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On 2/9/22 11:31 AM, olcott wrote:
On 2/9/2022 7:30 AM, Richard Damon wrote:
Ĥ applied to ⟨Ĥ⟩ is an entirely different sequence of configurations than embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ therefore embedded_H can transition to Ĥ.qn causing Ĥ to transition to Ĥ.qn without contradiction.
On 2/9/22 8:13 AM, olcott wrote:
On 2/9/2022 6:13 AM, Richard Damon wrote:IF H <H^> <H^> -> H.Qy which it is supposed to do if H^ <H^> Will Halt.
On 2/8/22 9:19 PM, olcott wrote:
On 2/8/2022 7:39 PM, Richard Damon wrote:
On 2/8/22 7:31 PM, olcott wrote:
On 2/8/2022 6:04 PM, Richard Damon wrote:
On 2/8/22 10:35 AM, olcott wrote:
On 2/8/2022 5:56 AM, Richard Damon wrote:
On 2/8/22 12:28 AM, olcott wrote:
On 2/7/2022 8:03 PM, Richard Damon wrote:
On 2/7/22 8:52 PM, 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:
When embedded_H matches this infinite pattern in the same three iterations:
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:
This is incomplete because it does not cover the case where the
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.
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
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.
If embedded_H DOES an infinite number of steps and doesn't reach a final state, then it shows its input never halts.
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.
Except that if H/embedded_H aborts its simulation and goes to H.Qn, then the CORRECT simulation of its input (that done by a REAL UTM) will show that it will go to H^.Qn.
All you have proven is that if H doesn't abort, and thus doesn't go to H.Qn, and thus fails to be a correct decider, then H^ applied to <H^> is non-halting.
You keep on thinking that a simulation that aborts its simulation is a 'correct' simulation. By the definition in Computation Theory, this is not true. If you think it is, it just proves that you don't understand the field.
If we know that we have a black cat then we know that we have a cat.
Except that if you DON'T have a black cat but think you do then you are wrong. If H aborts its simulation, it isn't a UTM and doesn't 'correctly' simulate.
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.
Except that is has been PROVEN that if H -> H.Qn then the pattern WILL reach the final state.
The fact that H can't ever reach that state proves just proves that if H is a UTM, which don't abort, then H^ will be non-halting, but H is still wrong for not answering. If H does abort, then it hasn't proven anything, and it has been proven that it is wrong.
You are either not bright enough to get this or dishonest.
I don't care which, I need to up my game to computer scientists.
So, can't refute what I say so you go to arguing by insults, classic Olcott logical fallicy.
Fundamentally you seem to lack the intellectual capacity to understand what I am saying. This is proven on the basis that what I am saying can be verified as true entirely on the basis of the meaning of its words.
Except that it has been shown that you keep on using the WRONG definitions of the words.
A UTM can NEVER abort its simulation as BY DEFINITION, a UTM EXACTLY repoduces the behavior of its input (so if it is non-halting, so will the UTM). Also you think that there can be a 'Correct Simulation' by something that is NOT actully a UTM.
Care to show anywhere where your misdefinitions are support in the field fo Computation Theory.
That just PROVES that you aren't actually working on the Halting Problem of Computation Theory.
Face it, you are just WRONG about your assertions, maybe because you just don't know the field, so don't have any idea what is legal or not.
Also note, you keep talking about needing 'Computer Scientists' to understand, that is really incorrect, you need to be able to explain it to someone who understands Computation Theory, which is a fairly specialized branch of Mathematics. Yes, it is part of the foundation of Computer Science, but isn't the sort of thing that a normal Computer Scientist will deal with day to day.
I need someone to analyze what I am saying on the deep meaning of what I am saying instead of mere rote memorized meanings from textbooks.
No, you need to learn that words have PRECISE meanings, and you aren't allowed to change them, no mwtter how much it 'makes sense' to do so.
The key mistake that my reviewers are making is that they believe that the halt decider is supposed to evaluate its input on the basis of some proxy for the actual behavior of this actual input rather than the actual behavior specified by this actual input.
Just proves you aren't working on the Halting Problem, as the DEFINITION of the Halting problems says that it is, because you don't actually understand the meaning of 'actual behavior'.
From Linz, H applied to wM w needs to go to H.Qy IFF M applied to w halts, and to H,Qn if M applied to w will never halt.
If you are supposed to report when Bill arrives at your house and Sam arrives at you house and you really really believe that Sam's arrival is a valid proxy for Bill's arrival then when I ask you did Bill arrive at your house? you say "yes" even though correct the answer is "no".
You really like to make you Herrings Red, don't you.
REMEMBER, the DEFINTION of a Halt Decider is that H applied to wM w is based on the behavior of M applied to w.
YOU are the one making the wrong report.
When anyone in the universe defines something besides the actual behavior specified by the input to embedded_H as the only correct halt status criterion measure that might as well say that cats are not animals.
Just shows your problem in comprehension, doesn't it. You just refuse to accept the definition because it doesn't match your idea of what you need.
Note, 'The Actual Behavior specifeid by the input' IS precisly defined, and it IS the behavior that the input specifes, The input to the decider is the description of a computation, and the actual behavior sepecified by the input is by defintion the behavior of that computation that the input describes.
YOU are the one that wants to change it to not be the behavior specified by the input, but the behavior of the program that is processing the input. YOUR definition of the behavior has the problem that the behavior is no longer just specified by 'the input' but is also a function of what program you give that input to.
Your logic is just not sound, and sometimes I wonder how sound your mind is.
This statement of your just shows how you have lost touch with the reality of the situation. You seem to think the Univese must be wrong because it doesn't match your expectations. THAT is a sign of mental illness.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qnIF H <H^> <H^> -> H.Qn which it is supposed to do if H^ <H^> will never Halt.
you keep forgetting the conditions, which are important.
Bing a pathological liar seems to have made you lose your sense of what is true.
While H^ applied to <H^> IS a different computation then H applied to <H^> <H^> the former uses the latter to determine its behavior.
The issue isn't a 'contradiction' between the behavior of the two machines but the contradiction between the behavior of these two machines and the concept that H is correct.
Like the guard that is only accountable for guarding the front door simulating halt decider embedded_H is only accountable for reporting whether or not its simulated input can possibly reach its own final state ⟨Ĥ⟩.qn.
Again, you pathological lying has blinded you to the actual fact.
H/embedded_H IS responsible for its answer match the the ACTUAL 'Behavior of its input', which is DEFINED as the behavior of the ACTUAL MACHINE the input represents.
| Concise refutation of halting problem proofs V62 [ Linz Proof ]|
By: olcott on Sun, 6 Feb 2022