Re: HHH maps its input to the behavior specified by it --- key error in all the proofs --- Mike --- basis

On 8/15/2024 3:46 AM, Mikko wrote:

On 2024-08-14 13:42:33 +0000, olcott said:
 

On 8/14/2024 2:30 AM, Mikko wrote:

On 2024-08-13 13:30:08 +0000, olcott said:
>

On 8/13/2024 6:23 AM, Richard Damon wrote:

On 8/12/24 11:45 PM, olcott wrote:

>
void DDD()
{
   HHH(DDD);
   return;
}
>
*DDD correctly emulated by HHH cannot possibly reach its*
*own "return" instruction final halt state, thus never halts*
>

>
Which is only correct if HHH actuallly does a complete and correct emulation, or the behavior DDD (but not the emulation of DDD by HHH) will reach that return.
>

>
A complete emulation of a non-terminating input has always
been a contradiction in terms.
>
HHH correctly predicts that a correct and unlimited emulation
of DDD by HHH cannot possibly reach its own "return" instruction
final halt state.

>
That is not a meaningful prediction because a complete and unlimited
emulation of DDD by HHH never happens.

>
A complete emulation is not required to correctly
predict that a complete emulation would never halt.

 Nice to see that you don't disagree.
 

In other words you agree with the first two of these?
A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is necessarily correct.
A correct simulation of N instructions of DDD by HHH is
sufficient to correctly predict the behavior of an unlimited
simulation.
Termination analyzers / halt deciders are only required
to correctly predict the behavior of their inputs.
Termination analyzers / halt deciders are only required
to correctly predict the behavior of their inputs, thus
the behavior of non-inputs is outside of their domain.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer