/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 U11 : [o * o] --> o U21 : [o * o * o] --> o U31 : [o] --> o U41 : [o * o * o] --> o activate : [o] --> o and : [o * o] --> o isNat : [o] --> o n!6220!62200 : [] --> o n!6220!6220isNat : [o] --> o n!6220!6220plus : [o * o] --> o n!6220!6220s : [o] --> o n!6220!6220x : [o * o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o x : [o * o] --> o U11(tt, X) => activate(X) U21(tt, X, Y) => s(plus(activate(Y), activate(X))) U31(tt) => 0 U41(tt, X, Y) => plus(x(activate(Y), activate(X)), activate(Y)) and(tt, X) => activate(X) isNat(n!6220!62200) => tt isNat(n!6220!6220plus(X, Y)) => and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) isNat(n!6220!6220s(X)) => isNat(activate(X)) isNat(n!6220!6220x(X, Y)) => and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) plus(X, 0) => U11(isNat(X), X) plus(X, s(Y)) => U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) x(X, 0) => U31(isNat(X)) x(X, s(Y)) => U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) 0 => n!6220!62200 plus(X, Y) => n!6220!6220plus(X, Y) isNat(X) => n!6220!6220isNat(X) s(X) => n!6220!6220s(X) x(X, Y) => n!6220!6220x(X, Y) activate(n!6220!62200) => 0 activate(n!6220!6220plus(X, Y)) => plus(activate(X), activate(Y)) activate(n!6220!6220isNat(X)) => isNat(X) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220x(X, Y)) => x(activate(X), activate(Y)) activate(X) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): U11(tt, X) >? activate(X) U21(tt, X, Y) >? s(plus(activate(Y), activate(X))) U31(tt) >? 0 U41(tt, X, Y) >? plus(x(activate(Y), activate(X)), activate(Y)) and(tt, X) >? activate(X) isNat(n!6220!62200) >? tt isNat(n!6220!6220plus(X, Y)) >? and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) isNat(n!6220!6220s(X)) >? isNat(activate(X)) isNat(n!6220!6220x(X, Y)) >? and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) plus(X, 0) >? U11(isNat(X), X) plus(X, s(Y)) >? U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) x(X, 0) >? U31(isNat(X)) x(X, s(Y)) >? U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) 0 >? n!6220!62200 plus(X, Y) >? n!6220!6220plus(X, Y) isNat(X) >? n!6220!6220isNat(X) s(X) >? n!6220!6220s(X) x(X, Y) >? n!6220!6220x(X, Y) activate(n!6220!62200) >? 0 activate(n!6220!6220plus(X, Y)) >? plus(activate(X), activate(Y)) activate(n!6220!6220isNat(X)) >? isNat(X) activate(n!6220!6220s(X)) >? s(activate(X)) activate(n!6220!6220x(X, Y)) >? x(activate(X), activate(Y)) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[U21(x_1, x_2, x_3)]] = U21(x_3, x_2, x_1) [[U41(x_1, x_2, x_3)]] = U41(x_3, x_2, x_1) [[activate(x_1)]] = x_1 [[tt]] = _|_ We choose Lex = {U21, U41, n!6220!6220plus, n!6220!6220x, plus, x} and Mul = {0, U11, U31, and, isNat, n!6220!62200, n!6220!6220isNat, n!6220!6220s, s}, and the following precedence: U41 = n!6220!6220x = x > U31 > 0 = n!6220!62200 > U21 = n!6220!6220plus = plus > isNat = n!6220!6220isNat = n!6220!6220s = s > U11 > and Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: U11(_|_, X) >= X U21(_|_, X, Y) > s(plus(Y, X)) U31(_|_) >= 0 U41(_|_, X, Y) > plus(x(Y, X), Y) and(_|_, X) >= X isNat(n!6220!62200) >= _|_ isNat(n!6220!6220plus(X, Y)) > and(isNat(X), n!6220!6220isNat(Y)) isNat(n!6220!6220s(X)) >= isNat(X) isNat(n!6220!6220x(X, Y)) > and(isNat(X), n!6220!6220isNat(Y)) plus(X, 0) >= U11(isNat(X), X) plus(X, s(Y)) >= U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) x(X, 0) > U31(isNat(X)) x(X, s(Y)) >= U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) 0 >= n!6220!62200 plus(X, Y) >= n!6220!6220plus(X, Y) isNat(X) >= n!6220!6220isNat(X) s(X) >= n!6220!6220s(X) x(X, Y) >= n!6220!6220x(X, Y) n!6220!62200 >= 0 n!6220!6220plus(X, Y) >= plus(X, Y) n!6220!6220isNat(X) >= isNat(X) n!6220!6220s(X) >= s(X) n!6220!6220x(X, Y) >= x(X, Y) X >= X With these choices, we have: 1] U11(_|_, X) >= X because [2], by (Star) 2] U11*(_|_, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] U21(_|_, X, Y) > s(plus(Y, X)) because [5], by definition 5] U21*(_|_, X, Y) >= s(plus(Y, X)) because U21 > s and [6], by (Copy) 6] U21*(_|_, X, Y) >= plus(Y, X) because U21 = plus, [7], [8], [9] and [10], by (Stat) 7] X >= X by (Meta) 8] Y >= Y by (Meta) 9] U21*(_|_, X, Y) >= Y because [8], by (Select) 10] U21*(_|_, X, Y) >= X because [7], by (Select) 11] U31(_|_) >= 0 because [12], by (Star) 12] U31*(_|_) >= 0 because U31 > 0, by (Copy) 13] U41(_|_, X, Y) > plus(x(Y, X), Y) because [14], by definition 14] U41*(_|_, X, Y) >= plus(x(Y, X), Y) because U41 > plus, [15] and [16], by (Copy) 15] U41*(_|_, X, Y) >= x(Y, X) because U41 = x, [7], [8], [16] and [17], by (Stat) 16] U41*(_|_, X, Y) >= Y because [8], by (Select) 17] U41*(_|_, X, Y) >= X because [7], by (Select) 18] and(_|_, X) >= X because [19], by (Star) 19] and*(_|_, X) >= X because [20], by (Select) 20] X >= X by (Meta) 21] isNat(n!6220!62200) >= _|_ by (Bot) 22] isNat(n!6220!6220plus(X, Y)) > and(isNat(X), n!6220!6220isNat(Y)) because [23], by definition 23] isNat*(n!6220!6220plus(X, Y)) >= and(isNat(X), n!6220!6220isNat(Y)) because isNat > and, [24] and [28], by (Copy) 24] isNat*(n!6220!6220plus(X, Y)) >= isNat(X) because isNat in Mul and [25], by (Stat) 25] n!6220!6220plus(X, Y) > X because [26], by definition 26] n!6220!6220plus*(X, Y) >= X because [27], by (Select) 27] X >= X by (Meta) 28] isNat*(n!6220!6220plus(X, Y)) >= n!6220!6220isNat(Y) because [29], by (Select) 29] n!6220!6220plus(X, Y) >= n!6220!6220isNat(Y) because [30], by (Star) 30] n!6220!6220plus*(X, Y) >= n!6220!6220isNat(Y) because n!6220!6220plus > n!6220!6220isNat and [31], by (Copy) 31] n!6220!6220plus*(X, Y) >= Y because [32], by (Select) 32] Y >= Y by (Meta) 33] isNat(n!6220!6220s(X)) >= isNat(X) because [34], by (Star) 34] isNat*(n!6220!6220s(X)) >= isNat(X) because [35], by (Select) 35] n!6220!6220s(X) >= isNat(X) because n!6220!6220s = isNat, n!6220!6220s in Mul and [36], by (Fun) 36] X >= X by (Meta) 37] isNat(n!6220!6220x(X, Y)) > and(isNat(X), n!6220!6220isNat(Y)) because [38], by definition 38] isNat*(n!6220!6220x(X, Y)) >= and(isNat(X), n!6220!6220isNat(Y)) because [39], by (Select) 39] n!6220!6220x(X, Y) >= and(isNat(X), n!6220!6220isNat(Y)) because [40], by (Star) 40] n!6220!6220x*(X, Y) >= and(isNat(X), n!6220!6220isNat(Y)) because n!6220!6220x > and, [41] and [43], by (Copy) 41] n!6220!6220x*(X, Y) >= isNat(X) because n!6220!6220x > isNat and [42], by (Copy) 42] n!6220!6220x*(X, Y) >= X because [36], by (Select) 43] n!6220!6220x*(X, Y) >= n!6220!6220isNat(Y) because n!6220!6220x > n!6220!6220isNat and [44], by (Copy) 44] n!6220!6220x*(X, Y) >= Y because [32], by (Select) 45] plus(X, 0) >= U11(isNat(X), X) because [46], by (Star) 46] plus*(X, 0) >= U11(isNat(X), X) because plus > U11, [47] and [48], by (Copy) 47] plus*(X, 0) >= isNat(X) because plus > isNat and [48], by (Copy) 48] plus*(X, 0) >= X because [8], by (Select) 49] plus(X, s(Y)) >= U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) because [50], by (Star) 50] plus*(X, s(Y)) >= U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) because plus = U21, [8], [51], [53], [55] and [58], by (Stat) 51] s(Y) > Y because [52], by definition 52] s*(Y) >= Y because [7], by (Select) 53] plus*(X, s(Y)) >= and(isNat(Y), n!6220!6220isNat(X)) because plus > and, [54] and [57], by (Copy) 54] plus*(X, s(Y)) >= isNat(Y) because plus > isNat and [55], by (Copy) 55] plus*(X, s(Y)) >= Y because [56], by (Select) 56] s(Y) >= Y because [52], by (Star) 57] plus*(X, s(Y)) >= n!6220!6220isNat(X) because plus > n!6220!6220isNat and [58], by (Copy) 58] plus*(X, s(Y)) >= X because [8], by (Select) 59] x(X, 0) > U31(isNat(X)) because [60], by definition 60] x*(X, 0) >= U31(isNat(X)) because x > U31 and [61], by (Copy) 61] x*(X, 0) >= isNat(X) because x > isNat and [62], by (Copy) 62] x*(X, 0) >= X because [8], by (Select) 63] x(X, s(Y)) >= U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) because [64], by (Star) 64] x*(X, s(Y)) >= U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) because x = U41, [8], [51], [65], [70] and [69], by (Stat) 65] x*(X, s(Y)) >= and(isNat(Y), n!6220!6220isNat(X)) because x > and, [66] and [68], by (Copy) 66] x*(X, s(Y)) >= isNat(Y) because [67], by (Select) 67] s(Y) >= isNat(Y) because s = isNat, s in Mul and [7], by (Fun) 68] x*(X, s(Y)) >= n!6220!6220isNat(X) because x > n!6220!6220isNat and [69], by (Copy) 69] x*(X, s(Y)) >= X because [8], by (Select) 70] x*(X, s(Y)) >= Y because [56], by (Select) 71] 0 >= n!6220!62200 because 0 = n!6220!62200, by (Fun) 72] plus(X, Y) >= n!6220!6220plus(X, Y) because plus = n!6220!6220plus, [73] and [74], by (Fun) 73] X >= X by (Meta) 74] Y >= Y by (Meta) 75] isNat(X) >= n!6220!6220isNat(X) because isNat = n!6220!6220isNat, isNat in Mul and [76], by (Fun) 76] X >= X by (Meta) 77] s(X) >= n!6220!6220s(X) because s = n!6220!6220s, s in Mul and [76], by (Fun) 78] x(X, Y) >= n!6220!6220x(X, Y) because x = n!6220!6220x, [73] and [74], by (Fun) 79] n!6220!62200 >= 0 because n!6220!62200 = 0, by (Fun) 80] n!6220!6220plus(X, Y) >= plus(X, Y) because n!6220!6220plus = plus, [81] and [82], by (Fun) 81] X >= X by (Meta) 82] Y >= Y by (Meta) 83] n!6220!6220isNat(X) >= isNat(X) because n!6220!6220isNat = isNat, n!6220!6220isNat in Mul and [76], by (Fun) 84] n!6220!6220s(X) >= s(X) because n!6220!6220s = s, n!6220!6220s in Mul and [85], by (Fun) 85] X >= X by (Meta) 86] n!6220!6220x(X, Y) >= x(X, Y) because n!6220!6220x = x, [81] and [82], by (Fun) 87] X >= X by (Meta) We can thus remove the following rules: U21(tt, X, Y) => s(plus(activate(Y), activate(X))) U41(tt, X, Y) => plus(x(activate(Y), activate(X)), activate(Y)) isNat(n!6220!6220plus(X, Y)) => and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) isNat(n!6220!6220x(X, Y)) => and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) x(X, 0) => U31(isNat(X)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): U11(tt, X) >? activate(X) U31(tt) >? 0 and(tt, X) >? activate(X) isNat(n!6220!62200) >? tt isNat(n!6220!6220s(X)) >? isNat(activate(X)) plus(X, 0) >? U11(isNat(X), X) plus(X, s(Y)) >? U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) x(X, s(Y)) >? U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) 0 >? n!6220!62200 plus(X, Y) >? n!6220!6220plus(X, Y) isNat(X) >? n!6220!6220isNat(X) s(X) >? n!6220!6220s(X) x(X, Y) >? n!6220!6220x(X, Y) activate(n!6220!62200) >? 0 activate(n!6220!6220plus(X, Y)) >? plus(activate(X), activate(Y)) activate(n!6220!6220isNat(X)) >? isNat(X) activate(n!6220!6220s(X)) >? s(activate(X)) activate(n!6220!6220x(X, Y)) >? x(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: 0 = 0 U11 = \y0y1.y0 + 2y1 U21 = \y0y1y2.y0 + y1 + y2 U31 = \y0.3 + 3y0 U41 = \y0y1y2.y0 + y1 + y2 activate = \y0.2y0 and = \y0y1.y0 + 2y1 isNat = \y0.y0 n!6220!62200 = 0 n!6220!6220isNat = \y0.y0 n!6220!6220plus = \y0y1.2 + y1 + 3y0 n!6220!6220s = \y0.2 + 2y0 n!6220!6220x = \y0y1.1 + y1 + 3y0 plus = \y0y1.2 + y1 + 3y0 s = \y0.3 + 2y0 tt = 0 x = \y0y1.1 + y1 + 3y0 Using this interpretation, the requirements translate to: [[U11(tt, _x0)]] = 2x0 >= 2x0 = [[activate(_x0)]] [[U31(tt)]] = 3 > 0 = [[0]] [[and(tt, _x0)]] = 2x0 >= 2x0 = [[activate(_x0)]] [[isNat(n!6220!62200)]] = 0 >= 0 = [[tt]] [[isNat(n!6220!6220s(_x0))]] = 2 + 2x0 > 2x0 = [[isNat(activate(_x0))]] [[plus(_x0, 0)]] = 2 + 3x0 > 3x0 = [[U11(isNat(_x0), _x0)]] [[plus(_x0, s(_x1))]] = 5 + 2x1 + 3x0 > 2x1 + 3x0 = [[U21(and(isNat(_x1), n!6220!6220isNat(_x0)), _x1, _x0)]] [[x(_x0, s(_x1))]] = 4 + 2x1 + 3x0 > 2x1 + 3x0 = [[U41(and(isNat(_x1), n!6220!6220isNat(_x0)), _x1, _x0)]] [[0]] = 0 >= 0 = [[n!6220!62200]] [[plus(_x0, _x1)]] = 2 + x1 + 3x0 >= 2 + x1 + 3x0 = [[n!6220!6220plus(_x0, _x1)]] [[isNat(_x0)]] = x0 >= x0 = [[n!6220!6220isNat(_x0)]] [[s(_x0)]] = 3 + 2x0 > 2 + 2x0 = [[n!6220!6220s(_x0)]] [[x(_x0, _x1)]] = 