The Boxdot Conjecture and the Language of Essence and Accident

We show the Boxdot Conjecture holds for a limited but familiar range of Lemmon-Scott axioms. We re-introduce the language of essence and accident, first introduced by J. Marcos, and show how it aids our strategy. 1   &       In modal logic, the boxdot translation, t, is the following translation: tp = p t? = ? t(φ! ψ) = (tφ! tψ) t φ = ( tφ∧ tφ) Note that t φ = ( tφ∨ tφ) and t¬φ = ¬tφ. The name derives from φ as an abbreviation for φ∧φ in Boolos [1]. Where K is the minimal normal modal logic, let K φ be the smallest normal modal logic containing φ. Let L be some normal modal logic and let KT be K φ! φ. French and Humberstone [4] conjectured, if (8ψ)(KT ` ψ if and only if L ` tψ), then L KT . This is the Boxdot Conjecture. The conjecture was established for normal modal logics of the form K φ with φ of modal degree 1. Here we show the conjecture holds for extensions of K which include any instance of the following axiom schema, h p! j p An instance is given by a specific choice of h, i, j, k 2 {0, 1, 2, . . .}. We use φhijk to represent an arbitrary instance. This schema is a limited form of the more Christopher Steinsvold, “The Boxdot Conjecture and the Language of Essence and Accident”, Australasian Journal of Logic (10) 2011, 18–35 http://www.philosophy.unimelb.edu.au/ajl/2011 19 general Lemmon-Scott axiom schema, see Goldblatt [5]. Clearly, there are infinitely many φhijk which are theorems of KT , thus we show: for all φhijk / 2 KT, (9ψ)(K φhijk ` tψ and KT 6` ψ) This is our main result. We now begin to discuss our strategy using examples and build up to a discussion of the language of essence and accident which will aid our strategy. For the remainder of this article, assume q and p are distinct. Consider K Dc (i.e., K p! p), and the following, (¬p∧ p)! [(q! p)! (q! p)] Call this sentence S(Dc). K ` (¬p ∧ p) ! p, and it straightforward to show that K ` p ! [(q ! p) ! (q ! p)]. Thus, K Dc ` S(Dc), but KT 6` S(Dc), and we leave it to the reader to find a reflexive frame where S(Dc) fails. Significantly, S(Dc) is K-equivalent to its own translation, i.e. K ` tS(Dc)$ S(Dc) Though French and Humberstone already showed this using a different sentence, our example shows that the conjecture holds for K Dc. For p! p / 2 KT , and K Dc ` tS(Dc) and KT 6` S(Dc). Consider K5 (i.e. K p ! p), and the following, (¬p∧ p)! [( (q! p) ∨ (q! p))! ( (q! p) ∨ (q! p))] Call this sentence S(5). As with the previous example K5 ` S(5), but S(5) is not a theorem of KT (again, we leave it to the reader to find a reflexive frame where S(5) fails). One can show,