/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 adx : [o] --> o cons : [o * o] --> o head : [o] --> o incr : [o] --> o mark : [o] --> o nats : [] --> o nil : [] --> o ok : [o] --> o proper : [o] --> o s : [o] --> o tail : [o] --> o top : [o] --> o zeros : [] --> o active(incr(nil)) => mark(nil) active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) active(adx(nil)) => mark(nil) active(adx(cons(X, Y))) => mark(incr(cons(X, adx(Y)))) active(nats) => mark(adx(zeros)) active(zeros) => mark(cons(0, zeros)) active(head(cons(X, Y))) => mark(X) active(tail(cons(X, Y))) => mark(Y) active(incr(X)) => incr(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(s(X)) => s(active(X)) active(adx(X)) => adx(active(X)) active(head(X)) => head(active(X)) active(tail(X)) => tail(active(X)) incr(mark(X)) => mark(incr(X)) cons(mark(X), Y) => mark(cons(X, Y)) s(mark(X)) => mark(s(X)) adx(mark(X)) => mark(adx(X)) head(mark(X)) => mark(head(X)) tail(mark(X)) => mark(tail(X)) proper(incr(X)) => incr(proper(X)) proper(nil) => ok(nil) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(s(X)) => s(proper(X)) proper(adx(X)) => adx(proper(X)) proper(nats) => ok(nats) proper(zeros) => ok(zeros) proper(0) => ok(0) proper(head(X)) => head(proper(X)) proper(tail(X)) => tail(proper(X)) incr(ok(X)) => ok(incr(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) s(ok(X)) => ok(s(X)) adx(ok(X)) => ok(adx(X)) head(ok(X)) => ok(head(X)) tail(ok(X)) => ok(tail(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(incr(nil)) >? mark(nil) active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) active(adx(nil)) >? mark(nil) active(adx(cons(X, Y))) >? mark(incr(cons(X, adx(Y)))) active(nats) >? mark(adx(zeros)) active(zeros) >? mark(cons(0, zeros)) active(head(cons(X, Y))) >? mark(X) active(tail(cons(X, Y))) >? mark(Y) active(incr(X)) >? incr(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(s(X)) >? s(active(X)) active(adx(X)) >? adx(active(X)) active(head(X)) >? head(active(X)) active(tail(X)) >? tail(active(X)) incr(mark(X)) >? mark(incr(X)) cons(mark(X), Y) >? mark(cons(X, Y)) s(mark(X)) >? mark(s(X)) adx(mark(X)) >? mark(adx(X)) head(mark(X)) >? mark(head(X)) tail(mark(X)) >? mark(tail(X)) proper(incr(X)) >? incr(proper(X)) proper(nil) >? ok(nil) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(s(X)) >? s(proper(X)) proper(adx(X)) >? adx(proper(X)) proper(nats) >? ok(nats) proper(zeros) >? ok(zeros) proper(0) >? ok(0) proper(head(X)) >? head(proper(X)) proper(tail(X)) >? tail(proper(X)) incr(ok(X)) >? ok(incr(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) s(ok(X)) >? ok(s(X)) adx(ok(X)) >? ok(adx(X)) head(ok(X)) >? ok(head(X)) tail(ok(X)) >? ok(tail(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 adx = \y0.2y0 cons = \y0y1.2y0 + 2y1 head = \y0.2y0 incr = \y0.y0 mark = \y0.y0 nats = 0 nil = 0 ok = \y0.y0 proper = \y0.y0 s = \y0.y0 tail = \y0.2 + y0 top = \y0.y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(incr(nil))]] = 0 >= 0 = [[mark(nil)]] [[active(incr(cons(_x0, _x1)))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(cons(s(_x0), incr(_x1)))]] [[active(adx(nil))]] = 0 >= 0 = [[mark(nil)]] [[active(adx(cons(_x0, _x1)))]] = 4x0 + 4x1 >= 2x0 + 4x1 = [[mark(incr(cons(_x0, adx(_x1))))]] [[active(nats)]] = 0 >= 0 = [[mark(adx(zeros))]] [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(head(cons(_x0, _x1)))]] = 4x0 + 4x1 >= x0 = [[mark(_x0)]] [[active(tail(cons(_x0, _x1)))]] = 2 + 2x0 + 2x1 > x1 = [[mark(_x1)]] [[active(incr(_x0))]] = x0 >= x0 = [[incr(active(_x0))]] [[active(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(active(_x0), _x1)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(adx(_x0))]] = 2x0 >= 2x0 = [[adx(active(_x0))]] [[active(head(_x0))]] = 2x0 >= 2x0 = [[head(active(_x0))]] [[active(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[tail(active(_x0))]] [[incr(mark(_x0))]] = x0 >= x0 = [[mark(incr(_x0))]] [[cons(mark(_x0), _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(cons(_x0, _x1))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[adx(mark(_x0))]] = 2x0 >= 2x0 = [[mark(adx(_x0))]] [[head(mark(_x0))]] = 2x0 >= 2x0 = [[mark(head(_x0))]] [[tail(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(tail(_x0))]] [[proper(incr(_x0))]] = x0 >= x0 = [[incr(proper(_x0))]] [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] [[proper(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(adx(_x0))]] = 2x0 >= 2x0 = [[adx(proper(_x0))]] [[proper(nats)]] = 0 >= 0 = [[ok(nats)]] [[proper(zeros)]] = 0 >= 0 = [[ok(zeros)]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(head(_x0))]] = 2x0 >= 2x0 = [[head(proper(_x0))]] [[proper(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[tail(proper(_x0))]] [[incr(ok(_x0))]] = x0 >= x0 = [[ok(incr(_x0))]] [[cons(ok(_x0), ok(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[ok(cons(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[adx(ok(_x0))]] = 2x0 >= 2x0 = [[ok(adx(_x0))]] [[head(ok(_x0))]] = 2x0 >= 2x0 = [[ok(head(_x0))]] [[tail(ok(_x0))]] = 2 + x0 >= 2 + x0 = [[ok(tail(_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(tail(cons(X, Y))) => mark(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(incr(nil)) >? mark(nil) active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) active(adx(nil)) >? mark(nil) active(adx(cons(X, Y))) >? mark(incr(cons(X, adx(Y)))) active(nats) >? mark(adx(zeros)) active(zeros) >? mark(cons(0, zeros)) active(head(cons(X, Y))) >? mark(X) active(incr(X)) >? incr(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(s(X)) >? s(active(X)) active(adx(X)) >? adx(active(X)) active(head(X)) >? head(active(X)) active(tail(X)) >? tail(active(X)) incr(mark(X)) >? mark(incr(X)) cons(mark(X), Y) >? mark(cons(X, Y)) s(mark(X)) >? mark(s(X)) adx(mark(X)) >? mark(adx(X)) head(mark(X)) >? mark(head(X)) tail(mark(X)) >? mark(tail(X)) proper(incr(X)) >? incr(proper(X)) proper(nil) >? ok(nil) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(s(X)) >? s(proper(X)) proper(adx(X)) >? adx(proper(X)) proper(nats) >? ok(nats) proper(zeros) >? ok(zeros) proper(0) >? ok(0) proper(head(X)) >? head(proper(X)) proper(tail(X)) >? tail(proper(X)) incr(ok(X)) >? ok(incr(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) s(ok(X)) >? ok(s(X)) adx(ok(X)) >? ok(adx(X)) head(ok(X)) >? ok(head(X)) tail(ok(X)) >? ok(tail(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 adx = \y0.2y0 cons = \y0y1.2y0 + 2y1 head = \y0.y0 incr = \y0.y0 mark = \y0.y0 nats = 0 nil = 2 ok = \y0.y0 proper = \y0.y0 s = \y0.y0 tail = \y0.2y0 top = \y0.y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(incr(nil))]] = 2 >= 2 = [[mark(nil)]] [[active(incr(cons(_x0, _x1)))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(cons(s(_x0), incr(_x1)))]] [[active(adx(nil))]] = 4 > 2 = [[mark(nil)]] [[active(adx(cons(_x0, _x1)))]] = 4x0 + 4x1 >= 2x0 + 4x1 = [[mark(incr(cons(_x0, adx(_x1))))]] [[active(nats)]] = 0 >= 0 = [[mark(adx(zeros))]] [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(head(cons(_x0, _x1)))]] = 2x0 + 2x1 >= x0 = [[mark(_x0)]] [[active(incr(_x0))]] = x0 >= x0 = [[incr(active(_x0))]] [[active(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(active(_x0), _x1)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(adx(_x0))]] = 2x0 >= 2x0 = [[adx(active(_x0))]] [[active(head(_x0))]] = x0 >= x0 = [[head(active(_x0))]] [[active(tail(_x0))]] = 2x0 >= 2x0 = [[tail(active(_x0))]] [[incr(mark(_x0))]] = x0 >= x0 = [[mark(incr(_x0))]] [[cons(mark(_x0), _x1)]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(cons(_x0, _x1))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[adx(mark(_x0))]] = 2x0 >= 2x0 = [[mark(adx(_x0))]] [[head(mark(_x0))]] = x0 >= x0 = [[mark(head(_x0))]] [[tail(mark(_x0))]] = 2x0 >= 2x0 = [[mark(tail(_x0))]] [[proper(incr(_x0))]] = x0 >= x0 = [[incr(proper(_x0))]] [[proper(nil)]] = 2 >= 2 = [[ok(nil)]] [[proper(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(adx(_x0))]] = 2x0 >= 2x0 = [[adx(proper(_x0))]] [[proper(nats)]] = 0 >= 0 = [[ok(nats)]] [[proper(zeros)]] = 0 >= 0 = [[ok(zeros)]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(head(_x0))]] = x0 >= x0 = [[head(proper(_x0))]] [[proper(tail(_x0))]] = 2x0 >= 2x0 = [[tail(proper(_x0))]] [[incr(ok(_x0))]] = x0 >= x0 = [[ok(incr(_x0))]] [[cons(ok(_x0), ok(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[ok(cons(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[adx(ok(_x0))]] = 2x0 >= 2x0 = [[ok(adx(_x0))]] [[head(ok(_x0))]] = x0 >= x0 = [[ok(head(_x0))]] [[tail(ok(_x0))]] = 2x0 >= 2x0 = [[ok(tail(_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(adx(nil)) => mark(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]): active(incr(nil)) >? mark(nil) active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) >? mark(incr(cons(X, adx(Y)))) active(nats) >? mark(adx(zeros)) active(zeros) >? mark(cons(0, zeros)) active(head(cons(X, Y))) >? mark(X) active(incr(X)) >? incr(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(s(X)) >? s(active(X)) active(adx(X)) >? adx(active(X)) active(head(X)) >? head(active(X)) active(tail(X)) >? tail(active(X)) incr(mark(X)) >? mark(incr(X)) cons(mark(X), Y) >? mark(cons(X, Y)) s(mark(X)) >? mark(s(X)) adx(mark(X)) >? mark(adx(X)) head(mark(X)) >? mark(head(X)) tail(mark(X)) >? mark(tail(X)) proper(incr(X)) >? incr(proper(X)) proper(nil) >? ok(nil) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(s(X)) >? s(proper(X)) proper(adx(X)) >? adx(proper(X)) proper(nats) >? ok(nats) proper(zeros) >? ok(zeros) proper(0) >? ok(0) proper(head(X)) >? head(proper(X)) proper(tail(X)) >? tail(proper(X)) incr(ok(X)) >? ok(incr(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) s(ok(X)) >? ok(s(X)) adx(ok(X)) >? ok(adx(X)) head(ok(X)) >? ok(head(X)) tail(ok(X)) >? ok(tail(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 adx = \y0.y0 cons = \y0y1.y0 + 2y1 head = \y0.2y0 incr = \y0.y0 mark = \y0.y0 nats = 2 nil = 0 ok = \y0.y0 proper = \y0.y0 s = \y0.y0 tail = \y0.2y0 top = \y0.2y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(incr(nil))]] = 0 >= 0 = [[mark(nil)]] [[active(incr(cons(_x0, _x1)))]] = x0 + 2x1 >= x0 + 2x1 = [[mark(cons(s(_x0), incr(_x1)))]] [[active(adx(cons(_x0, _x1)))]] = x0 + 2x1 >= x0 + 2x1 = [[mark(incr(cons(_x0, adx(_x1))))]] [[active(nats)]] = 2 > 0 = [[mark(adx(zeros))]] [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(head(cons(_x0, _x1)))]] = 2x0 + 4x1 >= x0 = [[mark(_x0)]] [[active(incr(_x0))]] = x0 >= x0 = [[incr(active(_x0))]] [[active(cons(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons(active(_x0), _x1)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(adx(_x0))]] = x0 >= x0 = [[adx(active(_x0))]] [[active(head(_x0))]] = 2x0 >= 2x0 = [[head(active(_x0))]] [[active(tail(_x0))]] = 2x0 >= 2x0 = [[tail(active(_x0))]] [[incr(mark(_x0))]] = x0 >= x0 = [[mark(incr(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[mark(cons(_x0, _x1))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[adx(mark(_x0))]] = x0 >= x0 = [[mark(adx(_x0))]] [[head(mark(_x0))]] = 2x0 >= 2x0 = [[mark(head(_x0))]] [[tail(mark(_x0))]] = 2x0 >= 2x0 = [[mark(tail(_x0))]] [[proper(incr(_x0))]] = x0 >= x0 = [[incr(proper(_x0))]] [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] [[proper(cons(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(adx(_x0))]] = x0 >= x0 = [[adx(proper(_x0))]] [[proper(nats)]] = 2 >= 2 = [[ok(nats)]] [[proper(zeros)]] = 0 >= 0 = [[ok(zeros)]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(head(_x0))]] = 2x0 >= 2x0 = [[head(proper(_x0))]] [[proper(tail(_x0))]] = 2x0 >= 2x0 = [[tail(proper(_x0))]] [[incr(ok(_x0))]] = x0 >= x0 = [[ok(incr(_x0))]] [[cons(ok(_x0), ok(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[ok(cons(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[adx(ok(_x0))]] = x0 >= x0 = [[ok(adx(_x0))]] [[head(ok(_x0))]] = 2x0 >= 2x0 = [[ok(head(_x0))]] [[tail(ok(_x0))]] = 2x0 >= 2x0 = [[ok(tail(_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(nats) => mark(adx(zeros)) 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(incr(nil)) >? mark(nil) active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) >? mark(incr(cons(X, adx(Y)))) active(zeros) >? mark(cons(0, zeros)) active(head(cons(X, Y))) >? mark(X) active(incr(X)) >? incr(active(X)) active(cons(X, Y)) >? cons(active(X), Y) active(s(X)) >? s(active(X)) active(adx(X)) >? adx(active(X)) active(head(X)) >? head(active(X)) active(tail(X)) >? tail(active(X)) incr(mark(X)) >? mark(incr(X)) cons(mark(X), Y) >? mark(cons(X, Y)) s(mark(X)) >? mark(s(X)) adx(mark(X)) >? mark(adx(X)) head(mark(X)) >? mark(head(X)) tail(mark(X)) >? mark(tail(X)) proper(incr(X)) >? incr(proper(X)) proper(nil) >? ok(nil) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(s(X)) >? s(proper(X)) proper(adx(X)) >? adx(proper(X)) proper(nats) >? ok(nats) proper(zeros) >? ok(zeros) proper(0) >? ok(0) proper(head(X)) >? head(proper(X)) proper(tail(X)) >? tail(proper(X)) incr(ok(X)) >? ok(incr(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) s(ok(X)) >? ok(s(X)) adx(ok(X)) >? ok(adx(X)) head(ok(X)) >? ok(head(X)) tail(ok(X)) >? ok(tail(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 adx = \y0.2y0 cons = \y0y1.y0 + 2y1 head = \y0.1 + y0 incr = \y0.y0 mark = \y0.y0 nats = 0 nil = 0 ok = \y0.y0 proper = \y0.y0 s = \y0.y0 tail = \y0.2y0 top = \y0.2y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(incr(nil))]] = 0 >= 0 = [[mark(nil)]] [[active(incr(cons(_x0, _x1)))]] = x0 + 2x1 >= x0 + 2x1 = [[mark(cons(s(_x0), incr(_x1)))]] [[active(adx(cons(_x0, _x1)))]] = 2x0 + 4x1 >= x0 + 4x1 = [[mark(incr(cons(_x0, adx(_x1))))]] [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(head(cons(_x0, _x1)))]] = 1 + x0 + 2x1 > x0 = [[mark(_x0)]] [[active(incr(_x0))]] = x0 >= x0 = [[incr(active(_x0))]] [[active(cons(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons(active(_x0), _x1)]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[active(adx(_x0))]] = 2x0 >= 2x0 = [[adx(active(_x0))]] [[active(head(_x0))]] = 1 + x0 >= 1 + x0 = [[head(active(_x0))]] [[active(tail(_x0))]] = 2x0 >= 2x0 = [[tail(active(_x0))]] [[incr(mark(_x0))]] = x0 >= x0 = [[mark(incr(_x0))]] [[cons(mark(_x0), _x1)]] = x0 + 2x1 >= x0 + 2x1 = [[mark(cons(_x0, _x1))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[adx(mark(_x0))]] = 2x0 >= 2x0 = [[mark(adx(_x0))]] [[head(mark(_x0))]] = 1 + x0 >= 1 + x0 = [[mark(head(_x0))]] [[tail(mark(_x0))]] = 2x0 >= 2x0 = [[mark(tail(_x0))]] [[proper(incr(_x0))]] = x0 >= x0 = [[incr(proper(_x0))]] [[proper(nil)]] = 0 >= 0 = [[ok(nil)]] [[proper(cons(_x0, _x1))]] = x0 + 2x1 >= x0 + 2x1 = [[cons(proper(_x0), proper(_x1))]] [[proper(s(_x0))]] = x0 >= x0 = [[s(proper(_x0))]] [[proper(adx(_x0))]] = 2x0 >= 2x0 = [[adx(proper(_x0))]] [[proper(nats)]] = 0 >= 0 = [[ok(nats)]] [[proper(zeros)]] = 0 >= 0 = [[ok(zeros)]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(head(_x0))]] = 1 + x0 >= 1 + x0 = [[head(proper(_x0))]] [[proper(tail(_x0))]] = 2x0 >= 2x0 = [[tail(proper(_x0))]] [[incr(ok(_x0))]] = x0 >= x0 = [[ok(incr(_x0))]] [[cons(ok(_x0), ok(_x1))]] = x0 + 2x1 >= x0 + 2x1 = [[ok(cons(_x0, _x1))]] [[s(ok(_x0))]] = x0 >= x0 = [[ok(s(_x0))]] [[adx(ok(_x0))]] = 2x0 >= 2x0 = [[ok(adx(_x0))]] [[head(ok(_x0))]] = 1 + x0 >= 1 + x0 = [[ok(head(_x0))]] [[tail(ok(_x0))]] = 2x0 >= 2x0 = [[ok(tail(_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(head(cons(X, Y))) => mark(X) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] active#(incr(cons(X, Y))) =#> cons#(s(X), incr(Y)) 1] active#(incr(cons(X, Y))) =#> s#(X) 2] active#(incr(cons(X, Y))) =#> incr#(Y) 3] active#(adx(cons(X, Y))) =#> incr#(cons(X, adx(Y))) 4] active#(adx(cons(X, Y))) =#> cons#(X, adx(Y)) 5] active#(adx(cons(X, Y))) =#> adx#(Y) 6] active#(zeros) =#> cons#(0, zeros) 7] active#(incr(X)) =#> incr#(active(X)) 8] active#(incr(X)) =#> active#(X) 9] active#(cons(X, Y)) =#> cons#(active(X), Y) 10] active#(cons(X, Y)) =#> active#(X) 11] active#(s(X)) =#> s#(active(X)) 12] active#(s(X)) =#> active#(X) 13] active#(adx(X)) =#> adx#(active(X)) 14] active#(adx(X)) =#> active#(X) 15] active#(head(X)) =#> head#(active(X)) 16] active#(head(X)) =#> active#(X) 17] active#(tail(X)) =#> tail#(active(X)) 18] active#(tail(X)) =#> active#(X) 19] incr#(mark(X)) =#> incr#(X) 20] cons#(mark(X), Y) =#> cons#(X, Y) 21] s#(mark(X)) =#> s#(X) 22] adx#(mark(X)) =#> adx#(X) 23] head#(mark(X)) =#> head#(X) 24] tail#(mark(X)) =#> tail#(X) 25] proper#(incr(X)) =#> incr#(proper(X)) 26] proper#(incr(X)) =#> proper#(X) 27] proper#(cons(X, Y)) =#> cons#(proper(X), proper(Y)) 28] proper#(cons(X, Y)) =#> proper#(X) 29] proper#(cons(X, Y)) =#> proper#(Y) 30] proper#(s(X)) =#> s#(proper(X)) 31] proper#(s(X)) =#> proper#(X) 32] proper#(adx(X)) =#> adx#(proper(X)) 33] proper#(adx(X)) =#> proper#(X) 34] proper#(head(X)) =#> head#(proper(X)) 35] proper#(head(X)) =#> proper#(X) 36] proper#(tail(X)) =#> tail#(proper(X)) 37] proper#(tail(X)) =#> proper#(X) 38] incr#(ok(X)) =#> incr#(X) 39] cons#(ok(X), ok(Y)) =#> cons#(X, Y) 40] s#(ok(X)) =#> s#(X) 41] adx#(ok(X)) =#> adx#(X) 42] head#(ok(X)) =#> head#(X) 43] tail#(ok(X)) =#> tail#(X) 44] top#(mark(X)) =#> top#(proper(X)) 45] top#(mark(X)) =#> proper#(X) 46] top#(ok(X)) =#> top#(active(X)) 47] top#(ok(X)) =#> active#(X) Rules R_0: active(incr(nil)) => mark(nil) active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) => mark(incr(cons(X, adx(Y)))) active(zeros) => mark(cons(0, zeros)) active(incr(X)) => incr(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(s(X)) => s(active(X)) active(adx(X)) => adx(active(X)) active(head(X)) => head(active(X)) active(tail(X)) => tail(active(X)) incr(mark(X)) => mark(incr(X)) cons(mark(X), Y) => mark(cons(X, Y)) s(mark(X)) => mark(s(X)) adx(mark(X)) => mark(adx(X)) head(mark(X)) => mark(head(X)) tail(mark(X)) => mark(tail(X)) proper(incr(X)) => incr(proper(X)) proper(nil) => ok(nil) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(s(X)) => s(proper(X)) proper(adx(X)) => adx(proper(X)) proper(nats) => ok(nats) proper(zeros) => ok(zeros) proper(0) => ok(0) proper(head(X)) => head(proper(X)) proper(tail(X)) => tail(proper(X)) incr(ok(X)) => ok(incr(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) s(ok(X)) => ok(s(X)) adx(ok(X)) => ok(adx(X)) head(ok(X)) => ok(head(X)) tail(ok(X)) => ok(tail(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(X)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 20, 39 * 1 : 21, 40 * 2 : 19, 38 * 3 : 19, 38 * 4 : 20, 39 * 5 : 22, 41 * 6 : * 7 : 19, 38 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 * 9 : 20, 39 * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 * 11 : 21, 40 * 12 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 * 13 : 22, 41 * 14 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 * 15 : 23, 42 * 16 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 * 17 : 24, 43 * 18 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 * 19 : 19, 38 * 20 : 20, 39 * 21 : 21, 40 * 22 : 22, 41 * 23 : 23, 42 * 24 : 24, 43 * 25 : 19, 38 * 26 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 27 : 20, 39 * 28 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 29 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 30 : 21, 40 * 31 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 32 : 22, 41 * 33 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 34 : 23, 42 * 35 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 36 : 24, 43 * 37 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 38 : 19, 38 * 39 : 20, 39 * 40 : 21, 40 * 41 : 22, 41 * 42 : 23, 42 * 43 : 24, 43 * 44 : 44, 45, 46, 47 * 45 : 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37 * 46 : 44, 45, 46, 47 * 47 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 