On Wednesday, 2 February 2022 at 06:19:04 UTC+8, olcott wrote:
> On 2/1/2022 4:12 PM, wij wrote:
> > On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:
> >> On 2/1/2022 3:23 PM, wij wrote:
> >>> On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:
> >>>> On 2/1/2022 10:33 AM, wij wrote:
> >>>>> On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:
> >>>>>> On 1/31/2022 11:25 PM, Richard Damon wrote:
> >>>>>>>
> >>>>>>> On 1/31/22 11:42 PM, olcott wrote:
> >>>>>>>> On 1/31/2022 10:33 PM, Richard Damon wrote:
> >>>>>>>>>
> >>>>>>>>> On 1/31/22 11:24 PM, olcott wrote:
> >>>>>>>>>> On 1/31/2022 10:17 PM, Richard Damon wrote:
> >>>>>>>>>>> On 1/31/22 10:40 PM, olcott wrote:
> >>>>>>>>>>>> On 1/31/2022 6:41 PM, Richard Damon wrote:
> >>>>>>>>>>>>> On 1/31/22 3:24 PM, olcott wrote:
> >>>>>>>>>>>>>> On 1/31/2022 2:10 PM, Ben wrote:
> >>>>>>>>>>>>>>> On 1/31/2022 8:06 AM, olcott wrote:
> >>>>>>>>>>>>>>>> On 1/30/2022 8:20 PM, Richard Damon wrote:
> >>>>>>>>>>>>>>>>> On 1/30/22 9:05 PM, olcott wrote:
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> >>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> >>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>> These statements need the conditions, that H^ goes to
> >>>>>>>>>>>>>>>>>>> H^.Qy/H^.Qn iff H goes to that corresponding state.
> >>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H
> >>>>>>>>>>>>>>>>>> in the same way that (5,3) is syntactically specified as an
> >>>>>>>>>>>>>>>>>> input to Sum(5,3)
> >>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>> Right, and the
> >>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to
> >>>>>>>>>>>>>>>>>> embedded_H in the same way that (1,2) is NOT syntactically
> >>>>>>>>>>>>>>>>>> specified as an input to Sum(5,3)
> >>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>> Right, but perhaps you don't understand that from you above
> >>>>>>>>>>>>>>>>> statement the right answer is based on if UTM(<H^>,<H^>)
> >>>>>>>>>>>>>>>>> Halts which by the definition of a UTM means if H^ applied to
> >>>>>>>>>>>>>>>>> <H^> Halts.
> >>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> The biggest reason for your huge mistakes is that you cannot
> >>>>>>>>>>>>>>>> stay sharply focused on a single point. It is as if you either
> >>>>>>>>>>>>>>>> have attention deficit disorder ADD or are addicted to
> >>>>>>>>>>>>>>>> methamphetamine.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> >>>>>>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and
> >>>>>>>>>>>>>>>> Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>> After we have mutual agreement on this point we will move on
> >>>>>>>>>>>>>>>> to the points that logically follow from this one.
> >>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>> Holy shit try to post something that makes sense.
> >>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> >>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>> Richard does not accept that the input to the copy of Linz H
> >>>>>>>>>>>>>> embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>>
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> No, but apparently you can't understand actual English words.
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must
> >>>>>>>>>>>>> give is based on the behavior of H^ applied to <H^> BECAUSE OF
> >>>>>>>>>>>>> THE DEFINITION of H.
> >>>>>>>>>>>>
> >>>>>>>>>>>> In other words Sum(3,5) must return the value of Sum(7,8)?
> >>>>>>>>>>>
> >>>>>>>>>>> Don't know how you get that from what I said.
> >>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> Any moron knows that a function is only accountable for its actual
> >>>>>>>>>>>> inputs.
> >>>>>>>>>>>
> >>>>>>>>>>>
> >>>>>>>>>>> And the actual input to H is <H^> <H^> which MEANS by the
> >>>>>>>>>>> DEFINITION of the Halting Problem that H is being asked to decide
> >>>>>>>>>>> on the Halting Status of H^ applied to <H^>
> >>>>>>>>>> No that is not it. That is like saying "by definition" Sum(3,5) is
> >>>>>>>>>> being asked about Sum(7,8).
> >>>>>>>>>
> >>>>>>>>> Again your RED HERRING.
> >>>>>>>>>
> >>>>>>>>> H is being asked EXACTLY what it being asked
> >>>>>>>>>
> >>>>>>>>> H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't
> >>>>>>>>>
> >>>>>>>>> AGREED?
> >>>>>>>>>
> >>>>>>>>
> >>>>>>>> No that is wrong. embedded_H is being asked:
> >>>>>>>> Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?
> >>>>>>>>
> >>>>>>>
> >>>>>>> If you say 'No', then you aren't doing the halting problem, as the
> >>>>>>> requirement I stated is EXACTLY the requirement of the Halting Problem.
> >>>>>> The halting problem is vague on the definition of halting, it includes
> >>>>>> that a machine has stopped running and that a machine cannot reach its
> >>>>>> final state. My definition only includes the latter.
> >>>>>
> >>>>> Sounds like a NDTM.
> >>>> https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine
> >>>>
> >>>> It is not a NDTM, a Turing Machine only actually halts when it reaches
> >>>> its own final state. People not very familiar with this material may get
> >>>> confused and believe that a TM halts when its stops running because its
> >>>> simulation has been aborted. This key distinction is not typically
> >>>> specified in most halting problem proofs.
> >>>> computation that halts … the Turing machine will halt whenever it enters
> >>>> a final state. (Linz:1990:234)
> >>>
> >>> Where did Linz mention 'simulation' and 'abort'?
> >> I have shown how my system directly applies to the actual halting
> >> problem and it can be understood as correct by anyone that understands
> >> the halting problem at a much deeper level than rote memorization.
> >>
> >> The following simplifies the syntax for the definition of the Linz
> >> Turing machine Ĥ, it is now a single machine with a single start state.
> >> A copy of Linz H is embedded at Ĥ.qx.
> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
> >> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> >> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to
> >> ⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).
> >>> You are defining POOP [Richard Damon]
> >>> André had recommended many online sites for you to learn or test, I forget which posts it is.
> >>> But I think C program is more simpler.
> >>>
> >>>> Halting problem undecidability and infinitely nested simulation (V3)
> >>>>
> >>>> https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
> >>>> --
> >>>> Copyright 2021 Pete Olcott
> >>>>
> >>>> Talent hits a target no one else can hit;
> >>>> Genius hits a target no one else can see.
> >>>> Arthur Schopenhauer
> >>>
> >>>
> >>>
> >>
> >>
> >> --
> >> Copyright 2021 Pete Olcott
> >>
> >> Talent hits a target no one else can hit;
> >> Genius hits a target no one else can see.
> >> Arthur Schopenhauer
> >
> > André had recommended many online sites for you to learn or test, I forget which posts it is.
> > Type it into a TM simulator and prove your claim, your words are meaningless.
> I have already proved that I know one key fact about halt deciders that
> no one else here seems to know.
>
> No one here understands that because a halt decider is a decider that it
> must compute the mapping from its inputs to an accept of reject state on
> the basis of the actual behavior specified by these inputs.
> --
> Copyright 2021 Pete Olcott
>
> Talent hits a target no one else can hit;
> Genius hits a target no one else can see.
> Arthur Schopenhauer

There is no 'actual TM' until you it into a TM simulator, otherwise all empty talks.
(I would expect to see you 'reinterpret' again)