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".
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.
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
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.
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
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^>
On 2/6/22 10:04 PM, olcott wrote:When embedded_H matches this infinite pattern in the same three iterations:
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
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.
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
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 infiniteThis is incomplete because it does not cover the case where the >>>>>>>>> machine neither halts nor matches an "infinite behavior pattern". >>>>>>>>>
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.
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.
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.
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
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.This is incomplete because it does not cover the case where the >>>>>>>>>>> machine neither halts nor matches an "infinite behavior
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.
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.
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,
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
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.This is incomplete because it does not cover the case where >>>>>>>>>>>>> the
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 >>>>>>>>>>>>> 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.
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.
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: >>>>>>>>>>>>>>
H determines [halting] on the basis of matching infinite >>>>>>>>>>>>>> behavior patterns.This is incomplete because it does not cover the case where >>>>>>>>>>>>> the
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 >>>>>>>>>>>>> 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.
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^>
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. >>>>>>>>>
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.
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.
If we know that we have a black cat then we know that we have a cat.
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.
--
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 2/8/22 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:
three iterations:
On 2/6/22 10:04 PM, olcott wrote:
On 2/6/2022 3:39 PM, Richard Damon wrote:WRONG.
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
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. >>>>>>>>>>>
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
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.
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.
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:WRONG.
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 suchTo 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. >>>>>>>>>>>>>>>>>>
that every Turing machine either halts or matches one >>>>>>>>>>>>>>>>>>> of these
patterns.
But I feel sure that theorem is not true. >>>>>>>>>>>>>>>>>>
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. >>>>>>>>>>>> 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.
olcott wrote:
...
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.
You are starting the race with a dead horse.
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 suchTo 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.
that every Turing machine either halts or matches >>>>>>>>>>>>>>>>>>>>>>> one of these
patterns.
But I feel sure that theorem is not true. >>>>>>>>>>>>>>>>>>>>>>
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. >>>>>>>>>>>>>>>>>>>>
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^>
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. >>>>>>>>>>>>>>>>>>>
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.
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.
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.
FAIL.
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