XPost: comp.theory, sci.logic

*Completely rewritten rebuttal of the halting theorem*

A simulating halt decider (SHD) correctly predicts what the behavior of
its input would be if it never aborted the simulation of this input. It
does this by correctly recognizing several non-halting behavior patterns
in a finite number of steps of correct simulation. It must abort the
simulation of all non-terminating inputs so that it can report that they
are non-halting. Inputs that do terminate are simply simulated until
they complete.

When simulating halt decider H correctly predicts that directly executed
D(D) would remain stuck in recursive simulation (run forever) unless H
aborts its simulation of D this directly applies to the halting theorem
because H correctly determines:

from a description of an arbitrary computer program and an input,
whether the program will finish running, or continue to run forever.
https://en.wikipedia.org/wiki/Halting_problem

https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X

MIT Professor Michael Sipser has agreed that the following verbatim
paragraph is correct (he has not 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.

(b) is a necessary consequence of (a)

01 int D(int (*x)())
02 {
03 int Halt_Status = H(x, x);
04 if (Halt_Status)
05 HERE: goto HERE;
06 return Halt_Status;
07 }
08
09 void main()
10 {
11 D(D);
12 }

*Here is the sequence when H never aborts it simulation*
main() calls D(D) at line 11
D(D) calls H(D,D) that simulates D(D) at line 03
simulated D(D) calls simulated H(D,D) that simulates D(D) at line 03
simulated D(D) calls simulated H(D,D) that simulates D(D) ...
repeating ...

Because it is an easily verified fact that D(D) would never stop running
unless H aborts its simulation of D, H is necessarily correct to return
0 indicating this verified fact.

*This is every aspect of my whole proof*
(a), (a) → (b) ∴ H(D,D)==0 is correct

H(D,D) is fully operational in the x86utm operating system: https://github.com/plolcott/x86utm

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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XPost: comp.theory, sci.logic

On 3/8/23 1:16 PM, olcott wrote:

*Completely rewritten rebuttal of the halting theorem*

A simulating halt decider (SHD) correctly predicts what the behavior of
its input would be if it never aborted the simulation of this input. It
does this by correctly recognizing several non-halting behavior patterns
in a finite number of steps of correct simulation. It must abort the simulation of all non-terminating inputs so that it can report that they
are non-halting. Inputs that do terminate are simply simulated until
they complete.

So, the criteria for a SHD is NOT the criteria for the Halting Problem,
so unless you can actually PROVE them equivalent, you are just admitting
that your Simulating Halt Deciders ar *NOT* HALT DECIDERS!!!

When simulating halt decider H correctly predicts that directly executed
D(D) would remain stuck in recursive simulation (run forever) unless H
aborts its simulation of D this directly applies to the halting theorem because H correctly determines:

   from a description of an arbitrary computer program and an input,
   whether the program will finish running, or continue to run forever.
   https://en.wikipedia.org/wiki/Halting_problem

So, how do you connect your simulation to the actual behavior of the
machine in question?

https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X
MIT Professor Michael Sipser has agreed that the following verbatim
paragraph is correct (he has not 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

Which you H doesn't do according to the meaning of the words as
Professor Sipser would use.

His only definition of "Correctly determine that its simulated D would
never stop running" would be that H determines that the UTM simulation,
by an ACTUAL UTM, would never stop running.

Since UTM D,D will halt because D(D) calls H(D,D) which you are saying
will return 0, which will cause D(D) to Halt.

Therefore, eiher you H returned 0 by breaking its requirement, or it
juyst will never ever return and fail to be a decicer.

You have admitted (by failing to provide rebutting proof) that Main
calling H(D,D) and main calling D(D) calling H(D,D) will both return o

(b) H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.

(b) is a necessary consequence of (a)

01 int D(int (*x)())
02 {
03   int Halt_Status = H(x, x);
04   if (Halt_Status)
05     HERE: goto HERE;
06   return Halt_Status;
07 }
08
09 void main()
10 {
11   D(D);
12 }

*Here is the sequence when H never aborts it simulation*
   main() calls D(D) at line 11
   D(D) calls H(D,D) that simulates D(D) at line 03
   simulated D(D) calls simulated H(D,D) that simulates D(D) at line 03
   simulated D(D) calls simulated H(D,D) that simulates D(D) ...
repeating ...

Which only happens if H never aborts, and simce you latter say H WILL
abort, just shows you are doing unsound logic based on a false premise
and thus becoming an admitted LIAR and HYPOCRITE.

Because it is an easily verified fact that D(D) would never stop running unless H aborts its simulation of D, H is necessarily correct to return
0 indicating this verified fact.

But since H DOES abort its simulation, the above is just unsound logic,
and your continued insistace proves you are of unsound mind.

*This is every aspect of my whole proof*
(a), (a) → (b) ∴ H(D,D)==0 is correct

Nope.

H(D,D) is fully operational in the x86utm operating system: https://github.com/plolcott/x86utm

Right, and that program proves your logic to be unsound, as H does abort
its simulation and returns 0 to D and thus D does halt.

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