Re: Liar detector:Peter Olcott --- Honest Dialogue ?

On 7/21/24 9:50 AM, olcott wrote:

On 7/21/2024 4:38 AM, Mikko wrote:

On 2024-07-20 13:28:36 +0000, olcott said:
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On 7/20/2024 3:54 AM, Mikko wrote:

On 2024-07-19 14:39:25 +0000, olcott said:
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On 7/19/2024 3:51 AM, Mikko wrote:

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You apparently mean that no HHHᵢ can simulate the corresponding DDDᵢ to
its termination?

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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.

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*Until you quit lying about this we cannot have an honest dialog*

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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.

Except that with the code for HHH include, N can't be over 4.

 The subscripts to HHH and DDD pairs are each element of
the set of positive integers ℤ+

And every DDDi is a DIFFERENT input (since to be the actual description of that program, it inlcudes at least mplicitly, the code of tHHHi

 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.
 

No, you show that HHHi doesn't reach the end of DDDi in its PARTIAL emulation.
But HHHinf when given the DDDi for any finite i *WILL* reach that end, because it will trace DDDi calling HHHi(DDDi) which, by your own definition, will simulate i steps of DDDi then return and then DDDi will return, all seen by HHHinf.
Thus, you claim is juat a LIE, you just try to ignore the cases where you are wrong,
In fact, for every DDDi, where i is finite, there will be a j > i such that ALL HHHj(DDDi) will emulate DDDi to its end.
Thus you claim that NO HHH gets there is just a LIE done by only looking at one input for each HHHi.