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

On 5/8/2024 3:59 AM, Mikko wrote:

On 2024-05-07 19:05:54 +0000, olcott said:
 

On 5/7/2024 1:54 PM, Fred. Zwarts wrote:

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.

>
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).
>
No-one can possibly show one element of this set where D(D) reaches
past its own line 03.

 If H is a decider of any kind then the D build from it reaches its line
4 as numberd above. Whether the simulation of D by H reaches that line
is another question.
 

*My fully operational code proves otherwise*
I seems like you guys don't have a clue about how infinite
recursion works. You can run the code and see that I am correct.
I have one concrete instance as fully operational code.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
line 555 u32 HH(ptr P, ptr I) its input in on
line 932 int DD(int (*x)())
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer