Re: The philosophy of computation reformulates existing ideas on a new basis ---x86 code is a liar?

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, since they are the same program sharing the
same code. This also means that the aborting of the inner HHH called
by DDD is not simulated by the outer HHH, because it aborts before.
DDD, the counterexample to the claim that HHH can decide
on the halting status of every program, is constructed on HHH.
DDD is not a single program that can’t be decided by ANY decider,
rather a template for every claimed halting decider, thus changes
depending on it. If you fix DDD to call a particular "decider",
no claim is made as to its decidability by a different "decider".
Each decider has their own DDD-like that calls that very same
decider.

When DDD calls a simulator that aborts, that simulator returns to DDD,
which then halts.

It is not the same DDD as the DDD under test.

Yes it is: DDD(){HHH(DDD);}
What do you think it is?

--
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.