• Re: Rebutting the Sipser Halting Problem Proof

    From Richard Damon@21:1/5 to olcott on Sun Sep 15 12:55:36 2024
    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,
    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.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Sun Sep 15 21:10:25 2024
    On 9/15/24 3:07 PM, olcott wrote:
    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.

    No, the only correct emulation is an unaborted on, and, since H DOES
    abort, it isn;t correct, and to talk about the "correct emulation by H"
    is a contradiction, (not just an oxymoron, but does make you a regular
    moron).

    As has been shown, when THIS D (The D that calls the H that gives the
    answer you claim) is actually correctly simulated, it will reacn an end.

    Only your LIES that your H does a correct simulation that isn't actually
    a correct simulation becuase it thinks it is and aborts, point to what
    you want, and theu have been proven to be LIES.


    <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

    Right, and since a correct simulation of this input (the D that calls
    this H that will abort) will reach a final state, you are just proving
    your stupidity in claming H can correctly determine something to happen
    that doesn't actually happen.

    Thus showing how broken your logic is.

    Note, the biggest part of your problem is you somehow think it is ok to
    change the problem in the middle, in part because you don't understand
    what a "program" is.

    Sorry, you are just proving that you have no idea what you are talking
    about and think that lying is a valid form of logic.

    It appears that you are booking a perpetual ticket for a logic refresher course.


    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.




    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Fred. Zwarts@21:1/5 to All on Mon Sep 16 10:09:47 2024
    Op 15.sep.2024 om 16:23 schreef olcott:

    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


    The simulation invariant is that H fails to reach the relevant part of
    the simulation. It stops the simulation and never sees how the simulated
    H aborts and therefore never reaches the 'pathological' part of D (lines
    7 and 8).
    This is due to the bug in H, that decides that there is an infinite
    recursion when it sees only a finite recursion.
    Although it is programmed to see the 'special condition' to abort and
    halt the simulation, it does not realise that the simulated H is also programmed to see this 'special condition' to abort and halt the simulation.
    H cannot possibly simulate itself correctly up to the end.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Mon Sep 16 11:06:01 2024
    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.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Fred. Zwarts@21:1/5 to All on Mon Sep 16 13:21:29 2024
    Op 15.sep.2024 om 16:23 schreef olcott:

    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.

    Yes H makes a good start, but fails to complete the simulation, because
    of a bug in the code to recognise an infinite 'recursion'.

    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.

    H fails to see that in-between the two beginnings of D, there are many conditional branch instructions, namely in the H called by D.
    The simulation fails to properly simulate the call to H. This bug makes
    that H decides incorrectly that there is an infinite recursion.

    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.

    Yes, and anyone sufficiently technically competent in the x86 language
    will also see that this proves that H fails to correctly complete the simulation.
    The bug in the code decides after two cycles that there are an infinite
    number of cycles, because it misses the conditional branch instructions
    in these cycles, because of the incorrect simulation that deviates from
    the semantics of the x86 language, with an improper simulation of the H
    called by D.
    So, it misses the fact that the simulated H would also (incorrectly) see
    this 'special condition', after which it would also abort, after which
    it would return to D and D would halt.

    The string w describes only one program and the semantics of the x86
    language allows only one behaviour for this program.
    This behaviour is 'halting', as proven by direct execution, by the
    simulation by the world class simulator and other simulations.
    But H fails to reproduce this behaviour in its attempt to simulate
    itself. This proves that H deviates from the semantics of the x86
    language in its simulation.

    H cannot possibly simulate itself correctly up to the end.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Mikko@21:1/5 to olcott on Tue Sep 17 09:51:37 2024
    On 2024-09-16 12:03:12 +0000, olcott said:

    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.

    Which 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.

    That is not said in the Appendix.

    --
    Mikko

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)