Re: Concise refutation of halting problem proofs V52 [ Ignorant or Dishonest ]

[olcott]

On 1/22/2022 11:13 AM, Richard Damon wrote:
> On 1/22/22 11:47 AM, olcott wrote:
>> On 1/22/2022 10:39 AM, Richard Damon wrote:
>>> On 1/22/22 11:29 AM, olcott wrote:
>>>> On 1/22/2022 10:23 AM, Richard Damon wrote:
>>>>>
>>>>> On 1/22/22 10:48 AM, olcott wrote:
>>>>>> Halting problem undecidability and infinitely nested simulation (V3)
>>>>>
>>>>> Take FIFTY TWO, I think you are stuck in an apparent infinite loop.
>>>>>
>>>>> You keep on repeating the same basic mistakes.
>>>>>
>>>>>>
>>>>>> We define Linz H to base its halt status decision on the behavior
>>>>>> of its pure simulation of N steps of its input. N is either the
>>>>>> number of steps that it takes for its simulated input to reach its
>>>>>> final state or the number of steps required for H to match an
>>>>>> infinite behavior pattern proving that the simulated input would
>>>>>> never reach its own final state. In this case H aborts the
>>>>>> simulation of this input and transitions to H.qn.
>>>>>
>>>>> Such a pattern NOT existing for the <H^> <H^> pattern, thus your H
>>>>> can't correctly abort and becomes non-answering and thus FAILS to
>>>>> be a decider.
>>>>>
>>>>> The non-existance has been proven and you have ignored that proof,
>>>>> showing you have no counter for the proof.
>>>>>
>>>>> If you want to claim such a pattern exists, you MUST provide it or
>>>>> accept that your logic is just plain flawed as you are claiming the
>>>>> existance of something that is impossible.
>>>>>
>>>>> In effect, you are saying that if you have a halt decider for you
>>>>> halt decider to use, you can write a halt decider.
>>>>>
>>>>> FAIL.
>>>>>
>>>>>>
>>>>>> 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
>>>>>>
>>>>>> Because it is known that the UTM simulation of a machine is
>>>>>> computationally equivalent to the direct execution of this same
>>>>>> machine H can always form its halt status decision on the basis of
>>>>>> what the behavior of the UTM simulation of its inputs would be.
>>>>>>
>>>>>> When Ĥ applied to ⟨Ĥ⟩ has embedded_H simulate ⟨Ĥ⟩ ⟨Ĥ⟩ these steps
>>>>>> would keep repeating:
>>>>>> Ĥ copies its input ⟨Ĥ⟩ to ⟨Ĥ⟩ then embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩...
>>>>>
>>>>> But only if H never aborts, if H does abort, then the pattern is
>>>>> broken.
>>>>>
>>>>
>>>> YOU ARE EITHER TOO IGNORANT OR DISHONEST TO ACKNOWLEDGE THE TRUTH OF
>>>> THIS:
>>>>
>>>> The fact that there are no finite number of steps that the simulated
>>>> input to embedded_H would ever reach its final state conclusively
>>>> proves that embedded_H is correct to abort its simulation of this
>>>> input and transition to Ĥ.qn.
>>>>
>>>
>>> The problem is that any H that aborts after a finite number of steps,
>>> gives the wrong answer because it only looked to see if the input
>>> doesn't halt at some specific finite number.
>>
>> OK IGNORANT it is. When there exists no finite (or infinite) number of
>> steps such that the simulated input to embedded_H reaches its final
>> state then we know that this simulated input does not halt.
>
> And the only case when that happens is when H does not abort its
> simulation,
>
WRONG !!! That happens in every possible case. The simulated input to
embedded_H cannot possibly ever reach its final state NO MATTER WHAT !!!



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

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