/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 first : [o * o] --> o from : [o] --> o n!6220!6220first : [o * o] --> o n!6220!6220from : [o] --> o n!6220!6220s : [o] --> o nil : [] --> o s : [o] --> o sel : [o * o] --> o from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(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] first#(s(X), cons(Y, Z)) =#> activate#(Z) 1] sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) 2] sel#(s(X), cons(Y, Z)) =#> activate#(Z) 3] activate#(n!6220!6220from(X)) =#> from#(activate(X)) 4] activate#(n!6220!6220from(X)) =#> activate#(X) 5] activate#(n!6220!6220s(X)) =#> s#(activate(X)) 6] activate#(n!6220!6220s(X)) =#> activate#(X) 7] activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y)) 8] activate#(n!6220!6220first(X, Y)) =#> activate#(X) 9] activate#(n!6220!6220first(X, Y)) =#> activate#(Y) Rules R_0: from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(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 : 3, 4, 5, 6, 7, 8, 9 * 1 : 1, 2 * 2 : 3, 4, 5, 6, 7, 8, 9 * 3 : * 4 : 3, 4, 5, 6, 7, 8, 9 * 5 : * 6 : 3, 4, 5, 6, 7, 8, 9 * 7 : 0 * 8 : 3, 4, 5, 6, 7, 8, 9 * 9 : 3, 4, 5, 6, 7, 8, 9 This graph has the following strongly connected components: P_1: first#(s(X), cons(Y, Z)) =#> activate#(Z) activate#(n!6220!6220from(X)) =#> activate#(X) activate#(n!6220!6220s(X)) =#> activate#(X) activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y)) activate#(n!6220!6220first(X, Y)) =#> activate#(X) activate#(n!6220!6220first(X, Y)) =#> activate#(Y) P_2: sel#(s(X), cons(Y, Z)) =#> sel#(X, activate(Z)) 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(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_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). The formative rules of (P_1, R_0) are R_1 ::= from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) activate(X) => X 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 with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_1, R_1) are: from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) activate(X) => X It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: first#(s(X), cons(Y, Z)) >? activate#(Z) activate#(n!6220!6220from(X)) >? activate#(X) activate#(n!6220!6220s(X)) >? activate#(X) activate#(n!6220!6220first(X, Y)) >? first#(activate(X), activate(Y)) activate#(n!6220!6220first(X, Y)) >? activate#(X) activate#(n!6220!6220first(X, Y)) >? activate#(Y) from(X) >= cons(X, n!6220!6220from(n!6220!6220s(X))) first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) from(X) >= n!6220!6220from(X) s(X) >= n!6220!6220s(X) first(X, Y) >= n!6220!6220first(X, Y) activate(n!6220!6220from(X)) >= from(activate(X)) activate(n!6220!6220s(X)) >= s(activate(X)) activate(n!6220!6220first(X, Y)) >= first(activate(X), activate(Y)) activate(X) >= X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.y0 activate# = \y0.y0 cons = \y0y1.y1 first = \y0y1.1 + 2y0 + 2y1 first# = \y0y1.1 + 2y1 from = \y0.y0 n!6220!6220first = \y0y1.1 + 2y0 + 2y1 n!6220!6220from = \y0.y0 n!6220!6220s = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[first#(s(_x0), cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[activate#(_x2)]] [[activate#(n!6220!6220from(_x0))]] = x0 >= x0 = [[activate#(_x0)]] [[activate#(n!6220!6220s(_x0))]] = x0 >= x0 = [[activate#(_x0)]] [[activate#(n!6220!6220first(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x1 = [[first#(activate(_x0), activate(_x1))]] [[activate#(n!6220!6220first(_x0, _x1))]] = 1 + 2x0 + 2x1 > x0 = [[activate#(_x0)]] [[activate#(n!6220!6220first(_x0, _x1))]] = 1 + 2x0 + 2x1 > x1 = [[activate#(_x1)]] [[from(_x0)]] = x0 >= x0 = [[cons(_x0, n!6220!6220from(n!6220!6220s(_x0)))]] [[first(s(_x0), cons(_x1, _x2))]] = 1 + 2x0 + 2x2 >= 1 + 2x0 + 2x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] [[from(_x0)]] = x0 >= x0 = [[n!6220!6220from(_x0)]] [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] [[first(_x0, _x1)]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[n!6220!6220first(_x0, _x1)]] [[activate(n!6220!6220from(_x0))]] = x0 >= x0 = [[from(activate(_x0))]] [[activate(n!6220!6220s(_x0))]] = x0 >= x0 = [[s(activate(_x0))]] [[activate(n!6220!6220first(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[first(activate(_x0), activate(_x1))]] [[activate(_x0)]] = x0 >= x0 = [[_x0]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, minimal, formative) by (P_3, R_1, minimal, formative), where P_3 consists of: activate#(n!6220!6220from(X)) =#> activate#(X) activate#(n!6220!6220s(X)) =#> activate#(X) activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y)) Thus, the original system is terminating if (P_3, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_1, 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, 2 * 1 : 0, 1, 2 * 2 : This graph has the following strongly connected components: P_4: activate#(n!6220!6220from(X)) =#> activate#(X) activate#(n!6220!6220s(X)) =#> activate#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_3, R_1, m, f) by (P_4, R_1, m, f). Thus, the original system is terminating if (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_1, minimal, formative). We apply the subterm criterion with the following projection function: nu(activate#) = 1 Thus, we can orient the dependency pairs as follows: nu(activate#(n!6220!6220from(X))) = n!6220!6220from(X) |> X = nu(activate#(X)) nu(activate#(n!6220!6220s(X))) = n!6220!6220s(X) |> X = nu(activate#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_1, minimal, f) by ({}, R_1, 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.