/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. 2nd : [o] --> o active : [o] --> o cons : [o * o] --> o cons1 : [o * o] --> o from : [o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o s : [o] --> o top : [o] --> o active(2nd(cons1(X, cons(Y, Z)))) => mark(Y) active(2nd(cons(X, Y))) => mark(2nd(cons1(X, Y))) active(from(X)) => mark(cons(X, from(s(X)))) active(2nd(X)) => 2nd(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(from(X)) => from(active(X)) active(s(X)) => s(active(X)) active(cons1(X, Y)) => cons1(active(X), Y) active(cons1(X, Y)) => cons1(X, active(Y)) 2nd(mark(X)) => mark(2nd(X)) cons(mark(X), Y) => mark(cons(X, Y)) from(mark(X)) => mark(from(X)) s(mark(X)) => mark(s(X)) cons1(mark(X), Y) => mark(cons1(X, Y)) cons1(X, mark(Y)) => mark(cons1(X, Y)) proper(2nd(X)) => 2nd(proper(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(from(X)) => from(proper(X)) proper(s(X)) => s(proper(X)) proper(cons1(X, Y)) => cons1(proper(X), proper(Y)) 2nd(ok(X)) => ok(2nd(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) from(ok(X)) => ok(from(X)) s(ok(X)) => ok(s(X)) cons1(ok(X), ok(Y)) => ok(cons1(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(2nd(cons1(X, cons(Y, Z)))) >? mark(Y) active(2nd(cons(X, Y))) >? mark(2nd(cons1(X, Y))) active(from(X)) >? mark(cons(X, from(s(X)))) active(2nd(X)) >? 2nd(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) active(cons1(X, Y)) >? cons1(active(X), Y) active(cons1(X, Y)) >? cons1(X, active(Y)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) 2nd(ok(X)) >? ok(2nd(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) from(ok(X)) >? ok(from(X)) s(ok(X)) >? ok(s(X)) cons1(ok(X), ok(Y)) >? ok(cons1(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: 2nd = \y0.2y0 active = \y0.2y0 cons = \y0y1.y0 + y1 cons1 = \y0y1.y0 + y1 from = \y0.2y0 mark = \y0.y0 ok = \y0.1 + 2y0 proper = \y0.y0 s = \y0.y0 top = \y0.2y0 Using this interpretation, the requirements translate to: [[active(2nd(cons1(_x0, cons(_x1, _x2))))]] = 4x0 + 4x1 + 4x2 >= x1 = [[mark(_x1)]] [[active(2nd(cons(_x0, _x1)))]] = 4x0 + 4x1 >= 2x0 + 2x1 = [[mark(2nd(cons1(_x0, _x1)))]] [[active(from(_x0))]] = 4x0 >= 3x0 = [[mark(cons(_x0, from(s(_x0))))]] [[active(2nd(_x0))]] = 4x0 >= 4x0 = [[2nd(active(_x0))]] [[active(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 4x0 >= 4x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(cons1(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons1(active(_x0), _x1)]] [[active(cons1(_x0, _x1))]] = 2x0 + 2x1 >= x0 + 2x1 = [[cons1(_x0, active(_x1))]] [[2nd(mark(_x0))]] = 2x0 >= 2x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = 2x0 >= 2x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = 2x0 >= 2x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 2x0 >= 2x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons1(proper(_x0), proper(_x1))]] [[2nd(ok(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[ok(2nd(_x0))]] [[cons(ok(_x0), ok(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[ok(cons(_x0, _x1))]] [[from(ok(_x0))]] = 2 + 4x0 > 1 + 4x0 = [[ok(from(_x0))]] [[s(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(s(_x0))]] [[cons1(ok(_x0), ok(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[ok(cons1(_x0, _x1))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = 2 + 4x0 > 4x0 = [[top(active(_x0))]] We can thus remove the following rules: 2nd(ok(X)) => ok(2nd(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) from(ok(X)) => ok(from(X)) cons1(ok(X), ok(Y)) => ok(cons1(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(2nd(cons1(X, cons(Y, Z)))) >? mark(Y) active(2nd(cons(X, Y))) >? mark(2nd(cons1(X, Y))) active(from(X)) >? mark(cons(X, from(s(X)))) active(2nd(X)) >? 2nd(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) active(cons1(X, Y)) >? cons1(active(X), Y) active(cons1(X, Y)) >? cons1(X, active(Y)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) s(ok(X)) >? ok(s(X)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.1 + y0 active = \y0.3y0 cons = \y0y1.y0 + y1 cons1 = \y0y1.y0 + y1 from = \y0.2y0 mark = \y0.y0 ok = \y0.y0 proper = \y0.y0 s = \y0.2y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(2nd(cons1(_x0, cons(_x1, _x2))))]] = 3 + 3x0 + 3x1 + 3x2 > x1 = [[mark(_x1)]] [[active(2nd(cons(_x0, _x1)))]] = 3 + 3x0 + 3x1 > 1 + x0 + x1 = [[mark(2nd(cons1(_x0, _x1)))]] [[active(from(_x0))]] = 6x0 >= 5x0 = [[mark(cons(_x0, from(s(_x0))))]] [[active(2nd(_x0))]] = 3 + 3x0 > 1 + 3x0 = [[2nd(active(_x0))]] [[active(cons(_x0, _x1))]] = 3x0 + 3x1 >= x1 + 3x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 6x0 >= 6x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 6x0 >= 6x0 = [[s(active(_x0))]] [[active(cons1(_x0, _x1))]] = 3x0 + 3x1 >= x1 + 3x0 = [[cons1(active(_x0), _x1)]] [[active(cons1(_x0, _x1))]] = 3x0 + 3x1 >= x0 + 3x1 = [[cons1(_x0, active(_x1))]] [[2nd(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = 2x0 >= 2x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = 2x0 >= 2x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = 1 + x0 >= 1 + x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 2x0 >= 2x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons1(proper(_x0), proper(_x1))]] [[s(ok(_x0))]] = 2x0 >= 2x0 = [[ok(s(_x0))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] We can thus remove the following rules: active(2nd(cons1(X, cons(Y, Z)))) => mark(Y) active(2nd(cons(X, Y))) => mark(2nd(cons1(X, Y))) active(2nd(X)) => 2nd(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(from(X)) >? mark(cons(X, from(s(X)))) active(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) active(cons1(X, Y)) >? cons1(active(X), Y) active(cons1(X, Y)) >? cons1(X, active(Y)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) s(ok(X)) >? ok(s(X)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.y0 active = \y0.3y0 cons = \y0y1.y0 + y1 cons1 = \y0y1.y0 + 2y1 from = \y0.y0 mark = \y0.y0 ok = \y0.1 + y0 proper = \y0.y0 s = \y0.2y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(from(_x0))]] = 3x0 >= 3x0 = [[mark(cons(_x0, from(s(_x0))))]] [[active(cons(_x0, _x1))]] = 3x0 + 3x1 >= x1 + 3x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 3x0 >= 3x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 6x0 >= 6x0 = [[s(active(_x0))]] [[active(cons1(_x0, _x1))]] = 3x0 + 6x1 >= 2x1 + 3x0 = [[cons1(active(_x0), _x1)]] [[active(cons1(_x0, _x1))]] = 3x0 + 6x1 >= x0 + 6x1 = [[cons1(_x0, active(_x1))]] [[2nd(mark(_x0))]] = x0 >= x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = x0 >= x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = 2x0 >= 2x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = x0 >= x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = x0 >= x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = 2x0 >= 2x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons1(proper(_x0), proper(_x1))]] [[s(ok(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[ok(s(_x0))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] We can thus remove the following rules: s(ok(X)) => ok(s(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(from(X)) >? mark(cons(X, from(s(X)))) active(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) active(cons1(X, Y)) >? cons1(active(X), Y) active(cons1(X, Y)) >? cons1(X, active(Y)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.y0 active = \y0.2y0 cons = \y0y1.y0 + y1 cons1 = \y0y1.1 + y0 + 2y1 from = \y0.2y0 mark = \y0.y0 proper = \y0.y0 s = \y0.y0 top = \y0.2y0 Using this interpretation, the requirements translate to: [[active(from(_x0))]] = 4x0 >= 