Re: DDD correctly emulated by H0 --- Why Lie? -- Repeat until Closure

On 6/26/24 9:12 PM, olcott wrote:

On 6/26/2024 7:40 PM, Richard Damon wrote:

On 6/26/24 8:24 PM, olcott wrote:

On 6/26/2024 6:52 PM, Richard Damon wrote:
>
Mike understands how your use of static variables is cheating.
Others here will be baffled and confused.
>

>
But the rules are the rules, and you can't create rules you didn't state.
>

 If you really have such brain damage that you cannot remember
that we already discussed this and closed it I will start
praying for you.

And if I am allowed, or even required to remember that, then I can remember that your Hs are all designed to be Halt Deciders, and thus the only correct behavior for their input is that of the direct execution of the program the input represents.

 YOU CAN'T FREAKING USE ANY STATIC LOCAL VARIABLES TO
CHANGE THE BEHAVIOR AND YOU KNOW WHY YOU CAN'T USE
THEM SO FREAKING QUIT IT !!!
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 The call from DDD to H0(DDD) when DDD is correctly emulated
by x86 emulator H0 cannot possibly return.
 

But it does, it just isn't emulated by H0.
The "Correct Behavior" is not limited by the PARTIAL simulation done by H0.
If H0 stops simulating and returns, it does not complete the CORRECT EMULATION of the input as defined, and thus can make no comment of the future that it can not prove. And, since the complete and correct emulation of the input does return if H0 ever returns an answer, it is impossible for it to correct say it doesn't.
You just don't understand the meaning of the word you are using