Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT

On 6/16/2025 3:34 AM, Mikko wrote:

On 2025-06-15 20:10:32 +0000, olcott said:
 

void DDD()
{
   HHH(DDD);
   return;
}
>
When I challenge anyone to show the details of exactly
how DDD correctly simulated by ANY simulating termination
analyzer HHH can possibly reach its own simulated "return"
statement final halt state they ignore this challenge.
>
THAT FACT THAT NOT ONE PERSON HAS MET THIS CHALLENGE
IN SEVERAL YEARS IS VERY STRONG EVIDENCE THAT I AM CORRECT.
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
DDD is a simplified version of DD.

 HHH is not inherently correct or incorrect.

Wrong.
Every Simulating termination analyzer HHH(DDD) correctly
determines that
When one or more instructions of DDD are correctly
simulated by ANY simulating termination analyzer HHH
that DDD never reaches its simulated "return" statement
final halt state and halts.

To be correct simply means
that it does what it is required to do and doesn't do what it is
required not to do. Therefore any discussion of correcteness is a
discussion about the requirements. The only reference to the requirements
above is "any simulating termination analyzer" which means that the
requirements are that HHH be a termination analyzer and an additional
vague requirement that it simulates something.
 The value of a challenge is mainly its entertainment value, if any.
 

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer