Re: DDD simulated by HHH cannot possibly halt (Halting Problem)

On 4/8/25 11:17 AM, olcott wrote:

On 4/8/2025 2:51 AM, Mikko wrote:

On 2025-04-08 04:33:57 +0000, olcott said:
>

On 4/7/2025 3:16 AM, Mikko wrote:

On 2025-04-06 16:12:36 +0000, olcott said:
>

On 4/6/2025 5:27 AM, Mikko wrote:

On 2025-04-05 16:45:28 +0000, olcott said:
>

On 4/5/2025 2:05 AM, Mikko wrote:

On 2025-04-05 06:18:06 +0000, olcott said:
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On 4/4/2025 3:12 AM, Mikko wrote:

On 2025-04-04 01:27:15 +0000, olcott said:
>

void DDD()
{
    HHH(DDD);
    return;
}
>
Do you really think that anyone knowing the C
programming language is too stupid to see that
DDD simulated by HHH cannot possibly return?

>
Anyone knowing the C language can see that if DDD() does not halt
it means that HHH(DDD) does not halt. The knowledge that that
means that HHH is not a decider is possible but not required.
>

>
*Perpetually ignoring this is not any actual rebuttal at all*
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination. The
only rebuttal to this is rejecting the notion that
deciders must always halt.

>
Wrong, because a termination analyzer is not required to halt.

>
Why say things that you know are untrue?

>
The term "termination analyzer" is used about programs that do not halt
on every input. There is no strict derfiniton of the term so there is
no requirement about halting.
>
On the first page of https://www.cs.princeton.edu/~zkincaid/pub/ pldi21.pdf
in the first parapgraph of Introduction:
>
   For example, termination analyzers may themselves fail to terminate on
   some input programs, or ...
>

A termination analyzer that doesn't halt
would flunk every proof of total program correctness.

>
There are no total termination analyzers.

>
Total proof of correctness does not require a halt
decider, it only requires a termination analyzer
with inputs in its domain.

>
Depends on what one wants to prove correct.
>
Often there is no way to determine whether a pariticular termination
analyser can determine about a particular program.
>

>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.
>
Rational minds would agree that the above principle
is correct and directly applies to HHH(DD) rejecting
its input.

>
 From the meaning of the word "correct" obviously follows that it is
never correct to reject a terminating input as non-terminating.
>

 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 This stuff is simply over-your-head.
HHH(DD) meets the above: *Simulating termination analyzer Principle*
Anyone with sufficient competence with the C programming language
will understand this.
 

NO, because either:
DD isn't a program, and thus HHH can't decide if it halts as it can't be simulated by a pure function, which you admit is a requirement for a proper decider, or
DD includes the code for the HHH that it calls, and thus the fact that this HHH *WILL* abort its simulation and return (since you claim that is what the outer simulating HHH, at the same memory address does) and thus HHH can't say that it DD doesn't halt by itself.
Note, you can't actually change the HHH at that address, or you are just admitting to changing the input, and thus admitting you are working on a strawman.