/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. 0 : [] --> o a!6220!6220f : [o] --> o a!6220!6220p : [o] --> o cons : [o * o] --> o f : [o] --> o mark : [o] --> o p : [o] --> o s : [o] --> o a!6220!6220f(0) => cons(0, f(s(0))) a!6220!6220f(s(0)) => a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) => mark(X) mark(f(X)) => a!6220!6220f(mark(X)) mark(p(X)) => a!6220!6220p(mark(X)) mark(0) => 0 mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) a!6220!6220f(X) => f(X) a!6220!6220p(X) => p(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]): a!6220!6220f(0) >? cons(0, f(s(0))) a!6220!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) >? mark(X) mark(f(X)) >? a!6220!6220f(mark(X)) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 mark(cons(X, Y)) >? cons(mark(X), Y) mark(s(X)) >? s(mark(X)) a!6220!6220f(X) >? f(X) a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.2 + y0 a!6220!6220p = \y0.2y0 cons = \y0y1.y0 + y1 f = \y0.1 + y0 mark = \y0.2y0 p = \y0.2y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(0)]] = 2 > 1 = [[cons(0, f(s(0)))]] [[a!6220!6220f(s(0))]] = 2 >= 2 = [[a!6220!6220f(a!6220!6220p(s(0)))]] [[a!6220!6220p(s(_x0))]] = 2x0 >= 2x0 = [[mark(_x0)]] [[mark(f(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220f(mark(_x0))]] [[mark(p(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220p(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220f(_x0)]] = 2 + x0 > 1 + x0 = [[f(_x0)]] [[a!6220!6220p(_x0)]] = 2x0 >= 2x0 = [[p(_x0)]] We can thus remove the following rules: a!6220!6220f(0) => cons(0, f(s(0))) a!6220!6220f(X) => f(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]): a!6220!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) >? mark(X) mark(f(X)) >? a!6220!6220f(mark(X)) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 mark(cons(X, Y)) >? cons(mark(X), Y) mark(s(X)) >? s(mark(X)) a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.y0 a!6220!6220p = \y0.y0 cons = \y0y1.y0 + y1 f = \y0.3 + 3y0 mark = \y0.y0 p = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(s(0))]] = 0 >= 0 = [[a!6220!6220f(a!6220!6220p(s(0)))]] [[a!6220!6220p(s(_x0))]] = x0 >= x0 = [[mark(_x0)]] [[mark(f(_x0))]] = 3 + 3x0 > x0 = [[a!6220!6220f(mark(_x0))]] [[mark(p(_x0))]] = x0 >= x0 = [[a!6220!6220p(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[mark(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(mark(_x0), _x1)]] [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] [[a!6220!6220p(_x0)]] = x0 >= x0 = [[p(_x0)]] We can thus remove the following rules: mark(f(X)) => a!6220!6220f(mark(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]): a!6220!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) >? mark(X) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 mark(cons(X, Y)) >? cons(mark(X), Y) mark(s(X)) >? s(mark(X)) a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.y0 a!6220!6220p = \y0.2y0 cons = \y0y1.2 + y0 + y1 mark = \y0.2y0 p = \y0.2y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(s(0))]] = 0 >= 0 = [[a!6220!6220f(a!6220!6220p(s(0)))]] [[a!6220!6220p(s(_x0))]] = 2x0 >= 2x0 = [[mark(_x0)]] [[mark(p(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220p(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[mark(cons(_x0, _x1))]] = 4 + 2x0 + 2x1 > 2 + x1 + 2x0 = [[cons(mark(_x0), _x1)]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220p(_x0)]] = 2x0 >= 2x0 = [[p(_x0)]] We can thus remove the following rules: mark(cons(X, Y)) => cons(mark(X), 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]): a!6220!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) >? mark(X) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 mark(s(X)) >? s(mark(X)) a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.2y0 a!6220!6220p = \y0.y0 mark = \y0.2y0 p = \y0.y0 s = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(s(0))]] = 2 >= 2 = [[a!6220!6220f(a!6220!6220p(s(0)))]] [[a!6220!6220p(s(_x0))]] = 1 + 2x0 > 2x0 = [[mark(_x0)]] [[mark(p(_x0))]] = 2x0 >= 2x0 = [[a!6220!6220p(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[mark(s(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[s(mark(_x0))]] [[a!6220!6220p(_x0)]] = x0 >= x0 = [[p(_x0)]] We can thus remove the following rules: a!6220!6220p(s(X)) => mark(X) mark(s(X)) => s(mark(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]): a!6220!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.y0 a!6220!6220p = \y0.y0 mark = \y0.2 + 2y0 p = \y0.y0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(s(0))]] = 0 >= 0 = [[a!6220!6220f(a!6220!6220p(s(0)))]] [[mark(p(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220p(mark(_x0))]] [[mark(0)]] = 2 > 0 = [[0]] [[a!6220!6220p(_x0)]] = x0 >= x0 = [[p(_x0)]] We can thus remove the following rules: mark(0) => 0 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] a!6220!6220f#(s(0)) =#> a!6220!6220f#(a!6220!6220p(s(0))) 1] a!6220!6220f#(s(0)) =#> a!6220!6220p#(s(0)) 2] mark#(p(X)) =#> a!6220!6220p#(mark(X)) 3] mark#(p(X)) =#> mark#(X) Rules R_0: a!6220!6220f(s(0)) => a!6220!6220f(a!6220!6220p(s(0))) mark(p(X)) => a!6220!6220p(mark(X)) a!6220!6220p(X) => p(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 : * 1 : * 2 : * 3 : 2, 3 This graph has the following strongly connected components: P_1: mark#(p(X)) =#> mark#(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). 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(mark#) = 1 Thus, we can orient the dependency pairs as follows: nu(mark#(p(X))) = p(X) |> X = nu(mark#(X)) 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.