Re: HHH maps its input to the behavior specified by it ---

On 8/10/2024 12:51 PM, Richard Damon wrote:

On 8/10/24 1:37 PM, olcott wrote:

>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>
That is true in every case except when an input calls its
own simulating halt decider.

  Nope. No exeptions. That is just another of your "I made it up but can't prove it but it must be true" lies.
 

>
void DDD()
{
   HHH(DDD);
   return;
}
>
*The set of HHH x86 emulators are defined such that*
Each element of this set corresponds to one element of the set
of positive integers indicating the number of  x86 instructions
of DDD that it emulates.

 But every one that emulates for a finite number of steps, and then returns create a halting DDD, so you claim is just disproven.
 

Changing the title indicates a reply that I can ignore.
I am only replying to a single thread of communication
with you.
*Maybe your ADD has always been much worse than I ever expected*
We have been over all of these same points hundreds of times.
When each element of the outer-most directly executed SHH
corresponds to one element of the set of positive integers
indicating the number of x86 instructions of DDD that it
emulates none of the emulated instances of HHH ever returns.
THAT IS WHAT THE C SOURCE CODE MEANS
THAT IS WHAT THE X86 CODE MEANS
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer