/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. !plus : [o * o] --> o p1 : [] --> o p10 : [] --> o p2 : [] --> o p5 : [] --> o !plus(p1, p1) => p2 !plus(p1, !plus(p2, p2)) => p5 !plus(p5, p5) => p10 !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !plus(p1, !plus(p1, X)) => !plus(p2, X) !plus(p1, !plus(p2, !plus(p2, X))) => !plus(p5, X) !plus(p2, p1) => !plus(p1, p2) !plus(p2, !plus(p1, X)) => !plus(p1, !plus(p2, X)) !plus(p2, !plus(p2, p2)) => !plus(p1, p5) !plus(p2, !plus(p2, !plus(p2, X))) => !plus(p1, !plus(p5, X)) !plus(p5, p1) => !plus(p1, p5) !plus(p5, !plus(p1, X)) => !plus(p1, !plus(p5, X)) !plus(p5, p2) => !plus(p2, p5) !plus(p5, !plus(p2, X)) => !plus(p2, !plus(p5, X)) !plus(p5, !plus(p5, X)) => !plus(p10, X) !plus(p10, p1) => !plus(p1, p10) !plus(p10, !plus(p1, X)) => !plus(p1, !plus(p10, X)) !plus(p10, p2) => !plus(p2, p10) !plus(p10, !plus(p2, X)) => !plus(p2, !plus(p10, X)) !plus(p10, p5) => !plus(p5, p10) !plus(p10, !plus(p5, X)) => !plus(p5, !plus(p10, 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]): !plus(p1, p1) >? p2 !plus(p1, !plus(p2, p2)) >? p5 !plus(p5, p5) >? p10 !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !plus(p1, !plus(p1, X)) >? !plus(p2, X) !plus(p1, !plus(p2, !plus(p2, X))) >? !plus(p5, X) !plus(p2, p1) >? !plus(p1, p2) !plus(p2, !plus(p1, X)) >? !plus(p1, !plus(p2, X)) !plus(p2, !plus(p2, p2)) >? !plus(p1, p5) !plus(p2, !plus(p2, !plus(p2, X))) >? !plus(p1, !plus(p5, X)) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, !plus(p1, X)) >? !plus(p1, !plus(p5, X)) !plus(p5, p2) >? !plus(p2, p5) !plus(p5, !plus(p2, X)) >? !plus(p2, !plus(p5, X)) !plus(p5, !plus(p5, X)) >? !plus(p10, X) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, !plus(p1, X)) >? !plus(p1, !plus(p10, X)) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, !plus(p2, X)) >? !plus(p2, !plus(p10, X)) !plus(p10, p5) >? !plus(p5, p10) !plus(p10, !plus(p5, X)) >? !plus(p5, !plus(p10, X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.2 + y0 + y1 p1 = 0 p10 = 0 p2 = 0 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(p1, p1)]] = 2 > 0 = [[p2]] [[!plus(p1, !plus(p2, p2))]] = 4 > 0 = [[p5]] [[!plus(p5, p5)]] = 2 > 0 = [[p10]] [[!plus(!plus(_x0, _x1), _x2)]] = 4 + x0 + x1 + x2 >= 4 + x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] [[!plus(p1, !plus(p1, _x0))]] = 4 + x0 > 2 + x0 = [[!plus(p2, _x0)]] [[!plus(p1, !plus(p2, !plus(p2, _x0)))]] = 6 + x0 > 2 + x0 = [[!plus(p5, _x0)]] [[!plus(p2, p1)]] = 2 >= 2 = [[!plus(p1, p2)]] [[!plus(p2, !plus(p1, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p1, !plus(p2, _x0))]] [[!plus(p2, !plus(p2, p2))]] = 4 > 2 = [[!plus(p1, p5)]] [[!plus(p2, !plus(p2, !plus(p2, _x0)))]] = 6 + x0 > 4 + x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p1)]] = 2 >= 2 = [[!plus(p1, p5)]] [[!plus(p5, !plus(p1, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p2)]] = 2 >= 2 = [[!plus(p2, p5)]] [[!plus(p5, !plus(p2, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p2, !plus(p5, _x0))]] [[!plus(p5, !plus(p5, _x0))]] = 4 + x0 > 2 + x0 = [[!plus(p10, _x0)]] [[!plus(p10, p1)]] = 2 >= 2 = [[!plus(p1, p10)]] [[!plus(p10, !plus(p1, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p1, !plus(p10, _x0))]] [[!plus(p10, p2)]] = 2 >= 2 = [[!plus(p2, p10)]] [[!plus(p10, !plus(p2, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p2, !plus(p10, _x0))]] [[!plus(p10, p5)]] = 2 >= 2 = [[!plus(p5, p10)]] [[!plus(p10, !plus(p5, _x0))]] = 4 + x0 >= 4 + x0 = [[!plus(p5, !plus(p10, _x0))]] We can thus remove the following rules: !plus(p1, p1) => p2 !plus(p1, !plus(p2, p2)) => p5 !plus(p5, p5) => p10 !plus(p1, !plus(p1, X)) => !plus(p2, X) !plus(p1, !plus(p2, !plus(p2, X))) => !plus(p5, X) !plus(p2, !plus(p2, p2)) => !plus(p1, p5) !plus(p2, !plus(p2, !plus(p2, X))) => !plus(p1, !plus(p5, X)) !plus(p5, !plus(p5, X)) => !plus(p10, 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]): !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !plus(p2, p1) >? !plus(p1, p2) !plus(p2, !plus(p1, X)) >? !plus(p1, !plus(p2, X)) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, !plus(p1, X)) >? !plus(p1, !plus(p5, X)) !plus(p5, p2) >? !plus(p2, p5) !plus(p5, !plus(p2, X)) >? !plus(p2, !plus(p5, X)) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, !plus(p1, X)) >? !plus(p1, !plus(p10, X)) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, !plus(p2, X)) >? !plus(p2, !plus(p10, X)) !plus(p10, p5) >? !plus(p5, p10) !plus(p10, !plus(p5, X)) >? !plus(p5, !plus(p10, X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.1 + y1 + 2y0 p1 = 0 p10 = 0 p2 = 0 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(!plus(_x0, _x1), _x2)]] = 3 + x2 + 2x1 + 4x0 > 2 + x2 + 2x0 + 2x1 = [[!plus(_x0, !plus(_x1, _x2))]] [[!plus(p2, p1)]] = 1 >= 1 = [[!plus(p1, p2)]] [[!plus(p2, !plus(p1, _x0))]] = 2 + x0 >= 2 + x0 = [[!plus(p1, !plus(p2, _x0))]] [[!plus(p5, p1)]] = 1 >= 1 = [[!plus(p1, p5)]] [[!plus(p5, !plus(p1, _x0))]] = 2 + x0 >= 2 + x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p2)]] = 1 >= 1 = [[!plus(p2, p5)]] [[!plus(p5, !plus(p2, _x0))]] = 2 + x0 >= 2 + x0 = [[!plus(p2, !plus(p5, _x0))]] [[!plus(p10, p1)]] = 1 >= 1 = [[!plus(p1, p10)]] [[!plus(p10, !plus(p1, _x0))]] = 2 + x0 >= 2 + x0 = [[!plus(p1, !plus(p10, _x0))]] [[!plus(p10, p2)]] = 1 >= 1 = [[!plus(p2, p10)]] [[!plus(p10, !plus(p2, _x0))]] = 2 + x0 >= 2 + x0 = [[!plus(p2, !plus(p10, _x0))]] [[!plus(p10, p5)]] = 1 >= 1 = [[!plus(p5, p10)]] [[!plus(p10, !plus(p5, _x0))]] = 2 + x0 >= 2 + x0 = [[!plus(p5, !plus(p10, _x0))]] We can thus remove the following rules: !plus(!plus(X, Y), Z) => !plus(X, !plus(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]): !plus(p2, p1) >? !plus(p1, p2) !plus(p2, !plus(p1, X)) >? !plus(p1, !plus(p2, X)) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, !plus(p1, X)) >? !plus(p1, !plus(p5, X)) !plus(p5, p2) >? !plus(p2, p5) !plus(p5, !plus(p2, X)) >? !plus(p2, !plus(p5, X)) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, !plus(p1, X)) >? !plus(p1, !plus(p10, X)) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, !plus(p2, X)) >? !plus(p2, !plus(p10, X)) !plus(p10, p5) >? !plus(p5, p10) !plus(p10, !plus(p5, X)) >? !plus(p5, !plus(p10, X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.3y0 + 3y1 p1 = 3 p10 = 2 p2 = 3 p5 = 3 Using this interpretation, the requirements translate to: [[!plus(p2, p1)]] = 18 >= 18 = [[!plus(p1, p2)]] [[!plus(p2, !plus(p1, _x0))]] = 36 + 9x0 >= 36 + 9x0 = [[!plus(p1, !plus(p2, _x0))]] [[!plus(p5, p1)]] = 18 >= 18 = [[!plus(p1, p5)]] [[!plus(p5, !plus(p1, _x0))]] = 36 + 9x0 >= 36 + 9x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p2)]] = 18 >= 18 = [[!plus(p2, p5)]] [[!plus(p5, !plus(p2, _x0))]] = 36 + 9x0 >= 36 + 9x0 = [[!plus(p2, !plus(p5, _x0))]] [[!plus(p10, p1)]] = 15 >= 15 = [[!plus(p1, p10)]] [[!plus(p10, !plus(p1, _x0))]] = 33 + 9x0 > 27 + 9x0 = [[!plus(p1, !plus(p10, _x0))]] [[!plus(p10, p2)]] = 15 >= 15 = [[!plus(p2, p10)]] [[!plus(p10, !plus(p2, _x0))]] = 33 + 9x0 > 27 + 9x0 = [[!plus(p2, !plus(p10, _x0))]] [[!plus(p10, p5)]] = 15 >= 15 = [[!plus(p5, p10)]] [[!plus(p10, !plus(p5, _x0))]] = 33 + 9x0 > 27 + 9x0 = [[!plus(p5, !plus(p10, _x0))]] We can thus remove the following rules: !plus(p10, !plus(p1, X)) => !plus(p1, !plus(p10, X)) !plus(p10, !plus(p2, X)) => !plus(p2, !plus(p10, X)) !plus(p10, !plus(p5, X)) => !plus(p5, !plus(p10, 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]): !plus(p2, p1) >? !plus(p1, p2) !plus(p2, !plus(p1, X)) >? !plus(p1, !plus(p2, X)) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, !plus(p1, X)) >? !plus(p1, !plus(p5, X)) !plus(p5, p2) >? !plus(p2, p5) !plus(p5, !plus(p2, X)) >? !plus(p2, !plus(p5, X)) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, p5) >? !plus(p5, p10) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.2y0 + 2y1 p1 = 1 p10 = 0 p2 = 0 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(p2, p1)]] = 2 >= 2 = [[!plus(p1, p2)]] [[!plus(p2, !plus(p1, _x0))]] = 4 + 4x0 > 2 + 4x0 = [[!plus(p1, !plus(p2, _x0))]] [[!plus(p5, p1)]] = 2 >= 2 = [[!plus(p1, p5)]] [[!plus(p5, !plus(p1, _x0))]] = 4 + 4x0 > 2 + 4x0 = [[!plus(p1, !plus(p5, _x0))]] [[!plus(p5, p2)]] = 0 >= 0 = [[!plus(p2, p5)]] [[!plus(p5, !plus(p2, _x0))]] = 4x0 >= 4x0 = [[!plus(p2, !plus(p5, _x0))]] [[!plus(p10, p1)]] = 2 >= 2 = [[!plus(p1, p10)]] [[!plus(p10, p2)]] = 0 >= 0 = [[!plus(p2, p10)]] [[!plus(p10, p5)]] = 0 >= 0 = [[!plus(p5, p10)]] We can thus remove the following rules: !plus(p2, !plus(p1, X)) => !plus(p1, !plus(p2, X)) !plus(p5, !plus(p1, X)) => !plus(p1, !plus(p5, 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]): !plus(p2, p1) >? !plus(p1, p2) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, p2) >? !plus(p2, p5) !plus(p5, !plus(p2, X)) >? !plus(p2, !plus(p5, X)) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, p5) >? !plus(p5, p10) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.2y0 + 2y1 p1 = 0 p10 = 0 p2 = 1 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(p2, p1)]] = 2 >= 2 = [[!plus(p1, p2)]] [[!plus(p5, p1)]] = 0 >= 0 = [[!plus(p1, p5)]] [[!plus(p5, p2)]] = 2 >= 2 = [[!plus(p2, p5)]] [[!plus(p5, !plus(p2, _x0))]] = 4 + 4x0 > 2 + 4x0 = [[!plus(p2, !plus(p5, _x0))]] [[!plus(p10, p1)]] = 0 >= 0 = [[!plus(p1, p10)]] [[!plus(p10, p2)]] = 2 >= 2 = [[!plus(p2, p10)]] [[!plus(p10, p5)]] = 0 >= 0 = [[!plus(p5, p10)]] We can thus remove the following rules: !plus(p5, !plus(p2, X)) => !plus(p2, !plus(p5, 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]): !plus(p2, p1) >? !plus(p1, p2) !plus(p5, p1) >? !plus(p1, p5) !plus(p5, p2) >? !plus(p2, p5) !plus(p10, p1) >? !plus(p1, p10) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, p5) >? !plus(p5, p10) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.y0 + 2y1 p1 = 1 p10 = 0 p2 = 0 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(p2, p1)]] = 2 > 1 = [[!plus(p1, p2)]] [[!plus(p5, p1)]] = 2 > 1 = [[!plus(p1, p5)]] [[!plus(p5, p2)]] = 0 >= 0 = [[!plus(p2, p5)]] [[!plus(p10, p1)]] = 2 > 1 = [[!plus(p1, p10)]] [[!plus(p10, p2)]] = 0 >= 0 = [[!plus(p2, p10)]] [[!plus(p10, p5)]] = 0 >= 0 = [[!plus(p5, p10)]] We can thus remove the following rules: !plus(p2, p1) => !plus(p1, p2) !plus(p5, p1) => !plus(p1, p5) !plus(p10, p1) => !plus(p1, p10) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): !plus(p5, p2) >? !plus(p2, p5) !plus(p10, p2) >? !plus(p2, p10) !plus(p10, p5) >? !plus(p5, p10) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.2y0 + 3y1 p10 = 0 p2 = 1 p5 = 0 Using this interpretation, the requirements translate to: [[!plus(p5, p2)]] = 3 > 2 = [[!plus(p2, p5)]] [[!plus(p10, p2)]] = 3 > 2 = [[!plus(p2, p10)]] [[!plus(p10, p5)]] = 0 >= 0 = [[!plus(p5, p10)]] We can thus remove the following rules: !plus(p5, p2) => !plus(p2, p5) !plus(p10, p2) => !plus(p2, p10) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): !plus(p10, p5) >? !plus(p5, p10) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.y0 + 2y1 p10 = 0 p5 = 2 Using this interpretation, the requirements translate to: [[!plus(p10, p5)]] = 4 > 2 = [[!plus(p5, p10)]] We can thus remove the following rules: !plus(p10, p5) => !plus(p5, p10) 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.