Re: D(D) simulated by H cannot possibly reach past its own line 03

On 6/2/2024 4:50 AM, Fred. Zwarts wrote:

Op 03.mei.2024 om 15:40 schreef olcott:

00 int H(ptr x, ptr x)  // ptr is pointer to int function
01 int D(ptr x)
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 void main()
10 {
11   H(D,D);
12 }
 We are examining the behavior of D(D) for every possible H/D pair
where 1 to N steps of D(D) are simulated by H.
 *Execution Trace*
Line 11: main() invokes H(D,D) that simulates D(D)
 *keeps repeating* (unless aborted)
Line 01
Line 02
Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)
 *Simulation invariant*
D correctly simulated by H cannot possibly reach past its own line 03.

 The Linz proof is based on the pathological relation ship that D contradicts the result of H. This is expressed in lines 04, 05, 06 of D, above.
It is strange that olcott claims that the simulation never sees the pathological part of D. He now seems to shift the meaning of pathological to the mere fact that D calls H. Lines 04, 05, and 06 are completely irrelevant.
In fact, any function that calls H now become pathological.
 E.G. if we replace D with a function P that only returns its own Halt_Status:
 01 int P(ptr x)
02 {
03   int Halt_Status = H(x, x);
04
05
06   return Halt_Status;
07 }
 Then we would normally expect that, because H is required to halt, P would halt as well. But the simulation of P by H does not halt. So, P, when it halts, reports that it not halting.
 The problem here is, that H is unable to simulate itself to its final state. That has no relation with the Linz proof, it is a problem of H.
 So, there is another *Simulation invariant*
H correctly simulated by H cannot possibly reach its own return.
 

 Already addressed in another reply.