This graph has the following strongly connected components: P_1: active#(incr(X)) =#> active#(X) active#(cons(X, Y)) =#> active#(X) active#(s(X)) =#> active#(X) active#(adx(X)) =#> active#(X) active#(head(X)) =#> active#(X) active#(tail(X)) =#> active#(X) P_2: incr#(mark(X)) =#> incr#(X) incr#(ok(X)) =#> incr#(X) P_3: cons#(mark(X), Y) =#> cons#(X, Y) cons#(ok(X), ok(Y)) =#> cons#(X, Y) P_4: s#(mark(X)) =#> s#(X) s#(ok(X)) =#> s#(X) P_5: adx#(mark(X)) =#> adx#(X) adx#(ok(X)) =#> adx#(X) P_6: head#(mark(X)) =#> head#(X) head#(ok(X)) =#> head#(X) P_7: tail#(mark(X)) =#> tail#(X) tail#(ok(X)) =#> tail#(X) P_8: proper#(incr(X)) =#> proper#(X) proper#(cons(X, Y)) =#> proper#(X) proper#(cons(X, Y)) =#> proper#(Y) proper#(s(X)) =#> proper#(X) proper#(adx(X)) =#> proper#(X) proper#(head(X)) =#> proper#(X) proper#(tail(X)) =#> proper#(X) P_9: top#(mark(X)) =#> top#(proper(X)) top#(ok(X)) =#> top#(active(X)) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f), (P_3, R_0, m, f), (P_4, R_0, m, f), (P_5, R_0, m, f), (P_6, R_0, m, f), (P_7, R_0, m, f), (P_8, R_0, m, f) and (P_9, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_9, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_9, R_0, minimal, formative). The formative rules of (P_9, R_0) are R_1 ::= active(incr(nil)) => mark(nil) active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) => mark(incr(cons(X, adx(Y)))) active(zeros) => mark(cons(0, zeros)) active(incr(X)) => incr(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(s(X)) => s(active(X)) active(adx(X)) => adx(active(X)) active(head(X)) => head(active(X)) active(tail(X)) => tail(active(X)) incr(mark(X)) => mark(incr(X)) cons(mark(X), Y) => mark(cons(X, Y)) s(mark(X)) => mark(s(X)) adx(mark(X)) => mark(adx(X)) head(mark(X)) => mark(head(X)) tail(mark(X)) => mark(tail(X)) proper(incr(X)) => incr(proper(X)) proper(nil) => ok(nil) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(s(X)) => s(proper(X)) proper(adx(X)) => adx(proper(X)) proper(nats) => ok(nats) proper(zeros) => ok(zeros) proper(0) => ok(0) proper(head(X)) => head(proper(X)) proper(tail(X)) => tail(proper(X)) incr(ok(X)) => ok(incr(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) s(ok(X)) => ok(s(X)) adx(ok(X)) => ok(adx(X)) head(ok(X)) => ok(head(X)) tail(ok(X)) => ok(tail(X)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_9, R_0, minimal, formative) by (P_9, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_9, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_9, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: top#(mark(X)) >? top#(proper(X)) top#(ok(X)) >? top#(active(X)) active(incr(nil)) >= mark(nil) active(incr(cons(X, Y))) >= mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) >= mark(incr(cons(X, adx(Y)))) active(zeros) >= mark(cons(0, zeros)) active(incr(X)) >= incr(active(X)) active(cons(X, Y)) >= cons(active(X), Y) active(s(X)) >= s(active(X)) active(adx(X)) >= adx(active(X)) active(head(X)) >= head(active(X)) active(tail(X)) >= tail(active(X)) incr(mark(X)) >= mark(incr(X)) cons(mark(X), Y) >= mark(cons(X, Y)) s(mark(X)) >= mark(s(X)) adx(mark(X)) >= mark(adx(X)) head(mark(X)) >= mark(head(X)) tail(mark(X)) >= mark(tail(X)) proper(incr(X)) >= incr(proper(X)) proper(nil) >= ok(nil) proper(cons(X, Y)) >= cons(proper(X), proper(Y)) proper(s(X)) >= s(proper(X)) proper(adx(X)) >= adx(proper(X)) proper(nats) >= ok(nats) proper(zeros) >= ok(zeros) proper(0) >= ok(0) proper(head(X)) >= head(proper(X)) proper(tail(X)) >= tail(proper(X)) incr(ok(X)) >= ok(incr(X)) cons(ok(X), ok(Y)) >= ok(cons(X, Y)) s(ok(X)) >= ok(s(X)) adx(ok(X)) >= ok(adx(X)) head(ok(X)) >= ok(head(X)) tail(ok(X)) >= ok(tail(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 active = \y0.2y0 adx = \y0.3y0 cons = \y0y1.2y0 head = \y0.y0 incr = \y0.2y0 mark = \y0.3y0 nats = 3 nil = 2 ok = \y0.1 + 2y0 proper = \y0.3y0 s = \y0.y0 tail = \y0.2y0 top# = \y0.2y0 zeros = 3 Using this interpretation, the