Sujet : Re: Defining a correct simulating halt decider
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 05. Sep 2024, 18:22:46
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <70a0b7e4bd0a0129649d8e77cdc36339bd74d6a5@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Thu, 05 Sep 2024 12:17:01 -0500 schrieb olcott:
On 9/5/2024 11:56 AM, joes wrote:
Am Thu, 05 Sep 2024 11:52:04 -0500 schrieb olcott:
On 9/5/2024 11:34 AM, joes wrote:
Am Thu, 05 Sep 2024 11:10:40 -0500 schrieb olcott:
On 9/5/2024 10:57 AM, joes wrote:
Am Thu, 05 Sep 2024 08:24:20 -0500 schrieb olcott:
On 9/5/2024 2:34 AM, Mikko wrote:
On 2024-09-03 13:00:50 +0000, olcott said:
On 9/3/2024 5:25 AM, Mikko wrote:
On 2024-09-02 16:38:03 +0000, olcott said:
>
Show the details of how DDD emulated by HHH reaches its own
machine address 0000217f.
By HHH returning, which we are guaranteed from its definition as a
decider.
How the F--- Does the emulated HHH return?
I don’t know, you claim it’s a decider!
DDD emulated by HHH CANNOT POSSIBLY reach its own machine address
0000217f.
Only HHH can’t simulate it.
The directly executed HHH correctly determines that its emulated DDD
must be aborted because DDD keeps *THE EMULATED HHH* stuck in
recursive emulation.
Why doesn’t the simulated HHH abort?
The first HHH cannot wait for its HHH to abort which is waiting for its
HHH to abort on and on with no HHH ever aborting.
But why does HHH halt and return that itself doesn’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.
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