On 3/25/25 8:26 AM, olcott wrote:
On 3/25/2025 6:19 AM, Richard Damon wrote:
On 3/24/25 10:10 PM, olcott wrote:
I told you too damn many times that all this stuff
is in the same global memory space of the compiled
object file.
And thus either all the global memory space is what is defined to be
the input, and thus every case you think of is a different input,
_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]
Correctly emulated is defined as emulated according to the
semantics of the x86 language.
Which means this input need the actual code of the HHH that it calls to
be included with it.
The question does the machine code of DDD (the program under test)
reach is own "ret" instruction when correctly emulated by HHH?
So, are you asserting that your HHH ACTUALLY does a correct emulation of
the input, which includes the code for that HHH?
If so, then THAT DDD (which calls the actually correctly emulating HHH)
will not halt, but that HHH will not answer.
is not effected by this.
or your "decider" fails to meet the requirements of being the pure
function that you have admited to be a known base requirement.
We cannot move on to any other point while you continue
to deny the proven facts of the first point.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
This first point is this:
Would HHH be correct to reject DD as non-halting?
No. because if HHH does that, then DD is halting, as HHH must not do a
correct emulation to do that rejection, and thus we see that the actual
correct emulation of the input will halt.
Sorry, you are just proving that everything you say is likely a lie,
because you just can remember the meaning of the words so you create
your own, INCONSISTENT meaning as you go.
You cannot show any example of that above.
Sure, many times, the problem is you keep on trying to make HHH be two different programs at once, which it can't be, so your argument is just
built on a lie.
Sorry, you are just proving your ignorance.
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