Sujet : Re: The philosophy of computation reformulates existing ideas on a new basis ---x86 code is a liar?
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 12. Nov 2024, 18:01:14
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <b1867135333edca9932462ac93da6ade4eb12ec6@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 12 Nov 2024 10:07:11 -0600 schrieb olcott:
On 11/12/2024 9:46 AM, joes wrote:
Am Tue, 12 Nov 2024 08:49:55 -0600 schrieb olcott:
On 11/12/2024 8:23 AM, joes wrote:
Am Tue, 12 Nov 2024 07:58:03 -0600 schrieb olcott:
On 11/12/2024 1:12 AM, joes wrote:
Am Mon, 11 Nov 2024 10:35:57 -0600 schrieb olcott:
On 11/11/2024 10:25 AM, joes wrote:
Am Mon, 11 Nov 2024 08:58:02 -0600 schrieb olcott:
On 11/11/2024 4:54 AM, Mikko wrote:
On 2024-11-09 14:36:07 +0000, olcott said:
On 11/9/2024 7:53 AM, Mikko wrote:
>
The actual computation itself does involve HHH emulating itself
emulating DDD. To simply pretend that this does not occur seems
dishonest.
Which is what you are doing: you pretend that DDD calls some
other HHH that doesn’t abort.
DDD emulated by HHH does not reach its "return" instruction
whether HHH aborts its emulation or not.
No. When the HHH that simulates DDD aborts, it also means that the
HHH that DDD calls aborts,
In no case does DDD emulated by any HHH that aborts at some point or
not does the emulated DDD ever reach its "return" instruction.
That is wrong. When you abort simulating, you can’t tell if it maybe
would have halted later on.
Guessing „No HHH that may or may not abort simulating a DDD that calls
that (aborting or not) HHH can simulate DDD halting.”
*There is no guessing to it*
I was guessing what you meant with your incorrect grammar.
You can’t say that something didn’t halt just because you didn’t
simulate further than some fixed number of steps.
*This must just be over your head*
Are you saying I can say that an infinite loop terminates because
it can be simulated it for a finite number of steps?
-- 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.