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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Proof that DDD specifies non-halting behavior --- point by point
 --- in our head
Date: Wed, 14 Aug 2024 22:11:05 -0400
Organization: i2pn2 (i2pn.org)
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On 8/14/24 10:03 PM, olcott wrote:
> On 8/14/2024 6:40 PM, Richard Damon wrote:
>> On 8/14/24 9:34 AM, olcott wrote:
>>> On 8/14/2024 6:22 AM, Richard Damon wrote:
>>>> On 8/14/24 12:24 AM, olcott wrote:
>>>>> On 8/13/2024 11:04 PM, Richard Damon wrote:
>>>>>> On 8/13/24 11:48 PM, olcott wrote:
>>>>>>> On 8/13/2024 10:21 PM, Richard Damon wrote:
>>>>>>>> On 8/13/24 10:38 PM, olcott wrote:
>>>>>>>>> On 8/13/2024 9:29 PM, Richard Damon wrote:
>>>>>>>>>> On 8/13/24 8:52 PM, olcott wrote:
>>>>>>>>>>> void DDD()
>>>>>>>>>>> {
>>>>>>>>>>>    HHH(DDD);
>>>>>>>>>>>    return;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> _DDD()
>>>>>>>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>>>>>>>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>>>>>>>>>>> [00002175] 6872210000 push 00002172 ; push DDD
>>>>>>>>>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>>>>>>>>>> [0000217f] 83c404     add esp,+04
>>>>>>>>>>> [00002182] 5d         pop ebp
>>>>>>>>>>> [00002183] c3         ret
>>>>>>>>>>> Size in bytes:(0018) [00002183]
>>>>>>>>>>>
>>>>>>>>>>> A simulation of N instructions of DDD by HHH according to
>>>>>>>>>>> the semantics of the x86 language is necessarily correct.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Nope, it is just the correct PARTIAL emulation of the first N 
>>>>>>>>>> instructions of DDD, and not of all of DDD, 
>>>>>>>>>
>>>>>>>>> That is what I said dufuss.
>>>>>>>>
>>>>>>>> Nope. You didn't. I added clairifying words, pointing out why 
>>>>>>>> you claim is incorrect.
>>>>>>>>
>>>>>>>> For an emulation to be "correct" it must be complete, as partial 
>>>>>>>> emulations are only partially correct, so without the partial 
>>>>>>>> modifier, they are not correct.
>>>>>>>>
>>>>>>>
>>>>>>> A complete emulation of one instruction is
>>>>>>> a complete emulation of one instruction
>>>>>>
>>>>>>
>>>>>>
>>>>>>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> A correct simulation of N instructions of DDD by HHH is
>>>>>>>>>>> sufficient to correctly predict the behavior of an unlimited
>>>>>>>>>>> simulation.
>>>>>>>>>>
>>>>>>>>>> Nope, if a HHH returns to its caller, 
>>>>>>>>>
>>>>>>>>> *Try to show exactly how DDD emulated by HHH returns to its 
>>>>>>>>> caller*
>>>>>>>>> (the first one doesn't even have a caller)
>>>>>>>>> Use the above machine language instructions to show this.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Remember how English works:
>>>>>>>>
>>>>>>>> When you ask "How DDD emulated by HHH returns to its callers".
>>>>>>>
>>>>>>> Show the exact machine code trace of how DDD emulated
>>>>>>> by HHH (according to the semantics of the x86 language)
>>>>>>> reaches its own machine address 00002183
>>>>>>
>>>>>> No. The trace is to long, 
>>>>>
>>>>> Show the Trace of DDD emulated by HHH
>>>>> and show the trace of DDD emulated by HHH
>>>>> emulated by the executed HHH
>>>>> Just show the DDD code traces.
>>>>>
>>>>
>>>> First you need to make a DDD that meets the requirements, and that 
>>>> means that it calls an HHH that meets the requirements.
>>>>
>>>
>>> _DDD()
>>> [00002172] 55         push ebp      ; housekeeping
>>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>>> [00002175] 6872210000 push 00002172 ; push DDD
>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>> [0000217f] 83c404     add esp,+04
>>> [00002182] 5d         pop ebp
>>> [00002183] c3         ret
>>> Size in bytes:(0018) [00002183]
>>>
>>> The is a hypothetical mental exercise and can be
>>> accomplished even if the only DDD in the world
>>> was simply typed into a word processor and never run.
>>
>> But, must behave the rules of Computation Theory.
>>
>> That means DDD, to be a program, includes the code of HHH, and that 
>> HHH obeys the requirements of programs in computation theory, which 
>> means that it always produces the same answer to its caller for the 
>> same input.
>>
>>
>> Note, its "Behavior" is defined as what it would do when run, even if 
>> it never is,
>>
> 
> No that is the big mistake of comp theory where it violates
> its own rules.
> 

WHAT rule does it violate? And where do you get it from?


This is just your ignorance speaking.

Deciders need to try to compute the Mapping that is defined MATHEMATICALLY.

For Halting, that mapping is from the behavior of the program the input 
describes to whether it Halts or Not.

What rule does that break?

Nothing in the definition says that the decider needs to actually be 
able to compute that results, and in fact, the big question is if there 
exists a program that can.

Thus, "Non-computable" functions are a thing, and fully proper.

Halting turns out to be one of them.

The mapping that a given program produces shows what it does decide, if 
that doesn't match the mapping it is supposed to produce, it just isn't 
correct.

Something you don't seem to be able to understand, that things can be 
incorrect.