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Re: Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn is correct and forms no contradiction.
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olcott
Aug 11, 2021, 10:03:19 PM8/11/21
to
On 8/11/2021 8:20 PM, Ben Bacarisse wrote:
> olcott <No...@NoWhere.com> writes:
>
>> On 8/11/2021 6:04 PM, Ben Bacarisse wrote:
>>> olcott <No...@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>>>>> olcott <No...@NoWhere.com> writes:
>>>
>>>>>> I went point by point. If I am actually incorrect then you can go
>>>>>> point by point and point out each individual error step by step.
>>>>> You fail at the first hurdle. You can't hope to persuade anyone that an
>>>>> add function with add(2, 3) == 9 is operating as specified simply by
>>>>> detailing, step by step, exactly how you implement the wrong behaviour.
>>>>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>>>>> Your add function is entirely correct in that is does exactly what you
>>>>> intend it to do. As far as I am concerned there is no significant error
>>>>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>>>>> code should add numbers and not do whatever it is your code does with
>>>>> 100% correctness. Do you follow?
>>>>>
>>>>>> Of course
>>>>>> everyone knows that this is impossible if I am totally correct.
>>>>> Then stop wasting time and try to publish! You'll need lots of time to
>>>>> explain away why every editor simply laughs at the paper.
>>>>>
>>>>>> H.q0 WM w ⊢* Ĥ.qn
>>>>>> becomes
>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>>>>> specification (for this case -- you don't claim to have a halt decider).
>>>>>
>>>>>> Can you admit when you are wrong when you really are wrong?
>>>>> Yes. I am wrong all the time. In this case I'm having trouble working
>>>>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>>>>> removing the key text from Linz's explanations it might sink in?
>>>>>
>>>>
>>>> Pages of the Linz text to verify the above quotes in their full context:
>>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> I see you are in "paste the same text" again mode. If you think I can
>>> help in any way, do let me know.
>>
>> I just noticed that you acknowledged that M refers to the TM
>> represented by the first input parameter to Ĥ.qx wM wM
>
> Did I? Oh dear. It's garbage. Where did I "acknowledge" it? I'd like
> to go back and point out that it's mathematical junk.
>
>> so we can move to the next point:
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and
>
> This is just a metaphorical math poem. You've taken a mathematical
> statement and substituted for only some on the occurrences of a
> variable. That is a schoolboy error.
>
That seems to be a very stupid thing to say when it only clarifies and
corrects Linz.
> Maybe you confused agreement with "can't be bothered to go of on a
> tangent about a math poem". I certainly let many meaningless things you
> write slip by unremarked upon. There are not enough hours in the day to
> point out all your mistakes.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn // see highlighted portion of Linz text
>> to confirm:
>> if M applied to wM does not halt // M refers to the TM of the first wM
>> parameter to Ĥ.qx
>
> Linz requires that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.
>
Sure and the Ĥ being referred to is the one that the first param to Ĥ.qx
⟨Ĥ⟩ ⟨Ĥ⟩ specifies.
> Unlike your garbage, this makes sense because all occurrences of M and
> the derived wM have now been substituted for actual values -- your Ĥ and
> its encoding.
>
>> Here is the overview of what I am claiming:
>> Turing machine Ĥ is applied to its input ⟨Ĥ⟩.
>
> Ĥ.q0 ⟨Ĥ⟩
>
>> It copies this input such that this input and the copy of this input
>> become the first and second parameters to the simulating halt decider
>> at Ĥ.qx.
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ decides that the simulation of its first parameter
>> on the input of its second parameter never halt it correctly
>> transitions to its own final state of Ĥ.qn.
>
> ⊢* Ĥ.qn
>
> None of which (other than your poor use of technical terms) has been in
> doubt for months.
>
You sure acted like M only referred to the first Ĥ shown below:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.
>> It is easy to see that when the halt decider at Ĥ.qx is a simulating
>> halt decider that there is an infinite cycle from Ĥ.qx to Ĥ.q0 that
>> prevents the input from every reaching its final state.
>
> If Ĥ were not written as you have chosen to write it -- if it were
> instead a pure simulator, then
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* ∞
>
This: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
specifies an infinite cycle from Ĥ.qx to Ĥ.q0 while the simulating halt
decider at Ĥ.qx acts as a UTM.
