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computers / comp.ai.philosophy / Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard [ continue to lie ]

Re: My Dishonest reviewers: André, Ben, Mike, Dennis, Richard [ continue to lie ]

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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Wed, 20 Apr 2022 20:43 UTC

On 4/20/2022 3:33 PM, Malcolm McLean wrote:
> On Wednesday, 20 April 2022 at 20:49:35 UTC+1, olcott wrote:
>> On 4/20/2022 2:44 PM, olcott wrote:
>>> On 4/20/2022 2:30 PM, Malcolm McLean wrote:
>>>> On Wednesday, 20 April 2022 at 19:10:37 UTC+1, olcott wrote:
>>>>> On 4/20/2022 12:51 PM, Malcolm McLean wrote:
>>>>>> On Wednesday, 20 April 2022 at 18:35:43 UTC+1, olcott wrote:
>>>>>>> On 4/19/2022 9:49 AM, Ben wrote:
>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 4/18/2022 4:53 PM, Ben wrote:
>>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 4/17/2022 6:43 PM, Ben wrote:
>>>>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 4/17/2022 10:20 AM, Ben wrote:
>>>>>>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>>>>>> If I was wrong then the correct simulation of the 27 bytes
>>>>>>>>>>>>>>> of machine
>>>>>>>>>>>>>>> code...
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If you were right you'd publish the code.
>>>>>>>>>>>>>
>>>>>>>>>>>>> This is the only relevant code to the question does the
>>>>>>>>>>>>> correctly
>>>>>>>>>>>>> simulated input to H(P,P) halt?
>>>>>>>>>>>>
>>>>>>>>>>>> Clearly not. The code for P is not in doubt.
>>>>>>>>>>>
>>>>>>>>>>> So then you must agree that when H correctly simulates the
>>>>>>>>>>> input to
>>>>>>>>>>> H(P,P) that it would never reach its own final state.
>>>>>>>>>> Deflection. As I said, if you were right you'd publish the code.
>>>>>>>>>> You
>>>>>>>>>> pointlessly showed P which is not in doubt. I can only assume
>>>>>>>>>> you know
>>>>>>>>>> that if you publish H the game will be well and truly up. You
>>>>>>>>>> need to
>>>>>>>>>> trot out one of your excuses for keeping the flawed code secret.
>>>>>>>>>
>>>>>>>>> I am going to continue to present this same point until everyone
>>>>>>>>> reading this forum realizes that the only reason that you dodge
>>>>>>>>> it is
>>>>>>>>> because you already know that it is correct.
>>>>>>>>
>>>>>>>> Everyone here has seen me address it many times and has seen you
>>>>>>>> ignore
>>>>>>>> what I've said about it many times. Do you really think anyone will
>>>>>>>> change their option of me just because you keep typing the same vague
>>>>>>>> sentence? And, more to the point, why do you care what people
>>>>>>>> think of
>>>>>>>> me?
>>>>>>>>
>>>>>>>> What you need is someone you trust to tell you to do something useful
>>>>>>>> with your time, not some way to get "everyone reading this forum" to
>>>>>>>> form your preferred opinion of me. I am not that important.
>>>>>>>>
>>>>>>>>> When the input to H(P,P) is non halting then it is necessarily
>>>>>>>>> correct
>>>>>>>>> for H to report that the input to H(P,P) is non halting.
>>>>>>>>
>>>>>>>> On more time: H(P,P) has no input.
>>>>>>> According to functional notation the inputs to a function are its
>>>>>>> parameters.
>>>>>>>
>>>>>>>
>>>>>>> Function notation is a way to write functions that is easy to read and
>>>>>>> understand. Functions have dependent and independent variables, and
>>>>>>> when
>>>>>>> we use function notation the independent variable is commonly x,
>>>>>>> and the
>>>>>>> dependent variable is F(x).
>>>>>>> https://www.brightstorm.com/math/algebra/graphs-and-functions/function-notation/#:~:text=Function%20notation%20is%20a%20way,variable%20is%20F(x).
>>>>>>>
>>>>>>>
>>>>>>> When we construe H(P,P) as a computable function then H computes the
>>>>>>> mapping from its inputs/parameters to its own accept of reject state,
>>>>>>> thus you "rebuttal" is merely double-talk misdirection.
>>>>>>>> If you mean the two pointer
>>>>>>>> parameters, say so. That "input" -- those two pointers -- are neither
>>>>>>>> halting nor non halting. Maybe you mean calling the first with the
>>>>>>>> second as its only argument is non halting?
>>>>>>> The finite string of machine code pointed to by P.
>>>>>>>
>>>>>> Machine code is tree-like in structure. I particular, in your
>>>>>> system, the "string of
>>>>>> machine code pointed to by P" contains a call to H.
>>>>>> Now are you excluding that code in H from "the input to H(P,P)?".
>>>>> The only relevant point is that the correctly simulated P simulated by H
>>>>> cannot possibly reach its own final state under any condition
>>>>> what-so-ever thus conclusively fails to meet the Linz criteria of a
>>>>> halting computation thus is definitively determined to be a non-halting
>>>>> sequence of configurations.
>>>>>
>>>>> If {an X is a Y} "the input to H(P,P) is non-halting"
>>>>>
>>>>> then it is necessarily correct for Z to report that {an X is a Y} "H
>>>>> reports that its input is non-halting"
>>>>>
>>>> You say that P(P) halts, but the correctly simulated input to H(P,P) does
>>>> not halt. You've been pretty consistent on this point.
>>>
>>>
>>>> But I'm still mystified. I've made a suggestion that might resolve the
>>>> obvious
>>>> contradiction, but it hasn't been accepted.
>>>
>>> It is an empirically proved verified fact that the input to H(P,P) does
>>> not halt so anyone and anything disagreeing is necessarily incorrect.
>>>
>> If I smash a Boston cream pie in someones face and they have a
>> fundamental religious belief that there is no such thing as pies the pie
>> dripping from their face conclusively proves that they are incorrect.
>>
>> The verified fact that the input to H(P,P) is non-halting is the pie
>> dripping from the face.
>>
> You have never published H, so we can't verify anything.

This right here provides complete proof that the input to H(P,P) is
non-halting. If you are clueless about the x86 language that does not
mean that it is not complete proof.

The technical computer science term "halt" means that a program will
reach its last instruction technically called its final state. For P
this would be its machine address [000009f0].

The function named H continues to simulate its input using an x86
emulator until this input either halts on its own or H detects that it
would never halt. If its input halts H returns 1. If H detects that its
input would never halt H returns 0.

void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
return;
}

int main()
{ Output("Input_Halts = ", H((u32)P, (u32)P));
}

_P()
[000009d6](01) 55 push ebp
[000009d7](02) 8bec mov ebp,esp
[000009d9](03) 8b4508 mov eax,[ebp+08]
[000009dc](01) 50 push eax // push P
[000009dd](03) 8b4d08 mov ecx,[ebp+08]
[000009e0](01) 51 push ecx // push P
[000009e1](05) e840feffff call 00000826 // call H
[000009e6](03) 83c408 add esp,+08
[000009e9](02) 85c0 test eax,eax
[000009eb](02) 7402 jz 000009ef
[000009ed](02) ebfe jmp 000009ed
[000009ef](01) 5d pop ebp
[000009f0](01) c3 ret // Final state
Size in bytes:(0027) [000009f0]

The simulated input to H(P,P) cannot possibly reach its own final state
of [000009f0] it keeps repeating [000009d6] to [000009e1] until aborted.

Begin Local Halt Decider Simulation
....[000009d6][00211368][0021136c] 55 push ebp // enter P
....[000009d7][00211368][0021136c] 8bec mov ebp,esp
....[000009d9][00211368][0021136c] 8b4508 mov eax,[ebp+08]
....[000009dc][00211364][000009d6] 50 push eax // Push P
....[000009dd][00211364][000009d6] 8b4d08 mov ecx,[ebp+08]
....[000009e0][00211360][000009d6] 51 push ecx // Push P
....[000009e1][0021135c][000009e6] e840feffff call 00000826 // Call H
....[000009d6][0025bd90][0025bd94] 55 push ebp // enter P
....[000009d7][0025bd90][0025bd94] 8bec mov ebp,esp
....[000009d9][0025bd90][0025bd94] 8b4508 mov eax,[ebp+08]
....[000009dc][0025bd8c][000009d6] 50 push eax // Push P
....[000009dd][0025bd8c][000009d6] 8b4d08 mov ecx,[ebp+08]
....[000009e0][0025bd88][000009d6] 51 push ecx // Push P
....[000009e1][0025bd84][000009e6] e840feffff call 00000826 // Call H
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

Because the correctly simulated input to H(P,P) cannot possibly reach
its own final state at [000009f0] it is necessarily correct for H to
reject this input as non-halting.

> I do believe you
> when you say that P(P) halts and H(P,P) reports "non-halting". The question
> which everyone is asking is how you get from that fact to the claim that
> "The input to H(P,P) is non-halting".
>
> As I said, I suspect that the reason is confusion, and the root of the confusion
> is that you are using an x86 emulator to make statements about Turing machines.
> Using an x86 emulator instead a Turing machine isn't of itself fatal to your
> project, but it opens the way for all sorts of errors that using Turing machines
> themselves would avoid. So really I urge you to learn more about Turing
> machines.

The exact same reasoning applies to
embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would reach its own final
state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to H would never reach its own
final state of ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩.

When Ĥ is applied to ⟨Ĥ⟩
Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Then these steps would keep repeating: (unless their simulation is aborted)
Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩...

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

SubjectRepliesAuthor
o Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key

By: olcott on Sun, 3 Apr 2022

54olcott
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