On 2/1/2022 7:40 PM, Richard Damon wrote:
>
> On 2/1/22 8:03 PM, olcott wrote:
>> On 2/1/2022 6:25 PM, Richard Damon wrote:
>>> On 2/1/22 5:18 PM, 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.
>>>
>>>
>>> And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the
>>> behavior of H^ applied to <H^> which does Halt if H goes to H.Qn.
>>
>>
>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to
>> ⟨Ĥ⟩.qn ?
>
> Doesn't matter if embedded_H is not a ACTUAL UTM.
>

When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩

As soon as embedded_H correctly recognizes this as an infinite behavior
pattern:

Then these steps would keep repeating:
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩ ⟨Ĥ5⟩...

Then embedded_H can correctly abort the simulation of its input and
correctly transition to Ĥ.qn.

The above words can be verified as completely true entirely on the basis
of their meaning.

--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer