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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Liar detector: Fred, Richard, Joes and Alan --- Honest Dialogue ? --- Infinite set of HHH/DDD pairs
Date: Tue, 23 Jul 2024 10:14:15 +0300
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On 2024-07-22 14:13:33 +0000, olcott said:

> On 7/22/2024 3:01 AM, Mikko wrote:
>> On 2024-07-21 13:50:17 +0000, olcott said:
>> 
>>> On 7/21/2024 4:38 AM, Mikko wrote:
>>>> On 2024-07-20 13:28:36 +0000, olcott said:
>>>> 
>>>>> On 7/20/2024 3:54 AM, Mikko wrote:
>>>>>> On 2024-07-19 14:39:25 +0000, olcott said:
>>>>>> 
>>>>>>> On 7/19/2024 3:51 AM, Mikko wrote:
>>>>>>>> 
>>>>>>>> You apparently mean that no HHHᵢ can simulate the corresponding DDDᵢ to
>>>>>>>> its termination?
>>>>>>> 
>>>>>>> No I don't mean that at all that incorrectly allocates the error
>>>>>>> to the emulator.
>>>>>> 
>>>>>> Anyway you did not say that some HHHᵢ can simulate the corresponding DDDᵢ
>>>>>> to its termination. And each DDDᵢ does terminate, whether simulated or not.
>>>>> 
>>>>> *Until you quit lying about this we cannot have an honest dialog*
>>>> 
>>>> I don't believe that you can have a honest dialog, at least not without
>>>> a chairman who wants to and can keep the dialog honest.
>>>> 
>>> 
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>> 
>>> When N steps of DDD are emulated by pure function HHH according
>>> to the semantics of the x86 language then N steps are emulated correctly.
>>> 
>>> The subscripts to HHH and DDD pairs are each element of
>>> the set of positive integers ℤ+
>>> 
>>> When we examine the infinite set of every HHH/DDD pair
>>> such that:
>>> 
>>> HHH₁ one step of DDD₁ is correctly emulated by HHH₁.
>>> HHH₂ two steps of DDD₂ are correctly emulated by HHH₂.
>>> HHH₃ three steps of DDD₃ are correctly emulated by HHH₃.
>>> ...
>>> HHHₙ n steps of DDDₙ are correctly emulated by HHHₙ.
>>> 
>>> Then DDD correctly simulated by any pure function HHH cannot
>>> possibly reach its own return instruction and halt, therefore
>>> every HHH is correct to reject its DDD as non-halting.
>> 
>> That does not follow. It is never correct to reject a halting comoputation
>> as non-halting.
>> 
> 
> In each of the above instances DDD never reaches its return
> instruction and halts. This proves that HHH is correct to
> report that its DDD never halts.

The same reasoning "proves" that HHH called by DDD does not return
and therefore HHH is not decider. But the "proof" is not sound.

> 
> When every element of an infinite set of the DDD of HHH/DDD
> pairs never halt (all black cats are black) then HHH can
> report that its DDD never halts (all black cats are cats).
> 
> Most people here seems intentionally ridiculously stupid when
> it comes to hypothetical scenarios:
> 
> _DDD()
> [00002163] 55         push ebp      ; housekeeping
> [00002164] 8bec       mov ebp,esp   ; housekeeping
> [00002166] 6863210000 push 00002163 ; push DDD
> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
> [00002170] 83c404     add esp,+04
> [00002173] 5d         pop ebp
> [00002174] c3         ret
> Size in bytes:(0018) [00002174]
> 
> A correct simulation is defined as emulating the x86 instructions
> of DDD according to the semantics specified by these instructions.
> This includes emulating HHH emulating DDD according to the semantics
> of the x86 instructions of HHH.
> 
> Within the hypothetical scenario where DDD is correctly emulated
> by its HHH and this HHH never aborts its simulation neither DDD
> nor HHH ever stops running.
> 
> This conclusively proves that HHH is required to abort the
> simulation of its corresponding DDD as required by the design
> spec that every partial halt decider must halt and is otherwise
> not any kind of decider at all.
> 
> That HHH is required to abort its simulation of DDD conclusively
> proves that this DDD never halts.

That DDD halts is conclusively proven by a direct excution or DDD.

-- 
Mikko