> would be the case. Mathematicians would avoid this messy language and
> the possibility of confusion by not re-using the name Ĥ for both your
> partial decider and one based on pure simulation.
>
Since it is the very same machine in two different modes of operation
this seems really stupid.
This <is> the same Ĥ that Linz specifies with the extra detail of how
this wildcard state transition works: ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Also, you are wrong about the cycles of states, but that's because you
> have no clue how a TM would simulate another TM.
No one ever investigates this because their paper on the subject would
be millions of pages long if it included a full listing of the source
code. If we add the single feature of random access memory TMs would be
100,000-fold less tedious.
> It's a rabbit hole we
> shouldn't go down, because (a) I don't think you could ever understand
> how a UTM really works, and (b) it's irrelevant since I agree that there
> would be infinite execution if Ĥ were not as it is but were, instead, a
> pure simulator.
>
Great. This is a big breakthrough.
>> When we define halting as reaching a final state
>
> Ĥ applied to ⟨Ĥ⟩ halts. The computation reaches the final state Ĥ.qn.
>
Yes. The halting problem is not about programs that halt. It is about
inputs to a halt decider. The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.
>> so that we can unequivocally divide computations that complete from
>> those that are aborted,
>
> So this is today's equivocation. TM computations either halt or they
> don't halt. None of them "abort".
>
I am concurrently perfecting my understanding of the technical terms via
Linz. A TM can also be said to halt if there is no transition out of the
current state. Ultimately I only need a clear and unequivocal way to
divide computations that reach their natural conclusion from ones that
have had their simulation aborted.
>> then we know that the input to Ĥ.qx never
>> halts, thus making its halt status decison correct.
>
> Garbled. Inputs don't halt or not halt. Your Ĥ does not meet Linz's
> spec because Ĥ applied to ⟨Ĥ⟩ halts when it should not. You don't get
> to say it does because it halts in what you think of as special way.
>
That is like saying because we know the liar paradox: "this sentence is
not true" is not true then it must be true.
My purpose of creating the x86utm operating system was so that every
single detail of the halting problem counter-examples can be completely
examined.
It is perfectly analogous in every way to the Linz Ĥ.
We can see that int main() { P(P); } halts and that the input to H(P,P)
cannot possibly halt. That is a lot like finding a cat that barks.
No one besides me has ever bothered to examine every single detail of
H(P,P) to see that its input really cannot possibly halt.
They are all so egotistically caught up in their own opinion that the
fact that int main() { P(P); } halts proves that H(P,P)==0 must be
incorrect.
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
> olcott <No...@NoWhere.com> writes:
>
>> On 8/11/2021 6:04 PM, Ben Bacarisse wrote:
>>> olcott <No...@NoWhere.com> writes:
>>>
>>>> On 8/11/2021 10:04 AM, Ben Bacarisse wrote:
>>>>> olcott <No...@NoWhere.com> writes:
>>>
>>>>>> I went point by point. If I am actually incorrect then you can go
>>>>>> point by point and point out each individual error step by step.
>>>>> You fail at the first hurdle. You can't hope to persuade anyone that an
>>>>> add function with add(2, 3) == 9 is operating as specified simply by
>>>>> detailing, step by step, exactly how you implement the wrong behaviour.
>>>>> In your case, it's simply that Ĥ.q0 ⟨Ĥ⟩ transitions to Ĥ.qn (via Ĥ.qx
>>>>> ⟨Ĥ⟩ ⟨Ĥ⟩ as you pointlessly keep insisting) when Linz says it should not.
>>>>> Your add function is entirely correct in that is does exactly what you
>>>>> intend it to do. As far as I am concerned there is no significant error
>>>>> in how you arrive at add(2, 3) == 9. The problem is that Linz says your
>>>>> code should add numbers and not do whatever it is your code does with
>>>>> 100% correctness. Do you follow?
>>>>>
>>>>>> Of course
>>>>>> everyone knows that this is impossible if I am totally correct.
>>>>> Then stop wasting time and try to publish! You'll need lots of time to
>>>>> explain away why every editor simply laughs at the paper.
>>>>>
>>>>>> H.q0 WM w ⊢* Ĥ.qn
>>>>>> becomes
>>>>>> H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
>>>>> Yes, we all know that. It's exactly why your Ĥ does not meet Linz's
>>>>> specification (for this case -- you don't claim to have a halt decider).
