On 2/7/2022 7:26 PM, Richard Damon wrote:
> On 2/7/22 8:08 PM, olcott wrote:
>> On 2/7/2022 5:46 PM, Richard Damon wrote:
>>> On 2/7/22 9:59 AM, olcott wrote:
>>>> On 2/7/2022 5:47 AM, Richard Damon wrote:
>>>>> On 2/6/22 11:30 PM, olcott wrote:
>>>>>> On 2/6/2022 10:05 PM, Richard Damon wrote:
>>>>>>>
>>>>>>> On 2/6/22 10:04 PM, olcott wrote:
>>>>>>>> On 2/6/2022 3:39 PM, Richard Damon wrote:
>>>>>>>>>
>>>>>>>>> On 2/6/22 3:53 PM, olcott wrote:
>>>>>>>>>> On 2/6/2022 2:33 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/6/22 3:15 PM, olcott wrote:
>>>>>>>>>>>> On 2/6/2022 1:43 PM, dklei...@gmail.com wrote:
>>>>>>>>>>>>> On Sunday, February 6, 2022 at 8:31:41 AM UTC-8, olcott wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> H determines [halting] on the basis of matching infinite
>>>>>>>>>>>>>> behavior patterns.
>>>>>>>>>>>>>> When an infinite behavior pattern is matched H aborts its
>>>>>>>>>>>>>> simulation and
>>>>>>>>>>>>>> transitions to its final reject state. Otherwise H
>>>>>>>>>>>>>> transitions to its
>>>>>>>>>>>>>> accept state when its simulation ends.
>>>>>>>>>>>>>>
>>>>>>>>>>>>> This is incomplete because it does not cover the case where
>>>>>>>>>>>>> the
>>>>>>>>>>>>> machine neither halts nor matches an "infinite behavior
>>>>>>>>>>>>> pattern".
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> It covers the case that had previously been considered to be
>>>>>>>>>>>> proof that the halting problem is undecidable. That is all
>>>>>>>>>>>> that I need to refute these proofs.
>>>>>>>>>>>>
>>>>>>>>>>>>> You need to prove a theorem: There is a finite set of
>>>>>>>>>>>>> patterns such
>>>>>>>>>>>>> that every Turing machine either halts or matches one of these
>>>>>>>>>>>>> patterns.
>>>>>>>>>>>>>
>>>>>>>>>>>>> But I feel sure that theorem is not true.
>>>>>>>>>>>>
>>>>>>>>>>>> To solve the halting problem my program must be all knowing.
>>>>>>>>>>>> To refute the proofs I merely need to show that their
>>>>>>>>>>>> counter-example can be proved to never halt.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> And you just ignore the fact that if H applied to <H^> <H^>
>>>>>>>>>>> goes to H.Qn, then by construction H^ <H^> goes to H^.Qn, and
>>>>>>>>>>> halts, and since H, to be an accurate Halt Decider, must only
>>>>>>>>>>> go to H,Qn if the machine its input represents will never
>>>>>>>>>>> halt. They you also don't seem to understand that the
>>>>>>>>>>> computaton that <H^> <H^> represents IS H^ applied to <H^>.
>>>>>>>>>>> So, H was just wrong.
>>>>>>>>>>>
>>>>>>>>>>> So, you haven't actually proved the thing you claim youhave,
>>>>>>>>>>> but only that you have amassed an amazing pile of unsound
>>>>>>>>>>> logic based on wrong definitions that have hoodwinked
>>>>>>>>>>> yourself into thinking you have shown something useful.
>>>>>>>>>>>
>>>>>>>>>>> You are so good at doing this that you have gaslighted
>>>>>>>>>>> yourself so you can't actually understand what actual Truth is.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You simply do know know enough computer science to understand
>>>>>>>>>> that you are wrong and never will because you believe that you
>>>>>>>>>> are right.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> And you clearly don't know enough Computation Theory to talk
>>>>>>>>> about it.
>>>>>>>>>
>>>>>>>>> Since the is a Theorm in Computation Theory, using Computation
>>>>>>>>> Theory Deffinitions, that is your problem.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Because all simulating halt deciders are deciders they are
>>>>>>>>>> only accountable for computing the mapping from their input
>>>>>>>>>> finite strings to an accept or reject state on the basis of
>>>>>>>>>> whether or not their correctly simulated input could ever
>>>>>>>>>> reach its final state: ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* ⟨Ĥ⟩.qn.
>>>>>>>>>
>>>>>>>>> And if you are working on the Halting Problem of Computation
>>>>>>>>> Theory, BY DEFINITION, the meaning of 'correcty simulted' is
>>>>>>>>> simulation by a REAL UTM which BY DEFINITION exactly matches
>>>>>>>>> the behavior of Computation that it is representation of, which
>>>>>>>>> for <H^> <H^> is H^ applied to <H^>
>>>>>>>>>
>>>>>>>>
>>>>>>>> If an infinite number is steps is not enough steps for the
>>>>>>>> correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H to transition to
>>>>>>>> ⟨Ĥ⟩.qn then the input to embedded_H meets the Linz definition of
>>>>>>>> a sequence of configurations that never halts.
>>>>>>>
>>>>>>> WRONG.
>>>>>>>
>>>>>>> If embedded_H DOES an infinite number of steps and doesn't reach
>>>>>>> a final state, then it shows its input never halts.
>>>>>> When embedded_H matches this infinite pattern in the same three
>>>>>> iterations:
>>>>>>
>>>>>> 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⟩...
>>>>>>
>>>>>> that you agreed show the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H will
>>>>>> never reach ⟨Ĥ⟩.qn in any number of steps, which proves that this
>>>>>> input cannot possibly meet the Linz definition of halting:
>>>>>>
>>>>>> computation that halts … the Turing machine will halt whenever it
>>>>>> enters a final state. (Linz:1990:234)
>>>>>>
>>>>>
>>>>> OK, so the only computatiopn that you show that does not halt is H,
>>>>> so H can not be a decider.
>>>>
>>>> In the above example embedded_H simulates three iterations of nested
>>>> simulation to match the infinitely nested simulation pattern.
>>>> In reality it needs less than this to match this pattern.
>>>>
>>>>
>>>
>>> And if it doesn't do an infinite number, the H^ that is using it will
>>> Halt,
>>
>> embedded_H only examines the actual behavior of its inputs as if its
>> was a guard assigned to watch the front. If someone comes in the back
>> door (non-inputs) embedded_H is not even allowed to pay attention.
>>
>
> If the 'actual behavior' of the input <H^> <H^> is not the behavior of
> H^ applied to <H^> you are lying about doing the Halting Problem.
>

If it is true that the simulated input to embedded_H cannot possibly
ever reach its final state of ⟨Ĥ⟩.qn, then nothing in the universe can
possibly contradict the fact that the input specifies a non-halting
sequences of configurations. If God himself said otherwise then God
himself would be a liar.

If we know that we have a black cat then we know that we have a cat.

If we know that we have a sequence of configurations that cannot
possibly ever reach its final state then we know that we have a
non-halting sequence of configurations.

--
Copyright 2021 Pete Olcott

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