On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:
On 2/1/2022 3:23 PM, wij wrote:
On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:I have shown how my system directly applies to the actual halting
On 2/1/2022 10:33 AM, wij wrote:
On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine
On 1/31/2022 11:25 PM, Richard Damon wrote:
On 1/31/22 11:42 PM, olcott wrote:
On 1/31/2022 10:33 PM, Richard Damon wrote:
On 1/31/22 11:24 PM, olcott wrote:
On 1/31/2022 10:17 PM, Richard Damon wrote:
On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is >>>>>>>>>> being asked about Sum(7,8).
On 1/31/2022 6:41 PM, Richard Damon wrote:Don't know how you get that from what I said.
On 1/31/22 3:24 PM, olcott wrote:
On 1/31/2022 2:10 PM, Ben wrote:
On 1/31/2022 8:06 AM, olcott wrote:
On 1/30/2022 8:20 PM, Richard Damon wrote:
On 1/30/22 9:05 PM, olcott wrote:
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
These statements need the conditions, that H^ goes to >>>>>>>>>>>>>>>>>>> H^.Qy/H^.Qn iff H goes to that corresponding state. >>>>>>>>>>>>>>>>>>>
⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H
in the same way that (5,3) is syntactically specified as an >>>>>>>>>>>>>>>>>> input to Sum(5,3)
Right, and the
Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to >>>>>>>>>>>>>>>>>> embedded_H in the same way that (1,2) is NOT syntactically >>>>>>>>>>>>>>>>>> specified as an input to Sum(5,3)
Right, but perhaps you don't understand that from you above >>>>>>>>>>>>>>>>> statement the right answer is based on if UTM(<H^>,<H^>) >>>>>>>>>>>>>>>>> Halts which by the definition of a UTM means if H^ applied to >>>>>>>>>>>>>>>>> <H^> Halts.
The biggest reason for your huge mistakes is that you cannot >>>>>>>>>>>>>>>> stay sharply focused on a single point. It is as if you either >>>>>>>>>>>>>>>> have attention deficit disorder ADD or are addicted to >>>>>>>>>>>>>>>> methamphetamine.
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and
Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H. >>>>>>>>>>>>>>>>
After we have mutual agreement on this point we will move on >>>>>>>>>>>>>>>> to the points that logically follow from this one. >>>>>>>>>>>>>>>>
Holy shit try to post something that makes sense. >>>>>>>>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>>>>>>
Richard does not accept that the input to the copy of Linz H >>>>>>>>>>>>>> embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.
No, but apparently you can't understand actual English words. >>>>>>>>>>>>>
The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must >>>>>>>>>>>>> give is based on the behavior of H^ applied to <H^> BECAUSE OF >>>>>>>>>>>>> THE DEFINITION of H.
In other words Sum(3,5) must return the value of Sum(7,8)? >>>>>>>>>>>
Any moron knows that a function is only accountable for its actual >>>>>>>>>>>> inputs.
And the actual input to H is <H^> <H^> which MEANS by the >>>>>>>>>>> DEFINITION of the Halting Problem that H is being asked to decide >>>>>>>>>>> on the Halting Status of H^ applied to <H^>
Again your RED HERRING.
H is being asked EXACTLY what it being asked
H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't >>>>>>>>>
AGREED?
No that is wrong. embedded_H is being asked:
Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?
If you say 'No', then you aren't doing the halting problem, as the >>>>>>> requirement I stated is EXACTLY the requirement of the Halting Problem. >>>>>> The halting problem is vague on the definition of halting, it includes >>>>>> that a machine has stopped running and that a machine cannot reach its >>>>>> final state. My definition only includes the latter.
Sounds like a NDTM.
It is not a NDTM, a Turing Machine only actually halts when it reaches >>>> its own final state. People not very familiar with this material may get >>>> confused and believe that a TM halts when its stops running because its >>>> simulation has been aborted. This key distinction is not typically
specified in most halting problem proofs.
computation that halts … the Turing machine will halt whenever it enters >>>> a final state. (Linz:1990:234)
Where did Linz mention 'simulation' and 'abort'?
problem and it can be understood as correct by anyone that understands
the halting problem at a much deeper level than rote memorization.
The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state.
