Saturday, 25 June 2011

Roy's Fortnightly Puzzle: Volume 5

(A day early, to make up for the last one being a day late!)

Assume we are working within Peano arithmetic (and, for a formula F, F will be the Goedel code of F) supplemented with a new predicate (either "T(...)", "K(...)", or "B(...)"). We obtain the Liar paradox by adding the T-schema to PA:

T-schema: For any statement F:
T(F) <--> F
is a theorem.

[This rule is of course initially plausible if we read "T(...)" as a truth predicate.] The Paradox of the Knower (or Montague's paradox) proceeds by weakening this assumption, replacing the T-schema with the following three principles:

Necessitation: For any formula F, if F is a theorem, then:
K(F)
is a theorem

Closure: For any formulas F and G:
K(F --> G) --> (K(F) --> K(G))
is a theorem.

Factivity: For any formula F:
K(F) --> F
is a theorem.

[These rules are initially plausible if we read "K(...)" as an in principle knowability predicate.] Notably, both of these paradoxes require Factivity. But there are similar paradoxes that do not require that Factivity hold, in general, of the predicate in question. For example, let "B(...)" be a "warranted belief" predicate, and consider the following rules. First, we have versions of Necessitation and Closure for B(...). Second, we have:

Transparency of Disbelief: For any formula F:
B( ~ B(F)) --> ~ B(F)
is a theorem.

In other words: (i) We are warranted to believe theorems. (ii) If we are warranted to believe a conditional, and we are warranted to believe the antecedent, then we are warranted to believe the consequent. (iii) If we are warranted to believe that we aren't warranted in believing something, then we aren't warranted in believing that something.

We can derive a contradiction similar to the one obtained in the Liar and the Knower, using only diagonalization and the three principles above (derivation left to the reader!) This derivation, to emphasize the central point, does not require that warranted belief entails truth!

[Note: Given the last few decades of work in epistemology, the transparency condition might not seem all that plausible. But I think it would have been plausible pre-Gettier. More importantly, the fact that its failure is a matter of pure logic (or, perhaps, pure logic plus pure arithmetic) is, I think, novel and substantial!]

Some questions about this paradox:

(1) Is it novel?

[If so, I hereby dub it the Cook-St. Croix paradox, because I am an arrogant @$%& and because I got thinking about this again while doing an independent study on this stuff with Cat St. Croix (who is completely not an arrogant @$%&, by the way!), who helped me work out some of the stuff below]

(2) Is it Yablo-izable? Are there additional assumptions needed that are not needed for the Liar variant of Yablo?

(3) Is there a set-theoretic analogue of this paradox?

(4) Is it an instance of Priest's inclosure schema?

(5) Is the existence of a paradox that does not depend on factivity of any other philosophical significance?

(6) What is the connection between this paradox and Loeb's theorem (there are, of course, deep connections between Loeb's theorem and both the Liar and the Knower!)

Discuss.

3 comments:

  1. First fuzzy thoughts (Sunday morning hangover here...):

    Your transparency of disbelief schema is of course a special case of the more general factivity schema (F is instantiated with ~B(P)). So what this tells me is that you don't need unrestricted factivity to get the paradox going, but in a sense there *is* a form of factivity in your paradox, actually. It would be interesting to check how weak/strong a quasi-factivity principle would have to be to give rise to the paradox. Here you are showing that a rather restricted factivity principle already does that.

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  2. You are certainly right about the version I gave in the post. It turns out, however, that the paradox can be obtained without any explicit version of factivity (although, as you will see below, it still requires something factivity-like!)

    Let:
    B[1](F) = B(F)
    B[n+1](F) = B(B[n](F))

    (In other words, B[m](F) is just F preceded by m applications of predicate B.)

    Now, let a be any natural number greater than or equal to b and b > 0.

    Then combining Necessitation, Closure, and:

    Principle: B[a] ~ B[b](F) --> ~ B[b](F)

    provides a contradiction.

    (note: This amounts to a special instance of factivity for B[a], but not for B(..) itself!)

    Roy

    PS: The proof goes as follows. By diagonalization, obtain G such that G <--> B[a - b] ~ B[b](G). Then by logic and closure, we get B[b](G) <--> B[a] ~ B[b](G). Assume B[b]G. So B[a] ~ B[b]G. By our principle, this entails ~ B[b](G). Contradiction. So ~ B[b](G) is a theorem. By necessitation, B[a] ~ B[b](G) is a theorem. By diagonalization, B[b](G) is a theorem. Contradiction.

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  3. Wow, this one didn't get the interest I thought it would. So here's one last attempt:

    Priest's argument that the Liar is an instance of his inclosure schema runs roughly as follows. Let T be the class of truths, and f(X) the function that maps any set of statements X onto the statement "This sentence is not in X". It follows that, for any set of true statements S - that is, any set of statements S where S is a subset of T:

    (i) f(S) is not in S.
    (ii) F(S) is in T.

    Paradox looms if we substitute T for S.

    Now, the question is whether we can get the Cook-St. Croix paradox to fit the inclosure scheme. Here is an attempt that doesn't quite work: let W be the set of all statements we are warranted to believe, and for any set of statements X, let f(X) = "This statement is not in X". Now, in order to get the inclosure scheme, we need, for any subset S of W:

    (i) f(S) is not in S.
    (ii) f(S) is in W.

    These clearly don't follow merely from S being a subset of W. So what modifications are needed (either to the inclosure scheme itself, or to my set-up above) to get this to fit Priest's view?

    (Of course, Priest could just reject the transparency condition and hence reject the claim that there is any paradox here at all, but that seems fishy somehow).

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