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

On 2/20/25 4:31 PM, olcott wrote:

On 2/20/2025 2:38 AM, Mikko wrote:

On 2025-02-20 00:31:33 +0000, olcott said:
>

On 2/19/2025 3:01 AM, Mikko wrote:

On 2025-02-18 11:26:25 +0000, olcott said:
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On 2/18/2025 3:24 AM, Mikko wrote:

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

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DD  correctly simulated by HHH cannot possibly terminate normally.

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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.
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>
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);
}
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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.
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A very strange and invalid stipulation.

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

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Those two comments are not discussed below.
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Unless HHH(DD) aborts its simulation of DD itself cannot possibly terminate normally.

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That cannot be determined without examination of HHH, which is not in the
scope of OP.

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I have given everyone here all of the complete source
code for a few years

>
True but irrelevant. OP did not specify that HHH means that particular
code.
>

 Every post that I have been talking about for two or
more years has referred to variations of that same code.
 

And thus, you LIE when you claim that HHH does something different then that code;
Note, a given program doesn't have "variations" but is always exactly what it is.
If you have variations, you don't have a single defined program, and all your work is prove to be based on admitted lies.
EIther HHH is one specific program always and forever, or you need to specify which variation you are using at every instance, something you do not do as it would prove you are just lying.