Quine s Fluted Fragment is Non-elementary

Transcription

1 Quine s Fluted Fragment is Non-elementary Ian Pratt-Hartmann Wies law Szwast Lidia Tendera University of Manchester/University of Opole 25th EACSL Annual Conference on Computer Science Logic 1 st September, 2016

2 Outline

3 Fluted fragment: first identified by W.V.Quine in Order of quantification of variables matches order of appearance in predicates. Examples: No student admires every professor x 1 (student(x 1 ) x 2 (prof(x 2 ) admires(x 1, x 2 ))) No lecturer introduces any professor to every student x 1 (lecturer(x 1 ) x 2 (prof(x 2 ) x 3 (student(x 3 ) intro(x 1, x 2, x 3 )))).

4 The syntax of the fluted fragment is defined as follows. Let x 1, x 2,... be a fixed sequence of variables. The fluted fragment with k free variables, FL [k], is defined by simultaneous induction for all k: - any atom p(x l, x l+1,..., x k ) is in FL [k] ; - FL [k] is closed under Boolean operations; - FL [k] contains x k+1 ϕ and x k+1 ϕ for any ϕ FL [k+1]. The fluted fragment, FL is the union of all the FL [k]. For all m > 0, we define FL m, to be the set of fluted formulas containing at most the variables x 1,..., x m, free or bound. The allusion seems to be architectural:

5 History: Quine (1968): Skolem s proof for the monadic fragment of FOL generalizes to homogeneous m-adic formulas. Quine (1972): We can further generalize to the fluted fragment.

6 History: Quine (1968): Skolem s proof for the monadic fragment of FOL generalizes to homogeneous m-adic formulas. Quine (1972): We can further generalize to the fluted fragment. Noah (1980): Quine s further generalization invalidates Skolem s proof.

7 History: Quine (1968): Skolem s proof for the monadic fragment of FOL generalizes to homogeneous m-adic formulas. Quine (1972): We can further generalize to the fluted fragment. Noah (1980): Quine s further generalization invalidates Skolem s proof. Purdy (1996): FL has the finite model property; hence its satisfiability problem is decidable.

8 History: Quine (1968): Skolem s proof for the monadic fragment of FOL generalizes to homogeneous m-adic formulas. Quine (1972): We can further generalize to the fluted fragment. Noah (1980): Quine s further generalization invalidates Skolem s proof. Purdy (1996): FL has the finite model property; hence its satisfiability problem is decidable. Purdy (2002): FL has the exponential-sized model property; hence its satisfiability problem is in NExpTime.

9 History: Quine (1968): Skolem s proof for the monadic fragment of FOL generalizes to homogeneous m-adic formulas. Quine (1972): We can further generalize to the fluted fragment. Noah (1980): Quine s further generalization invalidates Skolem s proof. Purdy (1996): FL has the finite model property; hence its satisfiability problem is decidable. Purdy (2002): FL has the exponential-sized model property; hence its satisfiability problem is in NExpTime. The claims in Purdy 2002 are false.

10 In fact, satisfiable formulas of FL 2m force models of m-tuply exponential size, that is, of size bounded below by a function where p is a polynomial....2p( ϕ ) m 2 s 2 = t(m, p( ϕ )) Essentially the same construction shows that the satisfiability problem for FL 2m is m-nexptime-hard. On the other hand, we also show that any satisfiable formula of FL m has a model of m-tuply exponential size; hence, the satisfiability problem for FL m is in m-nexptime. Therefore, for m 1, the complexity of the satisfiability problem for FL m lies between m/2 -NExpTime-hard and m-nexptime.

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12 For any z 1, the canonical representation of any integer n (0 n < 2 z 1) is its conventional binary encoding s z 1,..., s 0. For n in this range, to compute n 1 mod 2 z : for all i (0 < i < z), flip the ith digit of n just in case all digits of n less significant than the ith are 0. Let int 1 and p 0,..., p n 1 be unary predicates. We refer to any object b satisfying int 1 (in some structure) as a 1-integer. Define val 1 (b) to be the integer s n 1,..., s 0, where, for all i (0 i < n), { 1 if A = p i [b]; s i = 0 otherwise.

13 We can easily force a binary predicate pred 1 to denote the predecessor relation for 1-integers: x 1 x 2 pred 1 (x 1, x 2 ) n 1 i=0 0 j<i p i (x 1 ) [p i (x 1 ) p i (x 2 )] Suppose a structure A also makes the following true: x 1 int 1 (x 1 ) x 1 (int 1 (x 1 ) x 2 (int 1 (x 2 ) pred 1 (x 1, x 2 ))). Then A contains 1-integers with all values in the range [0, 2 n 1], whence A 2 n.

