Re: DDD specifies recursive emulation to HHH and halting to HHH1

On 3/31/2025 5:54 PM, dbush wrote:

On 3/31/2025 6:30 PM, olcott wrote:

On 3/31/2025 5:17 PM, dbush wrote:

On 3/31/2025 6:12 PM, olcott wrote:

On 3/31/2025 3:44 PM, joes wrote:

Am Sun, 30 Mar 2025 21:13:09 -0500 schrieb olcott:

On 3/30/2025 7:32 PM, Richard Damon wrote:

On 3/30/25 7:59 PM, olcott wrote:

On 3/30/2025 5:50 PM, Richard Damon wrote:

On 3/30/25 5:53 PM, olcott wrote:

On 3/30/2025 4:01 PM, Richard Damon wrote:

On 3/30/25 3:42 PM, olcott wrote:

On 3/30/2025 8:50 AM, Fred. Zwarts wrote:

Op 30.mrt.2025 om 04:35 schreef olcott:

On 3/29/2025 8:12 PM, Richard Damon wrote:

On 3/29/25 6:44 PM, olcott wrote:

On 3/29/2025 5:08 PM, dbush wrote:

Is not what I asked about.  I asked about the behavior
of D when executed directly.

Off topic for this thread.

Yes, HHH is off the topic of deciding halting.

The behavior that these machine code bytes specify:
558bec6872210000e853f4ffff83c4045dc3 as an input to HHH is
different than these same bytes as input to HHH1 as a
verified fact.

What does "specify to" mean? Which behaviour is correct?

DDD EMULATED BY HHH DOES SPECIFY THAT IT CANNOT POSSIBLY REACH
ITS OWN FINAL HALT STATE.

How does HHH emulate the call to HHH instruction

The semantics of the x86 language.

Right, which were defined by INTEL, and requires the data
emulated to be part of the input.

It is part of the input in the sense that HHH must emulate itself
emulating DDD. HHH it the test program thus not the program-under-
test.

It is part of the program under test, being called by it. That's
what you call a pathological relationship.

HHH is not asking does itself halt?

Yes it is saying "I can't simulate this".

It was encoded to always halt for such inputs. HHH is asking does
this input specify that it reaches its own final halt state?

Which it does (except when simulated by HHH).

How we we determine that DDD emulated by HHH cannot possibly
reach its final halt state?
Two recursive emulations provide correct inductive proof.

Nope, because if you admit to the first two lies, your HHH never
was a valid decider,

>
It is ALWAYS CORRECT for any simulating termination analyzer to stop
simulating and reject any input that would otherwise prevent its own
termination.
>

Except when doing so changes the input, as is the case with HHH and
DDD.
>
Changing the input is not allowed.

>
I have already addressed your misconception that the input is changed.
>

No, it is YOUR misconception.  The algorithm DDD consists of the
function DDD, the function HHH, and everything that HHH calls down to
the OS level.
 

We have already been over this.
HHH(DDD) and HHH1(DDD) have the same inputs all the way down to the OS
level. The ONLY difference is that DDD does not call HHH1(DDD) in
recursive emulation.