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

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.
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It seems that neither your D or your H actual meet the normal definition of what a "Program" is.
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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
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>
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So, what is the defined "class" of the input to a Termination Analyzer.
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I am only talking about H and D. You always "read things in"
to what I say that I never said.

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So, what are "H" and "D", are they "Programs" per the standard definitions, or something else that you are stipulating?
>

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They are 100% completely defined in the complete source-code
that I just linked above.

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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.
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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.
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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.
>
>

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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:
Simulating Termination Analyzer H determines whether or not D(D) Simulated bu H will reach its final state at its own line 06 and halt, or if H decided to abort its simulation.
Since if H aborts its simulation, that it by definition trivially true, the problem statement is just a toy.

 (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.

But there is only one H so "Whether or not" is MEANINGLESS.
The ONE AND ONLY H will abort its simulation and thus by definition D(D) will not reach its final state

 I copied that to an off-line file so I can make it a canned
boiler-plate reply to every reply that tries to change the subject.

And by doing so, you are showing that you are not reading the reply, liokely because the ligic is just over your head.

 Quite a few experts in C confirmed that (a) is definitely true
entirely on the basis of the 12 lines of code shown above.
Maybe you are not very good at C?
 

Nope, it is true because if a simulator aborts its simulation, irrespective of the input given, that input will not reach the final state. By defining the problam space to just a single decider and a single input you have slashed the problem to be just a TOY.
IF it took you 20 years to create a toy non-refutation of the proof, you have a lot of work to go. (non-refutation as your D wasn't actually built by the Linz Template, since it doesn't have the required COPY of the decider)