Re: Who knows that DDD correctly simulated by HHH cannot possibly reach its own return instruction final state? BUT ONLY that DDD

On 8/3/24 7:36 PM, olcott wrote:

On 8/3/2024 5:51 PM, Richard Damon wrote:

On 8/3/24 6:15 PM, olcott wrote:

On 8/3/2024 5:07 PM, Richard Damon wrote:

>
The problem is that every one of those emulation is of a *DIFFERENT* input, so they don't prove anything together except that each one didn't go far enough.

>
void DDD()
{
   HHH(DDD);
   return;
}
>
When each HHH correctly emulates 0 to infinity steps of
its corresponding DDD and none of them reach the "return"
halt state of DDD then even the one that emulated infinite
steps of DDD did not emulate enough steps?
>
>

>
Just says lying YOU.
>
You got any source for that other than yourself?
>

 It is self-evident and you know it. I do have four
people (two with masters in CS) that attest to that.
*It is as simple as I can possibly make it*

Maybe to your mind filled with false facts, but it isn't true.

 I wonder how you think that you are not swearing your
allegiance to that father of lies?

Because, I know I speak the truth.
Why do you not think you are lying?

 Anyone that truly understands infinite recursion knows
that DDD correctly simulated by HHH cannot possibly reach
its own "return" final state.

Right, but for every other HHH, which the ones that answer are, it isn't a fact.

 Surpisingly (to me) Jeff Barnett set the record straight
on exactly what halting means.
 

No, there is one, and only one definition, it is a machine that reaches its final state.
Note, *a machine*, not a (partial) emulation of the machine
And it is the FULL machine, so for DDD, it includes the HHH that it was built on.