The key basis of my refutation of the halting theorem

[olcott]

This is the key basis of my refutation of the halting theorem:
The halting theorem counter-examples present infinitely nested
simulation (non-halting) behavior to every simulating halt decider.

I presented this to Ben more than four years ago and he successfully
changed the subject with various dishonest dodges so that it could not
be properly evaluated until now.

Infinitely Recursive input on HP Proofs
peteolcott Mar 11, 2017, 3:13:03 PM
https://groups.google.com/g/comp.theory/c/NcFS02hKs1U

I had to stop tolerating dishonest dodges that changed the subject
rather than directly addressed the point at hand. A dishonest dodge most
often is an example of the strawman error.

A straw man (sometimes written as strawman) is a form of argument and an
informal fallacy of having the impression of refuting an argument,
whereas the real subject of the argument was not addressed or refuted,
but instead replaced with a false one.
https://en.wikipedia.org/wiki/Straw_man

All of the "rebuttals" to the {key basis of my refutation} have taken
the form of the strawman error, here is the most common one:
The halting theorem does not specify a simulating halt decider.
This is no actual rebuttal to the original claim at all.

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation


--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

[olcott]

On 9/27/2021 3:52 PM, Ben Bacarisse wrote:

> olcott <No...@NoWhere.com> writes:
>
>> This is the key basis of my refutation of the halting theorem:
>> The halting theorem counter-examples present infinitely nested
>> simulation (non-halting) behavior to every simulating halt decider.
>

> Not a refutation of anything.

>
>> I presented this to Ben more than four years ago and he successfully
>> changed the subject with various dishonest dodges so that it could not
>> be properly evaluated until now.
>

> Liar.

>
>> Infinitely Recursive input on HP Proofs
>> peteolcott Mar 11, 2017, 3:13:03 PM
>> https://groups.google.com/g/comp.theory/c/NcFS02hKs1U
>

> This was a post about a different topic. You are confused even about
> what you were saying back then. I've explained in anther reply. You
> are now saying the same thing over and over, so I won't copy my reply
> out here.

>
>> All of the "rebuttals" to the {key basis of my refutation} have taken
>> the form of the strawman error, here is the most common

>> The halting theorem does not specify a simulating halt decider.
>

> The most common one is that false is the wrong answer for a halting
> computation. That's the error I see most commonly pointed out.
>
> You were clear, even then, that your "solution" or "rebuttal" or
> whatever was to redefine halting:
>
> "This definition of halting circumvents the pathological
> self-reference error for every simulating halt decider:
>
> An input is decided to be halting only if its simulation never needs
> to be stopped by any simulating halt decider anywhere in its entire
> invocation chain.
>
> On that basis:
> Ĥ(<Ĥ>) ⊢* Ĥ.qn
> H(<Ĥ>,<Ĥ>) ⊢* H.qn"
>
> There you are (May 17 2017) clearly stating that you've defined H
> rejecting a halting computation to be correct! I must say I'd forgotten
> how long you have been flogging this dead horse.
>
> (There is an error of logic where you think that being specific -- some
> special kind of decider -- gets round a proof that is about all TMs, but
> you present that error only every now and then.)
>

As I carefully think this through again and again I continue to get
deeper insights.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
// because the simulation of ⟨Ĥ⟩ applied to ⟨Ĥ⟩ by Ĥ.qx DOES NOT HALT

The computation of Ĥ applied to ⟨Ĥ⟩ is a distinctly different
computation than the computation of the simulation of ⟨Ĥ⟩ applied to ⟨Ĥ⟩
by Ĥ.qx.

If we want to specify a computation that is equivalent to Ĥ applied to
⟨Ĥ⟩ we must specify H applied to ⟨Ĥ⟩ ⟨Ĥ⟩. H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy

The difference is that the input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ invokes a copy of Ĥ.qx
whereas the input to H does not invoke a copy of H.

