Re: This function proves that only the outermost HHH examines the execution trace

Am Thu, 01 Aug 2024 11:00:04 -0500 schrieb olcott:

On 8/1/2024 10:46 AM, joes wrote:

Am Thu, 01 Aug 2024 07:56:27 -0500 schrieb olcott:

On 8/1/2024 7:52 AM, Fred. Zwarts wrote:

Op 01.aug.2024 om 14:03 schreef olcott:

On 8/1/2024 2:52 AM, Mikko wrote:

On 2024-07-31 17:33:38 +0000, olcott said:

On 7/31/2024 4:15 AM, Mikko wrote:

On 2024-07-30 23:40:21 +0000, olcott said:

On 7/30/2024 2:00 AM, Mikko wrote:

On 2024-07-29 16:50:53 +0000, olcott said:

On 7/28/2024 3:59 AM, Mikko wrote:

On 2024-07-27 20:05:31 +0000, olcott said:

 

When instructions are executed/emulated according to the semantics
of the x86 language then they are executed/emulated correctly.

But only those instructions. A halting program is simulated correctly
if no instructions are skipped.

Correctly and completely are not the same damn thing you freaking
moron.

Yes, an incomplete simulation is also incorrect.

When N steps of DDD are emulated correctly one can only say that N steps
were not emulated correctly when one is a liar.

Good thing I only said this: not all steps were simulated.

When a correct and complete emulation is impossible because the
computation has no end then it is pretty damn stupid to insist an a
complete emulation.

It is possible. It just doesn't end. The simulated machine isn't required
to halt either

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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.