By Edith Spaan

This can be a doctoral dissertation of Edith Spaan lower than the supervision of prof. Johan van Benthem.

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**Extra resources for Complexity of Modal Logics [PhD Thesis]**

**Sample text**

CLASSIFICATION 39 A / \ is a skeleton subframe of T a and a a --------- ------------------- *»• Wt a = aa w0 B ~ is a skeleton subframe of Ta- wr ^ is a skeleton subframe of T a and a Oa a Wo a wi C , ^ is a skeleton subframe of Ta- a = aaa wr is a skeleton subframe of T a and is a skeleton subframe of Ta rn wr D is a skeleton subframe of T a and ? 2. 2. 2, we need to show the following two requirements: • T \ and T 2 satisfiability are in NP, and CHAPTER 3. 2, it follows that , and , ht are not skeleton subframes of T \ and ^ Let T be a class of frames that does not contain these three frames as skeleton subframes.

Let M = (VF, R u R2, 7r) be such that: • M ,w \= pt w e WT x {iu0}, • M, (i,w0) |= ( depth = d) O depth(i) = d, where depth is a propositional vector as on page 16, • M, (z, wo) |= pieft • M, (z, wq ) |= p i is a left child, M t , z |= p for p in 0. And define R 8UCc to simulate the edges of the tree in the obvious way: if M ,w |= depth = d and M ,w \= puft £ for d £ {0, . . , n} and i £ {T, _L}, then w R 8UCCw' iff • M, w' |= pt A depth(j) = d + 1, • there exists an R a path from w to w' such that for every world w" on this path: M, w" |= or M, w" |= ( depth = d + 1) or M, w" |= ( depth = d) A (puft <-+ i) It is immediate that j is a child of z iff (i,wo)R8Ucc{j^o)- We define a modality cor responding to R 8Ucc in the following way: for ^ a propositional formula, let [succ}ip be defined as follows: (depth A = d) A (p,cft *-> t) -»• 0 < d < n , *e{T ,_ L } [ai ](-»pt V ( depth = d + 1) V ((depth = d) A (puft <-+ £)) V (depth = d 4- 1) V ((depth = d) A ( pieft k * ](P t A (deptfz = F rom th e d e fin itio n o f F succ, it fo llo w s th a t M , ^(-0) b e th e r esu lt o f rep la cin g e v er y □ by d+ -> |= [su cc]^ iff Vu>'(u;Fsuccu / => ^ ) .

In this specific case, it is easy to see how to do this: look at the following T \ 0 Tz frame: 2 I 2 I” ! 1 This certainly looks like a binary tree. Furthermore, trees of arbitrary depth can be encoded in this way. 1 as repeated above, which proves PSPACE hardness for this simple case. Note that all the information of this tree simulation is contained in the following T \ 0 Tz frame F: 2 2 W£ Wr Binary tree simulations can be viewed as being constructed from F frames by replacing each node i by a copy of F, and identifying the wi (wr) world of the frame belonging to i with the w0 world of the frame belonging to j for j the left (right) child of i.

### Complexity of Modal Logics [PhD Thesis] by Edith Spaan

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