requirements translate to: [[top#(mark(_x0))]] = 6x0 >= 6x0 = [[top#(proper(_x0))]] [[top#(ok(_x0))]] = 2 + 4x0 > 4x0 = [[top#(active(_x0))]] [[active(incr(nil))]] = 8 >= 6 = [[mark(nil)]] [[active(incr(cons(_x0, _x1)))]] = 8x0 >= 6x0 = [[mark(cons(s(_x0), incr(_x1)))]] [[active(adx(cons(_x0, _x1)))]] = 12x0 >= 12x0 = [[mark(incr(cons(_x0, adx(_x1))))]] [[active(zeros)]] = 6 >= 6 = [[mark(cons(0, zeros))]] [[active(incr(_x0))]] = 4x0 >= 4x0 = [[incr(active(_x0))]] [[active(cons(_x0, _x1))]] = 4x0 >= 4x0 = [[cons(active(_x0), _x1)]] [[active(s(_x0))]] = 2x0 >= 2x0 = [[s(active(_x0))]] [[active(adx(_x0))]] = 6x0 >= 6x0 = [[adx(active(_x0))]] [[active(head(_x0))]] = 2x0 >= 2x0 = [[head(active(_x0))]] [[active(tail(_x0))]] = 4x0 >= 4x0 = [[tail(active(_x0))]] [[incr(mark(_x0))]] = 6x0 >= 6x0 = [[mark(incr(_x0))]] [[cons(mark(_x0), _x1)]] = 6x0 >= 6x0 = [[mark(cons(_x0, _x1))]] [[s(mark(_x0))]] = 3x0 >= 3x0 = [[mark(s(_x0))]] [[adx(mark(_x0))]] = 9x0 >= 9x0 = [[mark(adx(_x0))]] [[head(mark(_x0))]] = 3x0 >= 3x0 = [[mark(head(_x0))]] [[tail(mark(_x0))]] = 6x0 >= 6x0 = [[mark(tail(_x0))]] [[proper(incr(_x0))]] = 6x0 >= 6x0 = [[incr(proper(_x0))]] [[proper(nil)]] = 6 >= 5 = [[ok(nil)]] [[proper(cons(_x0, _x1))]] = 6x0 >= 6x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(s(_x0))]] = 3x0 >= 3x0 = [[s(proper(_x0))]] [[proper(adx(_x0))]] = 9x0 >= 9x0 = [[adx(proper(_x0))]] [[proper(nats)]] = 9 >= 7 = [[ok(nats)]] [[proper(zeros)]] = 9 >= 7 = [[ok(zeros)]] [[proper(0)]] = 3 >= 3 = [[ok(0)]] [[proper(head(_x0))]] = 3x0 >= 3x0 = [[head(proper(_x0))]] [[proper(tail(_x0))]] = 6x0 >= 6x0 = [[tail(proper(_x0))]] [[incr(ok(_x0))]] = 2 + 4x0 >= 1 + 4x0 = [[ok(incr(_x0))]] [[cons(ok(_x0), ok(_x1))]] = 2 + 4x0 >= 1 + 4x0 = [[ok(cons(_x0, _x1))]] [[s(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(s(_x0))]] [[adx(ok(_x0))]] = 3 + 6x0 >= 1 + 6x0 = [[ok(adx(_x0))]] [[head(ok(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[ok(head(_x0))]] [[tail(ok(_x0))]] = 2 + 4x0 >= 1 + 4x0 = [[ok(tail(_x0))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_9, R_1, minimal, formative) by (P_10, R_1, minimal, formative), where P_10 consists of: top#(mark(X)) =#> top#(proper(X)) Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_10, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_10, R_1, minimal, formative). The formative rules of (P_10, R_1) are R_2 ::= active(incr(nil)) => mark(nil) active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) => mark(incr(cons(X, adx(Y)))) active(zeros) => mark(cons(0, zeros)) active(incr(X)) => incr(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(s(X)) => s(active(X)) active(adx(X)) => adx(active(X)) active(head(X)) => head(active(X)) active(tail(X)) => tail(active(X)) incr(mark(X)) => mark(incr(X)) cons(mark(X), Y) => mark(cons(X, Y)) s(mark(X)) => mark(s(X)) adx(mark(X)) => mark(adx(X)) head(mark(X)) => mark(head(X)) tail(mark(X)) => mark(tail(X)) proper(incr(X)) => incr(proper(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(s(X)) => s(proper(X)) proper(adx(X)) => adx(proper(X)) proper(head(X)) => head(proper(X)) proper(tail(X)) => tail(proper(X)) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_10, R_1, minimal, formative) by (P_10, R_2, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative), (P_8, R_0, minimal, formative) and (P_10, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_10, R_2, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_10, R_2) are: incr(mark(X)) => mark(incr(X)) cons(mark(X), Y) => mark(cons(X, Y)) s(mark(X)) => mark(s(X)) adx(mark(X)) => mark(adx(X)) head(mark(X)) => mark(head(X)) tail(mark(X)) => mark(tail(X)) proper(incr(X)) => incr(proper(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(s(X)) => s(proper(X)) proper(adx(X)) => adx(proper(X)) proper(head(X)) => head(proper(X)) proper(tail(X)) => tail(proper(X)) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: