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

Op 30.mei.2024 om 15:16 schreef olcott:

On 5/30/2024 5:14 AM, Fred. Zwarts wrote:

Op 28.mei.2024 om 17:10 schreef olcott:

On 5/28/2024 4:17 AM, Fred. Zwarts wrote:

Op 27.mei.2024 om 16:39 schreef olcott:

On 5/27/2024 9:10 AM, Richard Damon wrote:

On 5/27/24 9:52 AM, olcott wrote:

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*
*Here it is more clearly*
>
When we hypothesize that H is a pure simulator we see that D correctly
simulated by pure simulator H remains stuck in recursive simulation thus
never reaches its own simulated final state at its line 06 and halts. In
this case H does not halt, thus is neither a pure function nor a
decider.

>
But when you hypothesize that H is actually a "pure simulator" (presumably one that never aborts) then you are creating a D that uses that pure simulator, and are ONLY deriving conclusions for such a D.
>

>
When D correctly simulated by pure simulator H cannot possibly reach
its own simulated final state at line 06 and halt in an infinite number
of simulated steps we can conclude that less than an infinite number of
steps is also not enough steps for D to halt.
>

>
When H correctly simulated by pure simulator H cannot possibly reach
its own simulated final state and halt in an infinite number
of simulated steps we can conclude that less than an infinite number of
steps is also not enough steps for H to halt.
>

>
When D correctly simulated by pure simulator H cannot possibly reach
its own simulated final state at line 06 and halt in an infinite number
of simulated steps we can conclude that less than an infinite number of
steps are also not enough steps for D to halt.
>
This tells us that when 1 to ∞ steps of D are correctly simulated by
H that D correctly simulated by H never halts under any circumstances
what-so-ever
*CONCLUSIVELY PROVING THAT D CORRECTLY SIMULATED BY H DOES NOT HALT*
>
Once we have mutual agreement and complete closure on this we can
move on to STEP TWO that analyzes these exact same issues with the
Linz proof. The H/D pairs were needed so that people cannot cheat
on the meaning of correct simulation.
>
*Dozens of people have consistently insisted on cheating on this*
*Dozens of people have consistently insisted on cheating on this*
*Dozens of people have consistently insisted on cheating on this*
>

>
When H correctly simulated by pure simulator H cannot possibly reach
its own simulated final state and halt in an infinite number
of simulated steps we can conclude that less than an infinite number of
steps are also not enough steps for H to halt.
>
This tells us that when 1 to ∞ steps of H are correctly simulated by
H that H correctly simulated by H never halts under any circumstances
what-so-ever
*CONCLUSIVELY PROVING THAT H CORRECTLY SIMULATED BY H DOES NOT HALT*
>
By this proof, we see that this H contradicts the requirement that it should halt.
We cannot continue with this before we have agreement about it.

 You should put my words in quotes otherwise people will be thinking
that you agree with me word-for-word.
 Not at all. When we see that the infinite set of every possible D
simulated by any H cannot possibly reach its own simulated final state
and halt this remains true for D correctly simulated by simulating halt
decider H. H halts and correctly simulated D never halts.
 

It is clear for anyone with a little bit of C knowledge, that if H halts, then D continues with line 04.
So, if the claim is true that the simulation of D does not reach line 04 it means that the simulation of H does not halt.
Line 04, 05 and 06 are completely irrelevant, because the simulator never uses them, if the claim is true. So, D's only function is the parameter duplication, so that H simulates itself. The simulation of H never reaches H's final state, so olcott should have said:
"When we see that the infinite set of every possible H
simulated by this H cannot possibly reach its own simulated final state
and halt this remains true for H correctly simulated by simulating halt
decider H. H is required to halt but the correctly simulated H never halts."
This contradicts the requirement that H should halt.
That the directly executed H halts is irrelevant because the directly executed D also halts and olcott claims that it is irrelevant. Both D and H, when simulated, never reach their final state.
We can only continue if we have 100% agreement about this.
If not true, show how the simulated H reaches its final state.