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

On 3/15/2024 1:15 PM, Richard Damon wrote:

On 3/15/24 10:23 AM, olcott wrote:

On 3/15/2024 12:07 PM, immibis wrote:

On 15/03/24 17:20, 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/
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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.
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*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*
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*Proof that H(D,D) meets its abort criteria*
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int D(int (*x)())
{
   int Halt_Status = H(x, x);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
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int main()
{
   Output("Input_Halts = ", H(D,D));
}
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  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)
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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()
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*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).
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There are conditional branch instructions inside H(D,D). This is obvious. Why do you keep lying?

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(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
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It is true that D(D) would never stop running unless the
outermost H(D,D) aborts its simulation thus meeting the
above criteria.
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Nope, not if D is claimed to be the equivalent of Linz H^.
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If you are willing to DISAVOW any posssible conntection to that we can discuss that version, but then you are admitting this is just a time wasting change of topic.
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The issue becomes a definition of identity.
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IF we are in an equivalency to Linz H/H^, then the H that D calls is a seperate identity to the H that is simulating that D.
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Thus, the outer H doesn't NEED to abort,

 (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
 That is incorrect yet too difficult for you to understand
that it is incorrect until after you first understand that
D(D,D)==0 is correct for the above criteria.