/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 2ndsneg : [o * o] --> o 2ndspos : [o * o] --> o activate : [o] --> o cons : [o * o] --> o cons2 : [o * o] --> o from : [o] --> o n!6220!6220from : [o] --> o negrecip : [o] --> o pi : [o] --> o plus : [o * o] --> o posrecip : [o] --> o rcons : [o * o] --> o rnil : [] --> o s : [o] --> o square : [o] --> o times : [o * o] --> o from(X) => cons(X, n!6220!6220from(s(X))) 2ndspos(0, X) => rnil 2ndspos(s(X), cons(Y, Z)) => 2ndspos(s(X), cons2(Y, activate(Z))) 2ndspos(s(X), cons2(Y, cons(Z, U))) => rcons(posrecip(Z), 2ndsneg(X, activate(U))) 2ndsneg(0, X) => rnil 2ndsneg(s(X), cons(Y, Z)) => 2ndsneg(s(X), cons2(Y, activate(Z))) 2ndsneg(s(X), cons2(Y, cons(Z, U))) => rcons(negrecip(Z), 2ndspos(X, activate(U))) pi(X) => 2ndspos(X, from(0)) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(Y, times(X, Y)) square(X) => times(X, X) from(X) => n!6220!6220from(X) activate(n!6220!6220from(X)) => from(X) activate(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] 2ndspos#(s(X), cons(Y, Z)) =#> 2ndspos#(s(X), cons2(Y, activate(Z))) 1] 2ndspos#(s(X), cons(Y, Z)) =#> activate#(Z) 2] 2ndspos#(s(X), cons2(Y, cons(Z, U))) =#> 2ndsneg#(X, activate(U)) 3] 2ndspos#(s(X), cons2(Y, cons(Z, U))) =#> activate#(U) 4] 2ndsneg#(s(X), cons(Y, Z)) =#> 2ndsneg#(s(X), cons2(Y, activate(Z))) 5] 2ndsneg#(s(X), cons(Y, Z)) =#> activate#(Z) 6] 2ndsneg#(s(X), cons2(Y, cons(Z, U))) =#> 2ndspos#(X, activate(U)) 7] 2ndsneg#(s(X), cons2(Y, cons(Z, U))) =#> activate#(U) 8] pi#(X) =#> 2ndspos#(X, from(0)) 9] pi#(X) =#> from#(0) 10] plus#(s(X), Y) =#> plus#(X, Y) 11] times#(s(X), Y) =#> plus#(Y, times(X, Y)) 12] times#(s(X), Y) =#> times#(X, Y) 13] square#(X) =#> times#(X, X) 14] activate#(n!6220!6220from(X)) =#> from#(X) Rules R_0: from(X) => cons(X, n!6220!6220from(s(X))) 2ndspos(0, X) => rnil 2ndspos(s(X), cons(Y, Z)) => 2ndspos(s(X), cons2(Y, activate(Z))) 2ndspos(s(X), cons2(Y, cons(Z, U))) => rcons(posrecip(Z), 2ndsneg(X, activate(U))) 2ndsneg(0, X) => rnil 2ndsneg(s(X), cons(Y, Z)) => 2ndsneg(s(X), cons2(Y, activate(Z))) 2ndsneg(s(X), cons2(Y, cons(Z, U))) => rcons(negrecip(Z), 2ndspos(X, activate(U))) pi(X) => 2ndspos(X, from(0)) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(Y, times(X, Y)) square(X) => times(X, X) from(X) => n!6220!6220from(X) activate(n!6220!6220from(X)) => from(X) activate(X) => 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 : 2, 3 * 1 : 14 * 2 : 4, 5, 6, 7 * 3 : 14 * 4 : 6, 7 * 5 : 14 * 6 : 0, 1, 2, 3 * 7 : 14 * 8 : 0, 1 * 9 : * 10 : 10 * 11 : 10 * 12 : 11, 12 * 13 : 11, 12 * 14 : This graph has the following strongly connected components: P_1: 2ndspos#(s(X), cons(Y, Z)) =#> 2ndspos#(s(X), cons2(Y, activate(Z))) 2ndspos#(s(X), cons2(Y, cons(Z, U))) =#> 2ndsneg#(X, activate(U)) 2ndsneg#(s(X), cons(Y, Z)) =#> 2ndsneg#(s(X), cons2(Y, activate(Z))) 2ndsneg#(s(X), cons2(Y, cons(Z, U))) =#> 2ndspos#(X, activate(U)) P_2: plus#(s(X), Y) =#> plus#(X, Y) P_3: times#(s(X), Y) =#> times#(X, 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), (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(times#) = 1 Thus, we can orient the dependency pairs as follows: nu(times#(s(X), Y)) = s(X) |> X = nu(times#(X, Y)) 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). We apply the subterm criterion with the following projection function: nu(plus#) = 1 Thus, we can orient the dependency pairs as follows: nu(plus#(s(X), Y)) = s(X) |> X = nu(plus#(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(2ndsneg#) = 1 nu(2ndspos#) = 1 Thus, we can orient the dependency pairs as follows: nu(2ndspos#(s(X), cons(Y, Z))) = s(X) = s(X) = nu(2ndspos#(s(X), cons2(Y, activate(Z)))) nu(2ndspos#(s(X), cons2(Y, cons(Z, U)))) = s(X) |> X = nu(2ndsneg#(X, activate(U))) nu(2ndsneg#(s(X), cons(Y, Z))) = s(X) = s(X) = nu(2ndsneg#(s(X), cons2(Y, activate(Z)))) nu(2ndsneg#(s(X), cons2(Y, cons(Z, U)))) = s(X) |> X = nu(2ndspos#(X, activate(U))) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains: 2ndspos#(s(X), cons(Y, Z)) =#> 2ndspos#(s(X), cons2(Y, activate(Z))) 2ndsneg#(s(X), cons(Y, Z)) =#> 2ndsneg#(s(X), cons2(Y, activate(Z))) Thus, the original system is terminating if (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, 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 : 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.