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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 19:40:49 -0400
Organization: i2pn2 (i2pn.org)
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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,

> 
> HHH can be purely imaginary yet must emulate the
> above code and itself according to the semantics
> of the x86 language. In this case HHH is a pure
> emulator.
> 
No, it must emulate ALL the code of the PROGRAM DDD, which include the 
code of HHH.

That, or you never had a PROGRAM DDD to give to HHH and you whole 
hypothetical blows up as a LIE.


> On this basis we know that such an HHH would
> emulate the first four instructions of DDD.
> 
> This includes a calls from the emulated DDD
> to an emulated HHH(DDD).
> 
> This emulated HHH would emulate the first
> four instructions of DDD.


And the emulating HHH, needs to trace how that HHH does its emulation, 
not the results of the emulation, as that is what the code of "DDD"
> 
> We can do that all in our head never needing
> any actually existing HHH.

And eitehr HHH CONDITIONALLY emulates DDD, and thus can break out of the 
loop, or it can't.

If it can, and it will, then DDD is Halting.

If it can't, then HHH can NEVER answer, and thus isn't a decider.

> 
> All other points are moot and will simply be erased
> until we have mutual agreement on this point.
> 


No, you NEED to answer the problems shown, or you are just admitting 
that you idea is all just a pathetic lie.

This seems to be your fundamental problem (or as one friend of mine 
calls it, a FUNNY-MENTAL problem) that you don't understand the 
fundamental definitions of what you are talking about.

Halting is a property of FULL PROGRAMS, and by the rules of Compuation 
Theory programs include ALL of the code they use, and their behavior can 
only depend on that code and the data explicitly given as the input.

You "DDD" that contains only those byte specified, just isn't a program, 
so you lost the argument at your first paragraph.

The fact that you don't understand this, and refuse to learn or accept 
it does mean that you have a "funny-mental" problem that just proves you 
are incapable of dealing with the topic you are trying to talk about, 
and just failing horribly.