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

On 11/17/24 1:36 PM, olcott wrote:

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.
 

Except your DDD *CAN'T BE EMULTATED* by *ANY* HHH, as it is IMPOSSIBLE to emulate the Call HHH per the x86 language from your input, as the data isn't tnere.

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.
 

But since NO HHH can do the first part of your requirements, that of actually emulating the input, you are just proved to be an lying idiot.
Also, Computation Theory isn't interested in Subjective Non-semantic problems like yours, but only objective problems, especially those that are semantic (which means one that actually IS based on the FINAL behavior, after an unbounded number of steps) properties.
Your ignorance of that fact just shows your utter stupidity.
It doesn't what the emulation of HHH is, if it is only finite, only the unbounded emulation of that input (Which will be halting if that DDD is based on an HHH that answers after finite time).
THe HHH that DOES a semantic emulation unfortunately never answers, so fails to be the needed decider.
So, you just struck out twice.
You then keep on lying about it, which gets your ejected from the logic pool..
Sorry, that is just the facts, you are showing you are just too stupid to have your idea have any merits, and thus even if there was something to your ideas, you have probably successfully killed them.