On 6/8/2024 11:03 AM, Richard Damon wrote:
On 6/8/24 11:32 AM, olcott wrote:
On 6/8/2024 10:15 AM, Richard Damon wrote:
On 6/8/24 11:07 AM, olcott wrote:
On 6/8/2024 9:54 AM, Richard Damon wrote:
On 6/8/24 10:20 AM, olcott wrote:
On 6/8/2024 9:10 AM, Richard Damon wrote:
>
I HAVE pointed out what is missing, ANY set of truth-perserving operations from the accepted facts (which will of course need to name the fact they are working from) to your conclusion.
>
The accepted facts are here
(a) The x86 language
(b) The notion of an x86 emulator
>
{The proof that No DDD correctly emulated by any x86
emulator H can possibly reach its own [00001df6] instruction}
>
So, how do you show this claim?
>
Do you have a tracing of the full INFINITE SET of possible Hs?
>
>
Is the set of possible execution traces of DDD correctly
emulated by x86 emulator HH on the basis of the above
accepted facts.
>
Maybe you are just clueless about these technical details
are are trying to hide this with pure bluster.
>
_DDD()
[00001de2] 55 push ebp
[00001de3] 8bec mov ebp,esp
[00001de5] 8b4508 mov eax,[ebp+08]
[00001de8] 50 push eax ; push DD
[00001de9] 8b4d08 mov ecx,[ebp+08]
[00001dec] 51 push ecx ; push DD
[00001ded] e890f5ffff call 00001382 ; call HH
[00001df2] 83c408 add esp,+08
[00001df5] 5d pop ebp
[00001df6] c3 ret
Size in bytes:(0021) [00001df6]
>
You keep disagreeing with the fact that DDD correctly
emulated by x86 emulator HH only has one single correct
execution trace of repeating the fist seven lines until
out-of-memory error.
>
>
But that is an INCORRECT trace per your definition,
>
The call HH instruction MUST be simulated into HH because that IS the behavior of the x86 instruction.
>
Did I ever say that it is not?
For the above DDD correctly emulated by x86 emulator HH
the first seven instructions of DD keep repeating because
DDD keeps calling HH(DDD,DDD) to emulate itself again and
again until HH/DDD hits out-of-memory exception.
>
So the x86 emulation of the code must go into HH(DDD,DDD)
>
>
It is pretty stupid to assume otherwise when HH is
stipulated to be an x86 emulator.
Right, so why did you say otherwise?
I never said otherwise you simply "read" meanings that I didn't say.
this thread: [Should I quit Richard at this point?]
stands alone and should not be interpreted within the
context of anything else that I ever said.
>
The correct x86 emulation of the call to HH(DDD,DDD) will NEVER get to the sequence of instrucitions starting at 00001DE2, as the code will never jump there to just execute it.
>
>
Your are saying that incorrectly DDD correctly emulated by
x86 emulator HH cannot possibly reach it own machine address
of [00001df6].
I said nothing about that. You are just serving Herring with Red Sauce agian.
>>> The correct x86 emulation of the call to HH(DDD,DDD) will NEVER get
>>> to the sequence of instructions starting at 00001DE2
Wrong! DDD correctly simulated by HH will never reach its own
machine address of 00001df6 because DDD correctly simulated
by HH keeps starting over with another instance of itself at
00001DE2 after the prior instance calls HH(DDD,DDD) to simulate
itself again.
The CORRECT simulation of DDD can NEVER get back to the sequence of instructions at 00001DE2, as there is never a jump to that address, only the emulator starting an emuation of that address, and the correst simulation of a simulatore is NOT the code the simulated simulator is looking at, but the code of the simulator doing the simulation.
DDD correctly simulated by HH will never reach its own
machine address of 00001df6 because DDD correctly simulated
by HH keeps starting over with another instance of itself at
00001DE2 after the prior instance calls HH(DDD,DDD) to simulate
itself again.
>
By your code, the simulator will "Debug Step" those instructions.
