• Re: Who here is too stupid to know that DDD simulated by HHH cannot rea

    From joes@21:1/5 to All on Wed Aug 7 18:25:31 2024
    Am Wed, 07 Aug 2024 08:40:31 -0500 schrieb olcott:
    On 8/7/2024 2:22 AM, Mikko wrote:
    On 2024-08-05 15:00:12 +0000, olcott said:
    On 8/5/2024 2:44 AM, Mikko wrote:
    On 2024-08-04 13:11:56 +0000, olcott said:
    On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
    Op 03.aug.2024 om 17:20 schreef olcott:>>

    In another message you have said that when HHH simulates itself
    simulating DDD is does not simulate itself simulating itself
    simulating DDD. You have not told whether it makes a cup of coffee.
    Neither action can be seen in the traces you have shown.

    HHH and HH and the original H have proved that they simulate
    themselves simulating DDD, DD and P for three years now.

    Your trace don't show siulation of exectuion differently from
    simulation of simulation of execution.

    It does but it is too difficult to dig it out of emulations of emulators emulating inputs.
    You could make it clearer by prefixing the simulation level.

    As soon as the first HHH sees the second DDD about to invoke a third HHH
    it aborts the emulation. At this point DDD, the second HHH and the
    second DDD all immediately stop running and HHH returns 0 to main.
    And HHH concludes that the second invocation of itself would somehow
    not also abort but run forever, and then returns that itself wouldn't
    halt, and halts. ???

    --
    Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
    It is not guaranteed that n+1 exists for every n.

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  • From Richard Damon@21:1/5 to olcott on Wed Aug 7 21:01:08 2024
    On 8/7/24 4:50 PM, olcott wrote:
    On 8/7/2024 2:02 PM, olcott wrote:
    On 8/7/2024 1:25 PM, joes wrote:
    Am Wed, 07 Aug 2024 08:40:31 -0500 schrieb olcott:
    On 8/7/2024 2:22 AM, Mikko wrote:
    On 2024-08-05 15:00:12 +0000, olcott said:
    On 8/5/2024 2:44 AM, Mikko wrote:
    On 2024-08-04 13:11:56 +0000, olcott said:
    On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
    Op 03.aug.2024 om 17:20 schreef olcott:>>

    In another message you have said that when HHH simulates itself
    simulating DDD is does not simulate itself simulating itself
    simulating DDD. You have not told whether it makes a cup of coffee. >>>>>>> Neither action can be seen in the traces you have shown.

    HHH and HH and the original H have proved that they simulate
    themselves simulating DDD, DD and P for three years now.

    Your trace don't show siulation of exectuion differently from
    simulation of simulation of execution.

    It does but it is too difficult to dig it out of emulations of
    emulators
    emulating inputs.
    You could make it clearer by prefixing the simulation level.


    Maybe I can do this. it requires a static local, yet this
    is only for reporting purposes and does not have an effect
    on the computation.


    Why? This is all supposedly the output of the top level decideer, which
    know what level it is at.

    Or, are you admitting that you have been lying all the time about where
    the output comes from.


    None of this was ever really required. We have complete
    proof that the second HHH does emulate its DDD correctly
    by simply comparing the execution trace that it produces
    to the x86 source code of DDD.


    Only when you use the wrong definiton of emulating correctly, which
    means you don't actually get the results you need for your claims.



    We have had this complete proof for HH/DD and H/P for three
    years and everyone simply ignores it. This can only mean:
    (a) dishonesty
    (b) insufficient technical competence.

    Yep, YOU have both of those problems.

    Do you EVER intend to provide ANY of the "source" for any of your claims
    that you rely on?



    As soon as the first HHH sees the second DDD about to invoke a third
    HHH
    it aborts the emulation. At this point DDD, the second HHH and the
    second DDD all immediately stop running and HHH returns 0 to main.

    And HHH concludes that the second invocation of itself would somehow
    not also abort but run forever, and then returns that itself wouldn't
    halt, and halts. ???


    void DDD()
    {
       HHH(DDD);
       return;
    }

    Any expert in C can tell that DDD simulated by HHH cannot
    possibly ever reach its own "return" instruction final halt
    state.


