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

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Sujet : Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theory
Date : 16. Feb 2025, 21:02:08
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <3b8a5f4be53047b2a6c03f9678d0253e137d3c40@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
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:

This is the only topic that I will discuss and any
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);
}
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.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

Date Sujet#  Auteur
30 May 25 o 

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