On 2/3/22 9:15 AM, olcott wrote:
> On 2/3/2022 5:50 AM, Richard Damon wrote:
>> On 2/3/22 12:24 AM, olcott wrote:
>>> On 2/2/2022 11:09 PM, Richard Damon wrote:
>>>> On 2/2/22 11:50 PM, olcott wrote:
>>>>> On 2/2/2022 10:42 PM, Richard Damon wrote:
>>>>>> On 2/2/22 10:50 PM, olcott wrote:
>>>>>>> On 2/2/2022 9:30 PM, Richard Damon wrote:
>>>>>>>>
>>>>>>>> On 2/2/22 10:12 PM, olcott wrote:
>>>>>>>>> On 2/2/2022 8:57 PM, Richard Damon wrote:
>>>>>>>>>> On 2/2/22 9:31 PM, olcott wrote:
>>>>>>>>>>> On 2/2/2022 8:08 PM, Richard Damon wrote:
>>>>>>>>>>>> On 2/2/22 8:41 PM, olcott wrote:
>>>>>>>>>>>>> On 2/2/2022 6:22 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 2/2/22 10:20 AM, olcott wrote:
>>>>>>>>>>>>>>> On 2/1/2022 9:58 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>> On 2022-02-01 20:44, olcott wrote:
>>>>>>>>>>>>>>>>> On 2/1/2022 9:24 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>> On 2022-02-01 19:57, olcott wrote:
>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:48 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>> On 2022-02-01 19:37, olcott wrote:
>>>>>>>>>>>>>>>>>>>>> On 2/1/2022 8:08 PM, André G. Isaak wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> ⟨Ĥ⟩ applied to ⟨Ĥ⟩ is completely meaningless.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Sure and so is the "I am going to go to the" part of
>>>>>>>>>>>>>>>>>>>>> "I am going to go to the store to buy some ice cream."
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> When you don't cut off what I said in the middle of
>>>>>>>>>>>>>>>>>>>>> the sentence then it makes much more sense.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H
>>>>>>>>>>>>>>>>>>>>> possibly transition to ⟨Ĥ⟩.qn ?
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> That's just as meaningless. You can simulate Ĥ
>>>>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the
>>>>>>>>>>>>>>>>>>>> input to a simulator. You cannot simulate ⟨Ĥ⟩
>>>>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ anymore than you can apply ⟨Ĥ⟩ to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> So you are simply being nit picky about my use of
>>>>>>>>>>>>>>>>>>> terminology.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Yes, I insist on terminology being used correctly. And
>>>>>>>>>>>>>>>>>> any place where you attempt to publish your results
>>>>>>>>>>>>>>>>>> will be equally, if not more, nit picky.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> It is fine and good that you help correct my terminology.
>>>>>>>>>>>>>>>>> What is not fine and good is for you to reject the
>>>>>>>>>>>>>>>>> essence of the gist of what I am saying entirely on the
>>>>>>>>>>>>>>>>> basis that I did not say it exactly according to
>>>>>>>>>>>>>>>>> conventions. The is what Ben always did. He never paid
>>>>>>>>>>>>>>>>> any attention to the actual substance of what I was
>>>>>>>>>>>>>>>>> saying.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> When ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to simulating halt decider
>>>>>>>>>>>>>>>>>>> embedded_H and embedded_H correctly determines that
>>>>>>>>>>>>>>>>>>> its simulated input cannot possibly reach any final
>>>>>>>>>>>>>>>>>>> state then embedded_H is necessarily correct to
>>>>>>>>>>>>>>>>>>> transition to Ĥ.qn indicating that its simulated
>>>>>>>>>>>>>>>>>>> input never halts.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> But now you've just hidden your meaningless
>>>>>>>>>>>>>>>>>> terminological abuse. "Its simulated input" is only
>>>>>>>>>>>>>>>>>> meaningful when it is construed as meaning the
>>>>>>>>>>>>>>>>>> simulation of the computation REPRESENTED by the
>>>>>>>>>>>>>>>>>> input, i.e. the
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Not at all. A simulator simulates a finite string and
>>>>>>>>>>>>>>>>> the actual behavior of this simulated finite string is
>>>>>>>>>>>>>>>>> the ultimate basis of whether or not it specifies a
>>>>>>>>>>>>>>>>> finite sequence of configurations.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> No. A simulator simulates a Turing Machine applied to an
>>>>>>>>>>>>>>>> input. It takes as its input a finite string which
>>>>>>>>>>>>>>>> represents that Turing Machine/Input pair. It's
>>>>>>>>>>>>>>>> completely meaningless to talk about simulating a finite
>>>>>>>>>>>>>>>> string.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It is possible for Turing machines to have blank tapes.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The salient aspect for the Halting problem is whether or
>>>>>>>>>>>>>>> not the finite string machine description specifies a
>>>>>>>>>>>>>>> finite or infinite sequence of configurations. The
>>>>>>>>>>>>>>> ultimate basis for determining this is the actual
>>>>>>>>>>>>>>> behavior of the simulated finite string.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Since this equally applies to machines having inputs and
>>>>>>>>>>>>>>> machines not having inputs the distinction relative to
>>>>>>>>>>>>>>> inputs is moot.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> If the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
>>>>>>>>>>>>>>>>> cannot possibly reach ⟨Ĥ⟩.qn then it is necessarily
>>>>>>>>>>>>>>>>> correct for embedded_H to transition to Ĥ.qn and
>>>>>>>>>>>>>>>>> nothing else in the universe can possibly refute this.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Again, you're falling back on your belief that ⟨Ĥ⟩
>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ is both meaningful (it isn't) and somehow
>>>>>>>>>>>>>>>> distinct from H applied to ⟨Ĥ⟩.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The behavior of the simulated input when embedded_H
>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is the ultimate measure of the halt
>>>>>>>>>>>>>>> status of this input.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Which just proves you are not working on the Halting Problem,
>>>>>>>>>>>>> No it only proves that you and André don't understand that
>>>>>>>>>>>>> a halt decider computes the mapping from the inputs to an
>>>>>>>>>>>>> accept or reject state (here is the part that you two don't
>>>>>>>>>>>>> understand):
>>>>>>>>>>>>>
>>>>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>>>>> input.
>>>>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>>>>> input.
>>>>>>>>>>>>> On the basis of the actual behavior specified by the actual
>>>>>>>>>>>>> input.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Which is DEFINED by what a the machine the input represents
>>>>>>>>>>>> would do,
>>>>>>>>>>>
>>>>>>>>>>> These words prove themselves true on the basis of their meaning:
>>>>>>>>>>> The actual behavior of the correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
>>>>>>>>>>> embedded_H is the ultimate measure of the behavior specified
>>>>>>>>>>> by ⟨Ĥ⟩ ⟨Ĥ⟩.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> WRONG, which shows you do not actually know the meaning of the
>>>>>>>>>> words.
>>>>>>>>> When you disagree that the correct simulation of a machine
>>>>>>>>> description of a machine is the ultimate measure of the
>>>>>>>>> behavior specified by this machine description it is just like
>>>>>>>>> saying that a black cat is not a cat.
>>>>>>>>>
>>>>>>>>
>>>>>>>> The problem is that 'Correct Simulation of a machine
>>>>>>>> description' has an actual meaning, in that the simulation must
>>>>>>>> match the actual behavior of the machine whose description it is
>>>>>>>> simulating, RIGHT?
>>>>>>> It must only do exactly what it actually does, if this does not
>>>>>>> meet expectations then expectations must be incorrect.
>>>>>>>
>>>>>>> Here is what it actually does:
>>>>>>> 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⟩...
>>>>>>
>>>>>> And if that is what it actually does, then H NEVER aborts its
>>>>>> simulation and thus never give an answer.
>>>>>>
>>>>>
>>>>> When embedded_H correctly matches the above infinite sequence this
>>>>> conclusively proves that its correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot
>>>>> possibly reach ⟨Ĥ⟩.qn. (We don't even need to mention any UTM).
>>>>>
>>>>>
>>>>
>>>> Excepts as previously said, that pattern only exists if H never aborts.
>>>>
>>> That is not true. The pattern exists for at least any finite number
>>> of steps where it can be recognized. The three iterations shown above
>>> are plenty enough for it to be recogized.
>>>
>>
>> But if it only exists for a finite number of steps (till it is
>> recognized)
>
> Then embedded_H has conclusively proved that its simulated ⟨Ĥ⟩ ⟨Ĥ⟩
> cannot possibly ever reach ⟨Ĥ⟩.qn even in an infinite number of
> simulated steps thus meeting the Linz definition of a sequence of
> configurations that never halt.
>
> computation that halts … the Turing machine will halt whenever it enters
> a final state. (Linz:1990:234)
>

No, it hasn't, because I just showed you that if H -> H.Qn then the
computation H^ <H^> Halts.

Proven.

No attempt top refute.

So FAIL, you have just proved that all you are is a LIAR.

So, YOU LIE.