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

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

On 2025-02-18 11:26:25 +0000, olcott said:

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.
>

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.

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

I have given everyone here all of the complete source code for a few
years
918-1156 // All of the lines of termination analyzer HHH
https://github.com/plolcott/x86utm/blob/master/Halt7.c

Every expert in the C programming language can see this.

They can't when they can't see HHH and even then it is not obvious,
so the claim on the subject line is false.