olcott <NoOne@NoWhere.com> writes:
On 10/18/2021 5:59 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 7:47 PM, Ben Bacarisse wrote:if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
Quotes re-ordered for clarity...
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 10:03 AM, Ben Bacarisse wrote:
But you agree that
Ĥ.q0 ⟨Ĥ⟩ ⊦ Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦ Ĥ.qn
if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
is (one half of) what Linz means by Ĥ, yes? (I don't expect an >>>>>>> answer -- I expect you'll just repeat the waffle.)
No not at all, not in the least little bit
OK, so you are not using Ĥ to denote an exemplar of the same class of >>>>> TMs that Linz does. If you were, you would have to agree that Linz's >>>>> specification of when Ĥ transitions to qn is the correct one.
Oh, and you need to apologise for some very deceptive claims to the
contrary.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf page 319
That you do not understand the most basic point that Linz has adapted
TM H by prepending states and appending states such that the original
H "not" path of H remains in the middle is a necessary prerequisite.
I'll leave it to others to decide which one of use us is lacking basic understanding.
Just stop dishonestly removing the key criterion given in Linz, or admit
that you are not using this notation as Linz does.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM H
Linz says it better: if M applied to wM does not halt. Your silly
phrase does not state what it is that H is saying "no" to.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM H'
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
Linz says it better: if Ĥ applied to wM does not halt.
This leaves the "no" path of TM H at Linz Ĥq0
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
And Linz says exactly when this transitions sequence occurs: if Ĥ
applied to ⟨Ĥ⟩ does not halt.
The only excuse for dropping what Linz says about these transition
sequences is to deceive. Stop doing that.
olcott <NoOne@NoWhere.com> writes:
On 10/19/2021 11:55 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/18/2021 5:59 PM, Ben Bacarisse wrote:I'll leave it to others to decide which one of use us is lacking basic
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 7:47 PM, Ben Bacarisse wrote:if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
Quotes re-ordered for clarity...
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 10:03 AM, Ben Bacarisse wrote:
But you agree that
Ĥ.q0 ⟨Ĥ⟩ ⊦ Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦ Ĥ.qn >>>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. >>>>>>>>>
is (one half of) what Linz means by Ĥ, yes? (I don't expect an >>>>>>>>> answer -- I expect you'll just repeat the waffle.)
No not at all, not in the least little bit
OK, so you are not using Ĥ to denote an exemplar of the same class of >>>>>>> TMs that Linz does. If you were, you would have to agree that Linz's >>>>>>> specification of when Ĥ transitions to qn is the correct one.
Oh, and you need to apologise for some very deceptive claims to the >>>>>>> contrary.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf page 319
That you do not understand the most basic point that Linz has adapted
TM H by prepending states and appending states such that the original
H "not" path of H remains in the middle is a necessary prerequisite.
understanding.
Just stop dishonestly removing the key criterion given in Linz, or admit >>> that you are not using this notation as Linz does.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM HLinz says it better: if M applied to wM does not halt. Your silly
phrase does not state what it is that H is saying "no" to.
Linz is vague about the point in the execution trace that matters.
You replacement is useless. Linz's specification is clear. Obviously
it does not way what you want, but it's not vague.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM H'
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
Linz says it better: if Ĥ applied to wM does not halt.
The only point in the execution trace that matters is where the halt
decider makes its halting decision.
What matters is that if H is as specified in Linz, the corresponding Ĥ behaves as Linz says and you keep removing.
This leaves the "no" path of TM H at Linz Ĥq0And Linz says exactly when this transitions sequence occurs: if Ĥ
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
applied to ⟨Ĥ⟩ does not halt.
If he is referring to something besides the input to the halt decider
then Linz is incorrect.
I am happy to explain "if Ĥ applied to wM does not halt" when attached
to the line you keep leaving it off if you think it would help. What is
it you don't understand?
On 2021-10-20 09:25, olcott wrote:
On 10/20/2021 10:13 AM, André G. Isaak wrote:
On 2021-10-20 07:57, olcott wrote:
On 10/19/2021 8:21 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/19/2021 11:55 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/18/2021 5:59 PM, Ben Bacarisse wrote:I'll leave it to others to decide which one of use us is lacking >>>>>>> basic
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 7:47 PM, Ben Bacarisse wrote:if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
Quotes re-ordered for clarity...
