• Re: Overview of proof that DDD specifies non-halting behavior --- point

    From joes@21:1/5 to All on Tue Aug 13 22:01:06 2024
    Am Tue, 13 Aug 2024 11:45:42 -0500 schrieb olcott:
    On 8/13/2024 11:22 AM, joes wrote:
    Am Tue, 13 Aug 2024 08:58:09 -0500 schrieb olcott:
    On 8/13/2024 8:34 AM, Alan Mackenzie wrote:
    olcott <[email protected]> wrote:
    On 8/13/2024 2:29 AM, Fred. Zwarts wrote:
    Op 12.aug.2024 om 14:42 schreef olcott:
    On 8/11/2024 2:54 PM, Fred. Zwarts wrote:
    Op 11.aug.2024 om 13:45 schreef olcott:

    (b) Strawman-deception of changing what I said and rebutting that
    That's a lie, too. I've not seen anybody else apart from you doing
    this. Indeed you're doing this as a response to Fred's last post.
    Beautiful:
    Through something like mathematical induction we can directly see that
    DDD correctly emulated by any HHH cannot possibly reach its "return"
    instruction final halt state.
    HHH is only required to predict whether or not an unlimited emulation
    of DDD would ever halt.
    Not really. It should predict whether DDD *by itself* halts, not what
    any simulator does to its encoding. It can trivially predict what
    *itself* returns. By the way DDD's halting depends on HHH.

    It is objectively incorrect to disagree with the semantics of the x86 language when one is assessing whether or not an emulation of N
    instructions of an input is correct or incorrect.
    Again: what am I disagreeing with?
    An aborted simulation of a non-halting machine is incorrect.

    When the measure of the behavior of the input is based on examining N
    steps of DDD correctly emulated by HHH to correctly predict the behavior
    of an unlimited emulation
    and DDD correctly emulated by HHH cannot possibly reach its own "return" instruction final halt state then the input to HHH(DDD) specifies not
    halting behavior.
    HHH cannot simulate DDD (correctly).

    Thus when computing the behavior that this finite string specifies DDD
    never halts.
    DDD halts.
    And its description specifies its behaviour.

    DDD always has the exact same finite string of machine code bytes.
    This requires each HHH to always be at machine address 000015d2.
    It also requires HHH to make up its mind whether it will abort or not. Please respond.

    --
    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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