/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. A : [] --> o B : [] --> o C : [] --> o cons : [o * o] --> o f : [o * o] --> o f!450 : [o * o] --> o f!450!450 : [o] --> o foldf : [o * o] --> o g : [o] --> o nil : [] --> o triple : [o * o * o] --> o g(A) => A g(B) => A g(B) => B g(C) => A g(C) => B g(C) => C foldf(X, nil) => X foldf(X, cons(Y, Z)) => f(foldf(X, Z), Y) f(X, Y) => f!450(X, g(Y)) f!450(triple(X, Y, Z), C) => triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) f!450(triple(X, Y, Z), A) => f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) f!450!450(triple(X, Y, Z)) => foldf(triple(X, Y, nil), 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]): g(A) >? A g(B) >? A g(B) >? B g(C) >? A g(C) >? B g(C) >? C foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), B) >? f(triple(X, Y, Z), A) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) f!450!450(triple(X, Y, Z)) >? foldf(triple(X, Y, nil), Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 0 B = 1 C = 2 cons = \y0y1.y1 + 2y0 f = \y0y1.y0 + 2y1 f!450 = \y0y1.y0 + 2y1 f!450!450 = \y0.y0 foldf = \y0y1.y0 + y1 g = \y0.y0 nil = 0 triple = \y0y1y2.y1 + y2 + 2y0 Using this interpretation, the requirements translate to: [[g(A)]] = 0 >= 0 = [[A]] [[g(B)]] = 1 > 0 = [[A]] [[g(B)]] = 1 >= 1 = [[B]] [[g(C)]] = 2 > 0 = [[A]] [[g(C)]] = 2 > 1 = [[B]] [[g(C)]] = 2 >= 2 = [[C]] [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = x0 + x2 + 2x1 >= x0 + x2 + 2x1 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = 4 + x1 + x2 + 2x0 >= 4 + x1 + x2 + 2x0 = [[triple(_x0, _x1, cons(C, _x2))]] [[f!450(triple(_x0, _x1, _x2), B)]] = 2 + x1 + x2 + 2x0 > x1 + x2 + 2x0 = [[f(triple(_x0, _x1, _x2), A)]] [[f!450(triple(_x0, _x1, _x2), A)]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] [[f!450!450(triple(_x0, _x1, _x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[foldf(triple(_x0, _x1, nil), _x2)]] We can thus remove the following rules: g(B) => A g(C) => A g(C) => B f!450(triple(X, Y, Z), B) => f(triple(X, Y, Z), A) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): g(A) >? A g(B) >? B g(C) >? C foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) f!450!450(triple(X, Y, Z)) >? foldf(triple(X, Y, nil), Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 0 B = 1 C = 0 cons = \y0y1.y1 + 2y0 f = \y0y1.y0 + 2y1 f!450 = \y0y1.y0 + y1 f!450!450 = \y0.y0 foldf = \y0y1.y0 + y1 g = \y0.2y0 nil = 0 triple = \y0y1y2.y1 + y2 + 2y0 Using this interpretation, the requirements translate to: [[g(A)]] = 0 >= 0 = [[A]] [[g(B)]] = 2 > 1 = [[B]] [[g(C)]] = 0 >= 0 = [[C]] [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = x0 + x2 + 2x1 >= x0 + x2 + 2x1 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[triple(_x0, _x1, cons(C, _x2))]] [[f!450(triple(_x0, _x1, _x2), A)]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] [[f!450!450(triple(_x0, _x1, _x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[foldf(triple(_x0, _x1, nil), _x2)]] We can thus remove the following rules: g(B) => B We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): g(A) >? A g(C) >? C foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) f!450!450(triple(X, Y, Z)) >? foldf(triple(X, Y, nil), Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 2 C = 0 cons = \y0y1.y0 + y1 f = \y0y1.y0 + 2y1 f!450 = \y0y1.y0 + y1 f!450!450 = \y0.y0 foldf = \y0y1.y0 + 2y1 g = \y0.2y0 nil = 0 triple = \y0y1y2.y0 + 2y1 + 2y2 Using this interpretation, the requirements translate to: [[g(A)]] = 4 > 2 = [[A]] [[g(C)]] = 0 >= 0 = [[C]] [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[triple(_x0, _x1, cons(C, _x2))]] [[f!450(triple(_x0, _x1, _x2), A)]] = 2 + x0 + 2x1 + 2x2 >= 2 + x0 + 2x1 + 2x2 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] [[f!450!450(triple(_x0, _x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[foldf(triple(_x0, _x1, nil), _x2)]] We can thus remove the following rules: g(A) => A We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): g(C) >? C foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) f!450!450(triple(X, Y, Z)) >? foldf(triple(X, Y, nil), Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 0 C = 0 cons = \y0y1.1 + y1 + 2y0 f = \y0y1.2 + y0 + 2y1 f!450 = \y0y1.2 + y0 + 2y1 f!450!450 = \y0.1 + y0 foldf = \y0y1.y0 + 2y1 g = \y0.y0 nil = 0 triple = \y0y1y2.y0 + 2y1 + 2y2 Using this interpretation, the requirements translate to: [[g(C)]] = 0 >= 0 = [[C]] [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = 2 + x0 + 2x2 + 4x1 >= 2 + x0 + 2x1 + 2x2 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = 2 + x0 + 2x1 >= 2 + x0 + 2x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = 2 + x0 + 2x1 + 2x2 >= 2 + x0 + 2x1 + 2x2 = [[triple(_x0, _x1, cons(C, _x2))]] [[f!450(triple(_x0, _x1, _x2), A)]] = 2 + x0 + 2x1 + 2x2 >= 2 + x0 + 2x1 + 2x2 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] [[f!450!450(triple(_x0, _x1, _x2))]] = 1 + x0 + 2x1 + 2x2 > x0 + 2x1 + 2x2 = [[foldf(triple(_x0, _x1, nil), _x2)]] We can thus remove the following rules: f!450!450(triple(X, Y, Z)) => foldf(triple(X, Y, nil), 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]): g(C) >? C foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 0 C = 2 cons = \y0y1.y0 + y1 f = \y0y1.y0 + 2y1 f!450 = \y0y1.y0 + y1 f!450!450 = \y0.y0 foldf = \y0y1.y0 + 2y1 g = \y0.2y0 nil = 0 triple = \y0y1y2.y2 + 2y0 + 2y1 Using this interpretation, the requirements translate to: [[g(C)]] = 4 > 2 = [[C]] [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[triple(_x0, _x1, cons(C, _x2))]] [[f!450(triple(_x0, _x1, _x2), A)]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] We can thus remove the following rules: g(C) => C We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), C) >? triple(X, Y, cons(C, Z)) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 0 C = 2 cons = \y0y1.y0 + y1 f = \y0y1.y0 + 2y1 f!450 = \y0y1.y0 + 2y1 f!450!450 = \y0.y0 foldf = \y0y1.y0 + 2y1 g = \y0.y0 nil = 0 triple = \y0y1y2.y2 + 2y0 + 2y1 Using this interpretation, the requirements translate to: [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), C)]] = 4 + x2 + 2x0 + 2x1 > 2 + x2 + 2x0 + 2x1 = [[triple(_x0, _x1, cons(C, _x2))]] [[f!450(triple(_x0, _x1, _x2), A)]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] We can thus remove the following rules: f!450(triple(X, Y, Z), C) => triple(X, Y, cons(C, 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]): foldf(X, nil) >? X foldf(X, cons(Y, Z)) >? f(foldf(X, Z), Y) f(X, Y) >? f!450(X, g(Y)) f!450(triple(X, Y, Z), A) >? f!450!450(foldf(triple(cons(A, X), nil, Z), Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: A = 0 cons = \y0y1.1 + y0 + y1 f = \y0y1.2 + y0 + 2y1 f!450 = \y0y1.2 + y0 + y1 f!450!450 = \y0.y0 foldf = \y0y1.y0 + 3y1 g = \y0.y0 nil = 0 triple = \y0y1y2.y0 + y2 + 3y1 Using this interpretation, the requirements translate to: [[foldf(_x0, nil)]] = x0 >= x0 = [[_x0]] [[foldf(_x0, cons(_x1, _x2))]] = 3 + x0 + 3x1 + 3x2 > 2 + x0 + 2x1 + 3x2 = [[f(foldf(_x0, _x2), _x1)]] [[f(_x0, _x1)]] = 2 + x0 + 2x1 >= 2 + x0 + x1 = [[f!450(_x0, g(_x1))]] [[f!450(triple(_x0, _x1, _x2), A)]] = 2 + x0 + x2 + 3x1 > 1 + x0 + x2 + 3x1 = [[f!450!450(foldf(triple(cons(A, _x0), nil, _x2), _x1))]] We can thus remove the following rules: foldf(X, cons(Y, Z)) => f(foldf(X, Z), Y) f!450(triple(X, Y, Z), A) => f!450!450(foldf(triple(cons(A, X), nil, Z), 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]): foldf(X, nil) >? X f(X, Y) >? f!450(X, g(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0y1.3 + 3y0 + 3y1 f!450 = \y0y1.y0 + y1 foldf = \y0y1.3 + y0 + y1 g = \y0.y0 nil = 3 Using this interpretation, the requirements translate to: [[foldf(_x0, nil)]] = 6 + x0 > x0 = [[_x0]] [[f(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[f!450(_x0, g(_x1))]] We can thus remove the following rules: foldf(X, nil) => X f(X, Y) => f!450(X, g(Y)) 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.