1 + x1 + 3x0 >= 1 + x1 + 3x0 = [[n!6220!6220x(_x0, _x1)]] [[activate(n!6220!62200)]] = 0 >= 0 = [[0]] [[activate(n!6220!6220plus(_x0, _x1))]] = 4 + 2x1 + 6x0 > 2 + 2x1 + 6x0 = [[plus(activate(_x0), activate(_x1))]] [[activate(n!6220!6220isNat(_x0))]] = 2x0 >= x0 = [[isNat(_x0)]] [[activate(n!6220!6220s(_x0))]] = 4 + 4x0 > 3 + 4x0 = [[s(activate(_x0))]] [[activate(n!6220!6220x(_x0, _x1))]] = 2 + 2x1 + 6x0 > 1 + 2x1 + 6x0 = [[x(activate(_x0), activate(_x1))]] [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] We can thus remove the following rules: U31(tt) => 0 isNat(n!6220!6220s(X)) => isNat(activate(X)) plus(X, 0) => U11(isNat(X), X) plus(X, s(Y)) => U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) x(X, s(Y)) => U41(and(isNat(Y), n!6220!6220isNat(X)), Y, X) s(X) => n!6220!6220s(X) activate(n!6220!6220plus(X, Y)) => plus(activate(X), activate(Y)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220x(X, Y)) => x(activate(X), activate(Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): U11(tt, X) >? activate(X) and(tt, X) >? activate(X) isNat(n!6220!62200) >? tt 0 >? n!6220!62200 plus(X, Y) >? n!6220!6220plus(X, Y) isNat(X) >? n!6220!6220isNat(X) x(X, Y) >? n!6220!6220x(X, Y) activate(n!6220!62200) >? 0 activate(n!6220!6220isNat(X)) >? isNat(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 U11 = \y0y1.3 + 3y0 + 3y1 activate = \y0.2 + 3y0 and = \y0y1.3 + 3y0 + 3y1 isNat = \y0.2 + 2y0 n!6220!62200 = 0 n!6220!6220isNat = \y0.y0 n!6220!6220plus = \y0y1.y0 + y1 n!6220!6220x = \y0y1.y0 + y1 plus = \y0y1.3 + 3y0 + 3y1 tt = 1 x = \y0y1.3 + 3y0 + 3y1 Using this interpretation, the requirements translate to: [[U11(tt, _x0)]] = 6 + 3x0 > 2 + 3x0 = [[activate(_x0)]] [[and(tt, _x0)]] = 6 + 3x0 > 2 + 3x0 = [[activate(_x0)]] [[isNat(n!6220!62200)]] = 2 > 1 = [[tt]] [[0]] = 1 > 0 = [[n!6220!62200]] [[plus(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[n!6220!6220plus(_x0, _x1)]] [[isNat(_x0)]] = 2 + 2x0 > x0 = [[n!6220!6220isNat(_x0)]] [[x(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[n!6220!6220x(_x0, _x1)]] [[activate(n!6220!62200)]] = 2 > 1 = [[0]] [[activate(n!6220!6220isNat(_x0))]] = 2 + 3x0 >= 2 + 2x0 = [[isNat(_x0)]] [[activate(_x0)]] = 2 + 3x0 > x0 = [[_x0]] We can thus remove the following rules: U11(tt, X) => activate(X) and(tt, X) => activate(X) isNat(n!6220!62200) => tt 0 => n!6220!62200 plus(X, Y) => n!6220!6220plus(X, Y) isNat(X) => n!6220!6220isNat(X) x(X, Y) => n!6220!6220x(X, Y) activate(n!6220!62200) => 0 activate(X) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): activate(n!6220!6220isNat(X)) >? isNat(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.3 + 3y0 isNat = \y0.y0 n!6220!6220isNat = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[activate(n!6220!6220isNat(_x0))]] = 12 + 9x0 > x0 = [[isNat(_x0)]] We can thus remove the following rules: activate(n!6220!6220isNat(X)) => isNat(X) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.