3x0 = [[mark(cons(_x0, from(s(_x0))))]] [[active(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 4x0 >= 4x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(cons1(_x0, _x1))]] = 2 + 2x0 + 4x1 > 1 + 2x0 + 2x1 = [[cons1(active(_x0), _x1)]] [[active(cons1(_x0, _x1))]] = 2 + 2x0 + 4x1 > 1 + x0 + 4x1 = [[cons1(_x0, active(_x1))]] [[2nd(mark(_x0))]] = x0 >= x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = 2x0 >= 2x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = x0 >= x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 2x0 >= 2x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[cons1(proper(_x0), proper(_x1))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] We can thus remove the following rules: active(cons1(X, Y)) => cons1(active(X), Y) active(cons1(X, Y)) => cons1(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(from(X)) >? mark(cons(X, from(s(X)))) active(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.y0 active = \y0.2 + 3y0 cons = \y0y1.1 + y1 + 2y0 cons1 = \y0y1.y0 + y1 from = \y0.y0 mark = \y0.y0 proper = \y0.y0 s = \y0.y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(from(_x0))]] = 2 + 3x0 > 1 + 3x0 = [[mark(cons(_x0, from(s(_x0))))]] [[active(cons(_x0, _x1))]] = 5 + 3x1 + 6x0 >= 5 + x1 + 6x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[s(active(_x0))]] [[2nd(mark(_x0))]] = x0 >= x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = x0 >= x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = x0 >= x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = x0 >= x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons1(proper(_x0), proper(_x1))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] We can thus remove the following rules: active(from(X)) => mark(cons(X, from(s(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(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) active(s(X)) >? s(active(X)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.2y0 active = \y0.2y0 cons = \y0y1.y1 + 2y0 cons1 = \y0y1.y0 + y1 from = \y0.2y0 mark = \y0.y0 proper = \y0.y0 s = \y0.1 + y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(cons(_x0, _x1))]] = 2x1 + 4x0 >= x1 + 4x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 4x0 >= 4x0 = [[from(active(_x0))]] [[active(s(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[s(active(_x0))]] [[2nd(mark(_x0))]] = 2x0 >= 2x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = 2x0 >= 2x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = 2x0 >= 2x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 2x0 >= 2x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = 1 + x0 >= 1 + x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons1(proper(_x0), proper(_x1))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] We can thus remove the following rules: active(s(X)) => s(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(cons(X, Y)) >? cons(active(X), Y) active(from(X)) >? from(active(X)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.2y0 active = \y0.2y0 cons = \y0y1.1 + y1 + 2y0 cons1 = \y0y1.y0 + y1 from = \y0.2y0 mark = \y0.y0 proper = \y0.y0 s = \y0.y0 top = \y0.2y0 Using this interpretation, the requirements translate to: [[active(cons(_x0, _x1))]] = 2 + 2x1 + 4x0 > 1 + x1 + 4x0 = [[cons(active(_x0), _x1)]] [[active(from(_x0))]] = 4x0 >= 4x0 = [[from(active(_x0))]] [[2nd(mark(_x0))]] = 2x0 >= 2x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = 2x0 >= 2x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = 2x0 >= 2x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 2x0 >= 2x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons1(proper(_x0), proper(_x1))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] We can thus remove the following rules: active(cons(X, Y)) => cons(active(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(from(X)) >? from(active(X)) 2nd(mark(X)) >? mark(2nd(X)) cons(mark(X), Y) >? mark(cons(X, Y)) from(mark(X)) >? mark(from(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.2y0 active = \y0.y0 cons = \y0y1.y1 + 2y0 cons1 = \y0y1.2 + y0 + y1 from = \y0.2 + 3y0 mark = \y0.2 + y0 proper = \y0.y0 s = \y0.y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(from(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[from(active(_x0))]] [[2nd(mark(_x0))]] = 4 + 2x0 > 2 + 2x0 = [[mark(2nd(_x0))]] [[cons(mark(_x0), _x1)]] = 4 + x1 + 2x0 > 2 + x1 + 2x0 = [[mark(cons(_x0, _x1))]] [[from(mark(_x0))]] = 8 + 3x0 > 4 + 3x0 = [[mark(from(_x0))]] [[s(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = 4 + x0 + x1 >= 4 + x0 + x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = 4 + x0 + x1 >= 4 + x0 + x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = 2x0 >= 2x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = 2 + x0 + x1 >= 2 + x0 + x1 = [[cons1(proper(_x0), proper(_x1))]] [[top(mark(_x0))]] = 2 + x0 > x0 = [[top(proper(_x0))]] We can thus remove the following rules: 2nd(mark(X)) => mark(2nd(X)) cons(mark(X), Y) => mark(cons(X, Y)) from(mark(X)) => mark(from(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(from(X)) >? from(active(X)) s(mark(X)) >? mark(s(X)) cons1(mark(X), Y) >? mark(cons1(X, Y)) cons1(X, mark(Y)) >? mark(cons1(X, Y)) proper(2nd(X)) >? 2nd(proper(X)) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(from(X)) >? from(proper(X)) proper(s(X)) >? s(proper(X)) proper(cons1(X, Y)) >? cons1(proper(X), proper(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.y0 active = \y0.3y0 cons = \y0y1.1 + y0 + y1 cons1 = \y0y1.3 + 2y0 + 2y1 from = \y0.y0 mark = \y0.2 + y0 proper = \y0.3y0 s = \y0.3 + 2y0 Using this interpretation, the requirements translate to: [[active(from(_x0))]] = 3x0 >= 3x0 = [[from(active(_x0))]] [[s(mark(_x0))]] = 7 + 2x0 > 5 + 2x0 = [[mark(s(_x0))]] [[cons1(mark(_x0), _x1)]] = 7 + 2x0 + 2x1 > 5 + 2x0 + 2x1 = [[mark(cons1(_x0, _x1))]] [[cons1(_x0, mark(_x1))]] = 7 + 2x0 + 2x1 > 5 + 2x0 + 2x1 = [[mark(cons1(_x0, _x1))]] [[proper(2nd(_x0))]] = 3x0 >= 3x0 = [[2nd(proper(_x0))]] [[proper(cons(_x0, _x1))]] = 3 + 3x0 + 3x1 > 1 + 3x0 + 3x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(from(_x0))]] = 3x0 >= 3x0 = [[from(proper(_x0))]] [[proper(s(_x0))]] = 9 + 6x0 > 3 + 6x0 = [[s(proper(_x0))]] [[proper(cons1(_x0, _x1))]] = 9 + 6x0 + 6x1 > 3 + 6x0 + 6x1 = [[cons1(proper(_x0), proper(_x1))]] We can thus remove the following rules: s(mark(X)) => mark(s(X)) cons1(mark(X), Y) => mark(cons1(X, Y)) cons1(X, mark(Y)) => mark(cons1(X, Y)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(s(X)) => s(proper(X)) proper(cons1(X, Y)) => cons1(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(from(X)) >? from(active(X)) proper(2nd(X)) >? 2nd(proper(X)) proper(from(X)) >? from(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.y0 active = \y0.3y0 from = \y0.2 + y0 proper = \y0.3y0 Using this interpretation, the requirements translate to: [[active(from(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[from(active(_x0))]] [[proper(2nd(_x0))]] = 3x0 >= 3x0 = [[2nd(proper(_x0))]] [[proper(from(_x0))]] = 6 + 3x0 > 2 + 3x0 = [[from(proper(_x0))]] We can thus remove the following rules: active(from(X)) => from(active(X)) proper(from(X)) => from(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]): proper(2nd(X)) >? 2nd(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 2nd = \y0.1 + y0 proper = \y0.3y0 Using this interpretation, the requirements translate to: [[proper(2nd(_x0))]] = 3 + 3x0 > 1 + 3x0 = [[2nd(proper(_x0))]] We can thus remove the following rules: proper(2nd(X)) => 2nd(proper(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.