/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 active : [o] --> o mark : [o] --> o ok : [o] --> o plus : [o * o] --> o proper : [o] --> o s : [o] --> o top : [o] --> o tt : [] --> o active(U11(tt, X, Y)) => mark(U12(tt, X, Y)) active(U12(tt, X, Y)) => mark(s(plus(Y, X))) active(plus(X, 0)) => mark(X) active(plus(X, s(Y))) => mark(U11(tt, Y, X)) active(U11(X, Y, Z)) => U11(active(X), Y, Z) active(U12(X, Y, Z)) => U12(active(X), Y, Z) active(s(X)) => s(active(X)) active(plus(X, Y)) => plus(active(X), Y) active(plus(X, Y)) => plus(X, active(Y)) U11(mark(X), Y, Z) => mark(U11(X, Y, Z)) U12(mark(X), Y, Z) => mark(U12(X, Y, Z)) s(mark(X)) => mark(s(X)) plus(mark(X), Y) => mark(plus(X, Y)) plus(X, mark(Y)) => mark(plus(X, Y)) proper(U11(X, Y, Z)) => U11(proper(X), proper(Y), proper(Z)) proper(tt) => ok(tt) proper(U12(X, Y, Z)) => U12(proper(X), proper(Y), proper(Z)) proper(s(X)) => s(proper(X)) proper(plus(X, Y)) => plus(proper(X), proper(Y)) proper(0) => ok(0) U11(ok(X), ok(Y), ok(Z)) => ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) => ok(U12(X, Y, Z)) s(ok(X)) => ok(s(X)) plus(ok(X), ok(Y)) => ok(plus(X, Y)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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]): active(U11(tt, X, Y)) >? mark(U12(tt, X, Y)) active(U12(tt, X, Y)) >? mark(s(plus(Y, X))) active(plus(X, 0)) >? mark(X) active(plus(X, s(Y))) >? mark(U11(tt, Y, X)) active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) s(mark(X)) >? mark(s(X)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.2 + y2 + 2y0 + 2y1 U12 = \y0y1y2.2 + y2 + 2y0 + 2y1 active = \y0.y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y0 + 2y1 proper = \y0.y0 s = \y0.2 + y0 top = \y0.2y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U11(tt, _x0, _x1))]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[mark(U12(tt, _x0, _x1))]] [[active(U12(tt, _x0, _x1))]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[mark(s(plus(_x1, _x0)))]] [[active(plus(_x0, 0))]] = x0 >= x0 = [[mark(_x0)]] [[active(plus(_x0, s(_x1)))]] = 4 + x0 + 2x1 > 2 + x0 + 2x1 = [[mark(U11(tt, _x1, _x0))]] [[active(U11(_x0, _x1, _x2))]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 2 + x0 >= 2 + x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[mark(U12(_x0, _x1, _x2))]] [[s(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(s(_x0))]] [[plus(mark(_x0), _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[proper(U11(_x0, _x1, _x2))]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[U11(proper(_x0), proper(_x1), proper(_x2))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(U12(_x0, _x1, _x2))]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[U12(proper(_x0), proper(_x1), proper(_x2))]] [[proper(s(_x0))]] = 2 + x0 >= 2 + x0 = [[s(proper(_x0))]] [[proper(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = 2 + x2 + 2x0 + 2x1 >= 2 + x2 + 2x0 + 2x1 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 2 + x0 >= 2 + x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] We can thus remove the following rules: active(plus(X, s(Y))) => mark(U11(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]): active(U11(tt, X, Y)) >? mark(U12(tt, X, Y)) active(U12(tt, X, Y)) >? mark(s(plus(Y, X))) active(plus(X, 0)) >? mark(X) active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) s(mark(X)) >? mark(s(X)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) 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.y1 + y2 + 2y0 active = \y0.y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y0 + y1 proper = \y0.y0 s = \y0.y0 top = \y0.2y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U11(tt, _x0, _x1))]] = x0 + x1 >= x0 + x1 = [[mark(U12(tt, _x0, _x1))]] [[active(U12(tt, _x0, _x1))]] = x0 + x1 >= x0 + x1 = [[mark(s(plus(_x1, _x0)))]] [[active(plus(_x0, 0))]] = 1 + x0 > x0 = [[mark(_x0)]] [[active(U11(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[mark(U12(_x0, _x1, _x2))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[plus(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(plus(_x0, _x1))]] [[proper(U11(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U11(proper(_x0), proper(_x1), proper(_x2))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(U12(_x0, _x1, _x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[U12(proper(_x0), proper(_x1), proper(_x2))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 