• Re: Flat out dishonest or totally ignorant? Liar ??? For Olcott: YES to

    From Richard Damon@21:1/5 to olcott on Wed Jul 3 22:34:45 2024
    XPost: sci.logic

    On 7/3/24 10:05 PM, olcott wrote:
    On 7/3/2024 8:47 PM, Richard Damon wrote:
    On 7/3/24 9:26 PM, olcott wrote:
    On 7/3/2024 8:17 PM, Richard Damon wrote:
    On 7/3/24 8:40 PM, olcott wrote:
    On 7/3/2024 6:18 PM, Richard Damon wrote:
    On 7/3/24 10:19 AM, olcott wrote:
    On 7/3/2024 9:11 AM, joes wrote:
    Am Tue, 02 Jul 2024 22:55:12 -0500 schrieb olcott:
    On 7/2/2024 10:50 PM, joes wrote:
    Am Tue, 02 Jul 2024 14:46:38 -0500 schrieb olcott:
    On 7/2/2024 2:17 PM, Fred. Zwarts wrote:
    Op 02.jul.2024 om 21:00 schreef olcott:
    On 7/2/2024 1:42 PM, Fred. Zwarts wrote:
    Op 02.jul.2024 om 14:22 schreef olcott:
    On 7/2/2024 3:22 AM, Fred. Zwarts wrote:
    Op 02.jul.2024 om 03:25 schreef olcott:

    HHH repeats the process twice and aborts too soon.

    DDD is correctly emulated by any HHH that can exist which >>>>>>>>>>> calls this
    emulated HHH(DDD) to repeat the process until aborted (which >>>>>>>>>>> may be
    never).
    Whatever HHH does, it does not run forever but aborts.

    HHH halts on input DDD.
    DDD correctly simulated by HHH cannot possibly halt.
    WTF? It only calls HHH, which you just said halts.


    An aborted simulation does not count as halting.

    And doesn't show non-halting either.

    Reaching it own machine address 00002183 counts as halting.
    DDD correctly simulated by HHH cannot possibly do that.

    But HHH doesn't DO a "Correct Simulation" that can show that, it
    only does a PARTIAL simulation.


    <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

         H can abort its simulation of D and correctly report that D >>>>>      specifies a non-halting sequence of configurations.
    </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022> >>>>>
    until H correctly determines
    until H correctly determines
    until H correctly determines
    until H correctly determines
    until H correctly determines
    until H correctly determines
    until H correctly determines

    Which it doesn't.


    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT
    THUS STIPULATING THAT A PARTIAL SIMULATION IS CORRECT


    Nope, just double talk.

    H never CORRECTLY determined that a CORRECT SIMULATION (which means
    one that matchs the behavior of the machine represented by the
    input) would never halt, sinc ehta tmachine halts.


    _DDD()
    [00002172] 55         push ebp      ; housekeeping
    [00002173] 8bec       mov ebp,esp   ; housekeeping
    [00002175] 6872210000 push 00002172 ; push DDD
    [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
    [0000217f] 83c404     add esp,+04
    [00002182] 5d         pop ebp
    [00002183] c3         ret
    Size in bytes:(0018) [00002183]

    OK so it is not your ADD you continue to insist that
    you can disagree with the x86 language that conclusively
    proves that DDD correctly simulated by HHH cannot
    possibly get past machine instruction 0000217a.

    No, the claim is that it isn't the simulation by HHH that determines
    the actual behavior of the input, but the detailed semantics of the
    x86 instruciton set of the WHOLE input (which includes this particular
    HHH which you say DOES abort its simulation and return).

    You are not paying close enough attention to the exact words
    that I am saying. DDD cannot possibly reach past its own
    machine address 0000217a no matter what the Hell that HHH does.



    OF course it can. HHH might not be able to simulate it getting there,
    but if HHH returns to main as you claim, then the behavior defined by
    the x86 machine code presented (and include by refernce from HHH) says
    it will.

    You still seem to be stuck in your LIE that the partial emulation
    defines the behavior of DDD. The code for HHH defines what it does
    EVERYWHERE and when we combine that with the defined behaivor of DDD
    shown above, if HHH returns to the call to main, it returns to the call
    by DDD and DDD returns.

    Yes, you keep on trying to INCORRECTLY define the "behavior of the
    input" to be something it is not allowed to be, because it is something
    that depends on things that are not part of it, and thus is just a
    stupid lie.

    Your stupidity does not change the fundamental facts, it just makes you
    clearly stupid.

    Face it, you will just be remembered as the laughing stock of
    comp.theory (and the other groups you have also spread your ignorance.

    Even if some of your ideas of logic might have had some interesting
    points, you have poisioned them so baddly with this sort of argument,
    they will not be considered.

    YOU HAVE BURIED YOUR REPUTATION and it sounds like you will join it soon.

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