/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. c : [o * o] --> o d : [o] --> o f : [o] --> o g : [o] --> o s : [o] --> o g(c(X, s(Y))) => g(c(s(X), Y)) f(c(s(X), Y)) => f(c(X, s(Y))) f(f(X)) => f(d(f(X))) f(X) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): g(c(X, s(Y))) >? g(c(s(X), Y)) f(c(s(X), Y)) >? f(c(X, s(Y))) f(f(X)) >? f(d(f(X))) f(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: c = \y0y1.y0 + y1 d = \y0.y0 f = \y0.2 + 2y0 g = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[g(c(_x0, s(_x1)))]] = x0 + x1 >= x0 + x1 = [[g(c(s(_x0), _x1))]] [[f(c(s(_x0), _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[f(c(_x0, s(_x1)))]] [[f(f(_x0))]] = 6 + 4x0 >= 6 + 4x0 = [[f(d(f(_x0)))]] [[f(_x0)]] = 2 + 2x0 > x0 = [[_x0]] We can thus remove the following rules: f(X) => X 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] g#(c(X, s(Y))) =#> g#(c(s(X), Y)) 1] f#(c(s(X), Y)) =#> f#(c(X, s(Y))) 2] f#(f(X)) =#> f#(d(f(X))) 3] f#(f(X)) =#> f#(X) Rules R_0: g(c(X, s(Y))) => g(c(s(X), Y)) f(c(s(X), Y)) => f(c(X, s(Y))) f(f(X)) => f(d(f(X))) 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 : * 3 : 1, 2, 3 This graph has the following strongly connected components: P_1: g#(c(X, s(Y))) =#> g#(c(s(X), Y)) P_2: f#(c(s(X), Y)) =#> f#(c(X, s(Y))) P_3: f#(f(X)) =#> f#(X) 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#(f(X))) = f(X) |> X = nu(f#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, 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 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). This combination (P_2, R_0) has no formative rules! We will name the empty set of rules:R_1. By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: f#(c(s(X), Y)) >? f#(c(X, s(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: c = \y0y1.3y0 f# = \y0.2y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[f#(c(s(_x0), _x1))]] = 18 + 6x0 > 6x0 = [[f#(c(_x0, s(_x1)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_1) by ({}, R_1). 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). This combination (P_1, R_0) has no formative rules! This is exactly R_1. By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative). Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: g#(c(X, s(Y))) >? g#(c(s(X), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: c = \y0y1.3y1 g# = \y0.2y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[g#(c(_x0, s(_x1)))]] = 18 + 6x1 > 6x1 = [[g#(c(s(_x0), _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1). 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.