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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Liar detector: Peter Olcott lies
Date: Wed, 10 Jul 2024 20:01:50 -0400
Organization: i2pn2 (i2pn.org)
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On 7/10/24 9:37 AM, olcott wrote:
> On 7/10/2024 2:18 AM, Mikko wrote:
>> On 2024-07-09 14:14:16 +0000, olcott said:
>>
>>> On 7/9/2024 1:14 AM, Mikko wrote:
>>>> On 2024-07-08 17:36:58 +0000, olcott said:
>>>>
>>>>> On 7/8/2024 11:16 AM, Fred. Zwarts wrote:
>>>>>> Op 08.jul.2024 om 18:07 schreef olcott:
>>>>>>>
>>>>>>> Try to show how infinity is one cycle too soon.
>>>>>>>
>>>>>> You believe that two equals infinity.
>>>>>
>>>>> void Infinite_Loop()
>>>>> {
>>>>>    HERE: goto HERE;
>>>>> }
>>>>>
>>>>> void Infinite_Recursion()
>>>>> {
>>>>>    Infinite_Recursion();
>>>>> }
>>>>>
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>> }
>>>>>
>>>>> Two cycles is enough to correctly determine that none
>>>>> of the above functions correctly emulated by HHH can
>>>>> possibly halt.
>>>>>
>>>>> That you don't see this is ignorance or deception.
>>>>
>>>> There is an important detail that determines whether an infinite
>>>> execution can be inferred. That is best illustrated by the following
>>>> examples:
>>>>
>>>> void Finite_Loop()
>>>> {
>>>>   int x = 10000;
>>>> HERE:
>>>>   if (x > 0) {
>>>>     x--;
>>>>     goto HERE;
>>>>   }
>>>> }
>>>>
>>>> void Finite_Recursion(int n)
>>>> {
>>>>   if (n > 0) {
>>>>     Finite_Recursion(n + 1);
>>>>   }
>>>> }
>>>>
>>>> void DDD()
>>>> {
>>>>   HHH(DDD); // HHH detects recursive simulation and then simulates 
>>>> no more
>>>> }
>>>>
>>>> The important difference is that in my examples there is a conditional
>>>> instruction that can (and does) prevent infinite exectuion.
>>>>
>>>
>>> When we ask:
>>> Does the call from DDD emulated by HHH to HHH(DDD) return?
>>
>> Why would anyone ask that? A question should make clear its topic.
>> Instead one could ask whether HHH can fully emulate DDD if that is
>> what one wants to know. Or one may think that HHH and DDD are so
>> unimteresting that there is no point to ask anyting about them.
>>
> 
> A correct emulator can correctly any correct x86 instructions.
> When it emulates non-halting code then itself does not halt.
> 

Right, so if the input is non-halting a correct emulation of it can not 
halt, and thus HHH can not do both a correct emulation of the input and 
say the input is non-halting. That is just a contradiction of terms.

If HHH does abort its emulation, it no longer did a correct emulation 
and thus lost the ability to show that its input was non-halting. The 
only input you have shown to be non-halting is the input based on the 
non-aborting HHH, but that doesn't answer, and its DDD is DIFFERENT than 
the others, as DDD INCLUDES the HHH that it is calling. (something you 
try to gloss over as it proves you wrong).