On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
Op 13.jul.2024 om 01:19 schreef olcott:
On 7/12/2024 5:56 PM, Richard Damon wrote:
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.
*Here is the gist of my proof it is irrefutable*
When no DDD of every HHH/DDD that can possibly exist
halts then each HHH that rejects its DDD as non-halting
is necessarily correct.
*No double-talk and weasel words can overcome that*
This is double talk, because no HHH can possibly exist that simulates
itself correctly.
Your definition of correct contradicts the semantics of
the x86 language making it wrong.
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86 language.
You think that the x86 language specifies that skipping instructions
do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider actually
has, but look at the input that would have been given to a different
decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming. https://www.geeksforgeeks.org/context-switch-in-operating-system/
On 7/13/2024 9:14 AM, Richard Damon wrote:
On 7/13/24 9:27 AM, olcott wrote:
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have a wrong understanding of the semantics of the x86
language. You think that the x86 language specifies that skipping
instructions do not change the behaviour of a program.
You have the wrong understanding of a decider.
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider
actually has, but look at the input that would have been given to a
different decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
But every DDD that calls an HHH that aborts its simulation of a copy
of that DDD and returns is shown to be halting, not non-halting. It is
just that HHH can't see that behavior becuase it aborted its simulation.
"DDD" is the program, not the partial emulation of it, so it halts
even if HHHs PARTIAL simulaton of it ddn't reach thatpoint.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming.
https://www.geeksforgeeks.org/context-switch-in-operating-system/
Which is something I don't have problems with, since I have written my
own operating systems.
Your problem is you don't seem to understand is that all copies of a
given deterministic program act the same
By this same reasoning when you are hungry and eat until
you are full you are still hungry because you are still
yourself.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
On 7/13/2024 9:48 AM, Richard Damon wrote:
On 7/13/24 10:35 AM, olcott wrote:
On 7/13/2024 9:14 AM, Richard Damon wrote:
On 7/13/24 9:27 AM, olcott wrote:
On 7/13/2024 8:15 AM, Richard Damon wrote:
On 7/13/24 9:04 AM, olcott wrote:
On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
You have the wrong understanding of a decider.
You have a wrong understanding of the semantics of the x86
language. You think that the x86 language specifies that
skipping instructions do not change the behaviour of a program. >>>>>>>
All deciders are required to halt.
And are required to give the correct answer.
You seem to think it is ok for them to lie if they don't know the
right answer.
As soon as the decider correctly determines that itself
would never halt unless is aborts the simulation of its
input the decider is required to abort this simulation.
Which it never does, so it gives up and guesses.
YOU lie that it does correctly determines the answer, but that is
because you lie and don't look at the input that this decider
actually has, but look at the input that would have been given to
a different decider to show that one wrong.
*This proves that every rebuttal is wrong somewhere*
No DDD instance of each HHH/DDD pair of the infinite set of
every HHH/DDD pair ever reaches past its own machine address of
0000216b and halts thus proving that every HHH is correct to
reject its input DDD as non-halting.
But every DDD that calls an HHH that aborts its simulation of a copy
of that DDD and returns is shown to be halting, not non-halting. It
is just that HHH can't see that behavior becuase it aborted its
simulation.
"DDD" is the program, not the partial emulation of it, so it halts
even if HHHs PARTIAL simulaton of it ddn't reach thatpoint.
You seem to fail to understand the notion of differing
process contexts. It is a tricky notion for people that
have never done operating system level programming.
https://www.geeksforgeeks.org/context-switch-in-operating-system/
Which is something I don't have problems with, since I have written
my own operating systems.
Your problem is you don't seem to understand is that all copies of a
given deterministic program act the same
By this same reasoning when you are hungry and eat until
you are full you are still hungry because you are still
yourself.
Not at all, I am not a deterministic entity like HHH and DDD.
In other words when you are very hungry you have the
free will to decide that you are not hungry at all
and never eat anything ever again with no ill effects
to your health what-so-ever.
Try and use this free will to make a square circle.
After HHH has already aborted its simulation of DDD
and returns to the DDD that called it is not the same
behavior as DDD simulated by HHH that must be aborted.
Right, and the question is about the behavior of DDD,
the input finite string not an external process that HHH
has no access to.
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