/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] --> o U12 : [o * o * o] --> o a!6220!6220U11 : [o * o * o] --> o a!6220!6220U12 : [o * o * o] --> o a!6220!6220plus : [o * o] --> o mark : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> 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!6220plus(X, 0) => mark(X) a!6220!6220plus(X, s(Y)) => a!6220!6220U11(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(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) 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!6220plus(X, 0) >? mark(X) a!6220!6220plus(X, s(Y)) >? a!6220!6220U11(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(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 U11 = \y0y1y2.y0 + y2 + 2y1 U12 = \y0y1y2.y2 + 2y0 + 2y1 a!6220!6220U11 = \y0y1y2.y0 + y2 + 2y1 a!6220!6220U12 = \y0y1y2.y2 + 2y0 + 2y1 a!6220!6220plus = \y0y1.y0 + 2y1 mark = \y0.y0 plus = \y0y1.y0 + 2y1 s = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[a!6220!6220U12(tt, _x0, _x1)]] [[a!6220!6220U12(tt, _x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[s(a!6220!6220plus(mark(_x1), mark(_x0)))]] [[a!6220!6220plus(_x0, 0)]] = 2 + x0 > x0 = [[mark(_x0)]] [[a!6220!6220plus(_x0, s(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[a!6220!6220U11(tt, _x1, _x0)]] [[mark(U11(_x0, _x1, _x2))]] = x0 + x2 + 2x1 >= x0 + x2 + 2x1 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[a!6220!6220plus(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 + x2 + 2x1 >= x0 + x2 + 2x1 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: a!6220!6220plus(X, 0) => mark(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!6220plus(X, s(Y)) >? a!6220!6220U11(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(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.y0 + 2y1 + 2y2 U12 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220U11 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220U12 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220plus = \y0y1.2y0 + 2y1 mark = \y0.y0 plus = \y0y1.2y0 + 2y1 s = \y0.1 + y0 tt = 1 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[a!6220!6220U12(tt, _x0, _x1)]] [[a!6220!6220U12(tt, _x0, _x1)]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[s(a!6220!6220plus(mark(_x1), mark(_x0)))]] [[a!6220!6220plus(_x0, s(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[a!6220!6220U11(tt, _x1, _x0)]] [[mark(U11(_x0, _x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(tt)]] = 1 >= 1 = [[tt]] [[mark(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[a!6220!6220U11(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: a!6220!6220plus(X, s(Y)) => a!6220!6220U11(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))) 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(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 2 U11 = \y0y1y2.y0 + 2y1 + 2y2 U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220U11 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220U12 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220plus = \y0y1.y0 + y1 mark = \y0.2y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220U12(tt, _x0, _x1)]] [[a!6220!6220U12(tt, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[s(a!6220!6220plus(mark(_x1), mark(_x0)))]] [[mark(U11(_x0, _x1, _x2))]] = 2x0 + 4x1 + 4x2 >= 2x0 + 2x1 + 2x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[mark(0)]] = 4 > 2 = [[0]] [[a!6220!6220U11(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + x1 + x2 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: mark(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))) 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(tt) >? tt mark(s(X)) >? s(mark(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.y1 + y2 + 2y0 U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220U11 = \y0y1y2.y1 + y2 + 2y0 a!6220!6220U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220plus = \y0y1.y0 + y1 mark = \y0.y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 1 Using this interpretation, the requirements translate to: [[a!6220!6220U11(tt, _x0, _x1)]] = 2 + 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)))]] [[mark(U11(_x0, _x1, _x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[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(tt)]] = 1 >= 1 = [[tt]] [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] [[a!6220!6220U11(_x0, _x1, _x2)]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[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)]] We can thus remove the following rules: 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))) 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(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(tt) >? tt mark(s(X)) >? s(mark(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) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.y0 + 2y1 + 2y2 U12 = \y0y1y2.1 + y2 + 2y0 + 2y1 a!6220!6220U11 = \y0y1y2.y0 + 2y1 + 2y2 a!6220!6220U12 = \y0y1y2.2 + 2y0 + 2y1 + 2y2 a!6220!6220plus = \y0y1.y0 + y1 mark = \y0.2y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 2 Using this interpretation, the requirements translate to: [[mark(U11(_x0, _x1, _x2))]] = 2x0 + 4x1 + 4x2 >= 2x0 + 2x1 + 2x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = 2 + 2x2 + 4x0 + 4x1 >= 2 + 2x1 + 2x2 + 4x0 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(tt)]] = 4 > 2 = [[tt]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220U11(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220U12(_x0, _x1, _x2)]] = 2 + 2x0 + 2x1 + 2x2 > 1 + x2 + 2x0 + 2x1 = [[U12(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: mark(tt) => tt 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]): 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(s(X)) >? s(mark(X)) a!6220!6220U11(X, Y, Z) >? U11(X, Y, Z) a!6220!6220plus(X, Y) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.2 + y0 + y1 + y2 U12 = \y0y1y2.3 + y1 + y2 + 3y0 a!6220!6220U11 = \y0y1y2.3 + y0 + y1 + 2y2 a!6220!6220U12 = \y0y1y2.y0 + y1 + y2 a!6220!6220plus = \y0y1.y0 + y1 mark = \y0.2y0 plus = \y0y1.y0 + y1 s = \y0.y0 Using this interpretation, the requirements translate to: [[mark(U11(_x0, _x1, _x2))]] = 4 + 2x0 + 2x1 + 2x2 > 3 + x1 + 2x0 + 2x2 = [[a!6220!6220U11(mark(_x0), _x1, _x2)]] [[mark(U12(_x0, _x1, _x2))]] = 6 + 2x1 + 2x2 + 6x0 > x1 + x2 + 2x0 = [[a!6220!6220U12(mark(_x0), _x1, _x2)]] [[mark(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220U11(_x0, _x1, _x2)]] = 3 + x0 + x1 + 2x2 > 2 + x0 + x1 + x2 = [[U11(_x0, _x1, _x2)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: mark(U11(X, Y, Z)) => a!6220!6220U11(mark(X), Y, Z) mark(U12(X, Y, Z)) => a!6220!6220U12(mark(X), Y, Z) a!6220!6220U11(X, Y, Z) => U11(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(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(s(X)) >? s(mark(X)) a!6220!6220plus(X, Y) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220plus = \y0y1.y1 + 2y0 mark = \y0.2y0 plus = \y0y1.y1 + 2y0 s = \y0.2 + y0 Using this interpretation, the requirements translate to: [[mark(plus(_x0, _x1))]] = 2x1 + 4x0 >= 2x1 + 4x0 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(s(_x0))]] = 4 + 2x0 > 2 + 2x0 = [[s(mark(_x0))]] [[a!6220!6220plus(_x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, _x1)]] We can thus remove the following rules: mark(s(X)) => s(mark(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]): mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) a!6220!6220plus(X, Y) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220plus = \y0y1.1 + y0 + y1 mark = \y0.2y0 plus = \y0y1.1 + y0 + y1 Using this interpretation, the requirements translate to: [[mark(plus(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[a!6220!6220plus(_x0, _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: mark(plus(X, Y)) => a!6220!6220plus(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!6220plus(X, Y) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220plus = \y0y1.3 + 3y0 + 3y1 plus = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[a!6220!6220plus(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: a!6220!6220plus(X, Y) => plus(X, 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.