Re: D simulated by H never halts no matter what H does V3

Liste des GroupesRevenir à s logic 
Sujet : Re: D simulated by H never halts no matter what H does V3
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logic
Date : 28. Apr 2024, 02:21:49
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v0k4qt$laej$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
User-Agent : Mozilla Thunderbird
On 4/27/2024 6:44 PM, Richard Damon wrote:
On 4/27/24 7:29 PM, olcott wrote:
On 4/27/2024 6:09 PM, Richard Damon wrote:
On 4/27/24 6:56 PM, olcott wrote:
On 4/27/2024 5:45 PM, Richard Damon wrote:
On 4/27/24 6:29 PM, olcott wrote:
On 4/27/2024 5:19 PM, Richard Damon wrote:
On 4/27/24 6:02 PM, olcott wrote:
On 4/27/2024 4:45 PM, Richard Damon wrote:
On 4/27/24 5:36 PM, olcott wrote:
On 4/27/2024 4:24 PM, Richard Damon wrote:
On 4/27/24 3:48 PM, olcott wrote:
Simulating termination analyzer H determines whether or not
D(D) simulated by H can possibly reach its final state at its
own line 06 and halt whether or not H aborts its simulation.
>
We can resolve exactly what I mean by this as an aspect of
staying on this one point. We cannot move on to the slightest
trace of any nuance of any other point until AFTER we have
100% complete mutual agreement on this point.
>
(a) It is a verified fact that D(D) simulated by H cannot
possibly reach past line 03 of D(D) simulated by H whether H
aborts its simulation or not.
>
When we have 100% perfect mutual agreement on that point
then we can move on to the next aspect of the point of the
paragraph.
>
>
The problem is you don't seem to have a proper definition for a "program", as the input seems to change behavior as you analyize different options for what "different" H's might do.
>
It seems that neither your D or your H actual meet the normal definition of what a "Program" is.
>
>
I never even use the word "program"
*H and D are 100% completely specified right here*
https://github.com/plolcott/x86utm/blob/master/Halt7.c
>
>
>
So, what is the defined "class" of the input to a Termination Analyzer.
>
>
I am only talking about H and D. You always "read things in"
to what I say that I never said.
>
So, what are "H" and "D", are they "Programs" per the standard definitions, or something else that you are stipulating?
>
>
They are 100% completely defined in the complete source-code
that I just linked above.
>
So, if H is "defined" by its source code, then it can only do one thing, and thus your criteria of talking about "whether it aborts its simulation or not" is a MEANINGLESS Statement.
>
>
It <is> the mandatory prerequisite to proceeding to additional steps.
The end of these steps will show how it PRECISELY applies to the Linz
proof.
>
You cannot truthfully say that I have not defined H and D sufficiently
when I provide the full source code to their fully operational system.
>
>
>
Since there is now no "problem" left for H to be tested to see if it "solves" what can be agreed to?
>
01 int D(ptr x)  // ptr is pointer to int function
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 void main()
10 {
11   H(D,D);
12 }
>
Simulating termination analyzer H determines whether or not
D(D) simulated by H can possibly reach its final state at its
own line 06 and halt whether or not H aborts its simulation.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
 But, since you have stipulated that there only exist ONE H, the H provided (since you won't define the class of object that H is) this statement reduces to:
 
I cut-and-pasted the exact verbatim word-for-word post that
C experts found very easy to answer.
If you don't give me a straight answer I will assume that you
are dodging because you want to hide the fact that you are mostly
clueless about C.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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