XPost: comp.theory, sci.logic
On 7/12/24 10:56 AM, olcott wrote:
We stipulate that the only measure of a correct emulation is the
semantics of the x86 programming language.
Which means the only "correct emulation" that tells the behavior of the
program at the input is a non-aborted one.
_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]
When N steps of DDD are emulated by HHH according to the
semantics of the x86 language then N steps are emulated correctly.
And thus HHH that do that know only the first N steps of the behavior of
DDD, which continues per the definition of the x86 instruction set until
the COMPLETE emulation (or direct execution) reaches a terminal instruction.
When we examine the infinite set of every HHH/DDD pair such that:
HHH₁ one step of DDD is correctly emulated by HHH.
HHH₂ two steps of DDD are correctly emulated by HHH.
HHH₃ three steps of DDD are correctly emulated by HHH.
...
HHH∞ The emulation of DDD by HHH never stops running.
And thus, the subset that only did a finite number of steps and aborted
its emulation on a non-terminal instrucition only have partial knowledge
of the behavior of their DDD, and by returning to their caller, they
establish that behavior for ALL copies of that HHH, even the one that
DDD calls, which shows that DDD will be halting, even though HHH stopped
its observation of the input before it gets to that point.
The above specifies the infinite set of every HHH/DDD pair
where 1 to infinity steps of DDD are correctly emulated by HHH.
No DDD instance of each HHH/DDD pair ever reaches past its
own machine address of 0000216b and halts.
Wrong. EVERY DDD of an HHH that simulated its input for only a finite
number of steps WILL halt becuase it will reach its final return.
The HHH that simulated it for only a finite number of steps, only
learned that finite number of steps of the behaivor, and in EVERY case,
when we look at the behavior past that point, which DOES occur per the definition of the x86 instruction set, as we have not reached a
"termial" instruction that stops behavior, will see the HHH(DDD) that
DDD called continuing to simulate its input to the point that this one
was defined to stop, and then returns 0 to DDDD and then DDD returning
and ending the behavior.
You continue to stupidly confuse the PARTIAL observation that HHH does
of the behavior of DDD by its PARTIAL emulation with the ACTUAL FULL
behavior of DDD as defined by the full definition of the x86 insttuction
set.
Thus each HHH element of the above infinite set of HHH/DDD
pairs is necessarily correct to reject its DDD as non-halting.
Nope.
NONE Of them CORRECTLY rejected itS DDD as non-halting and you are shown
to be ignorant of what you are talking about.
The HHH that did a partial emulation got the wrong answer, because THEIR
DDD will halt. and the HHH that doen't abort never get around to
rejecting its DDD as non-halting.
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