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

On 2/10/2025 5:18 AM, joes wrote:

Am Sun, 09 Feb 2025 13:57:03 -0600 schrieb olcott:

On 2/9/2025 1:39 PM, joes wrote:

Am Sun, 09 Feb 2025 10:49:51 -0600 schrieb olcott:

On 2/9/2025 10:43 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 17:37 schreef olcott:

On 2/9/2025 9:53 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 16:15 schreef olcott:

On 2/9/2025 2:09 AM, Fred. Zwarts wrote:

Op 09.feb.2025 om 07:04 schreef olcott:

On 2/8/2025 3:49 PM, Fred. Zwarts wrote:

Op 08.feb.2025 om 15:43 schreef olcott:

On 2/8/2025 3:54 AM, Fred. Zwarts wrote:

Op 08.feb.2025 om 00:13 schreef olcott:

>

The input to HHH(DD) cannot possibly terminate normally.
Referring to some other DD does not change this verfied fact.

That DD halts is a verified fact.

The input to HHH(DD) DOES NOT HALT !!!

It is a verified fact that the finite string describes a halting
program. Du to a bug, HHH does not see that, because it
investigates only the first few instructions of DD. HHH is unable
to process the call from DD to HHH correctly.

DD simulated by HHH cannot possibly terminate normally. DD simulated
by HHH does specify the behavioral basis of the Boolean termination
value of the DD input to HHH.

DD terminates, and HHH can’t simulate it normally.

That is not the same DD as the input to HHH(DD). That DD has an
entirely different execution trace.

That is entirely due to how HHH chooses to missimulate it, namely by
not calling itself, but a different version that doesn’t abort.
Why do you not pass the same DD as an input to HHH?

The same machine address of DD is the only reference to DD that any
termination analyzer ever sees.