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On 9/16/2024 6:21 AM, Fred. Zwarts wrote:But the first seven instruction of D are never executed again, so that CAN'T be a part of a correct simulation of D, including the part where we look at the H that D calls simulatng D, because that should show the actions of H, not the results of the simulation.Op 15.sep.2024 om 16:23 schreef olcott:Then if you are not a damned liar you can see this>>
Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
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https://www.researchgate.net/ publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
>We can see that the first seven instructions of D emulated by H precisely match the first seven instructions of the x86 source-code of D. This conclusively proves that these instructions were emulated correctly.>
Yes H makes a good start, but fails to complete the simulation, because of a bug in the code to recognise an infinite 'recursion'.
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next part that you dishonestly erased.
D()
[0000218e] 55 push ebp ; begin D
[0000218f] 8bec mov ebp,esp
[00002191] 8b4508 mov eax,[ebp+08]
[00002194] 50 push eax ; push param
[00002195] 8b4d08 mov ecx,[ebp+08]
[00002198] 51 push ecx ; push param
[00002199] e8a0f2ffff call 0000143e ; call H
After D calls H(D, D) we can see that H correctly emulates itself emulating D because again we see that the first seven instructions of D emulated by the emulated H precisely match the first seven instructions of the x86 source-code of D. This conclusively proves that these instructions were emulated correctly.
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