/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 U11 : [o * o * o] --> o U12 : [o * o * o] --> o U21 : [o * o * o] --> o U22 : [o * o * o] --> o a!6220!6220U11 : [o * o * o] --> o a!6220!6220U12 : [o * o * o] --> o a!6220!6220U21 : [o * o * o] --> o a!6220!6220U22 : [o * o * o] --> o a!6220!6220plus : [o * o] --> o a!6220!6220x : [o * o] --> o mark : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o x : [o * o] --> o a!6220!6220U11(tt, X, Y) => a!6220!6220U12(tt, X, Y) a!6220!6220U12(tt, X, Y) => s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220U21(tt, X, Y) => a!6220!6220U22(tt, X, Y) a!6220!6220U22(tt, X, Y) => a!6220!6220plus(a!6220!6220x(mark(Y), mark(X)), mark(Y)) a!6220!6220plus(X, 0) => mark(X) a!6220!6220plus(X, s(Y)) => a!6220!6220U11(tt, Y, X) a!6220!6220x(X, 0) => 0 a!6220!6220x(X, s(Y)) => a!6220!6220U21(tt, Y, X) mark(U11(X, Y, Z)) => a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) => a!6220!6220U12(mark(X), Y, Z) mark(plus(X, Y)) => a!6220!6220plus(mark(X), mark(Y)) mark(U21(X, Y, Z)) => a!6220!6220U21(mark(X), Y, Z) mark(U22(X, Y, Z)) => a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) => a!6220!6220x(mark(X), mark(Y)) mark(tt) => tt mark(s(X)) => s(mark(X)) mark(0) => 0 a!6220!6220U11(X, Y, Z) => U11(X, Y, Z) a!6220!6220U12(X, Y, Z) => U12(X, Y, Z) a!6220!6220plus(X, Y) => plus(X, Y) a!6220!6220U21(X, Y, Z) => U21(X, Y, Z) a!6220!6220U22(X, Y, Z) => U22(X, Y, Z) a!6220!6220x(X, Y) => x(X, 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]): a!6220!6220U11(tt, X, Y) >? a!6220!6220U12(tt, X, Y) a!6220!6220U12(tt, X, Y) >? s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220U21(tt, X, Y) >? a!6220!6220U22(tt, X, Y) a!6220!6220U22(tt, X, Y) >? a!6220!6220plus(a!6220!6220x(mark(Y), mark(X)), mark(Y)) a!6220!6220plus(X, 0) >? mark(X) a!6220!6220plus(X, s(Y)) >? a!6220!6220U11(tt, Y, X) a!6220!6220x(X, 0) >? 0 a!6220!6220x(X, s(Y)) >? a!6220!6220U21(tt, Y, X) mark(U11(X, Y, Z)) >? a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) >? a!6220!6220U12(mark(X), Y, Z) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(U21(X, Y, Z)) >? a!6220!6220U21(mark(X), Y, Z) mark(U22(X, Y, Z)) >? a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) mark(0) >? 0 a!6220!6220U11(X, Y, Z) >? U11(X, Y, Z) a!6220!6220U12(X, Y, Z) >? U12(X, Y, Z) a!6220!6220plus(X, Y) >? plus(X, Y) a!6220!6220U21(X, Y, Z) >? U21(X, Y, Z) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[mark(x_1)]] = x_1 [[tt]] = _|_ We choose Lex = {} and Mul = {U11, U12, U21, U22, a!6220!6220U11, a!6220!6220U12, a!6220!6220U21, a!6220!6220U22, a!6220!6220plus, a!6220!6220x, plus, s, x}, and the following precedence: U21 = U22 = a!6220!6220U21 = a!6220!6220U22 = a!6220!6220x = x > U11 = U12 = a!6220!6220U11 = a!6220!6220U12 = a!6220!6220plus = plus > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: a!6220!6220U11(_|_, X, Y) >= a!6220!6220U12(_|_, X, Y) a!6220!6220U12(_|_, X, Y) >= s(a!6220!6220plus(Y, X)) a!6220!6220U21(_|_, X, Y) >= a!6220!6220U22(_|_, X, Y) a!6220!6220U22(_|_, X, Y) >= a!6220!6220plus(a!6220!6220x(Y, X), Y) a!6220!6220plus(X, _|_) >= X a!6220!6220plus(X, s(Y)) > a!6220!6220U11(_|_, Y, X) a!6220!6220x(X, _|_) >= _|_ a!6220!6220x(X, s(Y)) > a!6220!6220U21(_|_, Y, X) U11(X, Y, Z) >= a!6220!6220U11(X, Y, Z) U12(X, Y, Z) >= a!6220!6220U12(X, Y, Z) plus(X, Y) >= a!6220!6220plus(X, Y) U21(X, Y, Z) >= a!6220!6220U21(X, Y, Z) U22(X, Y, Z) >= a!6220!6220U22(X, Y, Z) x(X, Y) >= a!6220!6220x(X, Y) _|_ >= _|_ s(X) >= s(X) _|_ >= _|_ a!6220!6220U11(X, Y, Z) >= U11(X, Y, Z) a!6220!6220U12(X, Y, Z) >= U12(X, Y, Z) a!6220!6220plus(X, Y) >= plus(X, Y) a!6220!6220U21(X, Y, Z) >= U21(X, Y, Z) a!6220!6220U22(X, Y, Z) >= U22(X, Y, Z) a!6220!6220x(X, Y) >= x(X, Y) With these choices, we have: 1] a!6220!6220U11(_|_, X, Y) >= a!6220!6220U12(_|_, X, Y) because a!6220!6220U11 = a!6220!6220U12, a!6220!6220U11 in Mul, [2], [3] and [4], by (Fun) 2] _|_ >= _|_ by (Bot) 3] X >= X by (Meta) 4] Y >= Y by (Meta) 5] a!6220!6220U12(_|_, X, Y) >= s(a!6220!6220plus(Y, X)) because [6], by (Star) 6] a!6220!6220U12*(_|_, X, Y) >= s(a!6220!6220plus(Y, X)) because a!6220!6220U12 > s and [7], by (Copy) 7] a!6220!6220U12*(_|_, X, Y) >= a!6220!6220plus(Y, X) because a!6220!6220U12 = a!6220!6220plus, a!6220!6220U12 in Mul, [8] and [9], by (Stat) 8] X >= X by (Meta) 9] Y >= Y by (Meta) 10] a!6220!6220U21(_|_, X, Y) >= a!6220!6220U22(_|_, X, Y) because a!6220!6220U21 = a!6220!6220U22, a!6220!6220U21 in Mul, [2], [8] and [9], by (Fun) 11] a!6220!6220U22(_|_, X, Y) >= a!6220!6220plus(a!6220!6220x(Y, X), Y) because [12], by (Star) 12] a!6220!6220U22*(_|_, X, Y) >= a!6220!6220plus(a!6220!6220x(Y, X), Y) because a!6220!6220U22 > a!6220!6220plus, [13] and [14], by (Copy) 13] a!6220!6220U22*(_|_, X, Y) >= a!6220!6220x(Y, X) because a!6220!6220U22 = a!6220!6220x, a!6220!6220U22 in Mul, [8] and [9], by (Stat) 14] a!6220!6220U22*(_|_, X, Y) >= Y because [9], by (Select) 15] a!6220!6220plus(X, _|_) >= X because [16], by (Star) 16] a!6220!6220plus*(X, _|_) >= X because [9], by (Select) 17] a!6220!6220plus(X, s(Y)) > a!6220!6220U11(_|_, Y, X) because [18], by definition 18] a!6220!6220plus*(X, s(Y)) >= a!6220!6220U11(_|_, Y, X) because a!6220!6220plus = a!6220!6220U11, a!6220!6220plus in Mul, [9], [19] and [21], by (Stat) 19] s(Y) > _|_ because [20], by definition 20] s*(Y) >= _|_ by (Bot) 21] s(Y) > Y because [22], by definition 22] s*(Y) >= Y because [8], by (Select) 23] a!6220!6220x(X, _|_) >= _|_ by (Bot) 24] a!6220!6220x(X, s(Y)) > a!6220!6220U21(_|_, Y, X) because [25], by definition 25] a!6220!6220x*(X, s(Y)) >= a!6220!6220U21(_|_, Y, X) because a!6220!6220x = a!6220!6220U21, a!6220!6220x in Mul, [9], [19] and [21], by (Stat) 26] U11(X, Y, Z) >= a!6220!6220U11(X, Y, Z) because U11 = a!6220!6220U11, U11 in Mul, [27], [28] and [29], by (Fun) 27] X >= X by (Meta) 28] Y >= Y by (Meta) 29] Z >= Z by (Meta) 30] U12(X, Y, Z) >= a!6220!6220U12(X, Y, Z) because U12 = a!6220!6220U12, U12 in Mul, [27], [28] and [29], by (Fun) 31] plus(X, Y) >= a!6220!6220plus(X, Y) because plus = a!6220!6220plus, plus in Mul, [27] and [32], by (Fun) 32] Y >= Y by (Meta) 33] U21(X, Y, Z) >= a!6220!6220U21(X, Y, Z) because U21 = a!6220!6220U21, U21 in Mul, [27], [32] and [29], by (Fun) 34] U22(X, Y, Z) >= a!6220!6220U22(X, Y, Z) because U22 = a!6220!6220U22, U22 in Mul, [27], [32] and [29], by (Fun) 35] x(X, Y) >= a!6220!6220x(X, Y) because x = a!6220!6220x, x in Mul, [27] and [32], by (Fun) 36] _|_ >= _|_ by (Bot) 37] s(X) >= s(X) because s in Mul and [38], by (Fun) 38] X >= X by (Meta) 39] _|_ >= _|_ by (Bot) 40] a!6220!6220U11(X, Y, Z) >= U11(X, Y, Z) because a!6220!6220U11 = U11, a!6220!6220U11 in Mul, [27], [32] and [29], by (Fun) 41] a!6220!6220U12(X, Y, Z) >= U12(X, Y, Z) because a!6220!6220U12 = U12, a!6220!6220U12 in Mul, [27], [32] and [29], by (Fun) 42] a!6220!6220plus(X, Y) >= plus(X, Y) because a!6220!6220plus = plus, a!6220!6220plus in Mul, [27] and [32], by (Fun) 43] a!6220!6220U21(X, Y, Z) >= U21(X, Y, Z) because a!6220!6220U21 = U21, a!6220!6220U21 in Mul, [27], [32] and [29], by (Fun) 44] a!6220!6220U22(X, Y, Z) >= U22(X, Y, Z) because a!6220!6220U22 = U22, a!6220!6220U22 in Mul, [27], [32] and [29], by (Fun) 45] a!6220!6220x(X, Y) >= x(X, Y) because a!6220!6220x = x, a!6220!6220x in Mul, [27] and [32], by (Fun) We can thus remove the following rules: a!6220!6220plus(X, s(Y)) => a!6220!6220U11(tt, Y, X) a!6220!6220x(X, s(Y)) => a!6220!6220U21(tt, Y, 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]): a!6220!6220U11(tt, X, Y) >? a!6220!6220U12(tt, X, Y) a!6220!6220U12(tt, X, Y) >? s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220U21(tt, X, Y) >? a!6220!6220U22(tt, X, Y) a!6220!6220U22(tt, X, Y) >? a!6220!6220plus(a!6220!6220x(mark(Y), mark(X)), mark(Y)) a!6220!6220plus(X, 0) >? mark(X) a!6220!6220x(X, 0) >? 0 mark(U11(X, Y, Z)) >? a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) >? a!6220!6220U12(mark(X), Y, Z) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(U21(X, Y, Z)) >? a!6220!6220U21(mark(X), Y, Z) mark(U22(X, Y, Z)) >? a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) mark(0) >? 0 a!6220!6220U11(X, Y, Z) >? U11(X, Y, Z) a!6220!6220U12(X, Y, Z) >? U12(X, Y, Z) a!6220!6220plus(X, Y) >? plus(X, Y) a!6220!6220U21(X, Y, Z) >? U21(X, Y, Z) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 U11 = \y0y1y2.y0 + y1 + y2 U12 = \y0y1y2.y0 + y1 + y2 U21 = \y0y1y2.2y0 + 2y1 + 2y2 U22 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220U11 = \y0y1y2.y0 + y1 + y2 a!6220!6220U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220U21 = \y0y1y2.2y0 + 2y1 + 2y2 a!6220!6220U22 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220plus = \y0y1.y0 + y1 a!6220!6220x = \y0y1.y0 + 2y1 mark = \y0.y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 0 x = \y0y1.y0 + 2y1 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[a!6220!6220U12(tt, _x0, _x1)]] [[a!6220!6220U12(tt, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[s(a!6220!6220plus(mark(_x1), mark(_x0)))]] [[a!6220!6220U21(tt, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220U22(tt, _x0, _x1)]] [[a!6220!6220U22(tt, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220plus(a!6220!6220x(mark(_x1), mark(_x0)), mark(_x1))]] [[a!6220!6220plus(_x0, 0)]] = 1 + x0 > x0 = [[mark(_x0)]] [[a!6220!6220x(_x0, 0)]] = 2 + x0 > 1 = [[0]] [[mark(U11(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(U21(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[a!6220!6220U21(mark(_x0), _x1, _x2)]] [[mark(U22(_x0, _x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[a!6220!6220U22(mark(_x0), _x1, _x2)]] [[mark(x(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[a!6220!6220x(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] [[mark(0)]] = 1 >= 1 = [[0]] [[a!6220!6220U11(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[plus(_x0, _x1)]] [[a!6220!6220U21(_x0, _x1, _x2)]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[U21(_x0, _x1, _x2)]] [[a!6220!6220U22(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U22(_x0, _x1, _x2)]] [[a!6220!6220x(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[x(_x0, _x1)]] We can thus remove the following rules: a!6220!6220plus(X, 0) => mark(X) a!6220!6220x(X, 0) => 0 We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): a!6220!6220U11(tt, X, Y) >? a!6220!6220U12(tt, X, Y) a!6220!6220U12(tt, X, Y) >? s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220U21(tt, X, Y) >? a!6220!6220U22(tt, X, Y) a!6220!6220U22(tt, X, Y) >? a!6220!6220plus(a!6220!6220x(mark(Y), mark(X)), mark(Y)) mark(U11(X, Y, Z)) >? a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) >? a!6220!6220U12(mark(X), Y, Z) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(U21(X, Y, Z)) >? a!6220!6220U21(mark(X), Y, Z) mark(U22(X, Y, Z)) >? a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) mark(0) >? 0 a!6220!6220U11(X, Y, Z) >? U11(X, Y, Z) a!6220!6220U12(X, Y, Z) >? U12(X, Y, Z) a!6220!6220plus(X, Y) >? plus(X, Y) a!6220!6220U21(X, Y, Z) >? U21(X, Y, Z) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.y0 + y1 + y2 U12 = \y0y1y2.y0 + y1 + y2 U21 = \y0y1y2.y0 + y1 + 2y2 U22 = \y0y1y2.y0 + y1 + 2y2 a!6220!6220U11 = \y0y1y2.y0 + y1 + y2 a!6220!6220U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220U21 = \y0y1y2.y0 + y1 + 2y2 a!6220!6220U22 = \y0y1y2.y0 + y1 + 2y2 a!6220!6220plus = \y0y1.y0 + y1 a!6220!6220x = \y0y1.y0 + y1 mark = \y0.y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 1 x = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[a!6220!6220U12(tt, _x0, _x1)]] [[a!6220!6220U12(tt, _x0, _x1)]] = 1 + x0 + x1 > x0 + x1 = [[s(a!6220!6220plus(mark(_x1), mark(_x0)))]] [[a!6220!6220U21(tt, _x0, _x1)]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[a!6220!6220U22(tt, _x0, _x1)]] [[a!6220!6220U22(tt, _x0, _x1)]] = 1 + x0 + 2x1 > x0 + 2x1 = [[a!6220!6220plus(a!6220!6220x(mark(_x1), mark(_x0)), mark(_x1))]] [[mark(U11(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(U21(_x0, _x1, _x2))]] = x0 + x1 + 2x2 >= x0 + x1 + 2x2 = [[a!6220!6220U21(mark(_x0), _x1, _x2)]] [[mark(U22(_x0, _x1, _x2))]] = x0 + x1 + 2x2 >= x0 + x1 + 2x2 = [[a!6220!6220U22(mark(_x0), _x1, _x2)]] [[mark(x(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220x(mark(_x0), mark(_x1))]] [[mark(tt)]] = 1 >= 1 = [[tt]] [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[a!6220!6220U11(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[plus(_x0, _x1)]] [[a!6220!6220U21(_x0, _x1, _x2)]] = x0 + x1 + 2x2 >= x0 + x1 + 2x2 = [[U21(_x0, _x1, _x2)]] [[a!6220!6220U22(_x0, _x1, _x2)]] = x0 + x1 + 2x2 >= x0 + x1 + 2x2 = [[U22(_x0, _x1, _x2)]] [[a!6220!6220x(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[x(_x0, _x1)]] We can thus remove the following rules: a!6220!6220U12(tt, X, Y) => s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220U22(tt, X, Y) => a!6220!6220plus(a!6220!6220x(mark(Y), mark(X)), mark(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]): a!6220!6220U11(tt, X, Y) >? a!6220!6220U12(tt, X, Y) a!6220!6220U21(tt, X, Y) >? a!6220!6220U22(tt, X, Y) mark(U11(X, Y, Z)) >? a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) >? a!6220!6220U12(mark(X), Y, Z) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(U21(X, Y, Z)) >? a!6220!6220U21(mark(X), Y, Z) mark(U22(X, Y, Z)) >? a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) mark(0) >? 