Re: DDD correctly emulated by HHH cannot possibly halt

On 7/10/24 9:41 AM, olcott wrote:

On 7/10/2024 8:27 AM, joes wrote:

Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:

On 7/9/2024 11:01 PM, joes wrote:
  > That means that HHH doesn't return, in particular that it doesn't
  > abort.
DDD correctly emulated by any pure function HHH that correctly emulates
1 to ∞ steps of DDD can't make it past the above line of code no matter
what.

That line being the call to itself -> it can't simulate itself.
>

*DDD NEVER HALTS*

DDD ONLY calls HHH...
>

 void DDD()
{
   HHH(DDD);
   return;
}
 DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
 

Nope, DDD does if HHH(DDD) returns.
You mean the emulation by HHH never gets there which is something different.
The first is behavior in the real world of the programs.
The second is the observation of that behavion by HHH.
HHH only gets partial information before it stops its observation, and thus doesn't actually know the right answer.
You just don't understand the difference between truth and knowledge, which means you don't understand what truth actually is.