Simplified proof that DDD correctly simulated by HHH does not halt

On 6/9/2024 9:16 AM, Andy Walker wrote:

If anyone says something new, perhaps they could indicate this in the "Subject" line, as a service to humanity.
 

*You must know the C programming language to understand this*
typedef void (*ptr)(); // pointer to void function
void HHH(ptr P, ptr I)
{
   P(I);
   return;
}
void DDD(int (*x)())
{
   HHH(x, x);
   return;
}
int main()
{
   HHH(DDD,DDD);
}
In the above Neither DDD nor HHH ever reach their own return
statement thus never halt.
When HHH is a simulating halt decider then HHH sees that
DDD correctly simulated by HHH cannot possibly reach its
own return statement, AKA
    simulating halt decider HHH correctly simulates its input DDD
    until HHH correctly determines that its simulated DDD would never
    stop running unless aborted
*as defined here*
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
   If simulating halt decider H correctly simulates its input D
   until H correctly determines that its simulated D would never
   stop running unless aborted then
   H can abort its simulation of D and correctly report that D
   specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words10/13/2022>
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer