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

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.
Tolerating the dishonest dodge strawman deception
CHANGE-THE-SUBJECT Fake rebuttal ONLY TALKS IN CIRCLES.
We are almost done with the first step. After we are
done with the first step we can move on to the second
step. Talking in circles requires infinite time and
I have much less that that.

That is the most important step and if it is not
inferred correctly from one or two earlier sentences the other steps
don't matter. If no problem is found with the last step you can ask
about earlier steps.
 

When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
I formulated four key points that together prove that
H(D,D) is correct to report that its input does not halts
and embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ is equally correct for its input.
People that resort to dogma instead of reasoning will not
see it this way yet be unable to correctly show any error
in my reasoning.
Some of these people unable to find any actual error resort to
ad hominem personal attack instead of reasoning demonstrating
a pathetic degree of professional decorum that ruins their own
credibility.

Steps of a draft proof should not be numbered as you may need to add
more steps if you find that the proof is not clear enough. Instead,
you should use labels that do not imply any particular order.
 

*I have been at this same proof full time for three years*
It is exactly four steps. Steps ONE and TWO mutually support each other.
On 3/1/2024 12:41 PM, Mike Terry wrote:
Confirmed that the essence of of STEP THREE is computable.
STEP FOUR seems like nonsense until after steps ONE, TWO AND THREE
are established.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer