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

On 5/29/24 9:17 AM, olcott wrote:

On 5/29/2024 4:08 AM, Mikko wrote:

On 2024-05-28 14:37:15 +0000, olcott said:
>

On 5/28/2024 2:17 AM, Mikko wrote:

On 2024-05-27 13:52:09 +0000, olcott said:
>

On 5/27/2024 3:11 AM, Mikko wrote:

On 2024-05-26 16:50:21 +0000, olcott said:
>

>
<snip>
So that: *Usenet Article Lookup*
http://al.howardknight.net/
can see the whole message now that
*the Thai spammer killed Google Groups*
>
typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int D(ptr p)
02       {
03         int Halt_Status = H(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         H(D,D);
12         return 0;
13       }
>

When we see that D correctly simulated by pure simulator H would remain
stuck in recursive simulation then we also know that D never reaches its
own line 06 and halts in less than an infinite number of correctly
simulated steps.

>
Which means that H never terminates. You said that by your definition
a function that never terminates is not a pure function. Therefore
H, if it exists, is not a pure function, and the phrase "pure function
H" does not denote.

>
*I should have said that more clearly*
*That is why I need reviewers*

>
What is said here has little significance. You should ask reviewer's
comments about your working draft that you maintain in GitHub or some
other web site. And you should update that draft when a reviewer finds
some something wrong or unclear.
>

*This is STEP ONE of my four step proof*
STEP TWO applies these same ideas to the Peter Linz HP proof.
STEP THREE shows how the Linz Ĥ.H sees the behavior of its recursive
      simulations.
STEP FOUR shows why the behavior of the INPUT is the correct basis.

>
When discussing individual steps of a proof draft you shold start
at the last step.

>
I have been at this for twenty years.
Ben Bacarisse has spoken with me for 15 of those years.
>
What I found out is that exactly one point of one step
must be pursued until complete closure of that one point
of that one step.

>
How long do you think it takes to cover them all, starting
from the first?
>
How long do you think it would take, starting from the last?
>

 The first point was closed in less than one hour with each of
four people that were not playing head games.
 typedef int (*ptr)();  // ptr is pointer to int function in C
00       int H(ptr p, ptr i);
01       int D(ptr p)
02       {
03         int Halt_Status = H(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
08
09       int main()
10       {
11         H(D,D);
12         return 0;
13       }
 The above template refers to an infinite set of H/D pairs where D is
correctly simulated by either pure simulator H or pure function H.  This
was done because many reviewers used the shell game ploy to endlessly
switch which H/D pair was being referred to.
 H correctly simulates 1 to ∞ steps of D with either pure function H or
pure simulator H. In none of these cases does the correctly simulated D
ever reach its own simulated final state and halt.

No, DIFFERENT H's simulate different number of steps of DIFFERENT instantations of the template D.
Until you define how to simulate "a template' by your definition, when the template doesn't HAVE the code of H to simulate, you are just shown to be LYING.

 Correct Simulation Defined
    This is provided because many reviewers had a different notion of
    correct simulation that diverges from this notion.
     A simulator is an x86 emulator that correctly emulates 1 to N of the
    x86 instructions of D in the order specified by the x86 instructions
    of D. This may include M recursive emulations of H emulating itself
    emulating D.

But your "template
  D doesn HAVE any x86 instrucitons of H.
Only all the infinte number of DIFFERENT instantiations of it.
Of course, different instantiations might have different behavior, so you can't just interchange them.

 When we see that D correctly simulated by pure simulator H would remain
stuck in infinite recursive simulation then we also know that less than
an infinite number of steps is not enough steps for D correctly
simulated by pure function H to reach its own simulated final state at
line 06 and halt.
 

Nope. NONE of the D's are stuck in "infinite" recursion for ANY H that actually aborts its simulation, since that means the H that it calls will also do that and terminte the apparent infinite recursion.
The problem is you can't actuala use the longer simulation to show the shorter simulations won't end, as the inputs are NOT the same, since, to do the simulation, you have created DIFFERENT instantiations.