On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
https://en.wikipedia.org/wiki/Halting_problem
Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
All this says is that if P never reaches a final state P never halts.
That is just the definition of "halts".
PS: Nit to be fixed: there can be more than one final attae.
On 2021-10-25 17:11, olcott wrote:
On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
https://en.wikipedia.org/wiki/Halting_problemAll this says is that if P never reaches a final state P never halts.
Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
That is just the definition of "halts".
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Sure that seems to be totally obvious and no big deal until one
realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
determines that its input never halts. All of the conventional proofs
(and Linz himself) conclude that this is impossible.
Except that when we actually *run* Ĥ ⟨Ĥ⟩ it halts. Yes your decider claims it doesn't halt, but your decider is *wrong*.
André
On 10/25/2021 6:31 PM, André G. Isaak wrote:
On 2021-10-25 17:11, olcott wrote:
On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
https://en.wikipedia.org/wiki/Halting_problemAll this says is that if P never reaches a final state P never halts.
Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P >>>>> never reaches its final state (whether or not its simulation of P is >>>>> aborted) then H correctly decides that P never halts.
That is just the definition of "halts".
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Sure that seems to be totally obvious and no big deal until one
realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩
correctly determines that its input never halts. All of the
conventional proofs (and Linz himself) conclude that this is impossible.
Except that when we actually *run* Ĥ ⟨Ĥ⟩ it halts. Yes your decider
claims it doesn't halt, but your decider is *wrong*.
André
It can't possibly be wrong because it is a semantic tautology.
If we know that X is a black then then we know that X is a cat.
THIS STATEMENT IS NECESSARILY ALWAYS TRUE
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
It can't possibly be wrong because it is a semantic tautology.
If we know that X is a black then then we know that X is a cat.
On 2021-10-25 17:57, olcott wrote:
On 10/25/2021 6:31 PM, André G. Isaak wrote:
On 2021-10-25 17:11, olcott wrote:
On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
https://en.wikipedia.org/wiki/Halting_problemAll this says is that if P never reaches a final state P never halts. >>>>> That is just the definition of "halts".
Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P >>>>>> never reaches its final state (whether or not its simulation of P is >>>>>> aborted) then H correctly decides that P never halts.
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Sure that seems to be totally obvious and no big deal until one
realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ >>>> correctly determines that its input never halts. All of the
conventional proofs (and Linz himself) conclude that this is
impossible.
Except that when we actually *run* Ĥ ⟨Ĥ⟩ it halts. Yes your decider >>> claims it doesn't halt, but your decider is *wrong*.
André
It can't possibly be wrong because it is a semantic tautology.
If we know that X is a black then then we know that X is a cat.
We do? My TV remote is black. I guess it must be a cat...
THIS STATEMENT IS NECESSARILY ALWAYS TRUE
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
But that is *not* the problem which a halt decider addresses.
On 2021-10-25 18:25, olcott wrote:
On 10/25/2021 7:17 PM, André G. Isaak wrote:
On 2021-10-25 17:57, olcott wrote:
On 10/25/2021 6:31 PM, André G. Isaak wrote:
On 2021-10-25 17:11, olcott wrote:
On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
https://en.wikipedia.org/wiki/Halting_problemAll this says is that if P never reaches a final state P never
Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that
input P
never reaches its final state (whether or not its simulation of >>>>>>>> P is
aborted) then H correctly decides that P never halts.
halts.
That is just the definition of "halts".
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Sure that seems to be totally obvious and no big deal until one
realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ >>>>>> correctly determines that its input never halts. All of the
conventional proofs (and Linz himself) conclude that this is
impossible.
Except that when we actually *run* Ĥ ⟨Ĥ⟩ it halts. Yes your decider >>>>> claims it doesn't halt, but your decider is *wrong*.
André
It can't possibly be wrong because it is a semantic tautology.
If we know that X is a black then then we know that X is a cat.
We do? My TV remote is black. I guess it must be a cat...
THIS STATEMENT IS NECESSARILY ALWAYS TRUE
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
But that is *not* the problem which a halt decider addresses.
Sure it is. If H correctly decides that its input never halts then
nothing else in the universe can possibly show that H has decided its
input incorrectly.
Inputs don't halt or not halt.
And you *very* dishonestly cut the portion where I indicated the actual specification of the problem as defined by Linz.
André
olcott <NoOne@NoWhere.com> writes:
On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
https://en.wikipedia.org/wiki/Halting_problemAll this says is that if P never reaches a final state P never halts. >>> That is just the definition of "halts".
Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
Sure that seems to be totally obvious and no big deal
Cut off from the specification is seems to be no big deal. Put the
proper condition back:
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
if (and only of) Ĥ applied to ⟨Ĥ⟩ does not halt
and you can see why it is a big deal.
until one realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM >> description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
determines that its input never halts.
Declaring what Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ does "correct" does not make it so. What is
correct follows from how Linz defines Ĥ, and is the part you keep lying
by omitting.
All of the conventional proofs
(and Linz himself) conclude that this is impossible.
All the proofs (conventional or otherwise) conclude that no TM can
decide TM halting. Simply declaring that what Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ does is "correct" does not make it so.
On 2021-10-26 07:29, olcott wrote:
On 10/26/2021 12:55 AM, André G. Isaak wrote:
On 2021-10-25 23:03, olcott wrote:
On 10/25/2021 11:38 PM, André G. Isaak wrote:
On 2021-10-25 20:58, olcott wrote:
The problem with that way of saying it is that it is vague about
what point of the execution trace matters.
There's nothing remotely vague about this. It states *exactly*
under which circumstances Ĥq0 ⊢* Ĥqn is justified.
The halting problem isn't about execution traces. Its about whether
a particular TM halts when given a particular input string.
The precise way of saying that when we are assuming that Ĥq0 is a >>>>>> simulating halt decider is:
Assuming that Ĥq0 is a simulating halt decider makes not an iota of >>>>> difference to the criterion which Linz states. Ĥq0 is supposed to
transition to Ĥqn if and only if M applied to <M> does not halt.
Which in my encoding means that the ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by Ĥq0
No. It doesn't. It means exactly what it says. The specification of a
problem doesn't change depending on the implementation.
You can use whatever method you want internally, but if the final
answer you get doesn't correspond to the one specified by the problem
then your method simply isn't sound or you're answering the wrong
question.
Even if everyone else in the universe disagrees, it is always that
case that when a halt decider does correctly decide the halt status of
its input then this input did have its halt status correctly decided.
It seems a little psychotic that you disagree with this.
People *aren't* disagreeing with this conditional (though it is
horrendously worded).
People are disagreeing with your assertion that your decider actually
does decide correctly because it doesn't.
To decide correctly you must
actually match the condition specified by the problem, which is "does Ĥ applied to <Ĥ> halt". It does.
André
On 2021-10-26 08:50, olcott wrote:
On 10/26/2021 9:23 AM, André G. Isaak wrote:
On 2021-10-26 07:29, olcott wrote:
On 10/26/2021 12:55 AM, André G. Isaak wrote:
On 2021-10-25 23:03, olcott wrote:
On 10/25/2021 11:38 PM, André G. Isaak wrote:
On 2021-10-25 20:58, olcott wrote:
Which in my encoding means that the ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulatedThe problem with that way of saying it is that it is vague about >>>>>>>> what point of the execution trace matters.
There's nothing remotely vague about this. It states *exactly*
under which circumstances Ĥq0 ⊢* Ĥqn is justified.
The halting problem isn't about execution traces. Its about
whether a particular TM halts when given a particular input string. >>>>>>>
The precise way of saying that when we are assuming that Ĥq0 is >>>>>>>> a simulating halt decider is:
Assuming that Ĥq0 is a simulating halt decider makes not an iota >>>>>>> of difference to the criterion which Linz states. Ĥq0 is supposed >>>>>>> to transition to Ĥqn if and only if M applied to <M> does not halt. >>>>>>
by Ĥq0
No. It doesn't. It means exactly what it says. The specification of
a problem doesn't change depending on the implementation.
You can use whatever method you want internally, but if the final
answer you get doesn't correspond to the one specified by the
problem then your method simply isn't sound or you're answering the
wrong question.
Even if everyone else in the universe disagrees, it is always that
case that when a halt decider does correctly decide the halt status
of its input then this input did have its halt status correctly
decided.
It seems a little psychotic that you disagree with this.
People *aren't* disagreeing with this conditional (though it is
horrendously worded).
People are disagreeing with your assertion that your decider actually
does decide correctly because it doesn't.
Simulating Halt Decider Theorem (Olcott 2021):
Calling something a theorem doesn't actually make it one. You should
learn what 'theorem' means.
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
As soon as I show that Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ meets the above theorem then Linz
has been refuted. Only a fool would disagree.
Things don't 'meet a theorem'. That's gibberish. They satisfy a
criterion. And in the case of Linz's H/Ĥ the criterion is ALREADY STATED
as part of the problem (though you keep deleting it).
Linz CLEARLY states the criterion:
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn IFF Ĥ APPLIED TO ⟨Ĥ⟩ HALTS.
You can't replace the last bit with some other criterion and still claim
to be working on the same problem.
