Re: The execution trace of HHH1(DDD) shows the divergence

Am Sat, 07 Jun 2025 14:10:04 -0500 schrieb olcott:

On 6/7/2025 2:04 PM, joes wrote:

Am Sat, 07 Jun 2025 13:59:54 -0500 schrieb olcott:

On 6/7/2025 1:51 PM, dbush wrote:

On 6/7/2025 2:49 PM, olcott wrote:

On 6/7/2025 1:43 PM, dbush wrote:

On 6/7/2025 2:36 PM, olcott wrote:

On 6/7/2025 1:26 PM, dbush wrote:

On 6/7/2025 1:10 PM, olcott wrote:

On 6/7/2025 11:37 AM, dbush wrote:

On 6/7/2025 12:29 PM, olcott wrote:

On 6/7/2025 11:20 AM, dbush wrote:

On 6/7/2025 12:17 PM, olcott wrote:

On 6/7/2025 11:14 AM, dbush wrote:

On 6/7/2025 11:33 AM, olcott wrote:

On 6/7/2025 10:17 AM, dbush wrote:

On 6/7/2025 11:12 AM, olcott wrote:

On 6/7/2025 10:08 AM, dbush wrote:

On 6/7/2025 11:06 AM, olcott wrote:

On 6/7/2025 10:01 AM, dbush wrote:

On 6/7/2025 10:58 AM, olcott wrote:

On 6/7/2025 9:56 AM, dbush wrote:

On 6/7/2025 10:54 AM, olcott wrote:

On 6/7/2025 9:51 AM, dbush wrote:

False.
The next instruction of DDD that both HHH and
HHH1 emulates is at the machine address of
000015c3,

>
*That is not an instruction of DDD*

>
In other words, you're not operating on algorithms.

 

I didn't say that you didn't specify an algorithm.  I said your HHH
doesn't work with algorithms,
>

Termination analyzers are encoded algorithms that work with finite
strings that specify a precise sequence of configurations.

 
Yes, they should. Thus HHH is not a termination analyser if it doesn't
consider itself being part of the input. This is how computation works.
 

The directly executed DDD() and the directly executed HHH(DDD) NEVER
STOP RUNNING UNLESS HHH(DDD) ABORTS ITS EMULATION OF DDD.

The directly executed DDD can't be aborted because it is not being
simulated.

--
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.