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From: olcott <polcott2@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Proof that H(D,D) meets its abort criteria --Categorically
 Exhaustive Reasoning--
Date: Fri, 15 Mar 2024 22:41:25 -0500
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On 3/15/2024 10:11 PM, Richard Damon wrote:
> On 3/15/24 8:00 PM, olcott wrote:
>> On 3/15/2024 9:47 PM, Richard Damon wrote:
>>> On 3/15/24 7:18 PM, olcott wrote:
>>>> On 3/15/2024 8:22 PM, Richard Damon wrote:
>>>>> On 3/15/24 5:49 PM, olcott wrote:
>>>>>> On 3/15/2024 6:37 PM, Mike Terry wrote:
>>>>>>> On 15/03/2024 18:45, immibis wrote:
>>>>>>>> On 15/03/24 19:39, olcott wrote:
>>>>>>>>> On 3/15/2024 1:38 PM, immibis wrote:
>>>>>>>>>> On 15/03/24 18:52, olcott wrote:
>>>>>>>>>>> On 3/15/2024 12:36 PM, Richard Damon wrote:
>>>>>>>>>>>> On 3/15/24 9:20 AM, olcott wrote:
>>>>>>>>>>>>> Best selling author of Theory of Computation textbooks:
>>>>>>>>>>>>> *Introduction To The Theory Of Computation 3RD, by sipser*
>>>>>>>>>>>>> https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>>>>>>>>>>>>>
>>>>>>>>>>>>> Date 10/13/2022 11:29:23 AM
>>>>>>>>>>>>> *MIT Professor Michael Sipser agreed this verbatim 
>>>>>>>>>>>>> paragraph is correct*
>>>>>>>>>>>>> (He has neither reviewed nor agreed to anything else in 
>>>>>>>>>>>>> this paper)
>>>>>>>>>>>>> (a) If simulating halt decider H correctly simulates its 
>>>>>>>>>>>>> input D until H correctly determines that its simulated D 
>>>>>>>>>>>>> would never stop running unless aborted then
>>>>>>>>>>>>> (b) H can abort its simulation of D and correctly report 
>>>>>>>>>>>>> that D specifies a non-halting sequence of configurations.
>>>>>>>>>>>>>
>>>>>>>>>>>>> *When we apply the abort criteria* (elaborated above)
>>>>>>>>>>>>> Will you halt if you never abort your simulation?
>>>>>>>>>>>>> *Then H(D,D) is proven to meet this criteria*
>>>>>>>>>>>>>
>>>>>>>>>>>>> *Proof that H(D,D) meets its abort criteria*
>>>>>>>>>>>>>
>>>>>>>>>>>>> int D(int (*x)())
>>>>>>>>>>>>> {
>>>>>>>>>>>>>    int Halt_Status = H(x, x);
>>>>>>>>>>>>>    if (Halt_Status)
>>>>>>>>>>>>>      HERE: goto HERE;
>>>>>>>>>>>>>    return Halt_Status;
>>>>>>>>>>>>> }
>>>>>>>>>>>>>
>>>>>>>>>>>>> int main()
>>>>>>>>>>>>> {
>>>>>>>>>>>>>    Output("Input_Halts = ", H(D,D));
>>>>>>>>>>>>> }
>>>>>>>>>>>>>
>>>>>>>>>>>>>   machine   stack     stack     machine    assembly
>>>>>>>>>>>>>   address   address   data      code       language
>>>>>>>>>>>>>   ========  ========  ========  =========  =============
>>>>>>>>>>>>> [00001d22][00102fc9][00000000] 55         push ebp      ; 
>>>>>>>>>>>>> begin main()
>>>>>>>>>>>>> [00001d23][00102fc9][00000000] 8bec       mov ebp,esp
>>>>>>>>>>>>> [00001d25][00102fc5][00001cf2] 68f21c0000 push 00001cf2 ; 
>>>>>>>>>>>>> push DD
>>>>>>>>>>>>> [00001d2a][00102fc1][00001cf2] 68f21c0000 push 00001cf2 ; 
>>>>>>>>>>>>> push D
>>>>>>>>>>>>> [00001d2f][00102fbd][00001d34] e8eef7ffff call 00001522 ; 
>>>>>>>>>>>>> call H(D,D)
>>>>>>>>>>>>>
>>>>>>>>>>>>> H: Begin Simulation   Execution Trace Stored at:113075
>>>>>>>>>>>>> Address_of_H:1522
>>>>>>>>>>>>> [00001cf2][00113061][00113065] 55         push ebp       ; 
>>>>>>>>>>>>> enter D(D)
>>>>>>>>>>>>> [00001cf3][00113061][00113065] 8bec       mov ebp,esp
>>>>>>>>>>>>> [00001cf5][0011305d][00103031] 51         push ecx
>>>>>>>>>>>>> [00001cf6][0011305d][00103031] 8b4508     mov eax,[ebp+08]
>>>>>>>>>>>>> [00001cf9][00113059][00001cf2] 50         push eax       ; 
>>>>>>>>>>>>> push D
>>>>>>>>>>>>> [00001cfa][00113059][00001cf2] 8b4d08     mov ecx,[ebp+08]
>>>>>>>>>>>>> [00001cfd][00113055][00001cf2] 51         push ecx       ; 
>>>>>>>>>>>>> push D
>>>>>>>>>>>>> [00001cfe][00113051][00001d03] e81ff8ffff call 00001522  ; 
>>>>>>>>>>>>> call H(D,D)
>>>>>>>>>>>>> H: Recursive Simulation Detected Simulation Stopped
>>>>>>>>>>>>>                            H(D,D) returns 0 to main()
>>>>>>>>>>>>>
>>>>>>>>>>>>> *That was proof that H(D,D) meets its abort criteria*
>>>>>>>>>>>>> H(D,D) correctly determines that itself is being called 
>>>>>>>>>>>>> with its same inputs and there are no conditional branch 
>>>>>>>>>>>>> instructions between the invocation of D(D) and its call to 
>>>>>>>>>>>>> H(D,D).
