Re: Every D(D) simulated by H presents non-halting behavior to H

Op 07.mei.2024 om 17:40 schreef olcott:

On 5/7/2024 6:18 AM, Richard Damon wrote:

On 5/7/24 3:30 AM, Mikko wrote:

On 2024-05-06 18:28:37 +0000, olcott said:
>

On 5/6/2024 11:19 AM, Mikko wrote:

On 2024-05-05 17:02:25 +0000, olcott said:
>

The x86utm operating system: https://github.com/plolcott/x86utm enables
one C function to execute another C function in debug step mode.
Simulating Termination analyzer H simulates the x86 machine code of its
input (using libx86emu) in debug step mode until it correctly matches a
correct non-halting behavior pattern proving that its input will never
stop running unless aborted.
>
Can D correctly simulated by H terminate normally?
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 int main()
10 {
11   H(D,D);
12 }
>
*Execution Trace*
Line 11: main() invokes H(D,D);
>
*keeps repeating* (unless aborted)
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 above execution trace proves that (for every H/D pair of the
infinite set of H/D pairs) each D(D) simulated by the H that this D(D)
calls cannot possibly reach past its own line 03.

>
When you say "every H/D pair" you should specify which set of pairs
you are talking about. As you don't, your words don't mean anything.
>

>
Every H/D pair in the universe where D(D) is simulated by the
same H(D,D) that D(D) calls. This involves 1 to ∞ steps of D
and also includes zero to ∞ recursive simulations where H
H simulates itself simulating D(D).

>
"In the universe" is not a set. In typical set theories like ZFC there
is no universal set.

>

 This template defines an infinite set of finite string H/D pairs where each D(D) that is simulated by H(D,D) also calls this same H(D,D).
 These H/D pairs can be enumerated by the one to ∞ simulated steps of D and involve zero to ∞ recursive simulations of H simulating itself simulating D(D). Every time Lines 1,2,3 are simulated again defines
one more level of recursive simulation.
 1st element of H/D pairs 1 step  of D  is simulated by H
2nd element of H/D pairs 2 steps of D are simulated by H
3rd element of H/D pairs 3 steps of D are simulated by H
 4th element of H/D pairs 4 steps of D are simulated by H
this begins the first recursive simulation at line 01
 5th element of H/D pairs 5 steps of D are simulated by
next step of the first recursive simulation at line 02
 6th element of H/D pairs 6 steps of D are simulated by
last step of the first recursive simulation at line 03
 7th element of H/D pairs 7 steps of D are simulated by H
this begins the second recursive simulation at line 01

Is this the definition of the infinite set of H? We can think of many more simulations that only these. In particular since the H as presented is not a pure function, but uses hidden inputs. If hidden inputs are allowed, it is easy to construct very different H functions, e.g., H functions for which the number of steps differ at each simulation level.

 Can D correctly simulated by H terminate normally?
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 int main()
10 {
11   H(D,D);
12 }
 *Execution Trace*
Line 11: main() invokes H(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 key thing to note is that no D simulated by any H ever reaches
its own line 06 and halts. This means that the input to H(D,D) is
ALWAYS non-halting.
 

It seems that olcott has found a subset of the infinite set of possible H functions for which he can prove that none of them is able to perform a good simulation, so they have to abort and guess an answer, which happens to be wrong.
Not a surprising result, since for any H a D can be constructed for which it will be unable to decide correctly.