>>>>>
>>>>>> Can you admit when you are wrong when you really are wrong?
>>>>> Yes. I am wrong all the time. In this case I'm having trouble working
>>>>> out how I could be clearer about your Ĥ. Maybe if you didn't keep
>>>>> removing the key text from Linz's explanations it might sink in?
>>>>>
>>>>
>>>> Pages of the Linz text to verify the above quotes in their full context:
>>>> http://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>>
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>> I see you are in "paste the same text" again mode. If you think I can
>>> help in any way, do let me know.
>>
>> I just noticed that you acknowledged that M refers to the TM
>> represented by the first input parameter to Ĥ.qx wM wM
>
> Did I? Oh dear. It's garbage. Where did I "acknowledge" it? I'd like
> to go back and point out that it's mathematical junk.
>
>> so we can move to the next point:
>>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>> if M applied to wM halts, and
>
> This is just a metaphorical math poem. You've taken a mathematical
> statement and substituted for only some on the occurrences of a
> variable. That is a schoolboy error.
>
That seems to be a very stupid thing to say when it only clarifies and
corrects Linz.
> Maybe you confused agreement with "can't be bothered to go of on a
> tangent about a math poem". I certainly let many meaningless things you
> write slip by unremarked upon. There are not enough hours in the day to
> point out all your mistakes.
>
>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn // see highlighted portion of Linz text
>> to confirm:
>> if M applied to wM does not halt // M refers to the TM of the first wM
>> parameter to Ĥ.qx
>
> Linz requires that
>
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.
>
Sure and the Ĥ being referred to is the one that the first param to Ĥ.qx
⟨Ĥ⟩ ⟨Ĥ⟩ specifies.
> Unlike your garbage, this makes sense because all occurrences of M and
> the derived wM have now been substituted for actual values -- your Ĥ and
> its encoding.
>
>> Here is the overview of what I am claiming:
>> Turing machine Ĥ is applied to its input ⟨Ĥ⟩.
>
> Ĥ.q0 ⟨Ĥ⟩
>
>> It copies this input such that this input and the copy of this input
>> become the first and second parameters to the simulating halt decider
>> at Ĥ.qx.
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩
>
>> When Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ decides that the simulation of its first parameter
>> on the input of its second parameter never halt it correctly
>> transitions to its own final state of Ĥ.qn.
>
> ⊢* Ĥ.qn
>
> None of which (other than your poor use of technical terms) has been in
> doubt for months.
>
You sure acted like M only referred to the first Ĥ shown below:
> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> if, and only if, Ĥ applied to ⟨Ĥ⟩ does not halt.
>> It is easy to see that when the halt decider at Ĥ.qx is a simulating
>> halt decider that there is an infinite cycle from Ĥ.qx to Ĥ.q0 that
>> prevents the input from every reaching its final state.
>
> If Ĥ were not written as you have chosen to write it -- if it were
> instead a pure simulator, then
>
> ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* ∞
>
This: Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
specifies an infinite cycle from Ĥ.qx to Ĥ.q0 while the simulating halt
decider at Ĥ.qx acts as a UTM.
> would be the case. Mathematicians would avoid this messy language and
> the possibility of confusion by not re-using the name Ĥ for both your
> partial decider and one based on pure simulation.
>
Since it is the very same machine in two different modes of operation
this seems really stupid.
This <is> the same Ĥ that Linz specifies with the extra detail of how
this wildcard state transition works: ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
> Also, you are wrong about the cycles of states, but that's because you
> have no clue how a TM would simulate another TM.
No one ever investigates this because their paper on the subject would
be millions of pages long if it included a full listing of the source
code. If we add the single feature of random access memory TMs would be
100,000-fold less tedious.
> It's a rabbit hole we
> shouldn't go down, because (a) I don't think you could ever understand
> how a UTM really works, and (b) it's irrelevant since I agree that there
> would be infinite execution if Ĥ were not as it is but were, instead, a
> pure simulator.
>
Great. This is a big breakthrough.
>> When we define halting as reaching a final state
>
> Ĥ applied to ⟨Ĥ⟩ halts. The computation reaches the final state Ĥ.qn.
>
Yes. The halting problem is not about programs that halt. It is about
inputs to a halt decider. The input to Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ never halts.
>> so that we can unequivocally divide computations that complete from
>> those that are aborted,
>
> So this is today's equivocation. TM computations either halt or they
> don't halt. None of them "abort".
>
I am concurrently perfecting my understanding of the technical terms via
Linz. A TM can also be said to halt if there is no transition out of the
current state. Ultimately I only need a clear and unequivocal way to
divide computations that reach their natural conclusion from ones that
have had their simulation aborted.
>> then we know that the input to Ĥ.qx never
>> halts, thus making its halt status decison correct.
>
> Garbled. Inputs don't halt or not halt. Your Ĥ does not meet Linz's
> spec because Ĥ applied to ⟨Ĥ⟩ halts when it should not. You don't get
> to say it does because it halts in what you think of as special way.
>
That is like saying because we know the liar paradox: "this sentence is
not true" is not true then it must be true.
My purpose of creating the x86utm operating system was so that every
single detail of the halting problem counter-examples can be completely
examined.
It is perfectly analogous in every way to the Linz Ĥ.
We can see that int main() { P(P); } halts and that the input to H(P,P)
cannot possibly halt. That is a lot like finding a cat that barks.
No one besides me has ever bothered to examine every single detail of
H(P,P) to see that its input really cannot possibly halt.
They are all so egotistically caught up in their own opinion that the
fact that int main() { P(P); } halts proves that H(P,P)==0 must be
incorrect.
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
olcott
Aug 13, 2021, 9:30:42 AM8/13/21
to
On 8/13/2021 3:58 AM, Malcolm McLean wrote:
> On Thursday, 12 August 2021 at 16:50:19 UTC+1, olcott wrote:
>> On 8/12/2021 10:40 AM, Malcolm McLean wrote:
>>> On Thursday, 12 August 2021 at 13:24:46 UTC+1, olcott wrote:
>>>> On 8/12/2021 3:36 AM, Malcolm McLean wrote:
>>>>> On Thursday, 12 August 2021 at 06:11:30 UTC+1, Jeff Barnett wrote:
>>>>>> On 8/11/2021 5:40 PM, olcott wrote:
>>>>>>> On 8/11/2021 6:32 PM, Jeff Barnett wrote:
>>>>>>>> syndrome? That along with all of his other deficiencies will surely do
>>>>>>>> him in. Perhaps we can invent shortcuts a la LaTeX macros and font
>>>>>>>> switches to help him cut down the strain. He wont be insulted and will
>>>>>>>> take to it as a pig to mud. Just think, he'll have yet another
>>>>>>>> notation to misuse and abuse.
>>>>>>>
>>>>>>> Ben finally acknowledged the point that I was making:
>>>>>>> next point until the current point is accepted as correct.
>>>>>> In addition to carpal tunnel, you are delusional. I get the vague
>>>>>> impression reading the above that you are having a schizoid episode.
>>>>>> There is a complete and total discount between what others say to you
>>>>>> and your replies. I suggest you cut out some of the examples you like
>>>>>> best and show them to your therapist. You do have one, don't you? It may
>>>>>> lead to better, more effective treatment. We are all hoping you get
>>>>>> better but schizoid episodes are hard to suppress. I assume meds have
>>>>>> been prescribed for your condition. Are you taking them? Or are you
>>>>>> refusing medical help just as you refuse to allow Ben to heal your
>>>>>> ignorance. A little trust in your betters will go a long way towards
>>>>>> recovery and a cure. Start now before it's to late. God speed to you.
>>>>>> -
>>>>> YOu might be right, but I think it's just frustration. His execution trace shows
>>>>> what appears to be an infinitely recursive process aborted, and everyone is
>>>>> just ignoring that and focusing on the other reality that his H does not
>>>>> return the result that matches the behaviour of the input when run independently.
>>>>>
>>>>> The attempt to use formal math notation when he doesn't understand it
>>>>> doesn't help. That can create the impression of mental incoherence, but in fact
>>>>> it's not that.
>>>>>
>>>>> Despite a mass of posting, we still don't have a good idea how H works.
>>>> _P()
>>>> [00000d02](01) 55 push ebp
>>>> [00000d03](02) 8bec mov ebp,esp
>>>> [00000d05](03) 8b4508 mov eax,[ebp+08]
>>>> [00000d08](01) 50 push eax // push 2nd Param
>>>> [00000d09](03) 8b4d08 mov ecx,[ebp+08]
>>>> [00000d0c](01) 51 push ecx // push 1st Param
>>>> [00000d0d](05) e870feffff call 00000b82 // call H
>>>> [00000d12](03) 83c408 add esp,+08
>>>> [00000d15](02) 85c0 test eax,eax
>>>> [00000d17](02) 7402 jz 00000d1b
>>>> [00000d19](02) ebfe jmp 00000d19
>>>> [00000d1b](01) 5d pop ebp
>>>> [00000d1c](01) c3 ret
>>>> Size in bytes:(0027) [00000d1c]
>>>> machine stack stack machine assembly
>>>> address address data code language
>>>> ======== ======== ======== ========= =============
>>>> ...[00000d0d][00101829][00000d12] e870feffff call 00000b82 // call H
>>>>
>>>> Begin Local Halt Decider Simulation at Machine Address:d02
>>>> ...[00000d02][002118f1][002118f5] 55 push ebp
>>>> ...[00000d03][002118f1][002118f5] 8bec mov ebp,esp
>>>> ...[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
>>>> ...[00000d08][002118ed][00000d02] 50 push eax // push P
>>>> ...[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
>>>> ...[00000d0c][002118e9][00000d02] 51 push ecx // push P
>>>> ...[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
>>>>
>>>> ...[00000d02][0025c319][0025c31d] 55 push ebp
>>>> ...[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
>>>> ...[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
>>>> ...[00000d08][0025c315][00000d02] 50 push eax // push P
>>>> ...[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
>>>> ...[00000d0c][0025c311][00000d02] 51 push ecx // push P
>>>> ...[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
>>>> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>>>> If this means that you are telling me that:
>>>> (1) You know the x86 language very well.
>>>>
>>>> (2) Can't begin to understand why P would be stuck in infinitely nested
>>>> simulation while H acts as a pure simulator?
>>>>
>>> OK, so let's start.
>>> There's a line which says "Begin Local Halt Decider Simulation at Machine
>>> Address:d02". Is that part of the execution trace, or is it a commented added
>>> by hand? If it is part of the execution trace, why doesn't this line appear
>>> before the second line with the label 0d02?
>> The halt decider automatically writes this when it begins the simulation
>> that its halt analysis is based on.
>>
> OK, so why don't we get a second "Begin Local Halt Decider ..." line output?
Begin Local Halt Decider Simulation at Machine Address:d02
...[00000d02][002118f1][002118f5] 55 push ebp
...[00000d03][002118f1][002118f5] 8bec mov ebp,esp
...[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
...[00000d08][002118ed][00000d02] 50 push eax // push P
...[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
...[00000d0c][002118e9][00000d02] 51 push ecx // push P
...[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
...[00000d02][0025c319][0025c31d] 55 push ebp
...[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
...[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
...[00000d08][0025c315][00000d02] 50 push eax // push P
...[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
...[00000d0c][0025c311][00000d02] 51 push ecx // push P
...[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
The simulation of the input begins and continues until it ends.
>>>>
>>>> I would say that you must be a liar.
>>>>
>>> I have a PhD. PhD holder attach a very high value to honesty, in particular
>>> with respect to their subjects. But also more generally. I always draw attention
>>> if I'm given too much change at the pub, for example, not becaue I'm better
>>> than other people, but because I'm a doctor, and a doctor can't be seen to
>>> take money that doesn't belong to him.
>>>
>> So do (1) and (2) apply to you?
>>
> That was a demand for an apology, but you are too focused on your own
> interests to be sensitive to what other people are trying to communicate to you.
>
I still can't tell whether you are telling the truth or not because you
have not yet committed to the actual truth and rejected the falsehood.
If (1) and (2) do apply to you then you would be a liar.
The point that I am trying to make here is so damn simple that everyone
should have agreed to it in less than a week. It has been many months.
There is no freaking code in the body of P that can possibly escape
infinitely nested simulation while H remains a pure simulator. If you as
much as refrain from agreeing to that (and know the x86 language very
well) then you would be a liar.