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Except that D calling H(D,D) does NOT prove the required 
>>>>>>>>>>>> (a), since the simulated D WILL stop running because *ITS* H 
>>>>>>>>>>>> will abort *ITS* simulation and returm 0 so that simulated D 
>>>>>>>>>>>> will halt.
>>>>>>>>>>> You keep saying that H(D,D) never really needs to abort the
>>>>>>>>>>> simulation of its input because after H(D,D) has aborted the
>>>>>>>>>>> simulation of this input it no longer needs to be aborted.
>>>>>>>>>>>
>>>>>>>>>> You keep thinking there is more than one H(D,D) and then when 
>>>>>>>>>> it's convenient for you you think there is only one H(D,D). 
>>>>>>>>>> Why is that?
>>>>>>>>>
>>>>>>>>> The first H(D,D) to see that the abort criteria has been met
>>>>>>>>> (the outermost one) must abort the simulation of its input or
>>>>>>>>> none of them ever abort.
>>>>>>>>>
>>>>>>>>
>>>>>>>> that's wrong. They all abort, so if we prevent the first one 
>>>>>>>> from aborting, the second one will abort. If we prevent the 
>>>>>>>> first and second ones from aborting, the third one will abort. 
>>>>>>>
>>>>>>> Correct - but PO has the wrong understanding of "prevent".
>>>>>>>
>>>>>>> Correct understanding:  We're discussing a (putative) HD H 
>>>>>>> examining an input (P,I) representing some /fixed/ computation. 
>>>>>>> When we talk about "preventing" H from doing xxxxx (such as 
>>>>>>> aborting a simulation) we mean how an amended version H2 (like H 
>>>>>>> but without xxxxx) behaves in examining that /same input/ (P,I).
>>>>>>>
>>>>>>
>>>>>> *It can be construed that way, yet that is not it*
>>>>>> In software engineering the above is simply a pair of distinct
>>>>>> execution paths based on a conditional test within the same program.
>>>>>> In both cases D is simply a fixed constant string of machine-code 
>>>>>> bytes.
>>>>>
>>>>> Right D is a FIXED constant string, and thus the meaning doesn't 
>>>>> change if we hypothsize about changing an H.
>>>>>
>>>> It always calls whatever H is at the fixed machine address
>>>> that is encoded within D.
>>>
>>> In other words, it has NOTHING to do with Turing Machines, and thus 
>>> has NO application to the Linz proof, and you are admitting you are 
>>> LYING about it having application.
>>>
>>> As D isn't a "Computation" as it takes a "hidden input", namely thely 
>>> the contents of them memory now called "H". (It isn't PART OF D, and 
>>> isn't a declared parameter to D, so it is a "Hidden Input"
>>>
>>
>> I told you that you will not have the basis (prerequisite knowledge)
>> to see how this is isomorphic to H ⟨Ĥ⟩ ⟨Ĥ⟩ until after you see how
>> H(D,D) does correctly determine that it must abort its simulation.
>>
>>>>
>>>> This means that it DOES call H(D,D) in recursive simulation and
>>>> DOES NOT call H1(D,D) in recursive simulation.
>>>>
>>>
>>> And means you have been LYING that it has ANYTHING to do with the 
>>> HALTING PROBLEM
>>>
>>>> Thus H(D,D) must account for this difference and H1(D,D) can
>>>> ignore this difference.
>>>>
>>>>>>
>>>>>> When we use categorically exhaustive reasoning instead of locking
>>>>>> ourselves into the pathological thinking of Richard where H tries
>>>>>> to second guess itself such that anything that H(D,D) does can 
>>>>>> somehow
>>>>>> be construed as incorrect...
>>>>>
>>>>> Sounds like buzzwords.
>>>>>
>>>>> H doesn't try to second guess, H does what H does. PERIOD. That is 
>>>>> all it can do.
>>>>>
>>>>> You don't seem to understand how programs work.
>>>>>
>>>>>>
>>>>>> We as humans analyze everything that every encoding of H can possibly
>>>>>> do and find that categorically every H that never aborts its 
>>>>>> simulation
>>>>>> results in D(D) never halting.
>>>>>
>>>>> Right. But that doesn't mean that any of the H that DO abort is 
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