/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o plus : [o * o] --> o s : [o] --> o times : [o * o] --> o times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(X, s(Y)) => s(plus(X, Y)) 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] times#(X, plus(Y, s(Z))) =#> plus#(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 1] times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) 2] times#(X, plus(Y, s(Z))) =#> plus#(Y, times(s(Z), 0)) 3] times#(X, plus(Y, s(Z))) =#> times#(s(Z), 0) 4] times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) 5] times#(X, s(Y)) =#> plus#(times(X, Y), X) 6] times#(X, s(Y)) =#> times#(X, Y) 7] plus#(X, s(Y)) =#> plus#(X, Y) Rules R_0: times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(X, s(Y)) => s(plus(X, 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 : 7 * 1 : 0, 1, 2, 3, 4, 5, 6 * 2 : 7 * 3 : * 4 : 5, 6 * 5 : 7 * 6 : 0, 1, 2, 3, 4, 5, 6 * 7 : 7 This graph has the following strongly connected components: P_1: times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) =#> times#(X, s(Z)) times#(X, s(Y)) =#> times#(X, Y) P_2: plus#(X, s(Y)) =#> plus#(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) 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(plus#) = 2 Thus, we can orient the dependency pairs as follows: nu(plus#(X, s(Y))) = s(Y) |> Y = 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 will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: times#(X, plus(Y, s(Z))) >? times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) >? times#(X, s(Z)) times#(X, s(Y)) >? times#(X, Y) times(X, plus(Y, s(Z))) >= plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X) plus(X, 0) >= X plus(X, s(Y)) >= s(plus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: This leaves the following ordering requirements: times#(X, plus(Y, s(Z))) >= times#(X, plus(Y, times(s(Z), 0))) times#(X, plus(Y, s(Z))) > times#(X, s(Z)) times#(X, s(Y)) >= times#(X, Y) times(X, 0) >= 0 plus(X, 0) >= X plus(X, s(Y)) >= s(plus(X, Y)) The following interpretation satisfies the requirements: 0 = 0 plus = \y0y1.2 + y0 + 2y1 s = \y0.y0 times = \y0y1.0 times# = \y0y1.2y1 Using this interpretation, the requirements translate to: [[times#(_x0, plus(_x1, s(_x2)))]] = 4 + 2x1 + 4x2 >= 4 + 2x1 = [[times#(_x0, plus(_x1, times(s(_x2), 0)))]] [[times#(_x0, plus(_x1, s(_x2)))]] = 4 + 2x1 + 4x2 > 2x2 = [[times#(_x0, s(_x2))]] [[times#(_x0, s(_x1))]] = 2x1 >= 2x1 = [[times#(_x0, _x1)]] [[times(_x0, 0)]] = 0 >= 0 = [[0]] [[plus(_x0, 0)]] = 2 + x0 >= x0 = [[_x0]] [[plus(_x0, s(_x1))]] = 2 + x0 + 2x1 >= 2 + x0 + 2x1 = [[s(plus(_x0, _x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_3, R_0, minimal, formative), where P_3 consists of: times#(X, plus(Y, s(Z))) =#> times#(X, plus(Y, times(s(Z), 0))) times#(X, s(Y)) =#> times#(X, Y) 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 will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: times#(X, plus(Y, s(Z))) >? times#(X, plus(Y, times(s(Z), 0))) times#(X, s(Y)) >? times#(X, Y) times(X, plus(Y, s(Z))) >= plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) times(X, 0) >= 0 times(X, s(Y)) >= plus(times(X, Y), X) plus(X, 0) >= X plus(X, s(Y)) >= s(plus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: This leaves the following ordering requirements: times#(X, plus(Y, s(Z))) > times#(X, plus(Y, times(s(Z), 0))) times#(X, s(Y)) >= times#(X, Y) times(X, 0) >= 0 plus(X, 0) >= X plus(X, s(Y)) >= s(plus(X, Y)) The following interpretation satisfies the requirements: 0 = 0 plus = \y0y1.y0 + 2y1 s = \y0.2 + y0 times = \y0y1.0 times# = \y0y1.y1 Using this interpretation, the requirements translate to: [[times#(_x0, plus(_x1, s(_x2)))]] = 4 + x1 + 2x2 > x1 = [[times#(_x0, plus(_x1, times(s(_x2), 0)))]] [[times#(_x0, s(_x1))]] = 2 + x1 > x1 = [[times#(_x0, _x1)]] [[times(_x0, 0)]] = 0 >= 0 = [[0]] [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[plus(_x0, s(_x1))]] = 4 + x0 + 2x1 >= 2 + x0 + 2x1 = [[s(plus(_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_0) by ({}, R_0). 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.