/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 active : [o] --> o and : [o * o] --> o mark : [o] --> o ok : [o] --> o plus : [o * o] --> o proper : [o] --> o s : [o] --> o top : [o] --> o tt : [] --> o active(and(tt, X)) => mark(X) active(plus(X, 0)) => mark(X) active(plus(X, s(Y))) => mark(s(plus(X, Y))) active(and(X, Y)) => and(active(X), Y) active(plus(X, Y)) => plus(active(X), Y) active(plus(X, Y)) => plus(X, active(Y)) active(s(X)) => s(active(X)) and(mark(X), Y) => mark(and(X, Y)) plus(mark(X), Y) => mark(plus(X, Y)) plus(X, mark(Y)) => mark(plus(X, Y)) s(mark(X)) => mark(s(X)) proper(and(X, Y)) => and(proper(X), proper(Y)) proper(tt) => ok(tt) proper(plus(X, Y)) => plus(proper(X), proper(Y)) proper(0) => ok(0) proper(s(X)) => s(proper(X)) and(ok(X), ok(Y)) => ok(and(X, Y)) plus(ok(X), ok(Y)) => ok(plus(X, Y)) s(ok(X)) => ok(s(X)) 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(and(tt, X)) >? mark(X) active(plus(X, 0)) >? mark(X) active(plus(X, s(Y))) >? mark(s(plus(X, Y))) active(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) and(mark(X), Y) >? mark(and(X, Y)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) s(mark(X)) >? mark(s(X)) proper(and(X, Y)) >? and(proper(X), proper(Y)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(X)) 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 active = \y0.y0 and = \y0y1.1 + y1 + 2y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y1 + 2y0 proper = \y0.y0 s = \y0.y0 top = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[active(and(tt, _x0))]] = 1 + x0 > x0 = [[mark(_x0)]] [[active(plus(_x0, 0))]] = 2x0 >= x0 = [[mark(_x0)]] [[active(plus(_x0, s(_x1)))]] = x1 + 2x0 >= x1 + 2x0 = [[mark(s(plus(_x0, _x1)))]] [[active(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[and(mark(_x0), _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(and(_x0, _x1))]] [[plus(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[mark(plus(_x0, _x1))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[proper(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(plus(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: active(and(tt, X)) => 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(plus(X, 0)) >? mark(X) active(plus(X, s(Y))) >? mark(s(plus(X, Y))) active(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) and(mark(X), Y) >? mark(and(X, Y)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) s(mark(X)) >? mark(s(X)) proper(and(X, Y)) >? and(proper(X), proper(Y)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(X)) 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 active = \y0.y0 and = \y0y1.y1 + 2y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y0 + 2y1 proper = \y0.y0 s = \y0.1 + y0 top = \y0.2y0 tt = 0 Using this interpretation, the requirements translate to: [[active(plus(_x0, 0))]] = x0 >= x0 = [[mark(_x0)]] [[active(plus(_x0, s(_x1)))]] = 2 + x0 + 2x1 > 1 + x0 + 2x1 = [[mark(s(plus(_x0, _x1)))]] [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(active(_x0))]] [[and(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(and(_x0, _x1))]] [[plus(mark(_x0), _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[s(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(s(_x0))]] [[proper(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(plus(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = 1 + x0 >= 1 + x0 = [[ok(s(_x0))]] [[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(s(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(plus(X, 0)) >? mark(X) active(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) and(mark(X), Y) >? mark(and(X, Y)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) s(mark(X)) >? mark(s(X)) proper(and(X, Y)) >? and(proper(X), proper(Y)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(X)) 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 active = \y0.y0 and = \y0y1.y1 + 2y0 mark = \y0.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(plus(_x0, 0))]] = 1 + x0 > x0 = [[mark(_x0)]] [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[and(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(and(_x0, _x1))]] [[plus(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(plus(_x0, _x1))]] [[s(mark(_x0))]] = 2x0 >= 2x0 = [[mark(s(_x0))]] [[proper(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 1 >= 1 = [[ok(0)]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[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(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) and(mark(X), Y) >? mark(and(X, Y)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) s(mark(X)) >? mark(s(X)) proper(and(X, Y)) >? and(proper(X), proper(Y)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(X)) 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 active = \y0.y0 and = \y0y1.y1 + 2y0 mark = \y0.1 + y0 ok = \y0.y0 plus = \y0y1.2y0 + 2y1 proper = \y0.y0 s = \y0.2y0 top = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[and(mark(_x0), _x1)]] = 2 + x1 + 2x0 > 1 + x1 + 2x0 = [[mark(and(_x0, _x1))]] [[plus(mark(_x0), _x1)]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[mark(plus(_x0, _x1))]] [[s(mark(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[mark(s(_x0))]] [[proper(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[top(mark(_x0))]] = 1 + x0 > x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: and(mark(X), Y) => mark(and(X, Y)) plus(mark(X), Y) => mark(plus(X, Y)) plus(X, mark(Y)) => mark(plus(X, Y)) 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(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) proper(and(X, Y)) >? and(proper(X), proper(Y)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(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 active = \y0.y0 and = \y0y1.1 + 2y0 + 2y1 ok = \y0.y0 plus = \y0y1.y0 + y1 proper = \y0.2y0 s = \y0.y0 top = \y0.2y0 tt = 0 Using this interpretation, the requirements translate to: [[active(and(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[proper(and(_x0, _x1))]] = 2 + 4x0 + 4x1 > 1 + 4x0 + 4x1 = [[and(proper(_x0), proper(_x1))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]] [[proper(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = 1 + 2x0 + 2x1 >= 1 + 2x0 + 2x1 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[top(ok(_x0))]] = 2x0 >= 2x0 = [[top(active(_x0))]] We can thus remove the following rules: proper(and(X, Y)) => and(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(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(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 active = \y0.y0 and = \y0y1.y1 + 2y0 ok = \y0.y0 plus = \y0y1.y0 + y1 proper = \y0.2y0 s = \y0.y0 top = \y0.y0 tt = 2 Using this interpretation, the requirements translate to: [[active(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[proper(tt)]] = 4 > 2 = [[ok(tt)]] [[proper(plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = x0 + x1 >= x0 + x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: proper(tt) => ok(tt) 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(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(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 active = \y0.y0 and = \y0y1.1 + y0 + y1 ok = \y0.y0 plus = \y0y1.1 + y0 + 2y1 proper = \y0.2y0 s = \y0.2y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(and(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[proper(plus(_x0, _x1))]] = 2 + 2x0 + 4x1 > 1 + 2x0 + 4x1 = [[plus(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(s(_x0))]] = 4x0 >= 4x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: 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(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(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 = 2 active = \y0.2 + 2y0 and = \y0y1.3 + y0 + 3y1 ok = \y0.3 + 3y0 plus = \y0y1.3 + 2y0 + 2y1 proper = \y0.3 + 3y0 s = \y0.y0 top = \y0.3y0 Using this interpretation, the requirements translate to: [[active(and(_x0, _x1))]] = 8 + 2x0 + 6x1 > 5 + 2x0 + 3x1 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 8 + 4x0 + 4x1 > 7 + 2x1 + 4x0 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = 8 + 4x0 + 4x1 > 7 + 2x0 + 4x1 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[s(active(_x0))]] [[proper(0)]] = 9 >= 9 = [[ok(0)]] [[proper(s(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[s(proper(_x0))]] [[and(ok(_x0), ok(_x1))]] = 15 + 3x0 + 9x1 > 12 + 3x0 + 9x1 = [[ok(and(_x0, _x1))]] [[plus(ok(_x0), ok(_x1))]] = 15 + 6x0 + 6x1 > 12 + 6x0 + 6x1 = [[ok(plus(_x0, _x1))]] [[s(ok(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[ok(s(_x0))]] [[top(ok(_x0))]] = 9 + 9x0 > 6 + 6x0 = [[top(active(_x0))]] We can thus remove the following rules: active(and(X, Y)) => and(active(X), Y) active(plus(X, Y)) => plus(active(X), Y) active(plus(X, Y)) => plus(X, active(Y)) and(ok(X), ok(Y)) => ok(and(X, Y)) 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(s(X)) >? s(active(X)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) s(ok(X)) >? ok(s(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 active = \y0.2 + 3y0 ok = \y0.1 + y0 proper = \y0.2 + 3y0 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[active(s(_x0))]] = 11 + 6x0 > 7 + 6x0 = [[s(active(_x0))]] [[proper(0)]] = 11 > 4 = [[ok(0)]] [[proper(s(_x0))]] = 11 + 6x0 > 7 + 6x0 = [[s(proper(_x0))]] [[s(ok(_x0))]] = 5 + 2x0 > 4 + 2x0 = [[ok(s(_x0))]] We can thus remove the following rules: active(s(X)) => s(active(X)) proper(0) => ok(0) proper(s(X)) => s(proper(X)) s(ok(X)) => ok(s(X)) 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.