On 2/1/22 4:36 PM, 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).

But unless embedded_H actually IS a real UTM, that doesn't matter.

Yes, you can show that for the case where H/embedded_H is actually a
UTM, i.e. a simulator that simulates until its input reaches a halting
state, that H^ applied to <H^> is non-halting, but you can also show
that if H is defined that way, it also fails to answer the question
given by H applied to <H^> <H^>, and thus it fails to be a Halt Decider.

Change your H, then you HAVE changed H^, so that proof is no longer
valid, and H is wrong.

FAIL.
>
>
>> 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
>>
>>
>>
>
>