A copy of Linz H is embedded at Ĥ.qx.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to
⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt). >>> You are defining POOP [Richard Damon]
André had recommended many online sites for you to learn or test, I forget which posts it is.
But I think C program is more simpler.
Halting problem undecidability and infinitely nested simulation (V3)
https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
André had recommended many online sites for you to learn or test, I forget which posts it is.
Type it into a TM simulator and prove your claim, your words are meaningless.
On Wednesday, 2 February 2022 at 06:19:04 UTC+8, olcott wrote:
On 2/1/2022 4:12 PM, wij wrote:
On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:I have already proved that I know one key fact about halt deciders that
On 2/1/2022 3:23 PM, wij wrote:
On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:I have shown how my system directly applies to the actual halting
On 2/1/2022 10:33 AM, wij wrote:
On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine
On 1/31/2022 11:25 PM, Richard Damon wrote:
The halting problem is vague on the definition of halting, it includes >>>>>>>> that a machine has stopped running and that a machine cannot reach its >>>>>>>> final state. My definition only includes the latter.
On 1/31/22 11:42 PM, olcott wrote:
On 1/31/2022 10:33 PM, Richard Damon wrote:
On 1/31/22 11:24 PM, olcott wrote:
On 1/31/2022 10:17 PM, Richard Damon wrote:
On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" Sum(3,5) is >>>>>>>>>>>> being asked about Sum(7,8).
On 1/31/2022 6:41 PM, Richard Damon wrote:Don't know how you get that from what I said.
On 1/31/22 3:24 PM, olcott wrote:
On 1/31/2022 2:10 PM, Ben wrote:
On 1/31/2022 8:06 AM, olcott wrote:
On 1/30/2022 8:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 1/30/22 9:05 PM, olcott wrote:
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
These statements need the conditions, that H^ goes to >>>>>>>>>>>>>>>>>>>>> H^.Qy/H^.Qn iff H goes to that corresponding state. >>>>>>>>>>>>>>>>>>>>>
⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to embedded_H
in the same way that (5,3) is syntactically specified as an
input to Sum(5,3)
Right, and the
Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to >>>>>>>>>>>>>>>>>>>> embedded_H in the same way that (1,2) is NOT syntactically >>>>>>>>>>>>>>>>>>>> specified as an input to Sum(5,3)
Right, but perhaps you don't understand that from you above >>>>>>>>>>>>>>>>>>> statement the right answer is based on if UTM(<H^>,<H^>) >>>>>>>>>>>>>>>>>>> Halts which by the definition of a UTM means if H^ applied to
<H^> Halts.
The biggest reason for your huge mistakes is that you cannot >>>>>>>>>>>>>>>>>> stay sharply focused on a single point. It is as if you either
have attention deficit disorder ADD or are addicted to >>>>>>>>>>>>>>>>>> methamphetamine.
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to embedded_H and
Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H. >>>>>>>>>>>>>>>>>>
After we have mutual agreement on this point we will move on >>>>>>>>>>>>>>>>>> to the points that logically follow from this one. >>>>>>>>>>>>>>>>>>
Holy shit try to post something that makes sense. >>>>>>>>>>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>>>>>>>>
Richard does not accept that the input to the copy of Linz H >>>>>>>>>>>>>>>> embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it is Ĥ ⟨Ĥ⟩.
No, but apparently you can't understand actual English words. >>>>>>>>>>>>>>>
The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H must >>>>>>>>>>>>>>> give is based on the behavior of H^ applied to <H^> BECAUSE OF >>>>>>>>>>>>>>> THE DEFINITION of H.
In other words Sum(3,5) must return the value of Sum(7,8)? >>>>>>>>>>>>>
Any moron knows that a function is only accountable for its actual
inputs.
And the actual input to H is <H^> <H^> which MEANS by the >>>>>>>>>>>>> DEFINITION of the Halting Problem that H is being asked to decide >>>>>>>>>>>>> on the Halting Status of H^ applied to <H^>
Again your RED HERRING.
H is being asked EXACTLY what it being asked
H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't >>>>>>>>>>>
AGREED?
No that is wrong. embedded_H is being asked:
Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to ⟨Ĥ⟩.qn ?
If you say 'No', then you aren't doing the halting problem, as the >>>>>>>>> requirement I stated is EXACTLY the requirement of the Halting Problem.
