Liste des Groupes | Revenir à c theory |
On 5/31/2024 7:39 PM, Richard Damon wrote:Depends on the specification of HH, as has beeen shown.On 5/31/24 7:57 PM, olcott wrote:DD correctly simulated by HH cannot possibly reach its own simulatedOn 5/31/2024 6:33 PM, Richard Damon wrote:>On 5/31/24 6:54 PM, olcott wrote:>On 5/31/2024 5:46 PM, Richard Damon wrote:>On 5/31/24 6:08 PM, olcott wrote:>On 5/31/2024 4:36 PM, Richard Damon wrote:>On 5/31/24 10:10 AM, olcott wrote:>On 5/31/2024 6:16 AM, Richard Damon wrote:>On 5/30/24 11:27 PM, olcott wrote:>Try and show how HH using an x86 emulator can correctly emulate>
the following x86 machine code such that DD reaches its own
machine address 00001c47.
Why should I, since that isn't what I was saying.
>
*To me that looks like you know that*
*you have been busted in a lie and are backing down*
no, YOU are LYING RIGHT HERE AND NOW.
>
Prove that I said that the simulation by HH made it there, or admit to being a DAMNED LIAR.
>
What I have been saying is the the DIRECT EXDCUTION of DD, and the CORRECT (and complete) simulation of the input to HH by an actual UTM will get there.
>
That has always been the dishonest dodge strawman deception
CHANGE-THE-SUBJECT fake rebuttal regarding
the behavior of DD correctly simulated by pure function HH.
But it is your talking about the "correctly simulated by HH" that is the dishonest dodge,
Try and show how HH using an x86 emulator can correctly emulate
the following x86 machine code such that DD reaches its own
machine address 00001c47.
Never said it could. But haven't looked hard enough to be willing to say it can't, but then, who cares, it doesn't say a thing about the real halting problem, since H's simulation isn't "correct" by a definition that relates simulation to non-halting behavior,
>
"...the Turing machine will halt whenever it enters a final state."
Linz(1990:234)
Right, and that is talking about runnig the Turing Machine, not simulating a representation of it.
>
final state. This is conclusively proven beyond all possible doubt
by the x86 machine code of DD.
You can lie about this and try to get away with changing the subject.But I don't try to claim one side of the other, as it isn't relevent.
What you cannot do is show that it is not true.
*Showing that it is relevant is a whole other different subject that*But if it isn't relevent, why are we looking at it.
*we can get to as soon as you quit your deception on this subject*
*Showing that it is relevant is a whole other different subject that*But I DO beleive in UTMs, and know exact what they do.
*we can get to as soon as you quit your deception on this subject*
*Showing that it is relevant is a whole other different subject that*
*we can get to as soon as you quit your deception on this subject*
Trying to get away with saying that you don't "believe in" UTMs>Nope, Where did that definition say ANYTHING about one machine simulationg another.
*If DD correctly simulated by HH can't possibly reach its own*
*final state then DD correctly simulated by HH is non-halting*
>
can't possibly fool anyone that knows what UTMs are.
When Ĥ is applied to ⟨Ĥ⟩And then, while THAT H^ is non-halting, H isn't a decider.
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
When embedded_H <is> a UTM then ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H
cannot possibly reach its own simulated final state of ⟨Ĥ.qn⟩. In this
case embedded_H fails to be a decider, however it also proves:
that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach itsBut that only apply *if* embedded_H (and thus H) *IS* a UTM, if it aborts its simuliation, then its
own simulated final state of ⟨Ĥ.qn⟩ for any embedded_H based on a UTM
that only simulates some finite sequence of steps.
"...the Turing machine will halt whenever it enters a final state."And and embedded_H that actually correct simulates the input, by Computaiton Theory definitions is the UTM that, by definition, doesn't abort its simulation.
Linz(1990:234)
⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H cannot possibly reach its
own simulated final state of ⟨Ĥ.qn⟩
*The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*Only if embedded_H and H are ACTUALLY UTMs, and you aren't lying about just playing one on TV. If H (H^) (H^), and thus embedded_H (H^) (H^), go to Qn, then H^ (H^) goes to Qn too and halts, and thus the input specified HALTING behavior, as the behavior specified *IS* by the definiton of the problem, the behavior of the machine described by the input.
*The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*
*The input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H SPECIFIES non-halting behavior*
You can lie about this or try to get away with changing the subject.So, if H *IS* a UTM, H^ (H^) will be non-halting, but H isn't a decider.
What you cannot do it show that it is not true.
*Showing that it is relevant is a whole other different subject that*But until you do that. I just don't care about this part, and thus won't do anything to look at a problem that just isn't relevent.
*we can get to as soon as you quit your deception on this subject*
*Showing that it is relevant is a whole other different subject that*
*we can get to as soon as you quit your deception on this subject*
*Showing that it is relevant is a whole other different subject that*
*we can get to as soon as you quit your deception on this subject*
Les messages affichés proviennent d'usenet.