500
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On 6/26/2024 3:01 AM, Fred. Zwarts wrote:
> 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?)
> 

Simulated H0 is never aborted.
Only termination analyzers do that.

>> 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 is like I say 2 + 3 = 5 and you simply say no I am wrong.
My reasoning is that the first four lines of DDD necessary must
repeat for DDD correctly simulated by any x86 emulator at machine
address 0000217a.

If you reasoning is that a magic fairy sprinkles magic
fairy dust on line four then you are wrong.

> 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.

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