Re: Liar detector: Fred, Richard, Joes and Alan --- Honest Dialogue ? --- Infinite set of HHH/DDD pairs

On 7/23/2024 2:14 AM,   wrote:

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.
 

 HHH(DDD) does halt. Some of the data that HHH simulates
is not required to halt.