> On Thursday, 12 August 2021 at 16:50:19 UTC+1, olcott wrote:
>> On 8/12/2021 10:40 AM, Malcolm McLean wrote:
>>> On Thursday, 12 August 2021 at 13:24:46 UTC+1, olcott wrote:
>>>> On 8/12/2021 3:36 AM, Malcolm McLean wrote:
>>>>> On Thursday, 12 August 2021 at 06:11:30 UTC+1, Jeff Barnett wrote:
>>>>>> On 8/11/2021 5:40 PM, olcott wrote:
>>>>>>> On 8/11/2021 6:32 PM, Jeff Barnett wrote:
>>>>>>>> On 8/11/2021 5:04 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>
>>>>>>>>> I see you are in "paste the same text" again mode. If you think I can
>>>>>>>>> help in any way, do let me know.
>>>>>>>>
>>>>>>>> That's not nice. Aren't you afraid that he'll develop carpal tunnel
>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>> PROOF THAT M REFERS TO THE TURING MACHINE DESCRIPTION PARAMETER WM TO H
>>>>>>>>>
>>>>>>>>> I see you are in "paste the same text" again mode. If you think I can
>>>>>>>>> help in any way, do let me know.
>>>>>>>>
>>>>>>>> syndrome? That along with all of his other deficiencies will surely do
>>>>>>>> him in. Perhaps we can invent shortcuts a la LaTeX macros and font
>>>>>>>> switches to help him cut down the strain. He wont be insulted and will
>>>>>>>> take to it as a pig to mud. Just think, he'll have yet another
>>>>>>>> notation to misuse and abuse.
>>>>>>>
>>>>>>> Ben finally acknowledged the point that I was making:
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>> if M applied to wM halts, and
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qy ∞
>>>>>>> if M applied to wM halts, and
>>>>>>>
>>>>>>> Ĥ.q0 wM ⊢* Ĥ.qx wM wM ⊢* Ĥ.qn
>>>>>>> if M applied to wM does not halt
>>>>>>>
>>>>>>> // M refers to the TM of the first wM parameter to Ĥ.qx
>>>>>>>
>>>>>>> My proof requires a whole inference chain that cannot proceed to the
>>>>>>>
>>>>>>> // M refers to the TM of the first wM parameter to Ĥ.qx
>>>>>>>
>>>>>>> next point until the current point is accepted as correct.
>>>>>> In addition to carpal tunnel, you are delusional. I get the vague
>>>>>> impression reading the above that you are having a schizoid episode.
>>>>>> There is a complete and total discount between what others say to you
>>>>>> and your replies. I suggest you cut out some of the examples you like
>>>>>> best and show them to your therapist. You do have one, don't you? It may
>>>>>> lead to better, more effective treatment. We are all hoping you get
>>>>>> better but schizoid episodes are hard to suppress. I assume meds have
>>>>>> been prescribed for your condition. Are you taking them? Or are you
>>>>>> refusing medical help just as you refuse to allow Ben to heal your
>>>>>> ignorance. A little trust in your betters will go a long way towards
>>>>>> recovery and a cure. Start now before it's to late. God speed to you.
>>>>>> -
>>>>> YOu might be right, but I think it's just frustration. His execution trace shows
>>>>> what appears to be an infinitely recursive process aborted, and everyone is
>>>>> just ignoring that and focusing on the other reality that his H does not
>>>>> return the result that matches the behaviour of the input when run independently.
>>>>>
>>>>> The attempt to use formal math notation when he doesn't understand it
>>>>> doesn't help. That can create the impression of mental incoherence, but in fact
>>>>> it's not that.
>>>>>
>>>>> Despite a mass of posting, we still don't have a good idea how H works.
