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

Op 01.aug.2024 om 18:00 schreef 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.

Nobody claims it. It is just your dream.
When the last M steps of a halting DDD are skipped, one can only say that it is a correct simulation when one is a liar.
This is the reality.

 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.
 

Nobody claims it. It is just your dream.
When a correct and complete simulation of HHH by itself is not possible even when HHH halts, then it is pretty stupid to insist that a correct simulation of halting program shows that it does not halt.
That is the reality.
Dreams are no substitute for fact, not for logic.