Re: HHH(DDD)==0

On 10/9/2024 5:08 AM, Mikko wrote:

On 2024-10-09 03:47:10 +0000, olcott said:
 

On 10/8/2024 7:49 AM, Andy Walker wrote:
 > Richard -- no-one sane carries on an extended discussion with
 > someone they [claim to] consider a "stupid liar".  So which are you?
 > Not sane?  Or stupid enough to try to score points off someone who is
 > incapable of conceding them?  Or lying when you describe Peter?  You
 > must surely have better things to do.  Meanwhile, you surely noticed
 > that Peter is running rings around you.
>
I am incapable of conceding this self-evident truth:
>
   DDD emulated by each corresponding HHH that can possibly
   exist never returns

 That is not self-evident or even meaningful without a definition of
"each corresponding HHH".
 

An HHH/DDD pair such that DDD calls its own emulator.
HHH is an emulating termination analyzer that takes the machine
address of DDD as input then emulates the x86 machine language
of DDD until a non-terminating behavior pattern is recognized.
HHH recognizes this pattern when HHH emulates itself emulating DDD
void DDD()
{
   HHH(DDD);
   return;
}
One cannot simply ignore the actual behavior specified by the
finite string x86 machine language of DDD such that
DDD emulated by each corresponding HHH that can possibly
exist never returns
thus each of the directly executed HHH emulators that does
return 0 correctly reports the above non-terminating behavior.
https://github.com/plolcott/x86utm x86utm operating system
Every executed HHH that returns 0 correctly reports that
DDD emulated by its corresponding HHH never returns.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer