Re: Proof that DD correctly simulated by HH has different behavior than DD(DD) STEP(1)

On 6/10/2024 1:21 PM, joes wrote:

Am Mon, 10 Jun 2024 12:54:48 -0500 schrieb olcott:

On 6/10/2024 6:16 AM, Richard Damon wrote:

On 6/10/24 1:17 AM, olcott wrote:

On 6/9/2024 1:33 AM, Fred. Zwarts wrote:

Op 08.jun.2024 om 20:47 schreef olcott:

 

Try to show how this DD correctly simulated by any HH ever stops
running without having its simulation aborted by HH.

>
Stopping at your first error. So, we can focus on it. Your are asking
a question that contradicts itself.
A correct simulation of HH that aborts itself, should simulate up to
the point where the simulated HH aborts. That is logically
impossible. So, either it is a correct simulation and then we see
that the simulated HH aborts and returns, or the simulation is
incorrect, because it assumes incorrectly that things that happen
(abort) do not happen.

Why does it need to abort, when its recursive simulation does the same
and thus returns, not needing to be aborted?
 

void Infinite_Recursion(u32 N)
{
   Infinite_Recursion(N);
}
*D correctly simulated by H is more complex yet isomorphic to the above*
*DDD is the simplest possible version of D*
00   typedef void (*ptr)(); // pointer to void function
01
02   int HH(ptr P, ptr I);
03
04   void DDD(int (*x)())
05   {
06     HH(x, x);
07     return;
08   }
09
10   int main()
11   {
12     HH(DDD,DDD);
13   }
Line 12 main()
   invokes HH(DDD,DDD); that simulates DDD()
*REPEAT UNTIL outer HH aborts*
   Line 06 simulated DDD()
   invokes simulated HH(DDD,DDD); that simulates DDD()
--
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