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

On 2024-08-15 15:13:48 +0000, olcott said:

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?

I didn't say that you agree, I only said you did't disagree.

A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is necessarily correct.

There is no necessary. I can't agree because most of them are too
unimportant to be checked. But I havn't noticed any other errors
than omission of some (mainly unimportan) details without an
indication of omission and some lack of clarity that would be
easily avoided if the intent were not to deceive.
--
Mikko