On Thursday, February 3, 2022 at 8:47:24 PM UTC-6, richar...@gmail.com wrote:
> On 2/3/22 7:14 PM, 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)
> >
> > At that point right there embedded_H has complete proof that the correct
> > simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach ⟨Ĥ⟩.qn.
> >
> >
> IF you think you can prove something that isn't true, you are dumber
> than you seem.

On 2/1/2022 9:33 PM, André G. Isaak wrote:
> On 2022-02-01 20:08, olcott wrote:
>
>> embedded_H is exactly a UTM with extra features added.
>
> Apparently you don't know what 'exactly' means. embedded_H is not a UTM
> *at* *all*.
>
> If embedded_H were a UTM, then
>
> embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩

If ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by embedded_H cannot possibly ever transition to
⟨Ĥ⟩.qn then it fails to meet the Linz definition:

computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234) thus the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ does not halt.

This makes it necessarily correct for embedded_H to transition to Ĥ.qn.

--
Copyright 2021 Pete Olcott

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

On 2/6/22 12:13 AM, olcott wrote:
> On 2/1/2022 9:33 PM, André G. Isaak wrote:
>> On 2022-02-01 20:08, olcott wrote:
>>
>>> embedded_H is exactly a UTM with extra features added.
>>
>> Apparently you don't know what 'exactly' means. embedded_H is not a
>> UTM *at* *all*.
>>
>> If embedded_H were a UTM, then
>>
>> embedded_H would accept ⟨Ĥ⟩ ⟨Ĥ⟩ if Ĥ accepts ⟨Ĥ⟩
>
> If ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by embedded_H cannot possibly ever transition to
> ⟨Ĥ⟩.qn then it fails to meet the Linz definition:
>
> computation that halts … the Turing machine will halt whenever it enters
> a final state. (Linz:1990:234) thus the simulated ⟨Ĥ⟩ ⟨Ĥ⟩ does not halt.
>
> This makes it necessarily correct for embedded_H to transition to Ĥ.qn.
>
> MOSTLY WRONG.

Linz definition applies to the DIRECT EXECUTION of a machine.

A UTM, which can NEVER ABORT ITS SIMULATION, is an equivalent, so could
also be used.

IF embedded_H was a UTM, then your logic is sound, but if it is a UTM,
it can not transition to Qn, as it can never abort.

All you have 'proved' is that in the case that H never aborts, H^
applied to <H^> is indeed non-halting, but in that case H fails to be a
decider because it never gave an answer. So yes, IF H is a UTM, it WOULD
have been correct to transition to H.Qn, but it never can, so it wrong.

That is exactly like saying you have proved that cats bark by showing
that your dog barks. A Cat is not a Dog, just like your H that goes to
H.Qn is NOT a UTM.

FAIL.

You just continue to show off your ignorance of the topic.