Re: Who here understands that the last paragraph is Necessarily true?

On 7/20/2024 3:48 AM, Fred. Zwarts wrote:

Op 19.jul.2024 om 16:56 schreef olcott:

On 7/19/2024 1:40 AM, Fred. Zwarts wrote:

Op 18.jul.2024 om 16:18 schreef olcott:

On 7/18/2024 3:41 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 16:56 schreef olcott:

On 7/17/2024 9:32 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 16:20 schreef olcott:

On 7/17/2024 8:54 AM, Fred. Zwarts wrote:

Op 17.jul.2024 om 15:27 schreef olcott:

>
HHH is not allowed to report on the behavior of it actual self
in its own directly executed process. HHH is allowed to report on
the effect of the behavior of the simulation of itself simulating DDD.
>

>
But only on the effect of a correct simulation.

>
_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
>
*THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
DDD emulated by any pure function HHH according to the
semantic meaning of its x86 instructions never stops
running unless aborted.
>

>
It is self evident that a program that aborts will halt.
The semantics of the x86 code of a halting program is also self-evident: it halts.
So, the aborting HHH, when simulated correctly, stops.
Dreaming of a HHH that does not abort is irrelevant.
>

>
That is all the dishonest dodge of the strawman deception.
HHH is required to halt by its design spec.
>
_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
>
*THIS IS SELF EVIDENT THUS DISAGREEMENT IS INCORRECT*
DDD emulated by any pure function HHH according to the
semantic meaning of its x86 instructions never stops
running unless aborted.
>

>
Dreaming of a HHH that does not halt, when we are talking about a HHH that aborts and halts is irrelevant. Therefore, the 'unless aborted' is irrelevant. The semantics of the x86 instructions are self-evident: HHH halts.

>
When you are hungry you remain hungry until you eat.
   Before HHH(DDD) aborts its emulation the directly
   executed DDD() cannot possibly halt.

>
No, but HHH would have halted when not aborted, because that is how it is programmed. That is the semantics of its x86 code.
>

>
int main { DDD(); } calls HHH(DDD) that must abort the
emulation of its input or
HHH, emulated DDD and executed DDD never stop running.
>
>

 Whether HHH *must* abort, or *not abort* is irrelevant.

In other words you can't begin to understand hypothetical
possibilities, thus this material may simply be over your head.
Of every possible HHH that can possibly exist HHH must abort
its emulation of DDD or not. Not is the wrong answer thus
making abort the correct answer.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer