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:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> 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.
>>>>>>>
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> FAIL.
>>>>>>
>>>>>>> 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.
>>>>>>
>>>>>> FAIL
>>>>>
>>>>> 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.

FAIL.

>
> When we have a sequence of configurations that can not possibly reach
> their final state then we have a non-halting sequence because any an all
> sequences of configurations that cannot reach their final state are
> defined to be non-halting in the same way that a cat is defined to be an
> animal. If you disagree that a cat is an animal you are simply wrong.
>

Right, so the H that doesn't abort is non-halting and fails to be a decider.

If H does abort its simulation, and transitions to H.Qn, then the H^
that uses that H will also go to H^.Qn and Halt. So, I don't see how a
machine that gets to its final state represents a configuration that
never reaches it final state.

The only machine that your proof mentions that doesn't reach its final
state is H itsself, so that is the only one you have proved to be
non-halting.

> Refuting the halting problem proofs is only a sideline of mine, my
> actual goal is to mathematically formalize the notion of truth. This
> establishes the anchor for Davidson's truth conditional semantics.
>

Right, and you don't put the care into it to get it right. Your problem
is that as long as the Halting Problem Stands, your formalization is
shown to be flawed.

You can't just 'wish' for the Halting Problem to not be there.

FAIL.