Sounds like a NDTM.
It is not a NDTM, a Turing Machine only actually halts when it reaches >>>>>> its own final state. People not very familiar with this material may get >>>>>> confused and believe that a TM halts when its stops running because its >>>>>> simulation has been aborted. This key distinction is not typically >>>>>> specified in most halting problem proofs.
computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)
Where did Linz mention 'simulation' and 'abort'?
problem and it can be understood as correct by anyone that understands >>>> the halting problem at a much deeper level than rote memorization.
The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state. >>>> A copy of Linz H is embedded at Ĥ.qx.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to
⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt). >>>>> You are defining POOP [Richard Damon]
André had recommended many online sites for you to learn or test, I forget which posts it is.
But I think C program is more simpler.
Halting problem undecidability and infinitely nested simulation (V3) >>>>>>
https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
André had recommended many online sites for you to learn or test, I forget which posts it is.
Type it into a TM simulator and prove your claim, your words are meaningless.
no one else here seems to know.
No one here understands that because a halt decider is a decider that it
must compute the mapping from its inputs to an accept of reject state on
the basis of the actual behavior specified by these inputs.
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
There is no 'actual TM' until you it into a TM simulator,
otherwise all empty talks.
(I would expect to see you 'reinterpret' again)
On 2/1/22 5:18 PM, olcott wrote:
On 2/1/2022 4:12 PM, wij wrote:
On Wednesday, 2 February 2022 at 05:36:39 UTC+8, olcott wrote:
On 2/1/2022 3:23 PM, wij wrote:
On Wednesday, 2 February 2022 at 02:37:17 UTC+8, olcott wrote:I have shown how my system directly applies to the actual halting
On 2/1/2022 10:33 AM, wij wrote:
On Tuesday, 1 February 2022 at 23:22:32 UTC+8, olcott wrote:https://en.wikipedia.org/wiki/Nondeterministic_Turing_machine
On 1/31/2022 11:25 PM, Richard Damon wrote:
The halting problem is vague on the definition of halting, it
On 1/31/22 11:42 PM, olcott wrote:
On 1/31/2022 10:33 PM, Richard Damon wrote:
On 1/31/22 11:24 PM, olcott wrote:
On 1/31/2022 10:17 PM, Richard Damon wrote:
On 1/31/22 10:40 PM, olcott wrote:No that is not it. That is like saying "by definition" >>>>>>>>>>>> Sum(3,5) is
On 1/31/2022 6:41 PM, Richard Damon wrote:Don't know how you get that from what I said.
On 1/31/22 3:24 PM, olcott wrote:
On 1/31/2022 2:10 PM, Ben wrote:
On 1/31/2022 8:06 AM, olcott wrote:
On 1/30/2022 8:20 PM, Richard Damon wrote: >>>>>>>>>>>>>>>>>>> On 1/30/22 9:05 PM, olcott wrote:
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
These statements need the conditions, that H^ goes to >>>>>>>>>>>>>>>>>>>>> H^.Qy/H^.Qn iff H goes to that corresponding state. >>>>>>>>>>>>>>>>>>>>>
⟨Ĥ⟩ ⟨Ĥ⟩ is syntactically specified as an input to
embedded_H
in the same way that (5,3) is syntactically >>>>>>>>>>>>>>>>>>>> specified as an
input to Sum(5,3)
Right, and the
Ĥ ⟨Ĥ⟩ is NOT syntactically specified as an input to >>>>>>>>>>>>>>>>>>>> embedded_H in the same way that (1,2) is NOT >>>>>>>>>>>>>>>>>>>> syntactically
specified as an input to Sum(5,3)
Right, but perhaps you don't understand that from you >>>>>>>>>>>>>>>>>>> above
statement the right answer is based on if UTM(<H^>,<H^>) >>>>>>>>>>>>>>>>>>> Halts which by the definition of a UTM means if H^ >>>>>>>>>>>>>>>>>>> applied to
<H^> Halts.
The biggest reason for your huge mistakes is that you >>>>>>>>>>>>>>>>>> cannot
stay sharply focused on a single point. It is as if >>>>>>>>>>>>>>>>>> you either
have attention deficit disorder ADD or are addicted to >>>>>>>>>>>>>>>>>> methamphetamine.