1 >= 1 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(plus(_x0, _x1))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] We can thus remove the following rules: active(plus(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]): active(U11(tt, X, Y)) >? mark(U12(tt, X, Y)) active(U12(tt, X, Y)) >? mark(s(plus(Y, X))) active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) s(mark(X)) >? mark(s(X)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.1 + y0 + 2y1 + 2y2 U12 = \y0y1y2.y0 + 2y1 + 2y2 active = \y0.y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.2y0 + 2y1 proper = \y0.y0 s = \y0.y0 top = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U11(tt, _x0, _x1))]] = 1 + 2x0 + 2x1 > 2x0 + 2x1 = [[mark(U12(tt, _x0, _x1))]] [[active(U12(tt, _x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(s(plus(_x1, _x0)))]] [[active(U11(_x0, _x1, _x2))]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[mark(U12(_x0, _x1, _x2))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[plus(mark(_x0), _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[proper(U11(_x0, _x1, _x2))]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[U11(proper(_x0), proper(_x1), proper(_x2))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(U12(_x0, _x1, _x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[U12(proper(_x0), proper(_x1), proper(_x2))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: active(U11(tt, X, Y)) => mark(U12(tt, 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]): active(U12(tt, X, Y)) >? mark(s(plus(Y, X))) active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) s(mark(X)) >? mark(s(X)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.y2 + 2y0 + 2y1 U12 = \y0y1y2.1 + y0 + 2y1 + 2y2 active = \y0.y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y0 + y1 proper = \y0.y0 s = \y0.2y0 top = \y0.2y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U12(tt, _x0, _x1))]] = 1 + 2x0 + 2x1 > 2x0 + 2x1 = [[mark(s(plus(_x1, _x0)))]] [[active(U11(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[mark(U12(_x0, _x1, _x2))]] [[s(mark(_x0))]] = 2x0 >= 2x0 = [[mark(s(_x0))]] [[plus(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(plus(_x0, _x1))]] [[proper(U11(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[U11(proper(_x0), proper(_x1), proper(_x2))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(U12(_x0, _x1, _x2))]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[U12(proper(_x0), proper(_x1), proper(_x2))]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[proper(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = 1 + x0 + 2x1 + 2x2 >= 1 + x0 + 2x1 + 2x2 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(plus(_x0, _x1))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] We can thus remove the following rules: active(U12(tt, X, Y)) => mark(s(plus(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) s(mark(X)) >? mark(s(X)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) 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 active = \y0.y0 mark = \y0.1 + y0 ok = \y0.y0 plus = \y0y1.y0 + y1 proper = \y0.y0 s = \y0.2y0 top = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[mark(U12(_x0, _x1, _x2))]] [[s(mark(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[mark(s(_x0))]] [[plus(mark(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[mark(plus(_x0, _x1))]] [[proper(U11(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U11(proper(_x0), proper(_x1), proper(_x2))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(U12(_x0, _x1, _x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[U12(proper(_x0), proper(_x1), proper(_x2))]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[proper(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(plus(_x0, _x1))]] [[top(mark(_x0))]] = 1 + x0 > x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: s(mark(X)) => mark(s(X)) top(mark(X)) => top(proper(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.1 + y0 + y1 + y2 U12 = \y0y1y2.1 + y0 + y1 + y2 active = \y0.y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.1 + y0 + y1 proper = \y0.3y0 s = \y0.2y0 top = \y0.y0 tt = 1 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[mark(U12(_x0, _x1, _x2))]] [[plus(mark(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[mark(plus(_x0, _x1))]] [[proper(U11(_x0, _x1, _x2))]] = 3 + 3x0 + 3x1 + 3x2 > 1 + 3x0 + 3x1 + 3x2 = [[U11(proper(_x0), proper(_x1), proper(_x2))]] [[proper(tt)]] = 3 > 1 = [[ok(tt)]] [[proper(U12(_x0, _x1, _x2))]] = 3 + 3x0 + 3x1 + 3x2 > 1 + 3x0 + 3x1 + 3x2 = [[U12(proper(_x0), proper(_x1), proper(_x2))]] [[proper(s(_x0))]] = 6x0 >= 6x0 = [[s(proper(_x0))]] [[proper(plus(_x0, _x1))]] = 3 + 3x0 + 3x1 > 1 + 3x0 + 3x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[ok(plus(_x0, _x1))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: proper(U11(X, Y, Z)) => U11(proper(X), proper(Y), proper(Z)) proper(tt) => ok(tt) proper(U12(X, Y, Z)) => U12(proper(X), proper(Y), proper(Z)) proper(plus(X, Y)) => plus(proper(X), proper(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(s(X)) >? s(proper(X)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(ok(X)) >? top(active(X)) 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.y1 + y2 + 2y0 active = \y0.2y0 mark = \y0.2 + y0 ok = \y0.2y0 plus = \y0y1.2y0 + 2y1 proper = \y0.3y0 s = \y0.y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= x1 + x2 + 2x0 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 2x1 + 2x2 + 4x0 >= x1 + x2 + 4x0 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 4x0 + 4x1 >= 2x1 + 4x0 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 4x0 + 4x1 >= 2x0 + 4x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[mark(U11(_x0, _x1, _x2))]] [[U12(mark(_x0), _x1, _x2)]] = 4 + x1 + x2 + 2x0 > 2 + x1 + x2 + 2x0 = [[mark(U12(_x0, _x1, _x2))]] [[plus(mark(_x0), _x1)]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[proper(s(_x0))]] = 3x0 >= 3x0 = [[s(proper(_x0))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = 2x1 + 2x2 + 4x0 >= 2x1 + 2x2 + 4x0 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = 4x0 + 4x1 >= 4x0 + 4x1 = [[ok(plus(_x0, _x1))]] [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] We can thus remove the following rules: U12(mark(X), Y, Z) => mark(U12(X, Y, Z)) plus(mark(X), Y) => mark(plus(X, Y)) plus(X, mark(Y)) => mark(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) proper(s(X)) >? s(proper(X)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(ok(X)) >? top(active(X)) 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.y0 + y1 + y2 active = \y0.2y0 mark = \y0.y0 ok = \y0.2y0 plus = \y0y1.y0 + 2y1 proper = \y0.3y0 s = \y0.2y0 top = \y0.2y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 2x0 + 2x2 + 4x1 >= x2 + 2x0 + 2x1 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= x1 + x2 + 2x0 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 4x0 >= 4x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 2x0 + 4x1 >= 2x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 2x0 + 4x1 >= x0 + 4x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = x0 + x2 + 2x1 >= x0 + x2 + 2x1 = [[mark(U11(_x0, _x1, _x2))]] [[proper(s(_x0))]] = 6x0 >= 6x0 = [[s(proper(_x0))]] [[proper(0)]] = 3 > 2 = [[ok(0)]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = 2x0 + 2x2 + 4x1 >= 2x0 + 2x2 + 4x1 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = 2x0 + 2x1 + 2x2 >= 2x0 + 2x1 + 2x2 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 4x0 >= 4x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = 2x0 + 4x1 >= 2x0 + 4x1 = [[ok(plus(_x0, _x1))]] [[top(ok(_x0))]] = 4x0 >= 4x0 = [[top(active(_x0))]] We can thus remove the following rules: proper(0) => ok(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) proper(s(X)) >? s(proper(X)) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(ok(X)) >? top(active(X)) 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.y2 + 2y0 + 2y1 active = \y0.y0 mark = \y0.1 + y0 ok = \y0.y0 plus = \y0y1.1 + 2y0 + 2y1 proper = \y0.3y0 s = \y0.y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[plus(_x0, active(_x1))]] [[U11(mark(_x0), _x1, _x2)]] = 2 + x1 + x2 + 2x0 > 1 + x1 + x2 + 2x0 = [[mark(U11(_x0, _x1, _x2))]] [[proper(s(_x0))]] = 3x0 >= 3x0 = [[s(proper(_x0))]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = x1 + x2 + 2x0 >= x1 + x2 + 2x0 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = x2 + 2x0 + 2x1 >= x2 + 2x0 + 2x1 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: U11(mark(X), Y, Z) => mark(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) proper(s(X)) >? s(proper(X)) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.y0 + y1 + y2 U12 = \y0y1y2.2y0 + 2y1 + 2y2 active = \y0.2y0 ok = \y0.1 + 2y0 plus = \y0y1.y0 + y1 proper = \y0.y0 s = \y0.2y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 2x0 + 2x1 + 2x2 >= x1 + x2 + 2x0 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 4x0 + 4x1 + 4x2 >= 2x1 + 2x2 + 4x0 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 4x0 >= 4x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 2x0 + 2x1 >= x0 + 2x1 = [[plus(_x0, active(_x1))]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[U11(ok(_x0), ok(_x1), ok(_x2))]] = 3 + 2x0 + 2x1 + 2x2 > 1 + 2x0 + 2x1 + 2x2 = [[ok(U11(_x0, _x1, _x2))]] [[U12(ok(_x0), ok(_x1), ok(_x2))]] = 6 + 4x0 + 4x1 + 4x2 > 1 + 4x0 + 4x1 + 4x2 = [[ok(U12(_x0, _x1, _x2))]] [[s(ok(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[ok(s(_x0))]] [[plus(ok(_x0), ok(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[top(ok(_x0))]] = 1 + 2x0 > 2x0 = [[top(active(_x0))]] We can thus remove the following rules: U11(ok(X), ok(Y), ok(Z)) => ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) => ok(U12(X, Y, Z)) s(ok(X)) => ok(s(X)) plus(ok(X), ok(Y)) => ok(plus(X, Y)) top(ok(X)) => top(active(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) proper(s(X)) >? s(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.y0 + y1 + y2 U12 = \y0y1y2.y0 + y1 + y2 active = \y0.3y0 plus = \y0y1.3 + y0 + y1 proper = \y0.3y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 3x0 + 3x1 + 3x2 >= x1 + x2 + 3x0 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 3x0 + 3x1 + 3x2 >= x1 + x2 + 3x0 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 3x0 >= 3x0 = [[s(active(_x0))]] [[active(plus(_x0, _x1))]] = 9 + 3x0 + 3x1 > 3 + x1 + 3x0 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 9 + 3x0 + 3x1 > 3 + x0 + 3x1 = [[plus(_x0, active(_x1))]] [[proper(s(_x0))]] = 3x0 >= 3x0 = [[s(proper(_x0))]] We can thus remove the following rules: active(plus(X, Y)) => plus(active(X), Y) active(plus(X, Y)) => plus(X, active(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) proper(s(X)) >? s(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.y0 + y1 + y2 U12 = \y0y1y2.y0 + y1 + y2 active = \y0.3y0 proper = \y0.3y0 s = \y0.2 + y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 3x0 + 3x1 + 3x2 >= x1 + x2 + 3x0 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 3x0 + 3x1 + 3x2 >= x1 + x2 + 3x0 = [[U12(active(_x0), _x1, _x2)]] [[active(s(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[s(active(_x0))]] [[proper(s(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[s(proper(_x0))]] We can thus remove the following rules: active(s(X)) => s(active(X)) proper(s(X)) => s(proper(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]): active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) 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.y0 + y1 + y2 active = \y0.3y0 Using this interpretation, the requirements translate to: [[active(U11(_x0, _x1, _x2))]] = 6 + 3x0 + 3x1 + 3x2 > 2 + x1 + x2 + 3x0 = [[U11(active(_x0), _x1, _x2)]] [[active(U12(_x0, _x1, _x2))]] = 3x0 + 3x1 + 3x2 >= x1 + x2 + 3x0 = [[U12(active(_x0), _x1, _x2)]] We can thus remove the following rules: active(U11(X, Y, Z)) => U11(active(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]): active(U12(X, Y, Z)) >? U12(active(X), Y, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U12 = \y0y1y2.1 + y0 + y1 + y2 active = \y0.3y0 Using this interpretation, the requirements translate to: [[active(U12(_x0, _x1, _x2))]] = 3 + 3x0 + 3x1 + 3x2 > 1 + x1 + x2 + 3x0 = [[U12(active(_x0), _x1, _x2)]] We can thus remove the following rules: active(U12(X, Y, Z)) => U12(active(X), 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.