Re: DDD correctly emulated by HHH cannot possibly halt, thinks olcott, but it does.

On 7/10/24 8:24 PM, olcott wrote:

On 7/10/2024 7:01 PM, Richard Damon wrote:

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 have a dead cat in your driveway does not mean that
you have a peanut butter sandwich on your front porch.
It has taken you at least 1000 messages to see that.
>
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.

 WRONG, you don't seem to understand the difference between DDD and HHH's emualtion of it.