Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
https://www.researchgate.net/ publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
On 9/15/2024 11:55 AM, Richard Damon wrote:
On 9/15/24 10:23 AM, olcott wrote:
Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
https://www.researchgate.net/
publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
Nope, just proves you are a stupid ignorant liar that doesn't know
what he is talking about.
First, you title is incorrect, as it isn't "D" that needs to report
its halt status, but the "H" that "D" Calls.
Remember, in the problem H is, and only is, the machine that H is, and
D is, and only is, the machine that D is, as the code in the problem
presents.
Remember, you have been tole (but ignore) that the phrase:
until H correctly determines that its simulated D would never stop
running unless aborted
Means that a CORRECT UNABORTED SIMULATION of *THIS* D would not stop,
unless aborted, Therefore giving H is criterion measure.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D
until H correctly determines that its simulated D would never
stop running unless aborted then
but *THIS* D calls *THIS* H, which you admit *WILL* decide to abort
(and thus its simulation is not the "correct simulation" we are to
look at).
Since THIS H does return to its caller, since it DOES abort, this
means that the correct simulation of this D will halt, and thus this H
never got the "authorization" to abort, but did so anyway and
introduced its error into the system.
Note, when you say:
When H correctly simulates D it finds that D remains stuck in
recursive simulation
Line 13: main() invokes D(D)
Line 06: Invoked D calls H that simulates D(D)
Line 06: Simulated D calls simulated H that simulates D(D) (repeats
until aborted)
Simulation Invariant: D simulated by H never reaches Line 07 or Line 08.
We note that the first line is incorrect, as H DOESN'T correctly
simulate D, as it DOES abort, and that the "repeats until aborted"
talks about a condition that DOES happen, so if we look at the actual
CORRECT simulation of D, we see that main invokes D(D) that calls
H(D,D) that simulates its copy of D(D) for some time an then aborts
that simulation and returns to D which returns to main, and thus halts.
Thus, your "Simulation Invariant" is just a LIE.
Of course, that is because you stupidly keep on confusing the behavior
of D (simulated by H) with the behavior of the simulation of D by H.
You seem to do this because you, in your ignorance, can't keep track
of the difference between the TRUTH of the behavior of *THIS* D (the D
that calls the H that simulates it for a while and then aborts and
returns) with the knowledge that H gets by its PARTIAL simulation of
its input D, and the behavior of a totally different program D (with
the same code in the C functions D, but calling a different function
H, with different behavior).
These errors have been pointed out to you MANY times, and your
repeating the error either says you are mentally incapable of learning
facts, or you are just so brainwashed by yourself with your lies that
you just refuse to accept the facts and thus make your self just a
pathetic ignorant pathologically lying idiot.
Notd, at your end, you confuse the question, as you fill in the D / D
line with "Accept" even though this is supposed to be the mapping
function that H is computing, and BY YOUR ADMISSION, H REJECTED the
input.
Thus, you prove with that final comment that you just don't understand
what you are talking about and are nothing but a LYING IDIOT.
Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
https://www.researchgate.net/ publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
https://www.researchgate.net/publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
https://www.researchgate.net/ publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
We can see that the first seven instructions of D emulated by H precisely match the first seven instructions of the x86 source-code of D. This conclusively proves that these instructions were emulated correctly.
H detects that D is calling H twice in sequence with the same aguments. H also sees that thereare no conditional branch instructions from the beginning of D to its call to H that can possibly escape the repetition of this recursive simulation.
Anyone sufficiently technically competent in the x86 programming language will agree that the above execution trace of D emulated by H shows that D will never stop running unless H aborts its emulation of D.
On 9/16/2024 3:06 AM, Mikko wrote:
On 2024-09-15 14:23:38 +0000, olcott said:
Rebutting the Sipser Halting Problem Proof
D(D) correctly reports its own halt status
https://www.researchgate.net/
publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
That article does not prove its claim and therefore does not constiture
a refutation of anything.
The text agreed by Sipser is shown twice: first in the beginnig of
the article and again in the beginnig of the Appendix.
There are two execution traces: one on pare 3, another in the Appendix.
The relation between them is not explained. More generally, the roloe
or the purpose of the Appendix is not clear.
You have to actually carefully study the words.
There are not two execution traces.
One shows the execution trace of D emulated by H:
The essence of the halt status decision.
The other one shows the trace of the whole
program including the trace of D emulated by H.
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