XPost: comp.theory, sci.logic, sci.math

On 2/8/2022 4:55 AM, Don Stockbauer wrote:

On Monday, February 7, 2022 at 7:52:39 PM UTC-6, olcott wrote:

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

If someone is obsessed with halting how do you ever halt them?

I am next to 100% certain that I really did refute the halting problem
proofs and this has two key benefits for AI research:

(1) The previously understood absolute limit to the capabilities of
computation has been eliminated.

(2) Tarski Undefinability theorem is refuted by proxy thus enabling the
notion of truth to be formalized which provides Davidson's truth
conditional semantics to be anchored in a formal notion of truth.

--
Copyright 2021 Pete Olcott

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

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