## Re: [PrimeNumbers] Re: Goldbach Restatement

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• ... IF (big if ) GC is shown to be undecidable in standard NT then that would mean that it is true, but it would also mean that there is no proof of that it
Message 1 of 58 , Oct 31, 2001
At 02:44 AM 10/31/2001 -0500, Jack Brennen wrote:

>Then you agree that a proof that GC is undecidable is by its nature
>also a proof that GC is true. If GC is in fact undecidable, such a
>proof cannot be allowed to exist - it would decide GC to be true.

IF (big "if") GC is shown to be undecidable in standard NT then that would
mean that it is true, but it would also mean that there is no proof of that
it is true in NT. It would be like Godel's construction of a statement
that is true, but can't be proven in the system.

If GC is false then it is decidable (by showing the counterexample in a
finite number of steps). So if GC is shown to be undecidable in NT then it
can't be false.

To look at it a slightly different way, if X is shown to be undecidable in
a system, that says two things (1) there is no proof of X in the system,
(2) there is no proof of ~X in the system. So if GC is shown to be
undecidable, it can't be false since says that there is no counterexample
whereas a false GC says that there is a counterexample.

>I don't disagree with that.

It conflicts with what you said though. To be undecidable, it is only
necessary that at least one of X or ~X is undecidable. At most one of them
can be decidable if the problem is to be undecidable (this is called
partial decidability for the one that is decidable.)

+---------------------------------------------------------+
| Jud McCranie |
| |
| Programming Achieved with Structure, Clarity, And Logic |
+---------------------------------------------------------+
• ... IF (big if ) GC is shown to be undecidable in standard NT then that would mean that it is true, but it would also mean that there is no proof of that it
Message 58 of 58 , Oct 31, 2001
At 02:44 AM 10/31/2001 -0500, Jack Brennen wrote:

>Then you agree that a proof that GC is undecidable is by its nature
>also a proof that GC is true. If GC is in fact undecidable, such a
>proof cannot be allowed to exist - it would decide GC to be true.

IF (big "if") GC is shown to be undecidable in standard NT then that would
mean that it is true, but it would also mean that there is no proof of that
it is true in NT. It would be like Godel's construction of a statement
that is true, but can't be proven in the system.

If GC is false then it is decidable (by showing the counterexample in a
finite number of steps). So if GC is shown to be undecidable in NT then it
can't be false.

To look at it a slightly different way, if X is shown to be undecidable in
a system, that says two things (1) there is no proof of X in the system,
(2) there is no proof of ~X in the system. So if GC is shown to be
undecidable, it can't be false since says that there is no counterexample
whereas a false GC says that there is a counterexample.

>I don't disagree with that.

It conflicts with what you said though. To be undecidable, it is only
necessary that at least one of X or ~X is undecidable. At most one of them
can be decidable if the problem is to be undecidable (this is called
partial decidability for the one that is decidable.)

+---------------------------------------------------------+
| Jud McCranie |
| |
| Programming Achieved with Structure, Clarity, And Logic |
+---------------------------------------------------------+
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