Re: 195 page execution trace of DDD correctly simulated by HH0

On 6/20/24 6:45 PM, olcott wrote:

On 6/20/2024 5:37 PM, Richard Damon wrote:

On 6/20/24 10:12 AM, olcott wrote:

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.

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Why, you aren't?
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You seem to think you are God or something that gets to set the rules.
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YOU ARE NOT.
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Similarly, if a simulator which aborts after 2 cycles of recursive simulation of it self, it simulates only 1 of the 2 cycles of itself. So, it is incorrect, not because one instruction was simulated incorrectly, but because it did not simulate up to the final state of the simulated self.
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void Infinite_Loop()
{
   HERE: goto HERE;
}
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It also looks like you fail to comprehend that it is possible
for a simulating termination analyzer to recognize inputs that
would never terminate by recognizing the repeating state of
these inputs after a finite number of steps of correct simulation.

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Right, but they don't do it by "Correctly Simulating" the input, but by a PARTIAL simulation that provides the needed information to prove that an ACTUAL CORRECT (and complete) simulation of that input would not halt.
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 *This seems to be the first time that you told the truth about this*
*This seems to be the first time that you told the truth about this*
*This seems to be the first time that you told the truth about this*
 

And that works for Infinite_Loop, and Infinte_Recursion, but NOT for D or any of your variations of it.
You just don't understand how to build the required inductive argument, because it requires forming an actual proof.