>>>> _P()
>>>> [00000d02](01) 55 push ebp
>>>> [00000d03](02) 8bec mov ebp,esp
>>>> [00000d05](03) 8b4508 mov eax,[ebp+08]
>>>> [00000d08](01) 50 push eax // push 2nd Param
>>>> [00000d09](03) 8b4d08 mov ecx,[ebp+08]
>>>> [00000d0c](01) 51 push ecx // push 1st Param
>>>> [00000d0d](05) e870feffff call 00000b82 // call H
>>>> [00000d12](03) 83c408 add esp,+08
>>>> [00000d15](02) 85c0 test eax,eax
>>>> [00000d17](02) 7402 jz 00000d1b
>>>> [00000d19](02) ebfe jmp 00000d19
>>>> [00000d1b](01) 5d pop ebp
>>>> [00000d1c](01) c3 ret
>>>> Size in bytes:(0027) [00000d1c]
>>>> machine stack stack machine assembly
>>>> address address data code language
>>>> ======== ======== ======== ========= =============
>>>> ...[00000d0d][00101829][00000d12] e870feffff call 00000b82 // call H
>>>>
>>>> Begin Local Halt Decider Simulation at Machine Address:d02
>>>> ...[00000d02][002118f1][002118f5] 55 push ebp
>>>> ...[00000d03][002118f1][002118f5] 8bec mov ebp,esp
>>>> ...[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
>>>> ...[00000d08][002118ed][00000d02] 50 push eax // push P
>>>> ...[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
>>>> ...[00000d0c][002118e9][00000d02] 51 push ecx // push P
>>>> ...[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
>>>>
>>>> ...[00000d02][0025c319][0025c31d] 55 push ebp
>>>> ...[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
>>>> ...[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
>>>> ...[00000d08][0025c315][00000d02] 50 push eax // push P
>>>> ...[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
>>>> ...[00000d0c][0025c311][00000d02] 51 push ecx // push P
>>>> ...[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
>>>> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>>>> If this means that you are telling me that:
>>>> (1) You know the x86 language very well.
>>>>
>>>> (2) Can't begin to understand why P would be stuck in infinitely nested
>>>> simulation while H acts as a pure simulator?
>>>>
>>> OK, so let's start.
>>> There's a line which says "Begin Local Halt Decider Simulation at Machine
>>> Address:d02". Is that part of the execution trace, or is it a commented added
>>> by hand? If it is part of the execution trace, why doesn't this line appear
>>> before the second line with the label 0d02?
>> The halt decider automatically writes this when it begins the simulation
>> that its halt analysis is based on.
>>
> OK, so why don't we get a second "Begin Local Halt Decider ..." line output?
Begin Local Halt Decider Simulation at Machine Address:d02
...[00000d02][002118f1][002118f5] 55 push ebp
...[00000d03][002118f1][002118f5] 8bec mov ebp,esp
...[00000d05][002118f1][002118f5] 8b4508 mov eax,[ebp+08]
...[00000d08][002118ed][00000d02] 50 push eax // push P
...[00000d09][002118ed][00000d02] 8b4d08 mov ecx,[ebp+08]
...[00000d0c][002118e9][00000d02] 51 push ecx // push P
...[00000d0d][002118e5][00000d12] e870feffff call 00000b82 // call H
...[00000d02][0025c319][0025c31d] 55 push ebp
...[00000d03][0025c319][0025c31d] 8bec mov ebp,esp
...[00000d05][0025c319][0025c31d] 8b4508 mov eax,[ebp+08]
...[00000d08][0025c315][00000d02] 50 push eax // push P
...[00000d09][0025c315][00000d02] 8b4d08 mov ecx,[ebp+08]
...[00000d0c][0025c311][00000d02] 51 push ecx // push P
...[00000d0d][0025c30d][00000d12] e870feffff call 00000b82 // call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
The simulation of the input begins and continues until it ends.
>>>>
>>>> I would say that you must be a liar.
>>>>
>>> I have a PhD. PhD holder attach a very high value to honesty, in particular
>>> with respect to their subjects. But also more generally. I always draw attention
>>> if I'm given too much change at the pub, for example, not becaue I'm better
>>> than other people, but because I'm a doctor, and a doctor can't be seen to
>>> take money that doesn't belong to him.
>>>
>> So do (1) and (2) apply to you?
>>
> That was a demand for an apology, but you are too focused on your own
> interests to be sensitive to what other people are trying to communicate to you.
>
I still can't tell whether you are telling the truth or not because you
have not yet committed to the actual truth and rejected the falsehood.
If (1) and (2) do apply to you then you would be a liar.
The point that I am trying to make here is so damn simple that everyone
should have agreed to it in less than a week. It has been many months.
There is no freaking code in the body of P that can possibly escape
infinitely nested simulation while H remains a pure simulator. If you as
much as refrain from agreeing to that (and know the x86 language very
well) then you would be a liar.
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