On 2/1/2022 8:21 PM, Richard Damon wrote:
> On 2/1/22 9:14 PM, olcott wrote:
>> On 2/1/2022 7:55 PM, Richard Damon wrote:
>>> On 2/1/22 8:47 PM, olcott wrote:
>>>> 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.
>>>>
>>>
>>>
>>> Nope, proven otherwise.
>>>
>>
>> What I said above is true by logical necessity and you simply aren't
>> bright enough to understand this.
>>
>
> Then you can provide a step by step proof of it?
>
>> If X then Y and if Y then Z and X then Z. There is no way around this.
>
> And what are your X, Y and Z?
>
>
>>
>> If embedded_H correctly recognizes that its input specifies non
>> halting behavior then it is necessarily correct for embedded_H to
>> report this
>> non halting behavior.
>
> *IF* it correct recognizes. Since there is no pattern in H's simulation
> of <H^> <H^> THAT IS a proof of non-halting
You must be a liar.

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⟩...

--
Copyright 2021 Pete Olcott

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