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On 6/21/24 9:01 AM, olcott wrote:If you claim that I have made a mistake and cannot pointOn 6/21/2024 2:44 AM, Fred. Zwarts wrote:Which proves your duplicity, since no one is claiming that.Op 20.jun.2024 om 16:12 schreef olcott:>On 6/20/2024 3:09 AM, Fred. Zwarts wrote:>Op 20.jun.2024 om 02:00 schreef olcott:>This shows all of the steps of HH0 simulating DDD>
calling a simulated HH0 simulating DDD
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https://liarparadox.org/HH0_(DDD)_Full_Trace.pdf
*Some of the key instructions are color coded*
GREEN---DebugStep Address
RED-----HH Address
YELLOW--All of the DDD instructions
CYAN----Return from DebugStep to Decide_Halting_HH
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_DDD()
[000020a2] 55 push ebp ; housekeeping
[000020a3] 8bec mov ebp,esp ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404 add esp,+04 ; housekeeping
[000020b2] 5d pop ebp ; housekeeping
[000020b3] c3 ret ; never gets here
Size in bytes:(0018) [000020b3]
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Exactly which step of DDD emulated by H0 was emulated
incorrectly such that this emulation would be complete?
AKA DDD emulated by H0 reaches machine address [000020b3]
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If the simulation of a program with a loop of 5 iterations is aborted after 3 iterations, all instructions are correctly simulated. Nevertheless, it is an incorrect simulation, because it should simulate up to the final state of the program.
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It would be helpful if you answer the actual question being asked
right here and thus not answer some other question that was asked
somewhere else.
If you do not understand that I answered the question why the simulation is incorrect, it is hopeless. The question which instruction is incorrect is not the right question.
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If you say that something is incorrect and can't be specific
then your rebuttal is pure bluster with no actual basis.
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You need to show every single freaking step of exactly
DDD correctly emulated by HH0 reaches past its own
machine address [0000209b] or all you have is BULLSHIT!
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