/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o f : [o * o * o * o] --> o false : [] --> o if : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o perfectp : [o] --> o s : [o] --> o true : [] --> o minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] le#(s(X), s(Y)) =#> le#(X, Y) 2] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X)) 3] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) 4] f#(s(X), 0, Y, Z) =#> minus#(Y, s(X)) 5] f#(s(X), s(Y), Z, U) =#> if#(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 6] f#(s(X), s(Y), Z, U) =#> le#(X, Y) 7] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) 8] f#(s(X), s(Y), Z, U) =#> minus#(Y, X) 9] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) Rules R_0: minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 1 * 2 : 5, 6, 7, 8, 9 * 3 : 3, 4, 5, 6, 7, 8, 9 * 4 : 0 * 5 : * 6 : 1 * 7 : 3, 4, 5, 6, 7, 8, 9 * 8 : 0 * 9 : 3, 4, 5, 6, 7, 8, 9 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) P_3: f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(s(X), 0, Y, Z)) = s(X) |> X = nu(f#(X, Z, minus(Y, s(X)), Z)) nu(f#(s(X), s(Y), Z, U)) = s(X) = s(X) = nu(f#(s(X), minus(Y, X), Z, U)) nu(f#(s(X), s(Y), Z, U)) = s(X) |> X = nu(f#(X, U, Z, U)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains: f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_4, R_0) are: minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: f#(s(X), s(Y), Z, U) >? f#(s(X), minus(Y, X), Z, U) minus(0, X) >= 0 minus(s(X), 0) >= s(X) minus(s(X), s(Y)) >= minus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 f# = \y0y1y2y3.y1 minus = \y0y1.2y0 s = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[f#(s(_x0), s(_x1), _x2, _x3)]] = 2 + 2x1 > 2x1 = [[f#(s(_x0), minus(_x1, _x0), _x2, _x3)]] [[minus(0, _x0)]] = 2 >= 1 = [[0]] [[minus(s(_x0), 0)]] = 4 + 4x0 >= 2 + 2x0 = [[s(_x0)]] [[minus(s(_x0), s(_x1))]] = 4 + 4x0 >= 2x0 = [[minus(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(le#) = 1 Thus, we can orient the dependency pairs as follows: nu(le#(s(X), s(Y))) = s(X) |> X = nu(le#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.