André
On 2021-10-26 11:41, olcott wrote:
On 10/26/2021 12:22 PM, André G. Isaak wrote:
On 2021-10-26 08:50, olcott wrote:
On 10/26/2021 9:23 AM, André G. Isaak wrote:
On 2021-10-26 07:29, olcott wrote:
On 10/26/2021 12:55 AM, André G. Isaak wrote:
On 2021-10-25 23:03, olcott wrote:
On 10/25/2021 11:38 PM, André G. Isaak wrote:
On 2021-10-25 20:58, olcott wrote:
The problem with that way of saying it is that it is vague >>>>>>>>>> about what point of the execution trace matters.
There's nothing remotely vague about this. It states *exactly* >>>>>>>>> under which circumstances Ĥq0 ⊢* Ĥqn is justified.
The halting problem isn't about execution traces. Its about
whether a particular TM halts when given a particular input
string.
The precise way of saying that when we are assuming that Ĥq0 >>>>>>>>>> is a simulating halt decider is:
Assuming that Ĥq0 is a simulating halt decider makes not an >>>>>>>>> iota of difference to the criterion which Linz states. Ĥq0 is >>>>>>>>> supposed to transition to Ĥqn if and only if M applied to <M> >>>>>>>>> does not halt.
Which in my encoding means that the ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated
by Ĥq0
No. It doesn't. It means exactly what it says. The specification >>>>>>> of a problem doesn't change depending on the implementation.
You can use whatever method you want internally, but if the final >>>>>>> answer you get doesn't correspond to the one specified by the
problem then your method simply isn't sound or you're answering
the wrong question.
Even if everyone else in the universe disagrees, it is always that >>>>>> case that when a halt decider does correctly decide the halt
status of its input then this input did have its halt status
correctly decided.
It seems a little psychotic that you disagree with this.
People *aren't* disagreeing with this conditional (though it is
horrendously worded).
People are disagreeing with your assertion that your decider
actually does decide correctly because it doesn't.
Simulating Halt Decider Theorem (Olcott 2021):
Calling something a theorem doesn't actually make it one. You should
learn what 'theorem' means.
It is actually a self-evident truth, yet math, logic and computer
science people may have never heard of that term from philosophy so I
use its closest approximation: theorem
Anyone who has studied math or logic will have a far better grasp of philosophy than you do. I have no idea where you get the idea that
people in these fields are ignorant of the relevant philsophy.
And 'self-evident truth' isn't a notion that has played a significant
role in philosophy in at least the last century and a half. It's an antiquated notion and it doesn't even remotely approximate 'theorem'.
But the main point is that whatever you want to call this it isn't
actually the criterion which is specified by the problem at hand. If you
want to address the halting problem, you need to use the same criterion
as that problem.
a general proposition not self-evident but proved by a chain of
reasoning; a truth established by means of accepted truths.
In epistemology (theory of knowledge), a self-evident proposition is a
proposition that is known to be true by understanding its meaning
without proof. https://en.wikipedia.org/wiki/Self-evidence
A notion dating back to the Ancient Greeks which has proven to be
entirely inadequate. Euclid considered his fifth postulate to be self-evident[*]. But no one today would accept it as such.
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
As soon as I show that Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ meets the above theorem then Linz
has been refuted. Only a fool would disagree.
Things don't 'meet a theorem'. That's gibberish. They satisfy a
criterion. And in the case of Linz's H/Ĥ the criterion is ALREADY
STATED as part of the problem (though you keep deleting it).
Linz CLEARLY states the criterion:
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn IFF Ĥ APPLIED TO ⟨Ĥ⟩ HALTS.
If Linz says this then Linz is wrong. A halt decider is correct iff it
correctly decides the halt status of it inputs.
A halt decider decides the halting status of the *computation*
represented by its input. Which is exactly what Linz states. The input
itself is just a string. It has no halting status.
The halt decider is at Ĥq0 it is not at q0.
The inputs to the halt decider at Ĥq0 are: ⟨Ĥ⟩ ⟨Ĥ⟩.
Neither Ĥ nor Ĥq0 are halt deciders at all.
THE SIMULATION OFAs long as Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that its inputs never >> reach their final state then Ĥq0 is necessarily correct and impossibly
incorrect.
The inputs don't have final states.
As long as Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that
its inputs never
reach their final state then Ĥq0 is necessarily correct and
impossibly
incorrect.
The machine they represent does. And
its about that machine applied to the supplied input string that a halt decider is expect to decide. That's a simple matter of definition.
André
[*] Or at least he presented it as such, though there are reasons to
believe he wasn't overly satisfied with this and would rather have
derived this from the first four postulates, but this proved impossible.
Of course, he's dead, so we can't really ask him about this.
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