Re: HHH maps its input to the behavior specified by it --- never reaches its halt state

On 8/9/2024 4:22 AM, Mikko wrote:

On 2024-08-08 18:45:00 +0000, olcott said:
 

>
x86 machine code is a 100% exact and concrete specification.

 But has too many unimportant details.
 

*After we get agreement on this*
(a) The "return" instruction is the halt state of DDD.
(b) DDD correctly emulated by any HHH never reaches this state.
Then we move on to the last step we I prove how important
a 100% concrete specification is.

C is a little more vague and hides the details.

 With C it is possible to write unambigous programs. It is not
as easy as with many other languages because C is usually
used for purposes where implementation dependnet behaviour
need not be avoided.
 

typedef void (*ptr)();
int HHH(ptr P); // simulating termination analyzer
>
void DDD()
{
   HHH(DDD);
   return;
}
>
Each HHH of every HHH that can possibly exist definitely
emulates zero to infinity instructions of DDD correctly.
Every expert in the C language sees that this emulated DDD
cannot possibly reaches its own "return" instruction halt state.
>
Every rebuttal that anyone can possibly make is necessarily
erroneous because the above paragraph is a tautology.
>
HHH computes the mapping from its finite string of x86 machine
code to the actual behavior that this finite string specifies
which includes recursive emulation.

 It is not useful to repeat what has already been said. You should
always improve, at least the presentation. If you cant find any
other improvemnt you should at least correct the worst typo.
 

*Until we get agreement on this I am not talking about anything else*
(a) The "return" instruction is the halt state of DDD.
(b) DDD correctly emulated by any HHH never reaches this state.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer