*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
*This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation*
*This is a verified fact*
When simulating halt deciders always report on the behavior of
their simulated input from their own POV then when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.Hqn it is correct from its own POV.
*This is a verified fact*
When that occurs then H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to H.qy from
its own POV.
When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ report on the basis of their own
POV then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ reports incorrectly about the behavior of
Ĥ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ reports the behavior of Ĥ ⟨Ĥ⟩ correctly.
*Verified facts*
(a) It is a verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the simulation of its input to prevent its own infinite execution.
(b) It is a verified fact when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see that it
must abort its simulation then it would transition to Ĥ.Hqn
to reject this input as non-halting from its own POV.
On 3/9/2024 7:24 PM, immibis wrote:
On 10/03/24 01:30, olcott wrote:
On 3/9/2024 6:24 PM, immibis wrote:
On 10/03/24 01:22, olcott wrote:
On 3/9/2024 5:57 PM, immibis wrote:
On 10/03/24 00:26, olcott wrote:
On 3/9/2024 5:10 PM, immibis wrote:
On 9/03/24 23:22, olcott wrote:
On 3/9/2024 3:50 PM, immibis wrote:
On 9/03/24 22:34, olcott wrote:
What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
wrong answer to provide?
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that
H ⟨Ĥ⟩ uses.
Simulating halt deciders must make sure that they themselves >>>>>>>>> do not get stuck in infinite execution. This means that they >>>>>>>>> must abort every simulation that cannot possibly otherwise halt. >>>>>>>>>
This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that
H ⟨Ĥ⟩ uses.
*Only because Ĥ.H is embedded within Ĥ and H is not*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation. >>>>>>>
You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact
same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
The above is true no matter what criteria that is used
as long as H is a simulating halt decider.
Objective criteria cannot vary based on who the subject is. They are
objective. The answer to different people is the same answer if the
criteria are objective.
It is objectively true that Ĥ.H can get stuck in recursive
simulation because Ĥ copies its input thus never runs
out of params.
It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params.
Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do
better. Write the Olcott machine (not x86utm) code for Ĥ and I would
show you.
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
On 3/9/2024 3:49 PM, immibis wrote:
On 9/03/24 22:30, olcott wrote:
On 3/9/2024 3:17 PM, Richard Damon wrote:
On 3/9/24 10:33 AM, olcott wrote:Yes. (Notice that I am agreeing with you, yet never do that with me)
*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Specifications, not actual behavior until the existance of such an H is shown.
IF taken as actual behavior, then it is conditional on such an H existing. >>>>
Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process >>>>> *This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation*
It NEEDS to in order to meet its specification
It DOESN'T unless its algorithm says it does,Yes. (Notice that I am agreeing with you, yet never do that with me)
If it just fails to answer, then it has failed to be a correct Halt Decider.Yes. (Notice that I am agreeing with you, yet never do that with me)
The fact that you reach this conflict in actions, is the reason Halt
Deciding is uncomputable.
*No. We know that Ĥ ⟨Ĥ⟩ either halts or fails to halt*
If it halts then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn.
If Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn then H ⟨Ĥ⟩ transitioned to H.qn or
else Ĥ is the wrong Ĥ or you can't read instructions.
I generally agree that a pair of identical machines
must have the same behavior on the same input.
This may not apply when these machines having identical
states and identical inputs:
On 3/10/2024 8:33 AM, Mikko wrote:
On 2024-03-10 01:29:39 +0000, olcott said:
On 3/9/2024 7:24 PM, immibis wrote:
On 10/03/24 01:30, olcott wrote:
On 3/9/2024 6:24 PM, immibis wrote:
On 10/03/24 01:22, olcott wrote:
On 3/9/2024 5:57 PM, immibis wrote:
On 10/03/24 00:26, olcott wrote:
On 3/9/2024 5:10 PM, immibis wrote:
On 9/03/24 23:22, olcott wrote:
On 3/9/2024 3:50 PM, immibis wrote:
On 9/03/24 22:34, olcott wrote:
What criteria would you use so that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ knows what
wrong answer to provide?
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that
H ⟨Ĥ⟩ uses.
Simulating halt deciders must make sure that they themselves >>>>>>>>>>> do not get stuck in infinite execution. This means that they >>>>>>>>>>> must abort every simulation that cannot possibly otherwise halt. >>>>>>>>>>>
This requires Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation and does not
require H ⟨Ĥ⟩ ⟨Ĥ⟩ to abort its simulation when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ aborts
its simulation.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does simulate itself in recursive simulation H ⟨Ĥ⟩ ⟨Ĥ⟩
does not simulate itself in recursive simulation.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact same objective criteria that
H ⟨Ĥ⟩ uses.
*Only because Ĥ.H is embedded within Ĥ and H is not*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
You dishonestly ignored that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is stipulated to use the exact
same OBJECTIVE criteria that H ⟨Ĥ⟩ uses.
