Re: DDD emulated by HHH diverges from DDD emulated by HHH1--- BEST ONE

On 6/6/25 11:43 PM, olcott wrote:

On 6/6/2025 9:02 PM, Richard Damon wrote:

On 6/6/25 12:53 PM, olcott wrote:

On 6/6/2025 11:06 AM, Mike Terry wrote:

On 05/06/2025 05:27, olcott wrote:

On 6/4/2025 10:55 PM, Mike Terry wrote:

On 05/06/2025 02:39, olcott wrote:

On 6/4/2025 8:28 PM, dbush wrote:

On 6/4/2025 9:08 PM, olcott wrote:

On 6/4/2025 7:41 PM, dbush wrote:

On 6/4/2025 8:32 PM, olcott wrote:

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Show me this side-by-side trace and I will point out your mistake.

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See below, which shows that the simulations performed by HHH and HHH1 are identical up to the point that HHH aborts, as you have agreed on the record.
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False.  The correct trace is the one I posted, which shows all levels of emulation performed by HHH and HHH1.  See the corrections I made to your comments

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It is not supposed to do that.

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Are you saying it's not supposed to include /nested/ emulations? It is perfectly sensible to include nested emulations.
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It can include nested simulations yet nested
simulations are in a hierarchy thus not side-by-side.
A side-by-side analysis must be side-by-side.
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Hierarchies can be compared side-by-side.  In the case of these traces, the hierarchy can be "flattened" into one stream of nested simulations. You do this yourself every time you present one of your nested simulation traces.  Such a trace should include a simulation depth (or equivalent) for each entry.
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Two nested simulation traces can easily be presented side-by-side for comparisson.  You are just trying to divert attention from your own failings to properly understand the requirements.
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*From the execution trace of HHH1(DDD) shown below*
DDD emulated by HHH1              DDD emulated by HHH
[00002183] push ebp               [00002183] push ebp
[00002184] mov ebp,esp            [00002184] mov ebp,esp
[00002186] push 00002183 ; DDD    [00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH    [0000218b] call 000015c3 ; HHH
*HHH1 emulates DDD once then HHH emulates DDD once, these match*
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*Then HHH emulates itself emulating DDD, HHH1 NEVER DOES THIS*

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Because the correct emulation of the input doesn't call for this to be done, and the identity of the emulator doesn't affect the defintion of a correct emulation.
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That fact that NONE of your traces actually show a correct emulation,

 I have corrected you on this hundreds of times and
you keep "forgetting" what I said.
 

That you have an "excuse" doesn't change the fact that the traces shown are not correct.
\Just because YOU are to stupid to make sense of the long report, doesn't mean that it isn't factual, and NEEDED to make your point.
All you are doing is saying that logic needs to allow LYING for you to understand it, likely because you have an inherent misunderstanding about the nature of how truth and logic work.
Note, you can't even seem to keep seperate the trace of the execution of the outer HHH from the trace that it sees from its emulation, which is why the output is so long.
The problem is that you just don't understand what you are doing, and are just hacking at a program you jury-rigged from an good open-source prject that you hacked up to build your "proof" from, by figuing out how to "edit" its output to show what you want.