Re: Four Chatbots figure out on their own without prompting that HHH(DDD)==0

On 7/19/2025 3:57 PM, wij wrote:

On Sat, 2025-07-19 at 15:41 -0500, olcott wrote:

On 7/19/2025 3:14 PM, wij wrote:

>
HP is very simple: H(D)=1 if D halts, H(D)=0 if D does not halt.
>

>
The standard proof assumes a decider
H(M,x) that determines whether machine
M halts on input x.
>
But this formulation is flawed, because:

 Whatever the 'formulation' is, the HP result is a fact that no H can decide
the halting status of any given D.
 

And that is wrong because H(⟨D⟩) is correctly determined.
It has always been a type mismatch error when H(D) was
assumed.

Turing machines can only process finite encodings
(e.g. ⟨M⟩), not executable entities like M.
>
So the valid formulation must be
H(⟨M⟩,x), where ⟨M⟩ is a string.

 Halting Problem::= H(D)=1 if D halts, H(D)=0 if D does not halt.
The conclusion is, no such H exists.
 

And that is wrong because H(⟨D⟩) is correctly determined.
It has always been a type mismatch error when H(D) was
assumed.
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
DD correctly simulated by HHH cannot reach past
the "if" statement thus cannot reach the "return"
statement. This makes HHH(DD)==0 correct.

'formulation' does not really matter.
If 'formulation' matters, it is another problem.
 

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