Re: Anyone that claims this is not telling the truth

On 8/17/24 4:41 PM, olcott wrote:

On 8/17/2024 3:20 PM, Richard Damon wrote:

On 8/17/24 4:10 PM, olcott wrote:

On 8/17/2024 2:27 PM, Richard Damon wrote:
 > On 8/17/24 3:00 PM, olcott wrote:>> On 8/17/2024 1:50 PM, Richard Damon wrote:
 >>>>>> And thus ALL of memory is part of the input,

>
Any additional details have no effect what-so-ever on my claim.
>

>
Suure it does.
>
Since your argument tries to say that since DDD is the same to all of them, so its the behavior.
>
You are just admitting to being a LIAR.
>

>
*Calling me a liar admits that insults is all that you have*
*If I made a mistake then show that*

>
I did.
>

>
FOR THREE YEARS YOU ALWAYS CHEAT
BY CHANGING MY WORDS AND REBUTTING THESE CHANGED WORDS
>
*Everything that is not expressly stated below is*
*specified as unspecified*

>
Ok.
>

>
void DDD()
{
   HHH(DDD);
}
>
_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]
>
*It is a basic fact that DDD emulated by HHH according to*
*the semantics of the x86 language cannot possibly stop*
*running unless aborted* (out of memory error excluded)
>

>
No, DDD can NOT be emulated accoreding to the semantics of the x86 langauge, because the contents of the location 000015d2 is not provided to be emulated, and will need to be emulated after emulating the call instruction.
>

 Everything that is logically entailed by the above specification
is included by reference. The assumption that DDD and HHH were
not in the same memory space has always been ridiculous.
 

Then I guess you accept that every different HHH generates a DIFFERENT Input, as that input, BY LOGHICAL NECESSITY includes all the code of HHH so it can be emulated, and thus you claims that "All the DDDs have the same bytes" is just a blantent lie.
It also means that you are accepting as a logical necessity that all version of HHH will continue until they reach a terminal state as that IS part of the semantics of ths x86 language, and if you accepted that some HHH could do a partial emulation and return, it would also return to the DDD that called it, as "DDD emulated by HHH" refers to the behavior of the FULL PROGRAM DDD, as that is the subject, and NOT just the emulation of DDD by HHH since that is not, and isn't even mentioned in the sentence, and that behavior DOES stop, so obviously you are accepting that "emulation" means correct and complete emulation without aborting
Sorry, that *IS* what comes out of everythig that is logically entailed by your specified.
Note, it is NOT rediculous that the HHH doing the emulation is in a different memory space than the program it is emulating, and in fact that is the NORM for emulators and the only way you can get totally correct emulation of all programs.
It does mean that DDD needs to include its own copy of that HHH in its memory space, but that is why Linz and Sipser talk about the input using a COPY of the decider, and that removes a LOT of the logical problems you have, as then you can talk about changing the decider and not worry about such changes changing the input.
But then, you NEED that confusion to support your LIES.