Re: Simplified proof that DDD correctly simulated by HHH does not halt

Am Sun, 09 Jun 2024 19:02:12 -0500 schrieb olcott:

On 6/9/2024 2:13 PM, joes wrote:

Am Sun, 09 Jun 2024 13:23:04 -0500 schrieb olcott:

On 6/9/2024 12:59 PM, joes wrote:

Am Sun, 09 Jun 2024 11:07:19 -0500 schrieb olcott:

typedef void (*ptr)(); // pointer to void function 01   void HHH(ptr
P, ptr I)
02   {
03     P(I);
04     return;
05   }
06 07   void DDD(int (*x)())
08   {
09     HHH(x, x);
10     return;
11   }
12 13   int main()
14   {
15     HHH(DDD,DDD);
16   }
 
In the above Neither DDD nor HHH ever reach their own return
statement thus never halt.
Most of my reviewers incorrectly believe that when HH(DD,DD) aborts
its simulated input that this simulated input halts.

Please go back and prove that you understand what infinite recursion
is before proceeding.

Dude, I've got nothing to prove to you.

OK then we are done talking.

How could I even do that? What would convince you?
Ignoring me is not an argument.

You instead could explain how you can call a simulation that differs
from the direct execution "correct".
Or why H and HH are different.

I could but you refuse to go through the steps of the proof,
one-at-a-time with mutual agreement at each step.

Several times I have asked for the other steps.

I am not going to tolerate circular head games that never result in any
mutual agreement.

What head games?

--
joes