/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. !dot : [o * o] --> o 0 : [] --> o int : [o * o] --> o int!6220list : [o] --> o nil : [] --> o s : [o] --> o int(0, 0) => !dot(0, nil) int(0, s(X)) => !dot(0, int(s(0), s(X))) int(s(X), 0) => nil int(s(X), s(Y)) => int!6220list(int(X, Y)) int!6220list(nil) => nil int!6220list(!dot(X, Y)) => !dot(s(X), int!6220list(Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): int(0, 0) >? !dot(0, nil) int(0, s(X)) >? !dot(0, int(s(0), s(X))) int(s(X), 0) >? nil int(s(X), s(Y)) >? int!6220list(int(X, Y)) int!6220list(nil) >? nil int!6220list(!dot(X, Y)) >? !dot(s(X), int!6220list(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.y0 + y1 0 = 0 int = \y0y1.2 + 2y0 + 2y1 int!6220list = \y0.y0 nil = 0 s = \y0.y0 Using this interpretation, the requirements translate to: [[int(0, 0)]] = 2 > 0 = [[!dot(0, nil)]] [[int(0, s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[!dot(0, int(s(0), s(_x0)))]] [[int(s(_x0), 0)]] = 2 + 2x0 > 0 = [[nil]] [[int(s(_x0), s(_x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[int!6220list(int(_x0, _x1))]] [[int!6220list(nil)]] = 0 >= 0 = [[nil]] [[int!6220list(!dot(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!dot(s(_x0), int!6220list(_x1))]] We can thus remove the following rules: int(0, 0) => !dot(0, nil) int(s(X), 0) => nil We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): int(0, s(X)) >? !dot(0, int(s(0), s(X))) int(s(X), s(Y)) >? int!6220list(int(X, Y)) int!6220list(nil) >? nil int!6220list(!dot(X, Y)) >? !dot(s(X), int!6220list(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.y1 + 2y0 0 = 0 int = \y0y1.y0 + y1 int!6220list = \y0.2y0 nil = 2 s = \y0.2y0 Using this interpretation, the requirements translate to: [[int(0, s(_x0))]] = 2x0 >= 2x0 = [[!dot(0, int(s(0), s(_x0)))]] [[int(s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[int!6220list(int(_x0, _x1))]] [[int!6220list(nil)]] = 4 > 2 = [[nil]] [[int!6220list(!dot(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[!dot(s(_x0), int!6220list(_x1))]] We can thus remove the following rules: int!6220list(nil) => nil 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] int#(0, s(X)) =#> int#(s(0), s(X)) 1] int#(s(X), s(Y)) =#> int!6220list#(int(X, Y)) 2] int#(s(X), s(Y)) =#> int#(X, Y) 3] int!6220list#(!dot(X, Y)) =#> int!6220list#(Y) Rules R_0: int(0, s(X)) => !dot(0, int(s(0), s(X))) int(s(X), s(Y)) => int!6220list(int(X, Y)) int!6220list(!dot(X, Y)) => !dot(s(X), int!6220list(Y)) 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 : 1, 2 * 1 : 3 * 2 : 0, 1, 2 * 3 : 3 This graph has the following strongly connected components: P_1: int#(0, s(X)) =#> int#(s(0), s(X)) int#(s(X), s(Y)) =#> int#(X, Y) P_2: int!6220list#(!dot(X, Y)) =#> int!6220list#(Y) 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) and (P_2, R_0, m, f). 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(int!6220list#) = 1 Thus, we can orient the dependency pairs as follows: nu(int!6220list#(!dot(X, Y))) = !dot(X, Y) |> Y = nu(int!6220list#(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(int#) = 2 Thus, we can orient the dependency pairs as follows: nu(int#(0, s(X))) = s(X) = s(X) = nu(int#(s(0), s(X))) nu(int#(s(X), s(Y))) = s(Y) |> Y = nu(int#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by (P_3, R_0, minimal, f), where P_3 contains: int#(0, s(X)) =#> int#(s(0), s(X)) Thus, the original system is terminating if (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, 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 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. 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.