Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar

The "the mapping" on the subject line is not correct. The subject line
does not specify which mapping and there is no larger context that could
specify that. Therefore it should be "a mapping".
 On 2024-11-17 18:36:17 +0000, olcott said:
 

void DDD()
{
   HHH(DDD);
   return;
}
>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
DDD emulated by any encoding of HHH that emulates N
to infinity number of steps of DDD cannot possibly
reach its "return" instruction final halt state.

 Because it cannot reach the instructions before tha return.
Because it cannot reach the instruction after the HHH call.
Because it cannot reach return instruction of HHH.
 

This applies to every DDD emulated by any HHH no
matter the recursive depth of emulation. Thus it is
a verified fact that the input to HHH never halts.

 That is too vague to be regareded true or false. It is perfectly possibe
to define two programs and call them DDD and HHH

and an input that
specifies a halting program and give that to the program called to HHH.
Obviously the words "every DDD" and "any HHH" adn "the input to HHH" are
intended to be restricted to some smaller ranges but no restrictions are
specified.