Re: How the requirements that Professor Sipser agreed to are exactly met --- WDH

On 5/14/2025 6:20 AM, Richard Damon wrote:

On 5/13/25 9:48 PM, olcott wrote:

On 5/13/2025 8:31 PM, dbush wrote:

On 5/13/2025 9:27 PM, olcott wrote:

On 5/13/2025 8:07 PM, dbush wrote:

On 5/13/2025 5:30 PM, olcott wrote:

On 5/13/2025 6:43 AM, Richard Damon wrote:

On 5/13/25 12:52 AM, olcott wrote:

On 5/12/2025 11:05 PM, Richard Damon wrote:

On 5/12/25 10:53 PM, olcott wrote:

On 5/12/2025 8:27 PM, Richard Damon wrote:

On 5/12/25 2:17 PM, olcott wrote:

Introduction to the Theory of Computation 3rd Edition
by Michael Sipser (Author)
4.4 out of 5 stars    568 rating
>
https://www.amazon.com/Introduction-Theory-Computation- Michael- Sipser/ dp/113318779X
>
int DD()
  {
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
  }
>
DD correctly simulated by any pure simulator
named HHH cannot possibly terminate thus proving
that this criteria has been met:
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     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
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
  </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>

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Which your H doesn't do, as it can not correctly determine what doesn't happen.
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Any C programmer can correctly tell what doesn't happen.
What doesn't happen is DD reaching its "return" statement
final halt state.
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Sure they can, since that is the truth, as explained.
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Since your "logic" is based on lies and equivocation,

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If my logic was based on lies and equivocation
then you could provide actual reasoning that
corrects my errors.

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I hae.
>

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It is truism that simulating termination analyzers
must report on the behavior of their input as if
they themselves never aborted this simulation:

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Right, of the input actually given to them, which must include all their code, and that code is what is actually there, not created by this imaginary operation.
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In other words every single byte of HHH and DD are
100% totally identical except the hypothetical HHH
has its abort code commented out.

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In other words you changed the input.
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Changing the input is not allowed.
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>

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Thus, a HHH that aborts to return an answer, when looking at the DDD that calls it, must look at the unaborted emulation of THAT DDD, that calls the HHH that DOES abort and return an answer, as that is what the PROGRAM DDD is, If you can not create the HHH that does that without changing that input, that is a flaw in your system, not the problem.
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*simulated D would never stop running unless aborted*
or they themselves could become non-terminating.

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But you aren't simulating the same PROGRAM D that the original was given.
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It is not supposed to be the same program.

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So you *explicitly* admit to changing the input.
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The finite string of DD is specific sequence bytes.

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Which includes the specific sequence of bytes that is the finite string HHH
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No it does not. A function calls is not macro inclusion.
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The finite string of HHH is specific sequence bytes.
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The hypothetical HHH that does not abort its input
cannot have input that has changed because it never
comes into actual existence.

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But your HHH decides on that hypothetical non-input.
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The whole point here is not to critique the words
that professor Sipser agreed to.
>
The whole point here is to determine whether or
not HHH meets this spec. It is a verified fact
that it does meet this spec.

  And since the DD that HHH is simulating WILL HALT when fully simulated (an action that HHH doesn't do)

*NOT IN THE ACTUAL SPEC*
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     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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