/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. !dot : [o * o] --> o !plus!plus : [o * o] --> o make : [o] --> o nil : [] --> o rev : [o] --> o rev(nil) => nil rev(rev(X)) => X rev(!plus!plus(X, Y)) => !plus!plus(rev(Y), rev(X)) !plus!plus(nil, X) => X !plus!plus(X, nil) => X !plus!plus(!dot(X, Y), Z) => !dot(X, !plus!plus(Y, Z)) !plus!plus(X, !plus!plus(Y, Z)) => !plus!plus(!plus!plus(X, Y), Z) make(X) => !dot(X, nil) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): rev(nil) >? nil rev(rev(X)) >? X rev(!plus!plus(X, Y)) >? !plus!plus(rev(Y), rev(X)) !plus!plus(nil, X) >? X !plus!plus(X, nil) >? X !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) !plus!plus(X, !plus!plus(Y, Z)) >? !plus!plus(!plus!plus(X, Y), Z) make(X) >? !dot(X, nil) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.y1 + 2y0 !plus!plus = \y0y1.1 + y0 + y1 make = \y0.3 + 3y0 nil = 1 rev = \y0.2y0 Using this interpretation, the requirements translate to: [[rev(nil)]] = 2 > 1 = [[nil]] [[rev(rev(_x0))]] = 4x0 >= x0 = [[_x0]] [[rev(!plus!plus(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[!plus!plus(rev(_x1), rev(_x0))]] [[!plus!plus(nil, _x0)]] = 2 + x0 > x0 = [[_x0]] [[!plus!plus(_x0, nil)]] = 2 + x0 > x0 = [[_x0]] [[!plus!plus(!dot(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[!dot(_x0, !plus!plus(_x1, _x2))]] [[!plus!plus(_x0, !plus!plus(_x1, _x2))]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[!plus!plus(!plus!plus(_x0, _x1), _x2)]] [[make(_x0)]] = 3 + 3x0 > 1 + 2x0 = [[!dot(_x0, nil)]] We can thus remove the following rules: rev(nil) => nil rev(!plus!plus(X, Y)) => !plus!plus(rev(Y), rev(X)) !plus!plus(nil, X) => X !plus!plus(X, nil) => X make(X) => !dot(X, nil) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): rev(rev(X)) >? X !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) !plus!plus(X, !plus!plus(Y, Z)) >? !plus!plus(!plus!plus(X, Y), Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.y0 + y1 !plus!plus = \y0y1.3 + y0 + 3y1 rev = \y0.3 + y0 Using this interpretation, the requirements translate to: [[rev(rev(_x0))]] = 6 + x0 > x0 = [[_x0]] [[!plus!plus(!dot(_x0, _x1), _x2)]] = 3 + x0 + x1 + 3x2 >= 3 + x0 + x1 + 3x2 = [[!dot(_x0, !plus!plus(_x1, _x2))]] [[!plus!plus(_x0, !plus!plus(_x1, _x2))]] = 12 + x0 + 3x1 + 9x2 > 6 + x0 + 3x1 + 3x2 = [[!plus!plus(!plus!plus(_x0, _x1), _x2)]] We can thus remove the following rules: rev(rev(X)) => X !plus!plus(X, !plus!plus(Y, Z)) => !plus!plus(!plus!plus(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]): !plus!plus(!dot(X, Y), Z) >? !dot(X, !plus!plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.1 + y0 + y1 !plus!plus = \y0y1.y1 + 3y0 Using this interpretation, the requirements translate to: [[!plus!plus(!dot(_x0, _x1), _x2)]] = 3 + x2 + 3x0 + 3x1 > 1 + x0 + x2 + 3x1 = [[!dot(_x0, !plus!plus(_x1, _x2))]] We can thus remove the following rules: !plus!plus(!dot(X, Y), Z) => !dot(X, !plus!plus(Y, Z)) 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.