Re: How do simulating termination analyzers work? ---Truth Maker Maximalism FULL_TRACE

On 7/5/2025 2:07 PM, Alan Mackenzie wrote:

 You lie.  You don't have a proof.  Many people in this group have pointed
out lots of errors in various versions of your purported proof, which you

void DDD()
{
   HHH(DDD);
   return;
}
*They disagree with this truism* (that seems dishonest)
DDD simulated by HHH according to the semantics of the
C programming language cannot possibly reach its own
simulated "return" instruction final halt state.

just ignore.  The section in Professor Linz's book you used to be so fond
of citing will contain plenty of details, if only you would take the
trouble to understand it (assuming you're capable of such understanding).
 

I have addressed all of those details that you make sure
to ignore so that you can baselessly claim that I am wrong.
There cannot possibly be *AN ACTUAL INPUT* that does the
opposite of whatever its decider decides. All of the examples
of this have never been *ACTUAL INPUTS*
No Turing machine can possibly take another directly executing
Turing machine as in input, thus removing these from the
domain of every halt decider.
*Thus the requirement that HHH report on the behavior*
*of the directly executed DD has always been bogus*
Turing machine partial halt deciders compute the mapping
from their actual inputs to the actual behavior that these
inputs specify.
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
DD simulated by HHH according to the semantics of the C
programming language cannot possibly halt when halting
is defined as reaching its own simulated "return"
statement final halt state.
Because the directly executed DD() is outside of the
domain of HHH its behavior does not contradict HHH(DD)==0.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer