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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Liar detector: Fred, Richard, Joes and Alan --- Honest Dialogue ?
Date: Mon, 22 Jul 2024 11:01:20 +0300
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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.

-- 
Mikko