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 10:03 AM, Ben Bacarisse wrote:
But you agree that
Ĥ.q0 ⟨Ĥ⟩ ⊦ Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦ Ĥ.qn
if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
is (one half of) what Linz means by Ĥ, yes? (I don't >>>>>>>>>>>>> expect an
answer -- I expect you'll just repeat the waffle.)
No not at all, not in the least little bit
OK, so you are not using Ĥ to denote an exemplar of the same >>>>>>>>>>> class of
TMs that Linz does. If you were, you would have to agree >>>>>>>>>>> that Linz's
specification of when Ĥ transitions to qn is the correct one. >>>>>>>>>>>
Oh, and you need to apologise for some very deceptive claims >>>>>>>>>>> to the
contrary.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
page 319
That you do not understand the most basic point that Linz has
adapted
TM H by prepending states and appending states such that the
original
H "not" path of H remains in the middle is a necessary
prerequisite.
understanding.
Just stop dishonestly removing the key criterion given in Linz,
or admit
that you are not using this notation as Linz does.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM HLinz says it better: if M applied to wM does not halt. Your silly >>>>>>> phrase does not state what it is that H is saying "no" to.
Linz is vague about the point in the execution trace that matters.
You replacement is useless. Linz's specification is clear. Obviously >>>>> it does not way what you want, but it's not vague.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM H'Linz says it better: if Ĥ applied to wM does not halt.
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ >>>>>>>
The only point in the execution trace that matters is where the halt >>>>>> decider makes its halting decision.
What matters is that if H is as specified in Linz, the corresponding Ĥ >>>>> behaves as Linz says and you keep removing.
This leaves the "no" path of TM H at Linz Ĥq0And Linz says exactly when this transitions sequence occurs: if Ĥ >>>>>>> applied to ⟨Ĥ⟩ does not halt.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
If he is referring to something besides the input to the halt decider >>>>>> then Linz is incorrect.
I am happy to explain "if Ĥ applied to wM does not halt" when attached >>>>> to the line you keep leaving it off if you think it would help.
What is
it you don't understand?
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
if M applied to Wm does not halt
Linz cannot be correctly referring to q0 applied to Wm because there
is no halt decider at q0. If he is referring to that then he is wrong.
Nor is there a halt decider at Ĥq0. Both q0 and Ĥq0 refer to single
states in the execution of Ĥ. There isn't a halt decider anywhere in Ĥ. >>>
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
It is clear that the "no" path of the H halt decider from which Ĥ was
constructed begins at Ĥq0 and its input is Wm Wm. Pages 318 and 319
prove this.
??
Ĥ always begins in state q0, so all paths ultimately begin there.
Since it always passes through state Ĥq0, you can talk about the path beginning at Ĥq0 if you want, but that has no bearing on the point being made, which is that final state qn is only reached in cases where M
applied to Wm does not halt.
André
On 2021-10-20 09:51, olcott wrote:
On 10/20/2021 10:42 AM, André G. Isaak wrote:
On 2021-10-20 09:25, olcott wrote:
On 10/20/2021 10:13 AM, André G. Isaak wrote:
On 2021-10-20 07:57, olcott wrote:
On 10/19/2021 8:21 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/19/2021 11:55 AM, Ben Bacarisse wrote:You replacement is useless. Linz's specification is clear.
olcott <NoOne@NoWhere.com> writes:
On 10/18/2021 5:59 PM, Ben Bacarisse wrote:I'll leave it to others to decide which one of use us is
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 7:47 PM, Ben Bacarisse wrote:if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
Quotes re-ordered for clarity...
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 10:03 AM, Ben Bacarisse wrote:
No not at all, not in the least little bit
But you agree that
Ĥ.q0 ⟨Ĥ⟩ ⊦ Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦ Ĥ.qn
if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
is (one half of) what Linz means by Ĥ, yes? (I don't >>>>>>>>>>>>>>> expect an
answer -- I expect you'll just repeat the waffle.) >>>>>>>>>>>>>>
OK, so you are not using Ĥ to denote an exemplar of the >>>>>>>>>>>>> same class of
TMs that Linz does. If you were, you would have to agree >>>>>>>>>>>>> that Linz's
specification of when Ĥ transitions to qn is the correct one. >>>>>>>>>>>>>
Oh, and you need to apologise for some very deceptive >>>>>>>>>>>>> claims to the
contrary.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf >>>>>>>>>> page 319
That you do not understand the most basic point that Linz has >>>>>>>>>> adapted
TM H by prepending states and appending states such that the >>>>>>>>>> original
H "not" path of H remains in the middle is a necessary
prerequisite.
lacking basic
understanding.
