Re: Proof that DDD specifies non-halting behavior --- point by point

On 2024-08-16 12:42:22 +0000, olcott said:

On 8/16/2024 3:53 AM, Mikko wrote:

On 2024-08-15 15:25:07 +0000, olcott said:
 

On 8/15/2024 5:22 AM, Mikko wrote:

On 2024-08-14 13:06:27 +0000, olcott said:
 

On 8/14/2024 3:17 AM, Mikko wrote:

On 2024-08-14 00:52:36 +0000, olcott said:
 

void DDD()
{
   HHH(DDD);
   return;
}

 In order to prove that the above specifies a non-halting behavour
you must prove that HHH(DDD) does not terminate.

 Wrong.

 At least the proof that DDD does not terminate also proves as an
intermedate result or an obvious corollary that HHH does not halt.
 Non-halting means that an infinite number of instructions can be
executed without halting. That means that at least one instruction
is executed infinitely many times as there are only finitely many
instructions. But not instrunctions of DDD outside HHH is executed
infinitely many times.
 

 Wrong. Non-halting only means that when DDD is emulated
according to the semantics of the x86 language and this
emulation is unlimited that DDD would never reach its
own "return" instruction.

 If what I said is wrong then what you said is wrong, too,
as you say what I said.
 

 *You are getting the computer science incorrectly*
 On 8/2/2024 11:32 PM, Jeff Barnett wrote:
 > ...In some formulations, there are specific states
 >    defined as "halting states" and the machine only
 >    halts if either the start state is a halt state...

These formulations have the problem that it is possible that
a computation reaches a state where the computation shall not
halt and cannot be continued. Therefore most authors prefer a
simpler formulation (introduced by Post) where halting simply
means impossibility to continue. For x86 processors the best
definition is that a computation halts when the SP register
gets a value that is greater than its value at the start of
the computation (in x86 processors the stack grows downwards
to smaller addresses).
--
Mikko