The pathological self-reference error that I first discovered in 2004 is
the key to finally resolved these otherwise undecidable decision problem
inputs.

comp.theory
Halting Problem Final Conclusion
Peter Olcott Sep 5, 2004,

The Liar Paradox can be shown to be nothing more than
a incorrectly formed statement because of its pathological
self-reference. The Halting Problem can only exist because
of this same sort of pathological self-reference.
https://groups.google.com/g/comp.theory/c/RO9Z9eCabeE/m/Ka8-xS2rdEEJ

[olcott]

On 9/28/2021 8:15 PM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>

>> On 9/28/2021 10:13 AM, Ben Bacarisse wrote:
>>> olcott <No...@NoWhere.com> writes:
>>>
>>>> On 9/27/2021 8:52 PM, Ben Bacarisse wrote:
>>>>> olcott <No...@NoWhere.com> writes:
>>>>>
>>>>>> On 9/27/2021 3:52 PM, Ben Bacarisse wrote:
>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>
>>>>>>>> All of the "rebuttals" to the {key basis of my refutation} have taken
>>>>>>>> the form of the strawman error, here is the most common
>>>>>>>> The halting theorem does not specify a simulating halt decider.
>>>>>>>
>>>>>>> The most common one is that false is the wrong answer for a halting
>>>>>>> computation. That's the error I see most commonly pointed out.
>>>>>>>
>>>>>>> You were clear, even then, that your "solution" or "rebuttal" or
>>>>>>> whatever was to redefine halting:
>>>>>>>
>>>>>>> "This definition of halting circumvents the pathological
>>>>>>> self-reference error for every simulating halt decider:
>>>>>>>
>>>>>>> An input is decided to be halting only if its simulation never needs
>>>>>>> to be stopped by any simulating halt decider anywhere in its entire
>>>>>>> invocation chain.
>>>>>>>
>>>>>>> On that basis:
>>>>>>> Ĥ(<Ĥ>) ⊢* Ĥ.qn
>>>>>>> H(<Ĥ>,<Ĥ>) ⊢* H.qn"
>>>>>>>
>>>>>>> There you are (May 17 2017) clearly stating that you've defined H
>>>>>>> rejecting a halting computation to be correct! I must say I'd forgotten
>>>>>>> how long you have been flogging this dead horse.

>>> No answer to this of course.

>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>> if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts, and
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>> if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt
>>>>>

>>>>> Deceptive removal of Linz's annotations. I've told you before that you
>>>>> should not do that.
>>>>
>>>> I merely adapted his annotations to be more clear.
>>> "if Ĥ applied to ⟨Ĥ⟩ halts" and "if Ĥ applied to ⟨Ĥ⟩ does not halt" are
>>> perfectly clear. There are what the halting theorem is about --
>>> halting. Replacing then with your junk is dishonest. Stop it.
>>
>> The halt decider is at Ĥ.qx. The halt decider is not at Ĥ.q0.
>
> And Linz's annotations are the correct ones. If you don't understand
> why, ask. If you do (and you claim to simply ave clified them) leave
> them in place.
>
>> The halt decider does not decide that halt status of itself.
>
> And Linz's annotation are the correct ones.
>
>> The halt decider decides the halt status of its input.
>
> And Linz's annotation are still the correct ones.
>
>> The input to the halt decider at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩.
>
> And it is still a lie to remove Linz's correct annotations from the
> lines you write above.
>
>> Since you already know these things are true denying them is
>> dishonest.
>
> I have not denied these things. You have removed the correct
> descriptions about which lines apply in which situations and I telling
> you to stop being so dishonest.
>
> It's clear from the fact that you think what you've written here is
> relevant, that you don't know /why/ Linz says "if Ĥ applied to ⟨Ĥ⟩
> halts" about the first and "if Ĥ applied to ⟨Ĥ⟩ does not halt" about the
> second, but your inability to understand is no excuse for replacing them
> with your junk versions.
>

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 simulating halt decider is embedded at Ĥ.qx. It has been annotated so
that it only shows Ĥ applied to ⟨Ĥ⟩, converting the variables to constants.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

if the simulated input to Ĥ.qx ⟨Ĥ⟩ applied to ⟨Ĥ⟩ reaches its final
state. (Halts)

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

if the simulated input to Ĥ.qx ⟨Ĥ⟩ applied to ⟨Ĥ⟩ never reaches its
final state. (Does not halt)

[olcott]

On 9/30/2021 4:38 AM, zelos...@gmail.com wrote:

> it is not a refutation because if you have a machine that can solve the halting problem then we can construct a machine that the previous machine cannot determine if it holds or not. Ergo it has no solution
>

With my solution that does not work:

The halting theorem counter-examples present infinitely nested
simulation (non-halting) behavior to every simulating halt decider.

The pathological self-reference of the conventional halting problem
proof counter-examples is overcome.

The halt status of these examples is correctly determined. A simulating
halt decider remains in pure simulation mode until after it determines
that its input will never reach its final state.

This eliminates the conventional feedback loop where the behavior of the
halt decider effects the behavior of its input.