Re: Proof that H(D,D) meets its abort criteria

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?