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On 7/10/2024 1:12 PM, Alan Mackenzie wrote:But not a correct simulation of a program. The x86 language specifies that it is insufficient to interpret only one instruction of a program. For a correct simulation of a halting program, all instructions must be simulated.[ Followup-To: set ]>
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In comp.theory Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:
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>Proving that the simulation is incorrect. Because a correct simulation>
would not abort a halting program halfway its simulation.
Just for clarity, a correct simulation wouldn't abort a non-halting
program either, would it? Or have I misunderstood this correctness?
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Welcome back.
I stipulate that I am referring to 1 to ∞ steps of correct
emulation according to the semantics of the x86 language.
This means that when HHH does correctly emulate 1 step
that *it is a correct emulation* of this 1 step,
thus
making everyone that disagrees disagree with a tautology
making them look foolish.
We stipulate that the only measure of a correct emulationWhich proves that the simulation is incomplete and, therefore, incorrect.
is the semantics of the x86 programming language. By this
measure when 1 to ∞ steps of DDD are correctly emulated
by each pure function x86 emulator HHH (of the infinite
set of every HHH that can possibly exist) then DDD cannot
possibly reach past its own machine address of 0000216b
and halt.
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
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