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

On 2024-07-19 14:49:46 +0000, olcott said:

On 7/19/2024 4:14 AM, Mikko wrote:

On 2024-07-18 14:18:51 +0000, olcott said:
 

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.
 After you eat you are no longer hungry.
   After HHH(DDD) aborts its emulation the directly
   executed DDD() halts.

 If DDD does not halt it indicates that HHH is faulty. Therefore the
interesting question is whether DDD halts, not when DDD halts.

This time I made a typo that is much worse than my usual typos.
Pro "when DDD halts" lege "when HHH halts".

*By your same reasoning*
If Infinite_Loop() does not halt HHH is faulty.

No, the reasoning must be differ. Infinite_Loop can be proven to halt
by a simple inspection of a short code. Similar simple inspection of
DDD reveals that DDD does halt if HHH halts but not whether HHH halts.
Therefore the interesting question, needed to complete the proof, is
whether HHH halts. If that can be determined the question about DDD
is easy.

In other words if Infinite_Loop()  is an actual infinite
loop then this is all the fault of HHH.

Nothing is the fault of HHH. If a program is faulty it is the fault
of the author of the program. Usually an infinite loop is a fault
but that depends on the purpose and specification of the program.
Sometimes a program is faulty if it does terminate.
--
Mikko