/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 activate : [o] --> o cons : [o * o] --> o from : [o] --> o minus : [o * o] --> o n!6220!6220from : [o] --> o n!6220!6220zWquot : [o * o] --> o nil : [] --> o quot : [o * o] --> o s : [o] --> o sel : [o * o] --> o zWquot : [o * o] --> o from(X) => cons(X, n!6220!6220from(s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) minus(X, 0) => 0 minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) zWquot(X, nil) => nil zWquot(nil, X) => nil zWquot(cons(X, Y), cons(Z, U)) => cons(quot(X, Z), n!6220!6220zWquot(activate(Y), activate(U))) from(X) => n!6220!6220from(X) zWquot(X, Y) => n!6220!6220zWquot(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220zWquot(X, Y)) => zWquot(X, Y) 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] sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) 1] sel#(s(X), cons(Y, Z)) =#> activate#(Z) 2] minus#(s(X), s(Y)) =#> minus#(X, Y) 3] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 4] quot#(s(X), s(Y)) =#> minus#(X, Y) 5] zWquot#(cons(X, Y), cons(Z, U)) =#> quot#(X, Z) 6] zWquot#(cons(X, Y), cons(Z, U)) =#> activate#(Y) 7] zWquot#(cons(X, Y), cons(Z, U)) =#> activate#(U) 8] activate#(n!6220!6220from(X)) =#> from#(X) 9] activate#(n!6220!6220zWquot(X, Y)) =#> zWquot#(X, Y) Rules R_0: from(X) => cons(X, n!6220!6220from(s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) minus(X, 0) => 0 minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) zWquot(X, nil) => nil zWquot(nil, X) => nil zWquot(cons(X, Y), cons(Z, U)) => cons(quot(X, Z), n!6220!6220zWquot(activate(Y), activate(U))) from(X) => n!6220!6220from(X) zWquot(X, Y) => n!6220!6220zWquot(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220zWquot(X, Y)) => zWquot(X, Y) 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 : 0, 1 * 1 : 8, 9 * 2 : 2 * 3 : * 4 : 2 * 5 : 3, 4 * 6 : 8, 9 * 7 : 8, 9 * 8 : * 9 : 5, 6, 7 This graph has the following strongly connected components: P_1: sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) P_2: minus#(s(X), s(Y)) =#> minus#(X, Y) P_3: zWquot#(cons(X, Y), cons(Z, U)) =#> activate#(Y) zWquot#(cons(X, Y), cons(Z, U)) =#> activate#(U) activate#(n!6220!6220zWquot(X, Y)) =#> zWquot#(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). The formative rules of (P_3, R_0) are R_1 ::= from(X) => cons(X, n!6220!6220from(s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) minus(X, 0) => 0 minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) zWquot(cons(X, Y), cons(Z, U)) => cons(quot(X, Z), n!6220!6220zWquot(activate(Y), activate(U))) from(X) => n!6220!6220from(X) zWquot(X, Y) => n!6220!6220zWquot(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220zWquot(X, Y)) => zWquot(X, Y) activate(X) => X By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_3, R_0, minimal, formative) by (P_3, R_1, minimal, formative). 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_1, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_1, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_3, R_1) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: zWquot#(cons(X, Y), cons(Z, U)) >? activate#(Y) zWquot#(cons(X, Y), cons(Z, U)) >? activate#(U) activate#(n!6220!6220zWquot(X, Y)) >? zWquot#(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate# = \y0.2y0 cons = \y0y1.3 + 2y1 n!6220!6220zWquot = \y0y1.3 + 2y0 + 2y1 zWquot# = \y0y1.2y1 + 3y0 Using this interpretation, the requirements translate to: [[zWquot#(cons(_x0, _x1), cons(_x2, _x3))]] = 15 + 4x3 + 6x1 > 2x1 = [[activate#(_x1)]] [[zWquot#(cons(_x0, _x1), cons(_x2, _x3))]] = 15 + 4x3 + 6x1 > 2x3 = [[activate#(_x3)]] [[activate#(n!6220!6220zWquot(_x0, _x1))]] = 6 + 4x0 + 4x1 > 2x1 + 3x0 = [[zWquot#(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, 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 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(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_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(sel#) = 1 Thus, we can orient the dependency pairs as follows: nu(sel#(s(X), cons(Y, Z))) = s(X) |> X = nu(sel#(X, activate(Z))) 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.