Re: DDD emulated by HHH --- (does not refer to prior posts)

Am Wed, 28 Aug 2024 14:47:45 -0500 schrieb olcott:

On 8/28/2024 2:21 PM, joes wrote:

Am Wed, 28 Aug 2024 11:44:53 -0500 schrieb olcott:

On 8/28/2024 11:31 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 18:21 schreef olcott:

On 8/28/2024 11:11 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 17:13 schreef olcott:

On 8/28/2024 9:57 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 14:59 schreef olcott:

On 8/28/2024 7:46 AM, Fred. Zwarts wrote:

Op 28.aug.2024 om 14:12 schreef olcott:

On 8/28/2024 4:09 AM, Fred. Zwarts wrote:

Op 27.aug.2024 om 14:44 schreef olcott:

On 8/27/2024 3:38 AM, Fred. Zwarts wrote:

Op 27.aug.2024 om 04:33 schreef olcott:

 

When we assume that:
(a) HHH is an x86 emulator that is in the same memory
space as DDD.
(b) HHH emulates DDD according to the semantics of the x86
language.
then we can see that DDD emulated by HHH cannot possibly
get past its own machine address 0000217a.

>
Yes, we see. In fact DDD is not needed at all.

When 100% of the whole point is for HHH to correctly determine
whether or not DDD would stop running if not aborted *IT IS
RIDICULOUSLY STUPID TO SAY THAT DDD IS NOT NEEDED*

Like Fred has been saying for a month, what is HHH(HHH)?

^

When without DDD it is clear as crystal that HHH cannot
possibly simulate itself correctly:

You may repeat it many more times, but HHH violated the semantics
of the x86 language by skipping the last few instructions of a
halting program. This finite string, when given for direct
execution, shows a halting behaviour. This is the proof what the
semantics of the x86 language means for this finite string: a
halting program.

It is very telling to see where these exchanges peter out (haha).

A dishonest dodge way form the subject of DDD emulated by HHH.

To the subject of what?

And when the x86 string tells the computer that there is a
halting program and the simulator decides that there is a
non-halting program, this proves that the simulation is
incorrect.
Clear as crystal: the semantics of the x86 string is proved by
its direct execution.
This is shown in the example below, where the direct execution of
HHH halts, but HHH decides that it does not halt.
>

By this same reasoning that fact that you are no longer hungry
AFTER you have eaten proves that you never needed to eat.
The behavior of DDD before HHH aborts its simulation (before it
has eaten) it not the same behavior after DDD has been aborted
(after it has eaten).

I do not understand this. There is no „after having been aborted”.

The directly executed DDD() has different behavior than DDD emulated by
HHH because DDD() benefits from HHH having already aborted its emulation
of DDD. HHH itself does not receive this benefit.

„Already”? DDD and it’s simulation live in entirely different spaces.
There can be absolutely no influence.

If hungry stands for fear for infinite recursion

hungry stands for will not stop running unless aborted just like
will remain hungry until eating is always true whenever hungry

Your HHH will see a 'special condition' after a few recursions, abort
and halt.

It is a design requirement that HHH halts if it doesn't halt it is
wrong.

Then why does it report itself as nonterminating? (There is nothing
else in DDD that would cause that.)

How could it do that? IT MUST TERMINATE TO REPORT ANYTHING.

???

When DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own machine address of 00002183,
then HHH is correct to reject DDD as non-halting even of HHH does this
entirely by wild guess.

--
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.