Re: DDD correctly emulated by HHH is correctly rejected as non-halting.

On 7/11/24 9:05 AM, olcott wrote:

On 7/11/2024 2:25 AM, Fred. Zwarts wrote:

Op 10.jul.2024 om 22:53 schreef olcott:

On 7/10/2024 3:49 PM, olcott wrote:

On 7/10/2024 1:55 PM, Alan Mackenzie wrote:

Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:

Op 10.jul.2024 om 20:12 schreef Alan Mackenzie:

[ Followup-To: set ]

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In comp.theory Fred. Zwarts <F.Zwarts@hetnet.nl> wrote:

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[ .... ]

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Proving that the simulation is incorrect. Because a correct simulation
would not abort a halting program halfway its simulation.

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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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[ .... ]

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A non-halting program cannot be simulated correctly in a finite time.
So, it depends whether we can call it a correct simulation, when it does
not abort. But, for some meaning of 'correct', indeed, a simulator
should not abort a non-halting program either.

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OK, thanks!
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In other words he is saying that when you do
1 step correctly you did 0 steps correctly.
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In other words he is using deceitful weasel wording to
try to escape a truism.

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Why are you twisting my words? Is English such a difficult language for you?
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We stipulate that the only measure of a correct emulation 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.

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And since HHH is a program that halts, this

 DDD emulated by pure function x86 emulator HHH can't
possibly ever reach is own emulated machine address at
00002174 according to the correct semantics of the x86
language.

You are again LYIHG by using imprecise language.
The emulation of DDD by HHH doesn't reach that point.
The behavior of DDD, (which happens to be the DDD that HHH emulates) does reach that point, just after HHH stops its emulation.
You confuse the Observation by HHH for the reality of the behavior of the program, because you don't understand what Truth actually is, so you can't actually understand logic.

 

proves that the simulation was aborted halfway, which makes it incorrect.
Showing that the set of HHH that can correctly simulate itself is empty.
When a correct simulation needs N steps and you abort it halfway after M steps, with M much smaller than N, then the simulation of N steps is correct, but the simulation as a whole is incorrect, because the semantics of the x86 languages specifies that N steps must me simulated.
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_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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On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > But H determines (correctly) that D would not halt
 > if it were not halted.  That much is a truism.
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