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computers / comp.ai.philosophy / Re: H(P,P)==0 is proven to be correct thus refuting the halting problem proofs [ ignorance squared ]

Re: H(P,P)==0 is proven to be correct thus refuting the halting problem proofs [ ignorance squared ]

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Subject: Re: H(P,P)==0 is proven to be correct thus refuting the halting
problem proofs [ ignorance squared ]
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sun, 22 May 2022 03:24 UTC

On 5/21/2022 10:12 PM, Richard Damon wrote:
> On 5/21/22 10:34 PM, olcott wrote:
>> On 5/21/2022 9:27 PM, Richard Damon wrote:
>>> On 5/21/22 9:25 PM, olcott wrote:
>>>> On 5/21/2022 8:16 PM, Richard Damon wrote:
>>>>> On 5/21/22 9:02 PM, olcott wrote:
>>>>>> On 5/21/2022 7:55 PM, Richard Damon wrote:
>>>>>>> On 5/21/22 3:12 PM, olcott wrote:
>>>>>>>> On 5/21/2022 2:06 PM, Richard Damon wrote:
>>>>>>>>> On 5/21/22 3:01 PM, olcott wrote:
>>>>>>>>>> On 5/21/2022 1:58 PM, Richard Damon wrote:
>>>>>>>>>>> On 5/21/22 12:47 PM, olcott wrote:
>>>>>>>>>>>> On 5/21/2022 11:38 AM, Richard Damon wrote:
>>>>>>>>>>>>> On 5/21/22 12:28 PM, olcott wrote:
>>>>>>>>>>>>>> On 5/21/2022 11:22 AM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 5/21/22 11:47 AM, olcott wrote:
>>>>>>>>>>>>>>>> On 5/21/2022 10:14 AM, Richard Damon wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On 5/21/22 10:44 AM, olcott wrote:
>>>>>>>>>>>>>>>>>> It is an easily verified fact that the execution trace
>>>>>>>>>>>>>>>>>> provided by H(P,P) of the nested simulation of its
>>>>>>>>>>>>>>>>>> input exactly matches the behavior of the correctly
>>>>>>>>>>>>>>>>>> reverse-engineered nested execution trace would be.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> No, it is easy to verify that it does NOT.
>>>>>>>>>>>>>>>> You know that you are a liar so I challenge you to
>>>>>>>>>>>>>>>> provide the execution trace that a pure single level
>>>>>>>>>>>>>>>> nested emulation of the input to H(P,P) would be. Any
>>>>>>>>>>>>>>>> failure to provide this basis for your damned lies will
>>>>>>>>>>>>>>>> be considered direct admission that you know you are lying.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> _P()
>>>>>>>>>>>>>>>> [00001352](01)  55              push ebp
>>>>>>>>>>>>>>>> [00001353](02)  8bec            mov ebp,esp
>>>>>>>>>>>>>>>> [00001355](03)  8b4508          mov eax,[ebp+08]
>>>>>>>>>>>>>>>> [00001358](01)  50              push eax      // push P
>>>>>>>>>>>>>>>> [00001359](03)  8b4d08          mov ecx,[ebp+08]
>>>>>>>>>>>>>>>> [0000135c](01)  51              push ecx      // push P
>>>>>>>>>>>>>>>> [0000135d](05)  e840feffff      call 000011a2 // call H
>>>>>>>>>>>>>>>> [00001362](03)  83c408          add esp,+08
>>>>>>>>>>>>>>>> [00001365](02)  85c0            test eax,eax
>>>>>>>>>>>>>>>> [00001367](02)  7402            jz 0000136b
>>>>>>>>>>>>>>>> [00001369](02)  ebfe            jmp 00001369
>>>>>>>>>>>>>>>> [0000136b](01)  5d              pop ebp
>>>>>>>>>>>>>>>> [0000136c](01)  c3              ret
>>>>>>>>>>>>>>>> Size in bytes:(0027) [0000136c]
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Well, I wopuld need to have the code for H to do that,
>>>>>>>>>>>>>>> since that is PART of P.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It would begin as:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>      machine   stack     stack     machine    assembly
>>>>>>>>>>>>>>>      address   address   data      code       language
>>>>>>>>>>>>>>>      ========  ========  ========  =========  =============
>>>>>>>>>>>>>>> ...[00001352][0021233e][00212342] 55         push ebp //
>>>>>>>>>>>>>>> enter P
>>>>>>>>>>>>>>> ...[00001353][0021233e][00212342] 8bec       mov ebp,esp
>>>>>>>>>>>>>>> ...[00001355][0021233e][00212342] 8b4508     mov
>>>>>>>>>>>>>>> eax,[ebp+08]
>>>>>>>>>>>>>>> ...[00001358][0021233a][00001352] 50         push eax //
>>>>>>>>>>>>>>> push P
>>>>>>>>>>>>>>> ...[00001359][0021233a][00001352] 8b4d08     mov
>>>>>>>>>>>>>>> ecx,[ebp+08]
>>>>>>>>>>>>>>> ...[0000135c][00212336][00001352] 51         push ecx //
>>>>>>>>>>>>>>> push P
>>>>>>>>>>>>>>> ...[0000135d][00212332][00001362] e840feffff call
>>>>>>>>>>>>>>> 000011a2 // call H
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> At this point I don't have the data,
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The assumption is that H(P,P) correctly emulates its input.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Maybe a bad assumption!
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> It is stipulated that you must show what the execution trace
>>>>>>>>>>>> of the input to H(P,P) would be if H only simulated its
>>>>>>>>>>>> input. You must show this for one simulation and one nested
>>>>>>>>>>>> simulation. Failure to do this will be construed as a direct
>>>>>>>>>>>> admission that you know you are lying.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> So, give me a copy of H to trace,
>>>>>>>>>> You are required to provide a trace under the assumption that
>>>>>>>>>> H(P,P) only does a pure x86 emulation of its input for the
>>>>>>>>>> first emulation and the first nested emulation. Are you too
>>>>>>>>>> stupid to understand this?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You obviously have an reading problem.
>>>>>>>>>
>>>>>>>>> I said, for that I need the code of H, as that is what needs to
>>>>>>>>> be traced.
>>>>>>>>>
>>>>>>>>
>>>>>>>> If it is "given" that this code only performs a pure x86
>>>>>>>> emulation of its input (unless you have no idea what an x86
>>>>>>>> emulation is) there is no reason to see that the code derives a
>>>>>>>> pure x86 emulation of its input.
>>>>>>>
>>>>>>> The the trace of the emulation needs to show the actual steps of
>>>>>>> emulationg the input, like I mentioned.
>>>>>>>
>>>>>> If I showed either the source-code of H or the execution trace of
>>>>>> H people here would be so confused that I would never reach
>>>>>> closure in 50 years. If they can't comprehend a 14 line execution
>>>>>> trace then showing them much more than this would permanently
>>>>>> scramble their brains.
>>>>>
>>>>> No, people wouldn't be confused,
>>>>>
>>>>
>>>> That they don't understand that they don't need to see this
>>>> conclusively proves that they have woefully inadequate technical
>>>> skills to evaluate my work.
>>>>
>>>
>>> Just shows you are lying.
>>>
>> I coined the term "ignorance squared" decades ago to account for the
>> fact that people cannot possibly be directly aware of their own
>> ignorance.
>
> You seem tobe a good example of that.
>
>>
>> To be directly aware of their own ignorance requires them to contrast
>> their ignorance with the knowledge that they are missing.
>>
>> Since they don't have this missing knowledge they cannot become
>> directly aware that they are missing any knowledge. Their own
>> ignorance is simply perceived as disagreement.
>
> So, what actual FACT do you think I am missing?
That the next level execution trace of the input to H(P,P) must be
identical to the first level trace.

--
Copyright 2022 Pete Olcott

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

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o H(P,P)==0 is proven to be correct thus refuting the halting problem

By: olcott on Sat, 21 May 2022

20olcott
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