Re: Proof that DDD specifies non-halting behavior --- point by point

On 8/13/24 11:48 PM, olcott wrote:

On 8/13/2024 10:21 PM, Richard Damon wrote:

On 8/13/24 10:38 PM, olcott wrote:

On 8/13/2024 9:29 PM, Richard Damon wrote:

On 8/13/24 8:52 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]
>
A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is necessarily correct.
>

>
Nope, it is just the correct PARTIAL emulation of the first N instructions of DDD, and not of all of DDD,

>
That is what I said dufuss.

>
Nope. You didn't. I added clairifying words, pointing out why you claim is incorrect.
>
For an emulation to be "correct" it must be complete, as partial emulations are only partially correct, so without the partial modifier, they are not correct.
>

 A complete emulation of one instruction is
a complete emulation of one instruction

 

>

>

A correct simulation of N instructions of DDD by HHH is
sufficient to correctly predict the behavior of an unlimited
simulation.

>
Nope, if a HHH returns to its caller,

>
*Try to show exactly how DDD emulated by HHH returns to its caller*
(the first one doesn't even have a caller)
Use the above machine language instructions to show this.
>

>
Remember how English works:
>
When you ask "How DDD emulated by HHH returns to its callers".

 Show the exact machine code trace of how DDD emulated
by HHH (according to the semantics of the x86 language)
reaches its own machine address 00002183

No. The trace is to long, and since you HHH doesn't meet your requirements (since it isn't a pure function) you can't give me a compldte input to trace.
If the input DDD is just what you show below, your problem has a category error as that isn't a complete program, and thus not something that can be completely traced.

 _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]
 When DDD is emulated by HHH
HHH emulates the lines of DDD in this order:
[00002172]
[00002173]
[00002175]
[0000217a] calls HHH(DDD)

And here you show that you don't understand what a correct by x86 semantics means.
Following the call to HHH, must be the instuctions of HHH being emulated.
THAT *IS* the requirements of emulation by the semantics of the x86 language.
Since HHH(DDD) is know to return, if HHH actually WAS a pure function, then this call MUST return to DDD.
You are just shown how stuoid you are.

 which emulates the lines of DDD in this order
HHH emulates the lines of DDD in this order:
[00002172]
[00002173]
[00002175]
[0000217a]
 with 200 pages of HHH emulating itself emulating
DDD inbetween.
 

No, about 300 lines. (if we allow the skipping of the actual OS code that your trace skips). The 200 pages is the trace of the HHH deciding on the input, not the trace it DOES of the input. That has about 1-2 instructions emulated per page of trace.
Obviously you don't read the message, I guess you are just to stupid to understand long posts.
You are just PROVING you are just a self-made idiot that just doesn't know what he is talking about, and so stupid that he doesn't realize he is an idiot, which is the worse kind.
That, you are you no sense of ethics that make you want to even try to tell the truth, because you think lying is just fine.