>
>
The underlying details of one HH are irrelevant when I reference
an infinite set:\
But not for your definition of the simulation.
>
{The proof that No DDD correctly emulated by
any x86 emulator H
any x86 emulator H
any x86 emulator H
any x86 emulator H
any x86 emulator H
any x86 emulator H
can possibly reach its own [00001df6] instruction}
It seems that by your current analysis, it can't get past the instruction at 00001DED as there is nothing to simulate after that.
*A stupid thing to say*
>>>>>> [00001ded] e890f5ffff call 00001382 ; call HH
>>>>>> [00001df2] 83c408 add esp,+08
>>>>>> [00001df5] 5d pop ebp
>>>>>> [00001df6] c3 ret
>>>>>> Size in bytes:(0021) [00001df6]
Remember, your definition said to simulate the instructions in the strict order they were reached.
If we don't have the instruction at 00001382, the simulation has to stop, as we can't go on.
*A stupid thing to say*
>>>>>> _DDD()
>>>>>> [00001de2] 55 push ebp
>>>>>> [00001de3] 8bec mov ebp,esp
>>>>>> [00001de5] 8b4508 mov eax,[ebp+08]
>>>>>> [00001de8] 50 push eax ; push DD
>>>>>> [00001de9] 8b4d08 mov ecx,[ebp+08]
>>>>>> [00001dec] 51 push ecx ; push DD
>>>>>> [00001ded] e890f5ffff call 00001382 ; call HH
>>>>>> [00001df2] 83c408 add esp,+08
>>>>>> [00001df5] 5d pop ebp
>>>>>> [00001df6] c3 ret
>>>>>> Size in bytes:(0021) [00001df6]
It has been stipulated that there is ALWAYS an x86
emulator at the address 00001382.
>
>
By a pure emulator, that would mean translating the machine code into the operations it will perform, and then manipulating the virtual register set being kept by the emulator.
>
>
libx86emu does that.
And if HH is a pure emulator, it need to do it to (or let libx86emu do it for it). and HH can't interfear with that process by not following each instruction with the instruction that follows it.
This thread is only about what you have already agreed to dozens
of times and you fail to notice that because rebuttal is your
game and honest dialogues are most always off-the-table for you.
*Ben already agreed that the criteria that*
*Professor Sipser agreed to have been met*
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
> I don't think that is the shell game. PO really /has/ an H
> (it's trivial to do for this one case) that correctly determines
> that P(P) *would* never stop running *unless* aborted.
You want to always ignore that so you can remain in rebuttal
mode at the expense of truth.
So, does the libx86emu keep a seperate logging of the instructions that HH is emulating, as would be needed for HH to be able to examine that trace.
The previous trace you posted wasn't the simulation that HH was doing, but was a trace of the execution of HH iteself.
It seems that HH doesn't actual have a trace of what it did available (or at least you didn't show it).
LIAR LIAR PANTS ON FIRE
I HAVE SHOWN THIS TRACE AGAIN AND AGAIN FOR THREE YEARS
Begin Local Halt Decider Simulation Execution Trace Stored at:113075
[00001c22][00113061][00113065] 55 push ebp
[00001c23][00113061][00113065] 8bec mov ebp,esp
[00001c25][0011305d][00103031] 51 push ecx
[00001c26][0011305d][00103031] 8b4508 mov eax,[ebp+08]
[00001c29][00113059][00001c22] 50 push eax ; push DD
[00001c2a][00113059][00001c22] 8b4d08 mov ecx,[ebp+08]
[00001c2d][00113055][00001c22] 51 push ecx ; push DD
[00001c2e][00113051][00001c33] e80ff7ffff call 00001342 ; call HH
New slave_stack at:14da95
[00001c22][0015da89][0015da8d] 55 push ebp
[00001c23][0015da89][0015da8d] 8bec mov ebp,esp
[00001c25][0015da85][0014da59] 51 push ecx
[00001c26][0015da85][0014da59] 8b4508 mov eax,[ebp+08]
[00001c29][0015da81][00001c22] 50 push eax ; push DD
[00001c2a][0015da81][00001c22] 8b4d08 mov ecx,[ebp+08]
[00001c2d][0015da7d][00001c22] 51 push ecx ; push DD
[00001c2e][0015da79][00001c33] e80ff7ffff call 00001342 ; call HH
Local Halt Decider: Recursive Simulation Detected Simulation Stopped
First published complete execution trace proving that D is correctly
simulated by H and correctly simulated by simulated H. The proof of
this that everyone has ignored for three solid years is that the derived
execution traces exactly match the behavior specified by the x86
machine code of D.
On 5/29/2021 2:26 PM, olcott wrote:
[Would the simulation of D be infinitely nested unless simulating partial halt decider H terminated its simulation of D?]
Message-ID: <
YJKdnZg9v__rCC_9nZ2dnUU7-QXNnZ2d@giganews.com>
But then, you always seemed to have gotten the "levels" of simulation incorrect.
No you just lie about this.
>
If your "simulation" is ACTUALLY being done using the debug step hardware of the system (or simulating the actions of that hardware) then the instruction are executed, but not in sequence as they have all the steps of the debugger/tracing around them.
>
>
That is not how x86 emulators work.
But it is what "Debug Step" implies.
It <is> a Debug_Step though the emulated code.
You keep freaking forgetting what the Hell an emulator is.
Or, are you not familiar with that part of the x86 hardware.
Yes.
>
So, your claim of what happens just shows you don't understand what the program you are using actually is doing.
>
>
No it shows that you don't know how x86 emulators work.
So, what did I not understand?
You keep freaking forgetting what the Hell an emulator is.
>
That might explain why the trace you posted the other day wasn't actually the trace you claimed it was.
>
>
We are only focusing on this one thread and zero deflection
will be tolerated.
Ok, so why doesn't your HH here trace into HH
As I have said 5,000 times when people cannot possibly
understand 1/2 of one page of code mixing in another 250
pages cannot possibly help.
And why does it start to show a trace of instructions that are never actually directly simulated by the HH that we are talking about>
I have already proved THREE FREAKING YEARS AGO
that D is correctly simulated by H and D is correctly
simulated by simulated H.
That everyone has ignored this proof is probably
because they were trying to hide their own cluelessness.
>
>
This is the only post that I will reply to you on.
I need you to stay focused on this one single point
until you understand it.
>
Is that a promise? I think you will break it.
>
>
When you proved to break out of your "stuck in rebuttal mode"
nonsense and talked about closure I backed off this requirement
for a while.
>
*Even Ben admits that H does meet the Sipser criteria*
>
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
> I don't think that is the shell game. PO really /has/ an H
> (it's trivial to do for this one case) that correctly determines
> that P(P) *would* never stop running *unless* aborted.
>
Nope, he is just pointing out your circular logic.
LIAR LIAR PANTS ON FIRE
Yes, P(P) won't halt if the code of H won't abort it.
thus meeting the Sipser approved criteria.
But if the code of H does abort it, it will halt.
D is accountable for the behavior of its actual input and
not freaking accountable for any damn thing else in the
whole freaking universe.
That there is a consensus of opinion against this is
just like the consensus of opinion that the Earth was flat
prior to Pythagoras.
Halt deciders compute the mapping FROM THEIR INPUTS
BASED ON THE ACTUAL BEHAVIOR THAT THIS INPUT SPECIFIES.
This input specifies the behavior of DD correctly
simulated by HH.
That everyone wants to avoid looking at my conclusive
proof that DD correctly simulated by HH has provably
different behavior than the direct executed DD(DD) is
dishonest.
Thus, if the question actually WAS "Will the template P(P) Halt if the decider it is pair with doesn't abort its simulation", then he agreed you could answer No.
But that isn't the Halting Question, so it isn't the correct answer to the question that H is supposed to answer.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Why does Olcott care about simulation, anyway?
Re: Why does Olcott care about simulation, anyway?