0 a!6220!6220U11(X, Y, Z) >? U11(X, Y, Z) a!6220!6220U12(X, Y, Z) >? U12(X, Y, Z) a!6220!6220plus(X, Y) >? plus(X, Y) a!6220!6220U21(X, Y, Z) >? U21(X, Y, Z) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 U11 = \y0y1y2.2 + y0 + 2y1 + 2y2 U12 = \y0y1y2.2 + y0 + y1 + y2 U21 = \y0y1y2.1 + y0 + y1 + y2 U22 = \y0y1y2.y0 + y1 + y2 a!6220!6220U11 = \y0y1y2.3 + y0 + 2y1 + 2y2 a!6220!6220U12 = \y0y1y2.3 + y0 + y1 + y2 a!6220!6220U21 = \y0y1y2.1 + y0 + y1 + 2y2 a!6220!6220U22 = \y0y1y2.y0 + y1 + y2 a!6220!6220plus = \y0y1.1 + y0 + y1 a!6220!6220x = \y0y1.y0 + y1 mark = \y0.2y0 plus = \y0y1.1 + y0 + y1 s = \y0.y0 tt = 0 x = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = 3 + 2x0 + 2x1 >= 3 + x0 + x1 = [[a!6220!6220U12(tt, _x0, _x1)]] [[a!6220!6220U21(tt, _x0, _x1)]] = 1 + x0 + 2x1 > x0 + x1 = [[a!6220!6220U22(tt, _x0, _x1)]] [[mark(U11(_x0, _x1, _x2))]] = 4 + 2x0 + 4x1 + 4x2 > 3 + 2x0 + 2x1 + 2x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = 4 + 2x0 + 2x1 + 2x2 > 3 + x1 + x2 + 2x0 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(U21(_x0, _x1, _x2))]] = 2 + 2x0 + 2x1 + 2x2 > 1 + x1 + 2x0 + 2x2 = [[a!6220!6220U21(mark(_x0), _x1, _x2)]] [[mark(U22(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= x1 + x2 + 2x0 = [[a!6220!6220U22(mark(_x0), _x1, _x2)]] [[mark(x(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220x(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[mark(0)]] = 2 > 1 = [[0]] [[a!6220!6220U11(_x0, _x1, _x2)]] = 3 + x0 + 2x1 + 2x2 > 2 + x0 + 2x1 + 2x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = 3 + x0 + x1 + x2 > 2 + x0 + x1 + x2 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[plus(_x0, _x1)]] [[a!6220!6220U21(_x0, _x1, _x2)]] = 1 + x0 + x1 + 2x2 >= 1 + x0 + x1 + x2 = [[U21(_x0, _x1, _x2)]] [[a!6220!6220U22(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U22(_x0, _x1, _x2)]] [[a!6220!6220x(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[x(_x0, _x1)]] We can thus remove the following rules: a!6220!6220U21(tt, X, Y) => a!6220!6220U22(tt, X, Y) mark(U11(X, Y, Z)) => a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) => a!6220!6220U12(mark(X), Y, Z) mark(plus(X, Y)) => a!6220!6220plus(mark(X), mark(Y)) mark(U21(X, Y, Z)) => a!6220!6220U21(mark(X), Y, Z) mark(0) => 0 a!6220!6220U11(X, Y, Z) => U11(X, Y, Z) a!6220!6220U12(X, Y, Z) => U12(X, Y, Z) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): a!6220!6220U11(tt, X, Y) >? a!6220!6220U12(tt, X, Y) mark(U22(X, Y, Z)) >? a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) a!6220!6220plus(X, Y) >? plus(X, Y) a!6220!6220U21(X, Y, Z) >? U21(X, Y, Z) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U21 = \y0y1y2.y0 + y1 + y2 U22 = \y0y1y2.y0 + y1 + y2 a!6220!6220U11 = \y0y1y2.3 + 2y1 + 2y2 + 3y0 a!6220!6220U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220U21 = \y0y1y2.3 + y0 + y1 + y2 a!6220!6220U22 = \y0y1y2.y0 + y1 + y2 a!6220!6220plus = \y0y1.3 + y0 + 2y1 a!6220!6220x = \y0y1.y0 + y1 mark = \y0.2y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 0 x = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = 3 + 2x0 + 2x1 > x0 + x1 = [[a!6220!6220U12(tt, _x0, _x1)]] [[mark(U22(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= x1 + x2 + 2x0 = [[a!6220!6220U22(mark(_x0), _x1, _x2)]] [[mark(x(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220x(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220plus(_x0, _x1)]] = 3 + x0 + 2x1 > x0 + x1 = [[plus(_x0, _x1)]] [[a!6220!6220U21(_x0, _x1, _x2)]] = 3 + x0 + x1 + x2 > x0 + x1 + x2 = [[U21(_x0, _x1, _x2)]] [[a!6220!6220U22(_x0, _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U22(_x0, _x1, _x2)]] [[a!6220!6220x(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[x(_x0, _x1)]] We can thus remove the following rules: a!6220!6220U11(tt, X, Y) => a!6220!6220U12(tt, X, Y) a!6220!6220plus(X, Y) => plus(X, Y) a!6220!6220U21(X, Y, Z) => U21(X, Y, Z) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): mark(U22(X, Y, Z)) >? a!6220!6220U22(mark(X), Y, Z) mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U22 = \y0y1y2.1 + y1 + y2 + 2y0 a!6220!6220U22 = \y0y1y2.1 + 2y0 + 2y1 + 2y2 a!6220!6220x = \y0y1.y0 + y1 mark = \y0.2y0 s = \y0.y0 tt = 0 x = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[mark(U22(_x0, _x1, _x2))]] = 2 + 2x1 + 2x2 + 4x0 > 1 + 2x1 + 2x2 + 4x0 = [[a!6220!6220U22(mark(_x0), _x1, _x2)]] [[mark(x(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220x(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220U22(_x0, _x1, _x2)]] = 1 + 2x0 + 2x1 + 2x2 >= 1 + x1 + x2 + 2x0 = [[U22(_x0, _x1, _x2)]] [[a!6220!6220x(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[x(_x0, _x1)]] We can thus remove the following rules: mark(U22(X, Y, Z)) => a!6220!6220U22(mark(X), Y, Z) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): mark(x(X, Y)) >? a!6220!6220x(mark(X), mark(Y)) mark(tt) >? tt mark(s(X)) >? s(mark(X)) a!6220!6220U22(X, Y, Z) >? U22(X, Y, Z) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U22 = \y0y1y2.y0 + y1 + y2 a!6220!6220U22 = \y0y1y2.3 + y0 + y2 + 2y1 a!6220!6220x = \y0y1.1 + y0 + y1 mark = \y0.2y0 s = \y0.y0 tt = 0 x = \y0y1.1 + y0 + y1 Using this interpretation, the requirements translate to: [[mark(x(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[a!6220!6220x(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220U22(_x0, _x1, _x2)]] = 3 + x0 + x2 + 2x1 > x0 + x1 + x2 = [[U22(_x0, _x1, _x2)]] [[a!6220!6220x(_x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[x(_x0, _x1)]] We can thus remove the following rules: mark(x(X, Y)) => a!6220!6220x(mark(X), mark(Y)) a!6220!6220U22(X, Y, Z) => U22(X, Y, Z) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): mark(tt) >? tt mark(s(X)) >? s(mark(X)) a!6220!6220x(X, Y) >? x(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220x = \y0y1.3 + y0 + y1 mark = \y0.3 + 3y0 s = \y0.y0 tt = 0 x = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[mark(tt)]] = 3 > 0 = [[tt]] [[mark(s(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[s(mark(_x0))]] [[a!6220!6220x(_x0, _x1)]] = 3 + x0 + x1 > x0 + x1 = [[x(_x0, _x1)]] We can thus remove the following rules: mark(tt) => tt a!6220!6220x(X, Y) => x(X, 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]): mark(s(X)) >? s(mark(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: mark = \y0.3y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[mark(s(_x0))]] = 3 + 3x0 > 1 + 3x0 = [[s(mark(_x0))]] We can thus remove the following rules: mark(s(X)) => s(mark(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.