On 2/1/2022 4:25 PM, wij wrote:
> 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)

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 ? (An answer of "no" means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt).

Key elements rewritten today:
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