Nope. you even posted the trace of this.
If H(P,P) returns 0, as you claim it does to main, it will do the same
to P(P) and that makes P(P) Halt.
Since P(P) Halts, a correct simulation of the input to H will also halt
since the input to H(P,P) IS the representation of P(P) and if you want
to claim otherwise, you have just declaired that P isn't writen right.
Yes, there IS a case of a design for H where the correctly simulated
input never terminates, and that is the case when H NEVER aborts its
simulation, and thus H(P,P) never returns to main and thus fails to be a
Decider.
Which H are you claiming, the one that gives the wrong answer or the one
that fails to be a decider?
>
>>>
>>> So it seems to be the case that for simulating halt deciders that are
>>> invoked with pathological arguments we have possibly a single
>>> exception to the rule where the correct simulation of an input does
>>> not match the behavior of the correct direct execution of this input.
>>>
>>> Because we know that a halt decider must base its halt status
>>> decision on the actual behavior that is actually specified by its
>>> actual input and we know that the correct simulation of this input by
>>> H(P,P) <is> the actual behavior of this input then we know that
>>> H(P,P)==0 is the correct halt status of this input.
>>
>> And the actual behavior of the input to H(P,P) must be the behavior of
>> P(P) since that comes from the definition of a Halt Decider and the
>> "impossible program" P. All your "proof' that the input to H(P,P) has
>> different behavior than P(P) does is show that your H is wrong.
>>
>>>
>>> Now we move on to whether or not H(P,P) is a pure function.
>>>
https://en.wikipedia.org/wiki/Pure_function
>>>
>>> The instance of H(P,P) that is called by the simulated P is
>>> essentially aborted before it is actually invoked. This seems to be
>>> another exception to the rule of conventional understandings.
>>
>> BUT, for your simulation to be "correct" its behavior must match what
>> the program actually does.
>>
>
> Because you know these things on the basis of dogma rather than
> reasoning you will never get this. If enough people of high enough
> authority proclaimed that all white cats are black dogs many people
> would simply accept that as true.
And because you are ignorant of what you talk about because you don't
know what anything means, you will never understand what yoy are writing
about.
I don't take these facts as "dogma", but becuase they ARE the definition.
H(M,x) must accept if M(x) halts, and renect if M(x) never halts.
By construction, P(P) will halt if H(P,P) rejects (returns 0), (and loop
forever if H(P,P) accepts) so H(P,P) rejecting is NEVER a correct answer.
What is the error in that arguement. Not how do you show something else,
what it the error.
Remember, if two contradictory statements can both be proved true, the
logic system is inconsistent. Thus you "proving" something contrary to
some other statement without showing the actual logical error in it just
shows tha tyou are using a inconsisten logic system.
>
>> Since P(P) calls H(P,P) and will get the 0 return, the simulation of
>> the call, to be "correct" needs to be compatible with that.
>>
>> H's analysis that this H will not return says that H is either using
>> incorrect logic, that it is a incorrect simulator, or that H isn't
>
> These are provably correct yet beyond your capacity to verify.
To "prove" something contrary to another prove without disproving the
original just proves that you logic system is inconsistent.
Since you system includes "rules" beyond the generally accepted theries,
all you have done it prove that you have added some inconsistent rule.
>
>> actually the pure function it needs to be and its simulation behaves
>> different then its actual execution.
>>
>> All three show H to be WRONG.
>>
>>>
>>> I can't think of any other example than a simulating halt decider
>>> that would not be totally stuck in infinite recursion on the same
>>> inputs no matter how it was initially invoked as long as one
>>> invocation on these same inputs is infinitely recursive.
>>
>> But it isn't an allowed exception. As I pointed out, since H decides
>> to abort its simulation, THIS H never gets stuck in infinite
>> recursion, so it doesn't matter what happens to the "other H" that
>> isn't actually present.
>>
>
> *It is already provable that it definitely has another exception*
> The provably correct simulation of the input to H(P,P) have provably
> correct different behavior that the provably correct behavior of the
> direct execution of P(P).
WHAT EXCEPTIONS. THERE ARE NONE. Your "provable correct simulation" has
had logical errors pointed out that you have not corrected. You are just
proving you incabablility to perform correct logic.
>
> The no one here has the technical understanding of the x86 language
> (except possibly Mike) to verify this is zero evidence what-so-ever that
> I am incorrect.
No, there is much evidence, you are just ignoring it, proving your
stupidity.
I would actually challange you to a duel of technical knowledge, but I
have a policy of not fighting an unarmed foe.
>
>>>
>>> The ultimate measure of whether or not it it Turing computable would
>>> be to translate it into a Turing equivalent architecture such as a
>>> RASP machine. The ultimate measure is definitely not the mere
>>> dogmatic words of any textbook(s).
>>>
>>
>> Note, RASP architectures have exactly the same traps as the x86 that
>> code fragments are not necessarily computations.
>>
>> Yes, you can translate the Halting Problem to such architectures, but
>> the wording needs to be much more detailed to spell out the
>> requirements that the pieces being used are actually computations.
>>
>>>
>>>
>>> The only reason that H(P,P) returns to main() is that the H in main
>>> can abort the simulation of its input otherwise the H in main() would
>>> never return from P.
>>>
>>
>> Right, and since it does, it will do the same when P(P) calls it, so
>> to be correct its simulstion needs to reflect that same behavior. ad
>> that means the behavior of the input is halting.
>>
>> ALL the H's can abort the simulation because ALL the H's need to be
>> the SAME ALGORIGTHM and compute the same function.
>
> All H's can abort the simulation, yet the only the outermost H sees the
> abort criteria first and when it aborts its simulation none of the inner
> H's see anything at all because they are never invoked.
You still don't understand do you. I think you are just to stupid to see
anything that doesn't match your erroneous believes.
There are TWO different "outermost" H's in view, one from the call to
main, and one in the test program where main calls P(P) which calls H.
Both of those should show that H(P,P) returns 0, and the second shows
that when this happens to the actual machine, P(P) Halts.
Since the DEFINITION of a correct simuation is one that matches that
actual machine, the correct simulation of the input to H(P,P) by a real
correct and complete simulator will also Halt.
The fact that H thinks it doesn't just proves that H is wrong. As I
said, it can be wrong serveral ways.
1) It can use incorrect logic, like your incorrect rule (3), to reach a
bad conclusion of its input.
2) It can be that H does an incorrect emulation of its input, this is
related to 1) as your emulator thinks that H(P,P) never returns, when we
have previous show that, as long as H is a pure function, that H(P,P)
will always return 0.
3) It could be that H just isn't a pure function and not qualified to be
a decider.
Since we don't have the code for H, we can't tell, but we KNOW that case
1 is at least partly true based on your claims, since you DO quote the
incorrect rule.