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
The single point is that ⟨Ĥ⟩ ⟨Ĥ⟩ is the input to >>>>>>>>>>>>>>>>>> embedded_H and
Ĥ ⟨Ĥ⟩ is the NOT the input to embedded_H. >>>>>>>>>>>>>>>>>>
After we have mutual agreement on this point we will >>>>>>>>>>>>>>>>>> move on
to the points that logically follow from this one. >>>>>>>>>>>>>>>>>>
Holy shit try to post something that makes sense. >>>>>>>>>>>>>>>>>
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ >>>>>>>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>>>>>>>>>>>>>>
Richard does not accept that the input to the copy of >>>>>>>>>>>>>>>> Linz H
embedded at Ĥ.qx is ⟨Ĥ⟩ ⟨Ĥ⟩. He keeps insisting that it
is Ĥ ⟨Ĥ⟩.
No, but apparently you can't understand actual English >>>>>>>>>>>>>>> words.
The INPUT to H is <H^> <H^> but the CORRECT ANSWER that H >>>>>>>>>>>>>>> must
give is based on the behavior of H^ applied to <H^> >>>>>>>>>>>>>>> BECAUSE OF
THE DEFINITION of H.
In other words Sum(3,5) must return the value of Sum(7,8)? >>>>>>>>>>>>>
Any moron knows that a function is only accountable for >>>>>>>>>>>>>> its actual
inputs.
And the actual input to H is <H^> <H^> which MEANS by the >>>>>>>>>>>>> DEFINITION of the Halting Problem that H is being asked to >>>>>>>>>>>>> decide
on the Halting Status of H^ applied to <H^>
being asked about Sum(7,8).
Again your RED HERRING.
H is being asked EXACTLY what it being asked
H wM w -> H.Qy if M applied to w Halts, and H.Qn if it doesn't >>>>>>>>>>>
AGREED?
No that is wrong. embedded_H is being asked:
Can the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ possibly transition to
⟨Ĥ⟩.qn ?
If you say 'No', then you aren't doing the halting problem, as the >>>>>>>>> requirement I stated is EXACTLY the requirement of the Halting >>>>>>>>> Problem.
includes
that a machine has stopped running and that a machine cannot
reach its
final state. My definition only includes the latter.
Sounds like a NDTM.
It is not a NDTM, a Turing Machine only actually halts when it
reaches
its own final state. People not very familiar with this material
may get
confused and believe that a TM halts when its stops running
because its
simulation has been aborted. This key distinction is not typically >>>>>> specified in most halting problem proofs.
computation that halts … the Turing machine will halt whenever it >>>>>> enters
a final state. (Linz:1990:234)
Where did Linz mention 'simulation' and 'abort'?
problem and it can be understood as correct by anyone that understands >>>> the halting problem at a much deeper level than rote memorization.
The following simplifies the syntax for the definition of the Linz
Turing machine Ĥ, it is now a single machine with a single start state. >>>> A copy of Linz H is embedded at Ĥ.qx.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
Can ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by embedded_H possibly transition to
⟨Ĥ⟩.qn ? (No means that ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt). >>>>> You are defining POOP [Richard Damon]
André had recommended many online sites for you to learn or test, I >>>>> forget which posts it is.
But I think C program is more simpler.
Halting problem undecidability and infinitely nested simulation (V3) >>>>>>
https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer
André had recommended many online sites for you to learn or test, I
forget which posts it is.
Type it into a TM simulator and prove your claim, your words are
meaningless.
I have already proved that I know one key fact about halt deciders
that no one else here seems to know.
No one here understands that because a halt decider is a decider that
it must compute the mapping from its inputs to an accept of reject
state on the basis of the actual behavior specified by these inputs.
And the ACTUAL BEHAVIOR of the input <H^> <H^> is EXACTLY the behavior
of H^ applied to <H^> which does Halt if H goes to H.Qn.
| Sysop: | Keyop |
|---|---|
| Location: | Huddersfield, West Yorkshire, UK |
| Users: | 297 |
| Nodes: | 16 (0 / 16) |
| Uptime: | 121:08:31 |
| Calls: | 6,662 |
| Files: | 12,212 |
| Messages: | 5,334,487 |