• Re: embedded_H applied to =?UTF-8?B?4p+oxKTin6kg4p+oxKTin6k=?= computes

    From joes@21:1/5 to All on Thu Aug 1 12:17:09 2024
    Am Thu, 01 Aug 2024 06:49:13 -0500 schrieb olcott:
    On 8/1/2024 2:44 AM, Mikko wrote:
    On 2024-07-31 17:27:33 +0000, olcott said:
    On 7/31/2024 2:32 AM, Mikko wrote:
    On 2024-07-30 14:16:20 +0000, olcott said:
    On 7/30/2024 1:37 AM, Mikko wrote:
    On 2024-07-29 16:16:13 +0000, olcott said:
    On 7/28/2024 3:02 AM, Mikko wrote:
    On 2024-07-27 14:08:10 +0000, olcott said:
    On 7/27/2024 2:21 AM, Mikko wrote:
    On 2024-07-26 14:08:11 +0000, olcott said:

    When we compute the mapping from the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
    to the behavior specified by this input we know that embedded_H is >>>>>>> correct to transition to Ĥ.qn.

    The meaning of "correct" in this context is that if the transition >>>>>> of embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn is correct if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions
    to H.qn but incorrect otherwise.

    No you are wrong.
    Which dictionary (or other authority) disagrees?

    The common knowledge that a decider computes the mapping from its
    input finite string...
    This is almost always the same as the direct execution of the machine
    represented by this finite string.
    Not "almost". Otherwise it is doing something different.

    None of above indicates any disagreement by any authority.
    Everyone (even Linz) has the wrong headed idea that a halt decider must report on the behavior of the computation that itself is contained
    within. This has always been wrong.
    Dude. The halting problem /specifically/ asks about a machine simulating itself.
    A halt decider must always report on the behavior that its finite string specifies. This is different only when an input invokes its own decider.
    Um, no? Then it is making a mistake.

    The one rare exception is shown above where Ĥ ⟨Ĥ⟩ halts and the input >>> to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach its own final state of
    ⟨Ĥ.qn⟩ when embedded_H acts as if it was a UTM.
    H is not an UTM, though.

    That is not supported by any anuthority.

    The authority says *given an input of the function domain it*
    *can return the corresponding output*
    Which authority? Not that that would be a valid argument.

    In other words all deciders compute the mapping from their input (finite string) to an accept or reject state.
    This means that they do not compute the mapping of the executing process
    of themselves.
    They do, if those happen to coincide.

    I am the first person in the world that noticed these two could be
    different. Everyone that has disagreed with me is disagreeing with the semantics of the x86 language.
    What are the semantics that you disagree about?

    --
    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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