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

On Sat, 2025-07-19 at 16:05 -0500, olcott wrote:

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.

Yes, there is type mismatch problems in nearly all discussions.
But I don't think you will understand what it is.

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;
}
 

A type mismatch: HHH(DD) or HHH(<DDD>)?

DD correctly simulated by HHH cannot reach past
the "if" statement thus cannot reach the "return"
statement. T

That is roughly what HP proof says.

his makes HHH(DD)==0 correct.

How is this statement from? HHH(DD) above shows it cannot return to report 0.
(I guess you might say something and doing another, again)

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