orthonormal comments on Decision Theories, Part 3.75: Hang On, I Think This Works After All - Less Wrong

23 Post author: orthonormal 06 September 2012 04:23PM

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Comment author: orthonormal 06 September 2012 05:07:00PM *  3 points [-]

Proof that a Löbian cycle works out (hat tip, Eliezer and co.):

|- □A -> B

|- □B -> A

then

|- □(□A->B)

|- □□A->□B

|- □□A->A

|- □(□□A->A)

|- □□□A->□A

|- □A->□□A

|- □□□A->□□A

(apply Löb)

|- □□A

|- A

Comment author: Benja 06 September 2012 10:01:09PM *  10 points [-]

How about just,

(simple facts about PA:)

|- □(A ∧ B) -> (□A ∧ □B)

(apply the assumptions:)

|- □(A ∧ B) -> (B ∧ A)

|- □(A ∧ B) -> (A ∧ B)

(apply Löb:)

|- A ∧ B

Comment author: Eliezer_Yudkowsky 06 September 2012 10:46:25PM 2 points [-]

Heh. Nice!

Comment author: wedrifid 06 September 2012 06:13:22PM 8 points [-]

Am I supposed to be seeing lots of boxes here?

Comment author: [deleted] 06 September 2012 06:27:10PM 23 points [-]

For once in Unicode's sad history, yes.

Comment author: orthonormal 06 September 2012 06:27:56PM 6 points [-]

Yeah, it's the syntax "□A" for "there is a Gödel-numbered proof of A", and "|-" for "the following is provable".

Or is your browser showing a lot of junk instead of things that cash out to "it is provable that (a Gödel-numbered proof of A) implies B"?

Comment author: wedrifid 07 September 2012 03:34:27AM 1 point [-]

Or is your browser showing a lot of junk instead of things that cash out to "it is provable that (a Gödel-numbered proof of A) implies B"?

Assuming the only operators you used were <box>, "|-" and "->" it all came through.

Comment author: orthonormal 06 September 2012 06:14:28PM 1 point [-]

Note the key step,

|- □A->□□A

and why it's true even though

|- A->□A

is false: the formal system can see how to take a Gödel-numbered proof and replace all of the tokens with more complicated ones to get a Gödel-numbered proof of a Gödel-numbered proof, but it has nothing to work with if it starts by assuming just an object-level result.

Comment author: Decius 08 September 2012 10:06:39PM *  0 points [-]

What is special about A being provable that makes |- □A -> B provable, given only that it is true? For example, assume some A and B such that:

|- □A

|- B

|- □A -> B is trivial, as is |- □B -> A. (it follows from |-□A->A)

but |- □B is false.

Edit: changed some variables to be more clear.

Comment author: orthonormal 09 September 2012 01:07:52AM 2 points [-]

Uh, I don't understand you. |- □A->□□A is a very special case.

it follows from |-□A->A

That statement isn't true for general A!

Comment author: Decius 09 September 2012 05:06:33AM 0 points [-]

□A->A, (true statement, assuming a sound system)

but not

|- (□A->A)

□(□A->A) (confirm these are false statements)

but

|- □□A->A

and

|- □A-> □□A (these are statements you have made)

Comment author: orthonormal 09 September 2012 05:08:44AM 2 points [-]

|- □□A->A follows from the assumptions |- □A -> B and |- □B -> A; it's not a theorem in general.

Comment author: Decius 09 September 2012 01:37:56PM *  1 point [-]

Does it apply to (All A for which exist a B such that |- □A -> B and |- □B -> A)?

Is it possible to construct a B in general that meets those descriptions for an arbitrary statement? For example, B=" (a proof that a numbered proof of A implies B) and (a proof of a numbered proof of B implies A)" (B="( |- □A -> B) and ( |- □B -> A) ), or is a self-referential statement prohibited in sound systems?

Does B even have to be true, much less provable?

Comment author: orthonormal 09 September 2012 01:48:38PM 1 point [-]

To the first question, yes. That's what the top comment in this sequence proved (and Benja's reply proved much more elegantly).

To the second question, self-referential statements can't be ruled out in sufficiently powerful formal systems- that's what Gödel's Incompleteness Theorem uses!

To the third, again, if any formal system proves □A -> B and □B -> A, then it proves A and B.

Comment author: Decius 09 September 2012 07:58:44PM 1 point [-]

Consider the statement A "□A -> B and □B -> A" and the statement B "□A -> B and □B -> A and false"

Why are those two statements not provable in every system in which they can be written?