Just stop dishonestly removing the key criterion given in Linz, >>>>>>>>> or admit
that you are not using this notation as Linz does.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM HLinz says it better: if M applied to wM does not halt. Your silly >>>>>>>>> phrase does not state what it is that H is saying "no" to.
Linz is vague about the point in the execution trace that matters. >>>>>>>
Obviously
it does not way what you want, but it's not vague.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM H'Linz says it better: if Ĥ applied to wM does not halt.
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ >>>>>>>>>
The only point in the execution trace that matters is where the >>>>>>>> halt
decider makes its halting decision.
What matters is that if H is as specified in Linz, the
corresponding Ĥ
behaves as Linz says and you keep removing.
This leaves the "no" path of TM H at Linz Ĥq0And Linz says exactly when this transitions sequence occurs: if Ĥ >>>>>>>>> applied to ⟨Ĥ⟩ does not halt.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
If he is referring to something besides the input to the halt
decider
then Linz is incorrect.
I am happy to explain "if Ĥ applied to wM does not halt" when
attached
to the line you keep leaving it off if you think it would help.
What is
it you don't understand?
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
if M applied to Wm does not halt
Linz cannot be correctly referring to q0 applied to Wm because
there is no halt decider at q0. If he is referring to that then he >>>>>> is wrong.
Nor is there a halt decider at Ĥq0. Both q0 and Ĥq0 refer to single >>>>> states in the execution of Ĥ. There isn't a halt decider anywhere
in Ĥ.
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf It is
clear that the "no" path of the H halt decider from which Ĥ was
constructed begins at Ĥq0 and its input is Wm Wm. Pages 318 and 319
prove this.
??
Ĥ always begins in state q0, so all paths ultimately begin there.
That is only the point in the execution trace of Ĥ that copies its input
So what? It's still the starting point of the computation. Ĥ is a Turing Machine which takes a *single* input string. The fact that it
subsequently duplicates that is an integral part of the computation
performed by Ĥ on a given input.
so that its halt decider at Ĥq0 has its required input pair.
There IS NO halt decider at Ĥq0. Nor anywhere else in Ĥ.
Since it always passes through state Ĥq0, you can talk about the path
beginning at Ĥq0 if you want, but that has no bearing on the point
being made, which is that final state qn is only reached in cases
where M applied to Wm does not halt.
André
As long as the simulating halt decider at Ĥq0 correctly decides that
its simulated input Wm applied to Wm never reaches the final state of
this simulated input (whether or not the simulating halt decider
aborts its simulation of this input) then this simulating halt decider
at Ĥq0 is correct when it transitions to qn.
But that's not the criterion given by Linz. He specifies that qn is
reached if M applied to Wm does not halt.
If you have to change that last bit, you're not talking about Linz's Ĥ.
André
On 2021-10-20 14:34, olcott wrote:
On 10/20/2021 3:06 PM, André G. Isaak wrote:
But it is *not* guaranteed to halt, so it is not a decider at all.
Not any kind of decider.
If I make any thingamajig Whatchamacallit that correctly determines
the halt status of the "impossible" halting problem counter-example
input then this is very significant.
But again, that's what your *H* is purported to do. It isn't what Ĥ is purported to do. I have been commenting on your insistence of calling Ĥ
(or Ĥq0) a halt decider. It is not one. Linz doesn't describe it as one.
If your going to make grandiose claims about having constructed a
(partial) halt decider, at least restrict them to your H, not to Ĥ which isn't presented by Linz as a decider of anything. It's simply a TM which performs some relatively meaningless computation but which plays a role
in his proof.
