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

On 08/05/2024 14:01, olcott wrote:

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

On 2024-05-07 19:05:54 +0000, olcott said:
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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:
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On 5/6/2024 11:19 AM, Mikko wrote:

On 2024-05-05 17:02:25 +0000, olcott said:
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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 }
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*Execution Trace*
Line 11: main() invokes H(D,D);
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*keeps repeating* (unless aborted)
Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)
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*Simulation invariant*
D correctly simulated by H cannot possibly reach past its own line 03.
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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.
>

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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.

>

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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).
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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.
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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
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4th element of H/D pairs 4 steps of D are simulated by H
this begins the first recursive simulation at line 01
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5th element of H/D pairs 5 steps of D are simulated by
next step of the first recursive simulation at line 02
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6th element of H/D pairs 6 steps of D are simulated by
last step of the first recursive simulation at line 03
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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)())

HH is completely broken - it uses a global variable which is allows HH to detect whether it is the outer HH or a nested (simulated) HH.  As a result, the nested HH behaves completely differently to the outer HH - I mean /completely/ differently: it goes through a totally separate "I am called in nested mode" code path!
You agreed this a few weeks ago.  Or have you fixed this code now?  I bet you haven't.
Until you fix the code you should not be referring to it as "fully operational" code, because it is simply WRONG and so does not demonstrate anything at all.
Regards,
Mike.