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

On 6/21/24 1:04 PM, olcott wrote:

On 6/21/2024 10:25 AM, Richard Damon wrote:

On 6/21/24 10:44 AM, olcott wrote:

On 6/21/2024 9:02 AM, Richard Damon wrote:

On 6/21/24 9:01 AM, olcott wrote:

On 6/21/2024 2:44 AM, Fred. Zwarts wrote:

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.

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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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Which proves your duplicity, since no one is claiming that.
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If you claim that I have made a mistake and cannot point
to the exact details of this mistake then everyone will
know that you are lying about there being any actual mistake.

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Sure
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First, NO ONE has said that *H* (or what every you are calling your decider today) can correct simulate the input to a final state.
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 Then you understand that H(D,D) is not even being
asked the question: Does D(D) halt?
 

Of course it is, at least if it is claimed to be a Halt Decider.
Nothing says that the decider has to actually be ABLE to answer the question, only that the answer exists. Uncomputable problems just can't be solved with a computation.
You just refuse to accept the only definition of that which is valid, because you logic can't handle it (and thus is wrong for the field).