On 1/22/2022 3:36 PM, Richard Damon wrote:
> On 1/22/22 4:25 PM, olcott wrote:
>> On 1/22/2022 3:20 PM, Richard Damon wrote:
>>>
>>> On 1/22/22 3:42 PM, olcott wrote:
>>>> On 1/22/2022 2:27 PM, Richard Damon wrote:
>>>>> On 1/22/22 1:53 PM, olcott wrote:
>>>>>> On 1/22/2022 12:42 PM, Richard Damon wrote:
>>>>>>> On 1/22/22 1:26 PM, olcott wrote:
>>>>>>>> On 1/22/2022 12:21 PM, Richard Damon wrote:
>>>>>>>>> On 1/22/22 12:53 PM, olcott wrote:
>>>>>>>>>> On 1/22/2022 11:33 AM, Richard Damon wrote:
>>>>>>>>>>> On 1/22/22 12:19 PM, olcott wrote:
>>>>>>>>>>>> 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 !!!
>>>>>>>>>>>
>>>>>>>>>>> But if H/embedded_H aborts its simulation, it doesn't matter
>>>>>>>>>>> if IT sees it or not, as it isn't then a UTM any longer.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> It remains true (invariant) that the simulated input to
>>>>>>>>>> embedded_H cannot possibly ever reach its final state no
>>>>>>>>>> matter what embedded_H does, thus conclusively proving that
>>>>>>>>>> this input never halts.
>>>>>>>>>>
>>>>>>>>>>> If H aborts its simulation and goes to H.Qn then H^ applied
>>>>>>>>>>> to <H^> and the UTM Simulation of <H^> <H^> will go to H^.Qn
>>>>>>>>>>> and Halt.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Because a halt decider is a decider embedded_H is only
>>>>>>>>>> accountable for computing the mapping from ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qy or
>>>>>>>>>> Ĥ.qn on the basis of the behavior specified by these inputs.
>>>>>>>>>> embedded_H is not accountable for the behavior of the
>>>>>>>>>> computation that it is contained within: Ĥ applied to ⟨Ĥ⟩
>>>>>>>>>> because this is not an actual input to embedded_H.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Right, and if H(<H^>,<H^>) -> H.Qn then the proper simulation
>>>>>>>>> by the computation UTM(<H^>,<H^>) will show that H^ also go to
>>>>>>>>> H^.Qn and Halt.
>>>>>>>>>
>>>>>>>>> THAT is the defined behavior of the actual input to H.
>>>>>>>>>
>>>>>>>>
>>>>>>>> You can define that a cat is a dog, yet the actual simulated
>>>>>>>> input to embedded_H cannot possibly reach its final state NO
>>>>>>>> MATTER WHAT.
>>>>>>>
>>>>>>> But it does. embeddd_H can't simuate to that point,
>>>>>>
>>>>>> If the simulated input cannot possibly every reach its final state
>>>>>> no matter what embedded_H does then this simulated input is
>>>>>> correctly determined to meet the Linz definition of never halting.
>>>>>>
>>>>>
>>>>> Except that the simulated input DOES reach its final state if
>>>>> H/embeded_H goes to H.Qn.
>>>>
>>>> Before embedded_H transitions to Ĥ.qn it has already aborted its
>>>> simulated input making it utterly impossible for any aspect of this
>>>> simulated input to do anything at all.
>>>
>>> Then you are just talking POOP and not halting.
>>>
>>
>> It is simply beyond your intellectual capacity:
>>
>> This is true for infinite loops, infinite recursion, infinitely nested
>> simulation and all other non halting inputs:
>>
>> When-so-ever any simulated input to any simulating halt decider would
>> never reach the final state of this simulated input in any finite
>> number of steps it is always correct for the simulating halt decider
>> to abort its simulation and transition to its reject state.
>>
>
> Can you PROVE that statement, or is this just one of your false 'self
> evident truth'. Does the proof include the posibility that the input
> includes a copy of the decider?
>
> The problem is that IF the simulating halt decider does abort its input
> based on some condition, then it is no longer a source of truth for the
> halting status of that input.
>
> It turns out that for H^, if H does abort its simulation, then it turns
> out that an actual simulation of the input proves that it will halt.
>

You really must have actual brain damage. That would not be your fault
I have told you the Linz definintion of halting many dozens of times:

computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)

Any you can't seem to rememeber that I ever said it once.

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

HALTING DOES NOT MEAN STOPS RUNNING
NOT HALTING MEANS CANNOT POSSIBLY REACH A FINAL STATE

--
Copyright 2021 Pete Olcott

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