14 The above formula defining pred 1 is not fluted; but we can make it so using a trick. We shadow the unary predicates p 0,..., p n 1 with binary predicates p0 1,..., p1 n 1, and write, for all i (0 i < n) x 1 (p i (x 1 ) x 2.p 1 i (x 1, x 2 )) x 1 ( p i (x 1 ) x 2 p 1 i (x 1, x 2 )) Now we can perform a simple replacement x 1 x 2 pred 1 (x 1, x 2 ) n 1 i=0 0 j<i p i (x 1 ) [p i (x 1 ) p i (x 2 )]

15 The above formula defining pred 1 is not fluted; but we can make it so using a trick. We shadow the unary predicates p 0,..., p n 1 with binary predicates p0 1,..., p1 n 1, and write, for all i (0 i < n) x 1 (p i (x 1 ) x 2.p 1 i (x 1, x 2 )) x 1 ( p i (x 1 ) x 2 p 1 i (x 1, x 2 )) Now we can perform a simple replacement x 1 x 2 pred 1 (x 1, x 2 ) n 1 i=0 0 j<i p i (x 1 ) [p i (x 1 ) p i (x 2 )]

16 The above formula defining pred 1 is not fluted; but we can make it so using a trick. We shadow the unary predicates p 0,..., p n 1 with binary predicates p0 1,..., p1 n 1, and write, for all i (0 i < n) x 1 (p i (x 1 ) x 2.p 1 i (x 1, x 2 )) x 1 ( p i (x 1 ) x 2 p 1 i (x 1, x 2 )) Now we can perform a simple replacement x 1 x 2 pred 1 (x 1, x 2 ) n 1 i=0 0 j<i pi 1 (x 1, x 2 ) [ pi 1 (x 1, x 2 ) p i (x 2 ) ]

17 By writing further fluted formulas, we can fix the predicates zero 1 and eq 1 such that, for all 1-integers b and b :1 A = zero 1 [b] val 1 (b ) = 0 A = eq 1 [b, b ] val 1 (b ) = val 1 (b). Actually, if eq 1,l is a l + 2-ary predicate, we can write fluted formulas ensuring A = eq 1,l [b, c 1,..., c l, b ] val 1 (b ) = val 1 (b). That is, we can insert semantically inert arguments into the equality predicate.

18 We introduce a unary predicate int 2 ; any object satisfying int 2 in a model of interest will be call a 2-integer. We also introduce a binary predicate in 1 relating 1- and 2-integers. a 2 n 1 val 2 (b) = val 1 (a i ) = i a 2 a 1 b a 0 For any 2-integer b, define val 2 (b) to be the integer in the range [0, 2 2n 1] canonically represented by the 2 n -element bit-string s 2 n 1,..., s 0, where, for all i (0 i < 2 n ), { 1 if A = in 1 [a, b] for some 1-integer a s.t. val 1 (a) = i; s i = 0 otherwise.

19 We have to be a bit careful here: we want any 2-integer b in some structure A to satisfy a harmony requirement: If a and a are 1-integers with val 1 (a) = val 1 (a ) then A = in 1 [a, b] iff A = in 1 [a, b]. a 2 n val 2 (b) = val 1 (a i ) = i a 2 a 1 b a 0 We can do this by writing the (non-fluted) formula x 1 x 1x 2 (eq 1 (x 1, x 1) (in 1 (x 1, x 2 ) in 1 (x 1, x 2 ))).

20 Can we harmonize 2-integers using fluted formulas? Yes, provided that we represent them as mirrored structures related to 1-integers via the binary predicates in 1 and out 1, satisfying the strengthened harmony requirement: For any 2-integer b and 1-integers a, a val 1 (a) = val 1 (a ) implies A = in 1 [a, b] A = out 1 [b, a ]. a 2 n 1 a 2 a 1 a 0. val 2 (b) = b in 1 out 1 a 2 n 1 a 2 a 1 a 0

21 This harmony requirement is easily enforced by fluted formulas. Remember eq 1,1 (x 1, x 2, x 3 ) is to be read as val 1 (x 1 ) = val 1 (x 3 ). We then write: x 1 (int 1 (x 1 ) x 2 (int 2 (x 2 ) in 1 (x 1, x 2 ) x 3 (int 1 (x 3 ) eq 1,1 (x 1, x 2, x 3 ) out 1 (x 2, x 3 )))). eq 1,1 b a a in 1 out 1

22 This harmony requirement is easily enforced by fluted formulas. Remember eq 1,1 (x 1, x 2, x 3 ) is to be read as val 1 (x 1 ) = val 1 (x 3 ). We then write: x 1 (int 1 (x 1 ) x 2 (int 2 (x 2 ) in 1 (x 1, x 2 ) x 3 (int 1 (x 3 ) eq 1,1 (x 1, x 2, x 3 ) out 1 (x 2, x 3 )))). eq 1,1 b a a in 1 out 1

23 These mirrored structures allow us to sat that the 2-integers b and b are equal, in a particular digit (represented by a 1-integer a): eq 1,2 b b a a in 1 out 1 and hence we can define an equality predicate eq 2 for 2-integers. (We need four variables.) And of course we can insert l semantically inert arguments eq 2,l b b a in a 1 out 1 (for which we need 4 + l variables).

24 Can we now define a decrement relation on 2-integers? We first introduce a binary predicate in 1(x 1, x 2 ), and say: if zero 1 (x 1 ) then in 1(x 1, x 2 ); otherwise, if pred 1 (x 1, x 1 ) then in 1(x 1, x 2 ) iff in 1(x 1, x 2) and in 1 (x 1, x 2). Now we can define pred 2 (x 1, x 2 ), by writing a fluted formula saying: for all 1-integers y, in 1 (y, x 1 ) and in 1 (y, x 2 ) differ in truth-value just in case in 1(y, x 1 ). By writing x 1.int 2 (x 1 ) x 1 (int 2 (x 1 ) x 2 (int 2 (x 2 ) pred 2 (x 1, x 2 ))), we can enforce models of size 2 2n = t(2, n).

25 This process can be continued, to build 3-integers, with values up to 2 22n 1:. 2 2n 1. val 3 (b) = n b in 2 out in 1 out 1. Thus, we can force models of size 2 22n appear to need six variables. = t(3, n). But we

26 In this way, we can construct, in time bounded by a polynomial function of m and n, a (finitely) satisfiable formula Φ m,n of FL 2m featuring int m, such that any model of ϕ has cardinality at least t(m, n). Hence, there is no elementary bound on the sizes of models forced by satisfiable formulas of FL. A simple modification of this argument shows that FL 2m is m-nexptime-hard. Hence, the satisfiability problem for FL is not elementary. Purdy s (2002) bound of NExpTime is wrong.

27 Outline

28 We employ a fluted analogue of Hintikka constituents. An atomic fluted m-type is a maximal consistent conjunction of fluted literals in FL [m], e.g.: p(x 3 ) q(x 2, x 3 ) r(x 1, x 2, x 3 ) is an atomic fluted 3-type over the signature p/1, q/2, r/3. A fluted (m, m)-constituent is an atomic fluted m-type. Let k satisfy m > k 0. A fluted (k, m)-constituent is a formula λ = t(x 1,..., x k ) x k+1.λ x k+1 λ, λ Λ λ Λ where t is an atomic fluted k-type and Λ is some set of fluted (k + 1, m)-constituents, and t is a atomic fluted k-type.

29 Fluted constituents have analogous properties to Hintikka constituents. In particular, in any structure A, for any m > 0, there exists a unique (0, m)-fluted constituent true in A, denoted fc A m[ ]. It is useful to think of a fluted constituent λ = t(x 1,..., x k ) x k+1.λ x k+1.λ, λ Λ λ Λ as a (labelled) tree in which t is the label of the root and the λ Λ are the top-level daughters.

30 The following property of fluted constituents is a bit less obvious. If λ is a fluted (k, m)-constituent and k > 0, define λ to be the formula obtained by deleting all literals of λ containing x 1 and then shifting remaining variables left, i.e. replacing each variable x i+1 by x i, for i > 1. Then λ is a fluted (k 1, m 1)-constituent. Call λ the pull-back of λ.

31 Thus, the fluted constituent fc A m[ ] forms a kind of double-tree whose vertices, at level k, are labelled by realized fluted (k, m)-constituents. fc A m[ ] level k 1 level k level k + 1 f w g g f... These two trees obey a confluence property: g(f 1 (w)) f 1 (g(w)).

32 Thus, the fluted constituent fc A m[ ] forms a kind of double-tree whose vertices, at level k, are labelled by realized fluted (k, m)-constituents. fc A m[ ] level k 1 level k level k + 1 f w g g f... These two trees obey a confluence property: g(f 1 (w)) = f 1 (g(w)). By duplicating nodes, we can ensure that every non-leaf vertex is at the end of some g-arrow.

33 Given any such double tree (with this property), we can generate a structure realizing exactly the given fluted (k, m)-constituents. The domain of this structure coinsists of the set of leaves of the tree and so is bounded by t(m, p( ϕ )). We can check from the double tree alone whether the corresponding structure satisfies ϕ. Any satisfiable formula of FL m has a model of m-tuply exponential size; hence, the satisfiability problem for FL m is in m-nexptime.

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35 The complexity of the satisfiability problem for FL m lies between m/2 -NExpTime-hard and m-nexptime. For m 3, we can improve the upper bound to (m 2)-NExpTime with a little effort. Also, FL 1 is in NPTime, and FL 2 is in NExpTime. Thus, for m 4, we have tight bounds of m/2 -NExpTime-complete; but still have a gap when m 5. Despite considerable effort, we have not been able to close this gap.