Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2 ---woefully mistaken rebuttal

Am Fri, 26 Jul 2024 10:20:58 -0500 schrieb olcott:

On 7/26/2024 10:13 AM, joes wrote:

Am Fri, 26 Jul 2024 08:54:32 -0500 schrieb olcott:

On 7/26/2024 3:50 AM, joes wrote:

Am Thu, 25 Jul 2024 23:25:59 -0500 schrieb olcott:

On 7/25/2024 10:35 PM, Mike Terry wrote:

On 26/07/2024 01:53, olcott wrote:

On 7/25/2024 4:03 PM, Mike Terry wrote:

On 25/07/2024 14:56, olcott wrote:

On 7/24/2024 10:29 PM, Mike Terry wrote:

On 23/07/2024 14:31, olcott wrote:

On 7/23/2024 1:32 AM, 0 wrote:

On 2024-07-22 13:46:21 +0000, olcott said:

On 7/22/2024 2:57 AM, Mikko wrote:

On 2024-07-21 13:34:40 +0000, olcott said:

On 7/21/2024 4:34 AM, Mikko wrote:

On 2024-07-20 13:11:03 +0000, olcott said:

On 7/20/2024 3:21 AM, Mikko wrote:

On 2024-07-19 14:08:24 +0000, olcott said:

Then we know that HHH can see the the first four instructions of DDD
have no conditional code that could prevent them from endlessly
repeating.

True, but HHH does have a conditional abort. It should be coded to
recognise that, because one knows that at compile time already.
 

The whole thing with the slave instances might well be where the
bug lies!  That would be slightly funny, as I pointed out that
problem on some completely unrelated post, and this could be a
follow-on issue where it has caused observable misbehavior in the
code.  (Needs a bit more investigation...)

There never is any actual bug with the simulation.

I bet my nonexistent soul that there are bugs left in libx86. Apart
from that, your use of the library may be buggy.

That is irrelevant. We can see by the execution trace of DDD emulated
by HHH that this emulation does precisely match the semantics of the
first four x86 machine language instructions of DDD.

But not what comes afterwards, and HHH makes the incorrect assumption
that another instance of itself wouldn't abort.

That it is beyond your technical competence to understand that unless
the first HHH aborts then none of them do is less than no rebuttal at
all.

Fuck your insults. I know that. Like you say, if the first one aborts,
all of them do, therefore the first one can't say it was simulating
something that didn't halt.

--
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.