On 8/19/24 8:08 AM, olcott wrote:
On 8/19/2024 2:30 AM, Mikko wrote:
On 2024-08-18 12:25:05 +0000, olcott said:
*Everything that is not expressly stated below is*
*specified as unspecified*
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]
*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)
X = DDD emulated by HHH∞ according to the semantics of the x86 language
Y = HHH∞ never aborts its emulation of DDD
Z = DDD never stops running
My claim boils down to this: (X ∧ Y) ↔ Z
void EEE()
{
HERE: goto HERE;
}
HHHn predicts the behavior of DDD the same
way that HHHn predicts the behavior of EEE.
That HHH <is> and x86 emulator <is> sufficient to
determine exactly what the behavior of DDD emulated by HHH
according to the semantics of the x86 language would be.
The last "would be" means that the clause is conterfactual.
But why would anybody care about the conterfactual behaviour?
It is not counter-factual.
Remember, you said: Everything that is not expressly stated below is*
specified as unspecified
Therefore HHHn can NOT correctly emulate DDD past the call HHH
instruction, because it doesn't HAVE the instruciton of the PROGRAM DDD
(which is what you emulate) since it doesn't have the instruction at
000015D2.
The contents of the memory at 000015D2 can not be accessable to HHHn, as
the input is described as DDD and not DDDn, so the input doesn't change
between instances, and thus CAN'T contain that memory that changes, and
thus is not valid to be part of the input.
Thus we also have that HHH∞ can not exist, so both your premises just
fail to be possible.
Sorry, you are just repeating your error because apparently you just
can't learn.
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