The above is true no matter what criteria that is used
as long as H is a simulating halt decider.
Objective criteria cannot vary based on who the subject is. They are >>>>>> objective. The answer to different people is the same answer if the >>>>>> criteria are objective.
It is objectively true that Ĥ.H can get stuck in recursive
simulation because Ĥ copies its input thus never runs
out of params.
It is objectively true that Ĥ cannot possibly get stuck
in recursive because H does not copy its input thus runs
out of params.
Wrong. Dead wrong. Stupidly wrong. So wrong that a dead monkey could do >>>> better. Write the Olcott machine (not x86utm) code for Ĥ and I would
show you.
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
*In other words you are denying these verified facts*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
"Ĥ.q0 ⟨Ĥ⟩ does not halt // Ĥ applied to ⟨Ĥ⟩ halts" is not a
verified fact.
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
"Ĥ.q0 ⟨Ĥ⟩ halts // Ĥ applied to ⟨Ĥ⟩ does not halt" is not a
verified fact.
Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
The earliest point when Turing machine Ĥ can detect the repeating
state of its input is when Ĥ reaches (c) a second time where its
input would begin simulating a copy of itself with a copy of its
input. It could detect this one execution trace earlier
[ when its input first reaches (c) ] if Ĥ was an Olcott machine.
On 3/10/2024 8:21 AM, Mikko wrote:
On 2024-03-09 22:15:20 +0000, olcott said:
On 3/9/2024 3:49 PM, immibis wrote:
On 9/03/24 22:30, olcott wrote:
On 3/9/2024 3:17 PM, Richard Damon wrote:
On 3/9/24 10:33 AM, olcott wrote:Yes. (Notice that I am agreeing with you, yet never do that with me) >>>>>
*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
Specifications, not actual behavior until the existance of such an H is shown.
IF taken as actual behavior, then it is conditional on such an H existing.
It NEEDS to in order to meet its specification
Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process >>>>>>> *This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation* >>>>>>
It DOESN'T unless its algorithm says it does,Yes. (Notice that I am agreeing with you, yet never do that with me) >>>>>
If it just fails to answer, then it has failed to be a correct Halt Decider.Yes. (Notice that I am agreeing with you, yet never do that with me) >>>>>
The fact that you reach this conflict in actions, is the reason Halt >>>>>> Deciding is uncomputable.
*No. We know that Ĥ ⟨Ĥ⟩ either halts or fails to halt*
If it halts then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn.
If Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn then H ⟨Ĥ⟩ transitioned to H.qn or
else Ĥ is the wrong Ĥ or you can't read instructions.
I generally agree that a pair of identical machines
must have the same behavior on the same input.
If two machines do not behave the same on the same input
they are not identical because that is what the word
means.
This may not apply when these machines having identical
states and identical inputs:
Meanings of the words do apply.
Because H ⟨Ĥ⟩ ⟨Ĥ⟩ is one execution trace ahead of Ĥ ⟨Ĥ⟩ it terminates it simulation one execution trace sooner than Ĥ ⟨Ĥ⟩ does.
On 3/10/2024 8:13 AM, Mikko wrote:
On 2024-03-09 18:33:50 +0000, olcott said:
*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn // Ĥ applied to ⟨Ĥ⟩ does not halt
In order to say something you need a main clause. Above does
not conform to the syntax of then sentences of Common Language.
Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
*This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation*
No, it does'n. It merely proves that Ĥ either does not halt
or aborts its simulation at some point.
*This is a verified fact*
When simulating halt deciders always report on the behavior of
their simulated input from their own POV then when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ >>> transitions to Ĥ.Hqn it is correct from its own POV.
If there is no simulationg halt deider then that is true but
hardly a "fact".
*This is a verified fact*
When that occurs then H ⟨Ĥ⟩ ⟨Ĥ⟩ would transition to H.qy from
its own POV.
No, it isn't. No statement of fact contains the word "would".
A clause containing the word "would" is counter-factual because
that is what "would" means.
When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ report on the basis of their own
POV then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ reports incorrectly about the behavior of >>> Ĥ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ reports the behavior of Ĥ ⟨Ĥ⟩ correctly.
Neither claim is proven.
*Verified facts*
(a) It is a verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort the
simulation of its input to prevent its own infinite execution.
A statement of fact cannot contain the word "must". Either
the fact is that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does abort the simulation
or the fact is that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ does not halt.
(b) It is a verified fact when Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can see that it
must abort its simulation then it would transition to Ĥ.Hqn
to reject this input as non-halting from its own POV.
A statement of fact cannot containt the word "must". The
word "must" means that the containing sentence is a requirement,
not a a fact.
If is a verified fact/ self-evident truth/ semantic tautology
that when you weigh 300 pounds that your weigh more than 200 pounds
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