    No, The Program DDD which is simualte by HHH will return, if and only if
    the subroutine HHH returns. If HHH ACTUALY does a correct emualtion of
    its input, and thus never abort to it reaches an end, will not return.

    But, YOUR HHH doesn't do that, because you claim that HHH is a decider,
    so it MUST answer, and thus can't be the correct emulation that you hypothocate, and thus you can't use the behavior of the DIFFERENT
    program DDD that uses a DIFFERENT program HHH as a subroutine for your analysis.


    Is it possible for someone to have a PhD in computer science
    and not have any clue about something as simple as this?


    Since you base your claims on the deceptive lies of changinge
    definitions, YOU are the stupid one.

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  • From Mikko@21:1/5 to olcott on Thu Aug 8 10:53:52 2024
    On 2024-08-07 20:50:15 +0000, olcott said:

    On 8/7/2024 2:02 PM, olcott wrote:
    On 8/7/2024 1:25 PM, joes wrote:
    Am Wed, 07 Aug 2024 08:40:31 -0500 schrieb olcott:
    On 8/7/2024 2:22 AM, Mikko wrote:
    On 2024-08-05 15:00:12 +0000, olcott said:
    On 8/5/2024 2:44 AM, Mikko wrote:
    On 2024-08-04 13:11:56 +0000, olcott said:
    On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
    Op 03.aug.2024 om 17:20 schreef olcott:>>

    In another message you have said that when HHH simulates itself
    simulating DDD is does not simulate itself simulating itself
    simulating DDD. You have not told whether it makes a cup of coffee. >>>>>>> Neither action can be seen in the traces you have shown.

    HHH and HH and the original H have proved that they simulate
    themselves simulating DDD, DD and P for three years now.

    Your trace don't show siulation of exectuion differently from
    simulation of simulation of execution.

    It does but it is too difficult to dig it out of emulations of emulators >>>> emulating inputs.
    You could make it clearer by prefixing the simulation level.


    Maybe I can do this. it requires a static local, yet this
    is only for reporting purposes and does not have an effect
    on the computation.


    None of this was ever really required. We have complete
    proof that the second HHH does emulate its DDD correctly
    by simply comparing the execution trace that it produces
    to the x86 source code of DDD.

    You may have the complete knowledge for the proof but you have never
    shown the proof.

    --
    Mikko

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  • From Mikko@21:1/5 to olcott on Thu Aug 8 10:52:20 2024
    On 2024-08-07 19:02:21 +0000, olcott said:

    On 8/7/2024 1:25 PM, joes wrote:
    Am Wed, 07 Aug 2024 08:40:31 -0500 schrieb olcott:
    On 8/7/2024 2:22 AM, Mikko wrote:
    On 2024-08-05 15:00:12 +0000, olcott said:
    On 8/5/2024 2:44 AM, Mikko wrote:
    On 2024-08-04 13:11:56 +0000, olcott said:
    On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
    Op 03.aug.2024 om 17:20 schreef olcott:>>

    In another message you have said that when HHH simulates itself
    simulating DDD is does not simulate itself simulating itself
    simulating DDD. You have not told whether it makes a cup of coffee. >>>>>> Neither action can be seen in the traces you have shown.

    HHH and HH and the original H have proved that they simulate
    themselves simulating DDD, DD and P for three years now.

    Your trace don't show siulation of exectuion differently from
    simulation of simulation of execution.

    It does but it is too difficult to dig it out of emulations of emulators >>> emulating inputs.
    You could make it clearer by prefixing the simulation level.


    Maybe I can do this. it requires a static local, yet this
    is only for reporting purposes and does not have an effect
    on the computation.

    No, it can be done without a static local. It does require that the
    emulator recognizes output functions as a special case but it already
    does that. The emulator should emulate an output call so that it
    prepends the text to be output with a special character, for example
    with '|'.

    --
    Mikko

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