• Re: HHH maps its input to the behavior specified by it --- HHH never re

    From Richard Damon@21:1/5 to olcott on Wed Aug 7 22:04:17 2024
    On 8/7/24 9:35 PM, olcott wrote:
    On 8/7/2024 8:22 PM, Richard Damon wrote:
    On 8/7/24 9:12 PM, olcott wrote:
    On 8/7/2024 8:03 PM, Richard Damon wrote:
    On 8/7/24 2:14 PM, olcott wrote:
    On 8/7/2024 1:02 PM, joes wrote:
    Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:
    On 8/7/2024 2:29 AM, Mikko wrote:
    On 2024-08-05 13:49:44 +0000, olcott said:

    I know what it means. But the inflected form "emulated" does not >>>>>>>> mean
    what you apparently think it means. You seem to think that "DDD >>>>>>>> emulated by HHH" means whatever HHH thinks DDD means but it does >>>>>>>> not.
    DDD means what it means whether HHH emulates it or not.

    In other words when DDD is defined to have a pathological
    relationship
    to HHH we can just close our eyes and ignore it and pretend that it >>>>>>> doesn't exist?
    It doesn't change anything about DDD. HHH was supposed to decide
    anything
    and can't fulfill that promise. That doesn't mean that DDD is somehow >>>>>> faulty, it's just a counterexample.


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

    *HHH is required to report on the behavior of DDD*
    Anyone that does not understand that HHH meets this criteria
    has insufficient understanding.

    But it doesn't, as a correct simulation of a DDD that calls an HHH
    that returns will stop running,

    I really think that you must be a liar here because
    you have known this for years:

    On 8/2/2024 11:32 PM, Jeff Barnett wrote:
    ...In some formulations, there are specific states
    ;    defined as "halting states" and the machine only
    ;    halts if either the start state is a halt state...

    ...these and many other definitions all have
    ;    equivalent computing prowess...

    Anyone that knows C knows that DDD correctly simulated
    by any HHH cannot possibly reach its "return" {halt state}.


    But the problem is that you HHH ODESN'T correctly emulate the DDD it
    is given, because it aborts its emulation.

    Each HHH of every HHH that can possibly exist definitely
    emulates zero to infinity instructions correctly. In
    none of these cases does the emulated DDD ever reach
    its "return" instruction halt state.

    *There are no double-talk weasel words around this*
    *There are no double-talk weasel words around this*
    *There are no double-talk weasel words around this*


    Then why do you use double-talk wasel words that conflate the behavior
    of the program DDD that calls the HHH that emulates its input for a
    finite number of steps, and then returns 0 to its caller (that is DDD)
    making DDD halt, with the PARTIAL (and thus INCORRECT) emulation that
    HHH did of the DDD it was given that aborts before it gets to that point
    in the behavior of the program.

    Yes, there is one class of HHH that actually do perform a correct
    emulation of their input, and never abort it, and satisfy your claim, unfortunately those HHH and the DDDs that call them never return to give
    an answer, and are a TOTALLY DISTINCT group of machines/inputs from the
    others, and becuase DDD does include its HHH as part of its definition
    (or it couldn't be emulated past 4 instructions) they are clearly
    DIFFERENT inputs so not relevent to your problem.


    I KNOW, because you need that lie to make your false claim sound like it
    might be correct.

    Sorry, but it seems everything you accuse others of doing tends to be
    something you actualy are doin gyour self.

    Just like your buddy Trump does. You know, you use a lot of the same
    logic he does, ignore the truth and just keep repeating your lies.

    --- SoupGate-Win32 v1.05
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