/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 div : [o * o] --> o plus : [o * o] --> o quot : [o * o * o] --> o s : [o] --> o times : [o * o] --> o plus(X, 0) => X plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(0), X) => X times(s(X), Y) => plus(Y, times(X, Y)) div(0, X) => 0 div(X, Y) => quot(X, Y, Y) quot(0, s(X), Y) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(div(X, s(Y))) div(div(X, Y), Z) => div(X, times(Y, Z)) 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] plus#(s(X), Y) =#> plus#(X, Y) 1] times#(s(X), Y) =#> plus#(Y, times(X, Y)) 2] times#(s(X), Y) =#> times#(X, Y) 3] div#(X, Y) =#> quot#(X, Y, Y) 4] quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) 5] quot#(X, 0, s(Y)) =#> div#(X, s(Y)) 6] div#(div(X, Y), Z) =#> div#(X, times(Y, Z)) 7] div#(div(X, Y), Z) =#> times#(Y, Z) Rules R_0: plus(X, 0) => X plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(0), X) => X times(s(X), Y) => plus(Y, times(X, Y)) div(0, X) => 0 div(X, Y) => quot(X, Y, Y) quot(0, s(X), Y) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(div(X, s(Y))) div(div(X, Y), Z) => div(X, times(Y, Z)) 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 : 0 * 2 : 1, 2 * 3 : 4, 5 * 4 : 4, 5 * 5 : 3, 6, 7 * 6 : 3, 6, 7 * 7 : 1, 2 This graph has the following strongly connected components: P_1: plus#(s(X), Y) =#> plus#(X, Y) P_2: times#(s(X), Y) =#> times#(X, Y) P_3: div#(X, Y) =#> quot#(X, Y, Y) quot#(s(X), s(Y), Z) =#> quot#(X, Y, Z) quot#(X, 0, s(Y)) =#> div#(X, s(Y)) div#(div(X, Y), Z) =#> div#(X, times(Y, 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), (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(div#) = 1 nu(quot#) = 1 Thus, we can orient the dependency pairs as follows: nu(div#(X, Y)) = X = X = nu(quot#(X, Y, Y)) nu(quot#(s(X), s(Y), Z)) = s(X) |> X = nu(quot#(X, Y, Z)) nu(quot#(X, 0, s(Y))) = X = X = nu(div#(X, s(Y))) nu(div#(div(X, Y), Z)) = div(X, Y) |> X = nu(div#(X, times(Y, Z))) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by (P_4, R_0, minimal, f), where P_4 contains: div#(X, Y) =#> quot#(X, Y, Y) quot#(X, 0, s(Y)) =#> div#(X, s(Y)) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. (P_4, R_0) has no usable rules. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: div#(X, Y) >? quot#(X, Y, Y) quot#(X, 0, s(Y)) >? div#(X, s(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 div# = \y0y1.1 + 2y1 quot# = \y0y1y2.2y1 s = \y0.0 Using this interpretation, the requirements translate to: [[div#(_x0, _x1)]] = 1 + 2x1 > 2x1 = [[quot#(_x0, _x1, _x1)]] [[quot#(_x0, 0, s(_x1))]] = 6 > 1 = [[div#(_x0, s(_x1))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0). 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(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_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(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_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.