/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 app : [o * o] --> o cons : [o * o] --> o minus : [o * o] --> o nil : [] --> o plus : [o * o] --> o quot : [o * o] --> o s : [o] --> o sum : [o] --> o minus(X, 0) => X 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))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) 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] minus#(s(X), s(Y)) =#> minus#(X, Y) 1] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 2] quot#(s(X), s(Y)) =#> minus#(X, Y) 3] plus#(s(X), Y) =#> plus#(X, Y) 4] minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z)) 5] minus#(minus(X, Y), Z) =#> plus#(Y, Z) 6] app#(cons(X, Y), Z) =#> app#(Y, Z) 7] sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z)) 8] sum#(cons(X, cons(Y, Z))) =#> plus#(X, Y) 9] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U))))) 10] sum#(app(X, cons(Y, cons(Z, U)))) =#> app#(X, sum(cons(Y, cons(Z, U)))) 11] sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(cons(Y, cons(Z, U))) Rules R_0: minus(X, 0) => X 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))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) 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, 4, 5 * 1 : 1, 2 * 2 : 0, 4, 5 * 3 : 3 * 4 : 0, 4, 5 * 5 : 3 * 6 : 6 * 7 : 7, 8 * 8 : 3 * 9 : 7, 8, 9, 10, 11 * 10 : 6 * 11 : 7, 8 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) minus#(minus(X, Y), Z) =#> minus#(X, plus(Y, Z)) P_2: quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) P_3: plus#(s(X), Y) =#> plus#(X, Y) P_4: app#(cons(X, Y), Z) =#> app#(Y, Z) P_5: sum#(cons(X, cons(Y, Z))) =#> sum#(cons(plus(X, Y), Z)) P_6: sum#(app(X, cons(Y, cons(Z, U)))) =#> sum#(app(X, sum(cons(Y, cons(Z, U))))) 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), (P_3, R_0, m, f), (P_4, R_0, m, f), (P_5, R_0, m, f) and (P_6, 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), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_6, R_0) are: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) => sum(app(X, sum(cons(Y, cons(Z, U))))) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: sum#(app(X, cons(Y, cons(Z, U)))) >? sum#(app(X, sum(cons(Y, cons(Z, U))))) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) app(nil, X) >= X app(X, nil) >= X app(cons(X, Y), Z) >= cons(X, app(Y, Z)) sum(cons(X, nil)) >= cons(X, nil) sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) sum(app(X, cons(Y, cons(Z, U)))) >= sum(app(X, sum(cons(Y, cons(Z, U))))) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: cons(x_1,x_2) = cons(x_2) This leaves the following ordering requirements: sum#(app(X, cons(Y, cons(Z, U)))) > sum#(app(X, sum(cons(Y, cons(Z, U))))) app(nil, X) >= X app(X, nil) >= X app(cons(X, Y), Z) >= cons(X, app(Y, Z)) sum(cons(X, nil)) >= cons(X, nil) sum(cons(X, cons(Y, Z))) >= sum(cons(plus(X, Y), Z)) The following interpretation satisfies the requirements: 0 = 0 app = \y0y1.y0 + y1 cons = \y0y1.1 + y1 nil = 0 plus = \y0y1.0 s = \y0.0 sum = \y0.1 sum# = \y0.2y0 Using this interpretation, the requirements translate to: [[sum#(app(_x0, cons(_x1, cons(_x2, _x3))))]] = 4 + 2x0 + 2x3 > 2 + 2x0 = [[sum#(app(_x0, sum(cons(_x1, cons(_x2, _x3)))))]] [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] [[app(_x0, nil)]] = x0 >= x0 = [[_x0]] [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] [[sum(cons(_x0, nil))]] = 1 >= 1 = [[cons(_x0, nil)]] [[sum(cons(_x0, cons(_x1, _x2)))]] = 1 >= 1 = [[sum(cons(plus(_x0, _x1), _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_6, 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), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_5, R_0) are: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: sum#(cons(X, cons(Y, Z))) >? sum#(cons(plus(X, Y), Z)) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + 2y0 + 2y1 plus = \y0y1.y1 s = \y0.0 sum# = \y0.y0 Using this interpretation, the requirements translate to: [[sum#(cons(_x0, cons(_x1, _x2)))]] = 9 + 2x0 + 4x1 + 4x2 > 3 + 2x1 + 2x2 = [[sum#(cons(plus(_x0, _x1), _x2))]] [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] [[plus(s(_x0), _x1)]] = x1 >= 0 = [[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_5, 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), (P_2, R_0, minimal, formative), (P_3, 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 apply the subterm criterion with the following projection function: nu(app#) = 1 Thus, we can orient the dependency pairs as follows: nu(app#(cons(X, Y), Z)) = cons(X, Y) |> Y = nu(app#(Y, Z)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, 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), (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(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_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). The formative rules of (P_2, R_0) are R_1 ::= minus(X, 0) => X 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))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) app(nil, X) => X app(X, nil) => X By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_2, R_1) are: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) minus(minus(X, Y), Z) => minus(X, plus(Y, Z)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: quot#(s(X), s(Y)) >? quot#(minus(X, Y), s(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) plus(0, X) >= X plus(s(X), Y) >= s(plus(X, Y)) minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. We consider usable_rules with respect to the following argument filtering: minus(x_1,x_2) = minus(x_1) This leaves the following ordering requirements: quot#(s(X), s(Y)) > quot#(minus(X, Y), s(Y)) minus(X, 0) >= X minus(s(X), s(Y)) >= minus(X, Y) minus(minus(X, Y), Z) >= minus(X, plus(Y, Z)) The following interpretation satisfies the requirements: 0 = 0 minus = \y0y1.y0 plus = \y0y1.0 quot# = \y0y1.y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[quot#(s(_x0), s(_x1))]] = 1 + x0 > x0 = [[quot#(minus(_x0, _x1), s(_x1))]] [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] [[minus(s(_x0), s(_x1))]] = 1 + x0 >= x0 = [[minus(_x0, _x1)]] [[minus(minus(_x0, _x1), _x2)]] = x0 >= x0 = [[minus(_x0, plus(_x1, _x2))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, 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 (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(minus#) = 1 Thus, we can orient the dependency pairs as follows: nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) nu(minus#(minus(X, Y), Z)) = minus(X, Y) |> X = nu(minus#(X, plus(Y, 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.