André
olcott <NoOne@NoWhere.com> writes:
On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/19/2021 8:21 PM, Ben Bacarisse wrote:This notation is hopeless in ASCII. Using the notation that has evolved >>> in these threads (not my preferred one, but I don't want to complicate
olcott <NoOne@NoWhere.com> writes:
On 10/19/2021 11:55 AM, Ben Bacarisse wrote:You replacement is useless. Linz's specification is clear. Obviously >>>>> it does not way what you want, but it's not vague.
olcott <NoOne@NoWhere.com> writes:
On 10/18/2021 5:59 PM, Ben Bacarisse wrote:Just stop dishonestly removing the key criterion given in Linz, or admit
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 7:47 PM, Ben Bacarisse wrote:if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
Quotes re-ordered for clarity...
olcott <NoOne@NoWhere.com> writes:
On 10/17/2021 10:03 AM, Ben Bacarisse wrote:
But you agree that
Ĥ.q0 ⟨Ĥ⟩ ⊦ Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦ Ĥ.qn >>>>>>>>>>>>> if (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt. >>>>>>>>>>>>>
is (one half of) what Linz means by Ĥ, yes? (I don't expect an >>>>>>>>>>>>> answer -- I expect you'll just repeat the waffle.)
No not at all, not in the least little bit
OK, so you are not using Ĥ to denote an exemplar of the same class of
TMs that Linz does. If you were, you would have to agree that Linz's
specification of when Ĥ transitions to qn is the correct one. >>>>>>>>>>>
Oh, and you need to apologise for some very deceptive claims to the >>>>>>>>>>> contrary.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf page 319 >>>>>>>>
That you do not understand the most basic point that Linz has adapted >>>>>>>> TM H by prepending states and appending states such that the original >>>>>>>> H "not" path of H remains in the middle is a necessary prerequisite. >>>>>>> I'll leave it to others to decide which one of use us is lacking basic >>>>>>> understanding.
that you are not using this notation as Linz does.
q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM HLinz says it better: if M applied to wM does not halt. Your silly >>>>>>> phrase does not state what it is that H is saying "no" to.
Linz is vague about the point in the execution trace that matters.
What matters is that if H is as specified in Linz, the corresponding Ĥ >>>>> behaves as Linz says and you keep removing.q0 Wm Wm ⊢* y1 qn y2 / The "no" path of TM H'Linz says it better: if Ĥ applied to wM does not halt.
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ >>>>>>>
The only point in the execution trace that matters is where the halt >>>>>> decider makes its halting decision.
I am happy to explain "if Ĥ applied to wM does not halt" when attached >>>>> to the line you keep leaving it off if you think it would help. What is >>>>> it you don't understand?This leaves the "no" path of TM H at Linz Ĥq0And Linz says exactly when this transitions sequence occurs: if Ĥ >>>>>>> applied to ⟨Ĥ⟩ does not halt.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
If he is referring to something besides the input to the halt decider >>>>>> then Linz is incorrect.
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ
if M applied to Wm does not halt
Linz cannot be correctly referring to q0 applied to Wm because there
is no halt decider at q0.
matters any more) Linz writes:
Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
if M applied to <M> does not halt.
And with these changes in notation you seem to be saying
Linz cannot be correctly referring to Ĥ applied to <M> because thereYes, there is no halt decider at Ĥ.q0, but he is correctly referring to >>> what Ĥ applied to <M> should do: Ĥ applied to <M> should transition to >>> Ĥ.qn if (and only if) M (the TM encoded in the string input) applied to >>> <M> (that exactly same string input) does not halt.
is no halt decider at q0.
Your next point helps explain how he comes to say this about Ĥ.
Linz can only be referring to Ĥ.qx <M> <M> the machine of the first
<M> being applied to the machine description of the second <M>.
The annotation dose not specifically refer to what this sub-computation
does. It simply gives the condition under which Ĥ will transition from >>> Ĥ.q0 to Ĥ.qn.
If a simulating halt decider H correctly determines that the
simulation of its input <M> applied to <M> never reaches its final
state whether or not this simulating halt decider aborts the
simulation of this input then this simulating halt decider does
correctly decide that this input never halts NO MATTER WHAT ELSE.
This is not what Linz is saying.
If I can help with any further
explanation, I'm happy to have a go but you need to stop repeating what
you are saying and try to focus on what Linz is saying. Once you get
what he is saying you will see that you can say anything you like about
a subclass (those that "simulate") of the empty class of TMs (the halt deciders) that Linz is talking about. Atatements about no TMs are
vacuously true (or irrelevant -- take your pick).
Anything and everything that is not input to this halt decider is 100%
totally irrelevant.
Technically, there is no halt decider in the TM you are talking about,
but that's a detail and you don't do details. Did you follow what I
wrote about how the string input to the almost-decider at Ĥ.qx is
entirely why Linz's annotation is correct? Here it is:
But you are correct in that the annotation /derives/ from what we know
about the modified copy of H embedded at Ĥ.qx. We know that
H.q0 <M> s ⊢* y1 H.qn y2
if M applied to s does not halt.
and since the TM at Ĥ.qx is (in this regard at least) exactly like H we >>> know that
Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
if M applied to <M> does not halt.
Now, because
Ĥ.q0 s ⊢* Ĥ.qx s s
for /any/ string s, we know, specifically, that
Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
if M applied to <M> does not halt.
Can you now see where this correctly annotated line comes from? Can I
explain any part of this some other way to make it clearer?
I suspect you simply didn't read it since I am agreeing with you about
the central role of the Ĥ.qx <M> <M> part in explaining why Linz is
correct.
olcott <NoOne@NoWhere.com> writes:
On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/20/2021 8:37 PM, Ben Bacarisse wrote:But until you understand what he is saying, you can't see why you are
olcott <NoOne@NoWhere.com> writes:
On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
This is not what Linz is saying.q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ >>>>>>>> if M applied to Wm does not haltThis notation is hopeless in ASCII. Using the notation that has evolved
Linz cannot be correctly referring to q0 applied to Wm because there >>>>>>>> is no halt decider at q0.
in these threads (not my preferred one, but I don't want to complicate >>>>>>> matters any more) Linz writes:
Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2
if M applied to <M> does not halt.
And with these changes in notation you seem to be saying
Linz cannot be correctly referring to Ĥ applied to <M> because there >>>>>>>> is no halt decider at q0.Yes, there is no halt decider at Ĥ.q0, but he is correctly referring to
what Ĥ applied to <M> should do: Ĥ applied to <M> should transition to
Ĥ.qn if (and only if) M (the TM encoded in the string input) applied to
<M> (that exactly same string input) does not halt.
Your next point helps explain how he comes to say this about Ĥ. >>>>>>>
Linz can only be referring to Ĥ.qx <M> <M> the machine of the first >>>>>>>> <M> being applied to the machine description of the second <M>. >>>>>>>The annotation dose not specifically refer to what this sub-computation >>>>>>> does. It simply gives the condition under which Ĥ will transition from
Ĥ.q0 to Ĥ.qn.
If a simulating halt decider H correctly determines that the
simulation of its input <M> applied to <M> never reaches its final >>>>>> state whether or not this simulating halt decider aborts the
simulation of this input then this simulating halt decider does
correctly decide that this input never halts NO MATTER WHAT ELSE.
Of course this is not what Linz is saying.
wrong.
That Linz did not consider a simulating halt decider does not make me
wrong.
Quite. You are wrong for the reasons I gave.
If any of this was wrong someone could point out an actual error:
The machine at Ĥq0 transitions to Ĥqn if the simulation of ⟨Ĥ⟩ applied
to ⟨Ĥ⟩ would never reach its final state whether or not this
simulation is aborted.
The error is that this is a statement about an sub-set of and empty set
of Turing machines.
Adapted from bottom of page 319 (definition of Ĥ)
⟨Ĥ⟩ indicates the Turing machine description of Ĥ.
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ∞
If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts
This is either the wrong annotation for Linz's Ĥ or a pointless
re-wording of it.
q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
If the simulated input to Ĥq0 ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt
And again, this is the wrong annotation for Linz's Ĥ. Linz's Ĥ
transitions to qn is (and only if) Ĥ applied to ⟨Ĥ⟩ does not halt.
You can't change the specification of what Ĥ does if you want to be
taken seriously. You have to accept the logical consequences of his definitions or define your own class of machines.
The halt decider must only correctly decide whether or not its input<See my recent relevant to this one> N.B. Compulsive repetition such as
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.
The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.
The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.
The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.
The halt decider must only correctly decide whether or not its input
halts on its input. As long as this decision is correct then it is
impossible for anything else to show that the halt decider is incorrect.
olcott <NoOne@NoWhere.com> writes:
On 10/24/2021 6:30 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/22/2021 6:57 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/21/2021 9:19 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/21/2021 4:45 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
On 10/20/2021 8:37 PM, Ben Bacarisse wrote:But until you understand what he is saying, you can't see why you are >>>>>>>>> wrong.
olcott <NoOne@NoWhere.com> writes:
On 10/20/2021 7:00 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:
q0 Wm ⊢* Ĥq0 Wm Wm ⊢* Ĥ y1 qn y2 // The "no" path of TM Ĥ >>>>>>>>>>>>>> if M applied to Wm does not haltThis notation is hopeless in ASCII. Using the notation that has evolved
Linz cannot be correctly referring to q0 applied to Wm because there
is no halt decider at q0.
in these threads (not my preferred one, but I don't want to complicate
matters any more) Linz writes:
Ĥ.q0 <M> ⊢* Ĥ.qx <M> <M> ⊢* y1 Ĥ.qn y2 >>>>>>>>>>>>> if M applied to <M> does not halt.
And with these changes in notation you seem to be saying >>>>>>>>>>>>>
Linz cannot be correctly referring to Ĥ applied to <M> because thereYes, there is no halt decider at Ĥ.q0, but he is correctly referring to
is no halt decider at q0.
what Ĥ applied to <M> should do: Ĥ applied to <M> should transition to
Ĥ.qn if (and only if) M (the TM encoded in the string input) applied to
<M> (that exactly same string input) does not halt.
Your next point helps explain how he comes to say this about Ĥ. >>>>>>>>>>>>>
Linz can only be referring to Ĥ.qx <M> <M> the machine of the firstThe annotation dose not specifically refer to what this sub-computation
<M> being applied to the machine description of the second <M>. >>>>>>>>>>>>>
does. It simply gives the condition under which Ĥ will transition from
Ĥ.q0 to Ĥ.qn.
If a simulating halt decider H correctly determines that the >>>>>>>>>>>> simulation of its input <M> applied to <M> never reaches its final >>>>>>>>>>>> state whether or not this simulating halt decider aborts the >>>>>>>>>>>> simulation of this input then this simulating halt decider does >>>>>>>>>>>> correctly decide that this input never halts NO MATTER WHAT ELSE. >>>>>>>>>>> This is not what Linz is saying.
Of course this is not what Linz is saying.
That Linz did not consider a simulating halt decider does not make me >>>>>>>> wrong.
Quite. You are wrong for the reasons I gave.
If any of this was wrong someone could point out an actual error: >>>>>>>>The error is that this is a statement about an sub-set of and empty set >>>>>>> of Turing machines.
The machine at Ĥq0 transitions to Ĥqn if the simulation of ⟨Ĥ⟩ applied
to ⟨Ĥ⟩ would never reach its final state whether or not this >>>>>>>> simulation is aborted.
You are simply making a false assumption. You are beginning with the >>>>>> premise that Linz is correct then concluding that Linz is correct on >>>>>> the basis of this premise.
A proved theorem is not a "false assumption".
Ah so you are back to your blasphemy. One can not rely on a proved
theorem as exactly equal to infallibility.
You are being dishonest again. Setting aside the silly religious
language, I am not asserting that any argument is infallible something
that was obvious from what I wrote immediate after this.
I am simply saying you have not shown any error in the argument. Until
you show that, for example, 2+2 is not equal to 4, it is entirely
reasonable to assert that it is. That is not blasphemy.
I don't see how this is so difficult for you:
(a) If it is claimed that X cannot possibly be done
(b) I show X being correctly done
(c) then I have refuted X cannot possibly be done
I don't think you honestly believe that the logic is difficult for me.
Neither do I think you have trouble knowing what part I am saying you
have failed at (it's (b) by the way).
But my objection to what you've been saying goes my deeper than a
failure to show anything of value. You are being dishonest in that you
claim to be talking about Linz's class of TMs, but you keep cutting key
facts about them while trying to replace those facts with waffle of you
your own invention.
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