Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

On 2/18/2025 6:25 AM, Richard Damon wrote:

On 2/18/25 6:26 AM, olcott wrote:

On 2/18/2025 3:24 AM, Mikko wrote:

On 2025-02-17 09:05:42 +0000, Fred. Zwarts said:
>

Op 16.feb.2025 om 23:51 schreef olcott:

On 2/16/2025 4:30 PM, joes wrote:

Am Sun, 16 Feb 2025 15:58:14 -0600 schrieb olcott:

On 2/16/2025 2:02 PM, joes wrote:

Am Sun, 16 Feb 2025 13:24:14 -0600 schrieb olcott:

On 2/16/2025 10:35 AM, joes wrote:

Am Sun, 16 Feb 2025 06:51:12 -0600 schrieb olcott:

On 2/15/2025 2:49 AM, Mikko wrote:

On 2025-02-14 12:40:04 +0000, olcott said:

On 2/14/2025 2:58 AM, Mikko wrote:

On 2025-02-14 00:07:23 +0000, olcott said:

On 2/13/2025 3:20 AM, Mikko wrote:

On 2025-02-13 04:21:34 +0000, olcott said:

On 2/12/2025 4:04 AM, Mikko wrote:

On 2025-02-11 14:41:38 +0000, olcott said:

>

DD  correctly simulated by HHH cannot possibly terminate normally.

>
That claim has already shown to be false. Nothing above shows that
HHH does not return 0. If it does DD also returns 0.

When we are referring to the above DD simulated by HHH and not
trying to get away with changing the subject to some other DD
somewhere else

such as one that calls a non-aborting version of HHH
>

then anyone with sufficient knowledge of C programming knows that no
instance of DD shown above simulated by any corresponding instance
of HHH can possibly terminate normally.

Well, then that corresponding (by what?) HHH isn’t a decider.

I am focusing on the isomorphic notion of a termination analyzer.

(There are other deciders that are not termination analysers.)
>

A simulating termination analyzer correctly rejects any input that
must be aborted to prevent its own non-termination.

Yes, in particular itself is not such an input, because we *know* that
it halts, because it is a decider. You can’t have your cake and eat it
too.

I am not even using the confusing term "halts".
Instead I am using in its place "terminates normally".
DD correctly simulated by HHH cannot possibly terminate normally.

What’s confusing about „halts”? I find it clearer as it does not imply
an ambiguous „abnormal termination”. How does HHH simulate DD
terminating abnormally, then? Why doesn’t it terminate abnormally
itself?
You can substitute the term: the input DD to HHH does not need to be
aborted, because the simulated decider terminates.
>

>
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);
}
>
Every simulated input that must be aborted to
prevent the non-termination of HHH is stipulated
to be correctly rejected by HHH as non-terminating.
>

A very strange and invalid stipulation.

>
It merely means that the words do not have their ordinary meaning.
>

>
Unless HHH(DD) aborts its simulation of DD itself cannot possibly terminate normally. Every expert in the C programming language
can see this. People that are not experts get confused by the loop
after the "if" statement.
>

 So? Since it does that, it needs to presume that the copy of itself it sees called does that.
 

Not at all. Perhaps your technical skill is much more woefully
deficient than I ever imagined.
Here is the point that you just missed Unless the first HHH
that sees the non-terminating pattern aborts its simulation
none of them do because they all have the exact same code.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer