Liste des Groupes | Revenir à c theory |
On 6/22/2024 11:03 AM, joes wrote:And thus, your emulation traces show that your "Simulating Halt Deciders" do not do a "Correct Simulation" as the ONLY correct simulation of the call instruction is the simulitation of the instructions of the routine called, which is NOT what you show.Am Sat, 22 Jun 2024 10:16:18 -0500 schrieb olcott:void DDD()On 6/22/2024 9:42 AM, Richard Damon wrote:>On 6/22/24 10:31 AM, olcott wrote:The D(D) that calls H(D,D) such that this call returns has provably[s/is/as?]
different behavior than D correctly simulated by H is measured by the
actual semantics of the x86 programming language.
No, H of course needs to simulate the call to itself like any other.
x86 has nothing to do with that. A correct simulation has identical
behaviour to the real thing. Why should H simulate something that is
not its input?
>Yes, DDD is coded to call the HHH0 deciding on it.The input is the finite string.
The MEANING of that finite string is defined by the PROBLEM.LIAR. You know that the meaning of the finite string is defined by the>
semantics of the x86 language.As Christ said as ye judge ye shall be judged so I do wish the sameThanks for the permission. Your minimum seems to be quite high.
thing upon myself. If I am on the wrong path then I sincerely wish for
the minimum adversity required to definitely set me on the right path.
Not sure I want to fulfill your wishes.
>WTF? What contradiction? How can "halting" even be incorrect?Halting DEFINES the meaning/behavior to be that of the directly runThat makes it contradict one of its own axioms, thus conclusively
program represented by the input.
proving that it is incorrect:
>This is just silly. The x86 code of DDD is defined to call HH0.The problem is that the "behavior" that the finite string DDD presentsLIAR. It is defined by the semantics of the x86 language.
to HH0, is DEFINED by the problem.
>Exactly. Like you say, it must follow the semantics of its input.And if that problem is the Halting Problem, that behavior is the
behavior of the machine the input represents. If HH0 treats the input
as having a different behavior, then HH0 just isn't a Halting Decider,
but something else.
If HH0 is supposed to be a Halting decider, but uses a method that
makes it see something other than that behavior, then it is just an
incorrect Halting Decider, and its algorithm just creates an incorrect
recreation of the property of the input it is supposed to be working
on.
>The input to HHH0(DDD) includes itself.Yes, both include HHH0. The second case is boring.
The input to HHH1(DDD) DOES NOT include itself.
>DDD correctly emulated by HHH0 correctly determines that the call from*incorrectly
the emulated DDD to HHH0 DOES NOT RETURN.DDD correctly emulated by HHH1 correctly determines that the call from>
the emulated DDD to HHH0 DOES RETURN.
{
HHH0(DDD);
}
The input to HHH0(DDD) includes itself.
The input to HHH1(DDD) DOES NOT include itself.
It is stipulated that correct emulation is defined by the
semantics of the x86 programming language and nothing else.
DDD correctly emulated by HHH0 correctly determines thatBut doesn't "Correctly Emulate" the input, so it claim is just invalid.
the call from the emulated DDD to HHH0 DOES NOT RETURN.
DDD correctly emulated by HHH1 correctly determines thatNope.
the call from the emulated DDD to HHH0 DOES RETURN.
The fact that DDD calls HHH0(DDD) and does not call HHH1(DDD)
changes the behavior of DDD correctly emulated by HHH0 relative
to DDD correctly emulated by HHH1.
_DDD()
[00002172] 55 push ebp
[00002173] 8bec mov ebp,esp
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH0
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
Les messages affichés proviennent d'usenet.