Re: The philosophy of computation reformulates existing ideas on a new basis ---SUCCINCT

On 11/14/2024 8:05 AM, joes wrote:

Am Thu, 14 Nov 2024 07:22:12 -0600 schrieb olcott:

On 11/14/2024 2:56 AM, joes wrote:

Am Wed, 13 Nov 2024 17:11:30 -0600 schrieb olcott:

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.

What, then, is the DDD "under test"?

The machine code address that is passed to HHH on the stack is the input
to HHH thus the code under test. It specifies that HHH emulates itself
emulating DDD.
The DDD executed in main() is never pushed onto the stack of any HHH
thus <is not> the input DDD.

 It starts at the same address, however. In what sense is the input not
the DDD with that entry point?
 

DDD emulated by HHH specifies that HHH emulates itself
emulating DDD. This requires HHH to abort this DDD to
prevent its own non-termination. No other instance of DDD
has this same result.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer