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:That is not true. The pattern exists for at least any finite number
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:
It must only do exactly what it actually does, if this does not >>>>>>>> meet expectations then expectations must be incorrect.
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:When you disagree that the correct simulation of a machine >>>>>>>>>> description of a machine is the ultimate measure of the
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: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 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: >>>>>>>>>>>>>>>>>>>>>That's just as meaningless. You can simulate Ĥ >>>>>>>>>>>>>>>>>>>>> applied to ⟨Ĥ⟩ or you can provide ⟨Ĥ⟩ ⟨Ĥ⟩ as the
⟨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 ? >>>>>>>>>>>>>>>>>>>>>
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,
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.
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?
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.
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.
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:That is not true. The pattern exists for at least any finite number of
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:
It must only do exactly what it actually does, if this does not
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:When you disagree that the correct simulation of a machine
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.
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?
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.
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)
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