Re: Proof that H(D,D) meets its abort criteria --counter-factual assumptions--

On 3/16/2024 12:32 PM, immibis wrote:

On 16/03/24 17:44, olcott wrote:

On 3/16/2024 11:32 AM, immibis wrote:

On 16/03/24 05:23, olcott wrote:

On 3/15/2024 10:55 PM, immibis wrote:

On 16/03/24 04:14, olcott wrote:

On 3/15/2024 6:26 PM, Richard Damon wrote:

On 3/15/24 3:41 PM, olcott wrote:

On 3/15/2024 5:10 PM, Richard Damon wrote:

On 3/15/24 2:13 PM, olcott wrote:

On 3/15/2024 3:27 PM, Richard Damon wrote:

On 3/15/24 12:38 PM, olcott wrote:

On 3/15/2024 2:30 PM, Richard Damon wrote:

On 3/15/24 12:14 PM, olcott wrote:

On 3/15/2024 2:06 PM, Richard Damon wrote:

On 3/15/24 11:39 AM, olcott wrote:

On 3/15/2024 1:38 PM, immibis wrote:

On 15/03/24 18:52, olcott wrote:

On 3/15/2024 12:36 PM, Richard Damon wrote:

On 3/15/24 9:20 AM, olcott wrote:

Best selling author of Theory of Computation textbooks:
*Introduction To The Theory Of Computation 3RD, by sipser*
https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
>
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running unless aborted then
(b) H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
>
*When we apply the abort criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then H(D,D) is proven to meet this criteria*
>
*Proof that H(D,D) meets its abort criteria*
>
int D(int (*x)())
{
   int Halt_Status = H(x, x);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   Output("Input_Halts = ", H(D,D));
}
>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[00001d22][00102fc9][00000000] 55         push ebp ; begin main()
[00001d23][00102fc9][00000000] 8bec       mov ebp,esp
[00001d25][00102fc5][00001cf2] 68f21c0000 push 00001cf2 ; push DD
[00001d2a][00102fc1][00001cf2] 68f21c0000 push 00001cf2 ; push D
[00001d2f][00102fbd][00001d34] e8eef7ffff call 00001522 ; call H(D,D)
>
H: Begin Simulation   Execution Trace Stored at:113075
Address_of_H:1522
[00001cf2][00113061][00113065] 55         push ebp ; enter D(D)
[00001cf3][00113061][00113065] 8bec       mov ebp,esp
[00001cf5][0011305d][00103031] 51         push ecx
[00001cf6][0011305d][00103031] 8b4508     mov eax,[ebp+08]
[00001cf9][00113059][00001cf2] 50         push eax ; push D
[00001cfa][00113059][00001cf2] 8b4d08     mov ecx,[ebp+08]
[00001cfd][00113055][00001cf2] 51         push ecx ; push D
[00001cfe][00113051][00001d03] e81ff8ffff call 00001522 ; call H(D,D)
H: Recursive Simulation Detected Simulation Stopped
                           H(D,D) returns 0 to main()
>
*That was proof that H(D,D) meets its abort criteria*
H(D,D) correctly determines that itself is being called with its same inputs and there are no conditional branch instructions between the invocation of D(D) and its call to H(D,D).
>
>

>
Except that D calling H(D,D) does NOT prove the required (a), since the simulated D WILL stop running because *ITS* H will abort *ITS* simulation and returm 0 so that simulated D will halt.

You keep saying that H(D,D) never really needs to abort the
simulation of its input because after H(D,D) has aborted the
simulation of this input it no longer needs to be aborted.
>

You keep thinking there is more than one H(D,D) and then when it's convenient for you you think there is only one H(D,D). Why is that?

>
The first H(D,D) to see that the abort criteria has been met
(the outermost one) must abort the simulation of its input or
none of them ever abort.
>

>
But since it does, which is your definition of H, the others

>
never begin to be simulated.

>
But D(D) started to be simulated, and we can know what D(D) actually does, which includes it using its version of H.
>

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We cannot reference the behavior of what D(D) does after H(D,D)
has already aborted the simulation of its input at the point
in time before H(D,D) aborts its input as any criterion measure
for this H(D,D).
>

>
WHy not?
>
That is what Correct Simulation refers to.
>
I guess you are just admiting to being a LIAR (or stupid).

>
*I am not a liar or stupid and you admitted your mistake*
*I am not a liar or stupid and you admitted your mistake*
*I am not a liar or stupid and you admitted your mistake*
*I am not a liar or stupid and you admitted your mistake*

>
So, do you admit that the definition of a "Correct Simulation" for the purposes of that criteria are the complete not-aborted simulation done by possibly some other simulator?
>

>
(a) If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
would never stop running unless aborted then
>
Not at all the words don't say anything like that.
"H correctly simulates its input D until"
specifically means a partial simulation.
>

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Means H uses a partial simulation to make its decision.
>

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Finally you get this.
>

The correctness of the decision is measured by the full simulation, even past where H simulated. Thus, is based on things H might not know.
>

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No. the correctness of the decision is essentially anchored
in something like mathematical induction that correctly
predicts that complete simulation would never end.

>
The correctness of the decision is anchored in whether D(D) halts or not.

>
A termination analyzer must have some way to predicate this.
H(D,D) can only predict what it actually sees and H(D,D)
sees that it must abort the simulation of its input.
>

A termination analyzer cannot always correctly predict this. Every time you make a termination analyzer it is easy to demonstrate one program which it predicts incorrectly.

>
Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) If simulating halt decider H correctly simulates its input D until H correctly determines that its simulated D would never stop running unless aborted then
(b) H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
>
To the best of my knowledge no one here has ever shown that
the above criteria is impossible to always meet.
>

 You did not read the message or did not understand it.

Or you spouted off a rebuttal based on counter-factual assumptions
that you did not understand were counter-factual.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer