[Olcott 2021 generic halt deciding principle]
Whenever the pure simulation of the input to simulating halt decider
H(x,y) never stops running unless H aborts its simulation H correctly
aborts this simulation and returns 0 for not halting.

Halting problem
In computability theory, the halting problem is the problem of
determining, from a description of an arbitrary computer program and an
input, whether the program will finish running, or continue to run
forever. https://en.wikipedia.org/wiki/Halting_problem

Now we construct a new Turing machine D with H as a subroutine.
This new TM calls H to determine what M does when the input to M
is its own description ⟨M⟩. Once D has determined this information,
it does the opposite. (Sipser:1997:165)

// Simplified Linz(1990) Ĥ
// and Strachey(1965) P
void P(ptr x)
{ if (H(x, x))
HERE: goto HERE;
}

The pathological feedback loop between the halt decider and its input
that otherwise makes inputs like the above impossible for H(P,P) to
decide has been eliminated.

Halting problem undecidability and infinitely nested simulation (V2)
https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2

--
Copyright 2021 Pete Olcott

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

On 12/4/21 4:26 PM, olcott wrote:
> [Olcott 2021 generic halt deciding principle]
> Whenever the pure simulation of the input to simulating halt decider
> H(x,y) never stops running unless H aborts its simulation H correctly
> aborts this simulation and returns 0 for not halting.
>

You have yet to PROVE that such a machine can exist, and still always
answer in finite time.

FAIL.

> Halting problem
> In computability theory, the halting problem is the problem of
> determining, from a description of an arbitrary computer program and an
> input, whether the program will finish running, or continue to run
> forever. https://en.wikipedia.org/wiki/Halting_problem
>
>      Now we construct a new Turing machine D with H as a subroutine.
>      This new TM calls H to determine what M does when the input to M
>      is its own description ⟨M⟩. Once D has determined this information,
>      it does the opposite.  (Sipser:1997:165)
>
> // Simplified Linz(1990) Ĥ
> // and Strachey(1965) P
> void P(ptr x)
> {
>   if (H(x, x))
>     HERE: goto HERE;
> }
>
> The pathological feedback loop between the halt decider and its input
> that otherwise makes inputs like the above impossible for H(P,P) to
> decide has been eliminated.

How? The answer to the Halting problem is based on the behavior of the
computaiton that is given to the decider via a representation.

H(P,P) asks if P(P) Halts or not.

If H(P,P) returns 0, then it is clear that P(P) will Halt.
If H(P,P) returns 1, then it is clear that P(P) will never Halt.
If H(P,P) fails to return an answer in finite time it fails to be a
decider, let alone correct Halt Decider.

Thus H can NOT be a correct Halt decider for H(P,P).

FAIL.

>
> Halting problem undecidability and infinitely nested simulation (V2)
> https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2
>
>
>

On 12/4/2021 3:26 PM, olcott wrote:
> [Olcott 2021 generic halt deciding principle]
> Whenever the pure simulation of the input to simulating halt decider
> H(x,y) never stops running unless H aborts its simulation H correctly
> aborts this simulation and returns 0 for not halting.
>

(¬∃N ∈ Z+ such that H(x,y) simulates N steps of x(y) and x stops
running) ⊢ H(x,y)==0

If there is no N in the set of positive integers such that H(x,y)
simulates N steps of x(y) and x stops running this proves H(x,y)==0.

> Halting problem
> In computability theory, the halting problem is the problem of
> determining, from a description of an arbitrary computer program and an
> input, whether the program will finish running, or continue to run
> forever. https://en.wikipedia.org/wiki/Halting_problem
>
>      Now we construct a new Turing machine D with H as a subroutine.
>      This new TM calls H to determine what M does when the input to M
>      is its own description ⟨M⟩. Once D has determined this information,
>      it does the opposite.  (Sipser:1997:165)
>
> // Simplified Linz(1990) Ĥ
> // and Strachey(1965) P
> void P(ptr x)
> {
>   if (H(x, x))
>     HERE: goto HERE;
> }
>
> The pathological feedback loop between the halt decider and its input
> that otherwise makes inputs like the above impossible for H(P,P) to
> decide has been eliminated.
>
> Halting problem undecidability and infinitely nested simulation (V2)
> https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2
>
>
>

--
Copyright 2021 Pete Olcott

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

On 12/4/21 7:57 PM, olcott wrote:
> On 12/4/2021 3:26 PM, olcott wrote:
>> [Olcott 2021 generic halt deciding principle]
>> Whenever the pure simulation of the input to simulating halt decider
>> H(x,y) never stops running unless H aborts its simulation H correctly
>> aborts this simulation and returns 0 for not halting.
>>
>
> (¬∃N ∈ Z+ such that H(x,y) simulates N steps of x(y) and x stops
> running) ⊢ H(x,y)==0
>
> If there is no N in the set of positive integers such that H(x,y)
> simulates N steps of x(y) and x stops running this proves H(x,y)==0.

The logic only works if x and y are not dependent on 'H' as H is no
longer a specific machine but a set of them.

You DO understand this?

Thus, if we chose a particular member of this set Hm and build Pm off of
that, we can show that for ALL finite m, that for sufficiently high n
(based on m) that all Hn(Pm,Pm) will see the final halting state of
Pm(Pm), thus Pm(Pm) is Halting, by definition.

Thus Hm is WRONG when it says Hm(Pm,Pm) == 0 is correct.

FAIL.

You argument is basically of the form that would say that given some
positive finite integer k, that N+k must be infinite since there exists
no N such that N > N+k.

>
>> Halting problem
>> In computability theory, the halting problem is the problem of
>> determining, from a description of an arbitrary computer program and
>> an input, whether the program will finish running, or continue to run
>> forever. https://en.wikipedia.org/wiki/Halting_problem
>>
>>       Now we construct a new Turing machine D with H as a subroutine.
>>       This new TM calls H to determine what M does when the input to M
>>       is its own description ⟨M⟩. Once D has determined this information,
>>       it does the opposite.  (Sipser:1997:165)
>>
>> // Simplified Linz(1990) Ĥ
>> // and Strachey(1965) P
>> void P(ptr x)
>> {
>>    if (H(x, x))
>>      HERE: goto HERE;
>> }
>>
>> The pathological feedback loop between the halt decider and its input
>> that otherwise makes inputs like the above impossible for H(P,P) to
>> decide has been eliminated.
>>
>> Halting problem undecidability and infinitely nested simulation (V2)
>> https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2
>>
>>
>>
>
>