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

On 11/15/24 6:34 PM, olcott wrote:

On 11/15/2024 5:23 AM, joes wrote:

Am Thu, 14 Nov 2024 17:53:38 -0600 schrieb olcott:

On 11/14/2024 3:09 AM, Mikko wrote:

On 2024-11-13 23:11:30 +0000, olcott said:

On 11/13/2024 4:58 AM, Mikko wrote:

On 2024-11-12 13:58:03 +0000, olcott said:

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 final
halt state whether HHH aborts its emulation or not.

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.

>
How do they differ?
>

If the DDD under the test is not the same as DDD then the test is
performed incorrectly and the test result is not valid.

The DDD under test IS THE INPUT DDD

I agree that there is only one DDD but above you said otherwise.

DDD emulated by HHH emulates itself emulating DDD and DDD emulated by
HHH1 *DOES NOT DO THAT*

Those are the same DDD. The difference lies with the simulators.
>

 It is ridiculously stupid to require an emulating termination
analyzer to get stuck in recursive simulation.
 Alternatively DDD emulated by HHH must be aborted by HHH.
 

It isn't required to do that. Just to get the right answer.
A fact that you don't seem to understand, likely because you don't undetstand the foundations of truth.