top#(mark(X)) >? top#(proper(X)) incr(mark(X)) >= mark(incr(X)) cons(mark(X), Y) >= mark(cons(X, Y)) s(mark(X)) >= mark(s(X)) adx(mark(X)) >= mark(adx(X)) head(mark(X)) >= mark(head(X)) tail(mark(X)) >= mark(tail(X)) proper(incr(X)) >= incr(proper(X)) proper(cons(X, Y)) >= cons(proper(X), proper(Y)) proper(s(X)) >= s(proper(X)) proper(adx(X)) >= adx(proper(X)) proper(head(X)) >= head(proper(X)) proper(tail(X)) >= tail(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: adx = \y0.y0 cons = \y0y1.y0 head = \y0.3y0 incr = \y0.3y0 mark = \y0.3 + 3y0 proper = \y0.0 s = \y0.3y0 tail = \y0.y0 top# = \y0.3y0 Using this interpretation, the requirements translate to: [[top#(mark(_x0))]] = 9 + 9x0 > 0 = [[top#(proper(_x0))]] [[incr(mark(_x0))]] = 9 + 9x0 >= 3 + 9x0 = [[mark(incr(_x0))]] [[cons(mark(_x0), _x1)]] = 3 + 3x0 >= 3 + 3x0 = [[mark(cons(_x0, _x1))]] [[s(mark(_x0))]] = 9 + 9x0 >= 3 + 9x0 = [[mark(s(_x0))]] [[adx(mark(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[mark(adx(_x0))]] [[head(mark(_x0))]] = 9 + 9x0 >= 3 + 9x0 = [[mark(head(_x0))]] [[tail(mark(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[mark(tail(_x0))]] [[proper(incr(_x0))]] = 0 >= 0 = [[incr(proper(_x0))]] [[proper(cons(_x0, _x1))]] = 0 >= 0 = [[cons(proper(_x0), proper(_x1))]] [[proper(s(_x0))]] = 0 >= 0 = [[s(proper(_x0))]] [[proper(adx(_x0))]] = 0 >= 0 = [[adx(proper(_x0))]] [[proper(head(_x0))]] = 0 >= 0 = [[head(proper(_x0))]] [[proper(tail(_x0))]] = 0 >= 0 = [[tail(proper(_x0))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_10, R_2) by ({}, R_2). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative), (P_7, R_0, minimal, formative) and (P_8, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_8, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(proper#) = 1 Thus, we can orient the dependency pairs as follows: nu(proper#(incr(X))) = incr(X) |> X = nu(proper#(X)) nu(proper#(cons(X, Y))) = cons(X, Y) |> X = nu(proper#(X)) nu(proper#(cons(X, Y))) = cons(X, Y) |> Y = nu(proper#(Y)) nu(proper#(s(X))) = s(X) |> X = nu(proper#(X)) nu(proper#(adx(X))) = adx(X) |> X = nu(proper#(X)) nu(proper#(head(X))) = head(X) |> X = nu(proper#(X)) nu(proper#(tail(X))) = tail(X) |> X = nu(proper#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_8, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative), (P_6, R_0, minimal, formative) and (P_7, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_7, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(tail#) = 1 Thus, we can orient the dependency pairs as follows: nu(tail#(mark(X))) = mark(X) |> X = nu(tail#(X)) nu(tail#(ok(X))) = ok(X) |> X = nu(tail#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_7, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative), (P_5, R_0, minimal, formative) and (P_6, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_6, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(head#) = 1 Thus, we can orient the dependency pairs as follows: nu(head#(mark(X))) = mark(X) |> X = nu(head#(X)) nu(head#(ok(X))) = ok(X) |> X = nu(head#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_6, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative), (P_4, R_0, minimal, formative) and (P_5, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(adx#) = 1 Thus, we can orient the dependency pairs as follows: nu(adx#(mark(X))) = mark(X) |> X = nu(adx#(X)) nu(adx#(ok(X))) = ok(X) |> X = nu(adx#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_5, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative), (P_3, R_0, minimal, formative) and (P_4, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(s#) = 1 Thus, we can orient the dependency pairs as follows: nu(s#(mark(X))) = mark(X) |> X = nu(s#(X)) nu(s#(ok(X))) = ok(X) |> X = nu(s#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_4, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(cons#) = 1 Thus, we can orient the dependency pairs as follows: nu(cons#(mark(X), Y)) = mark(X) |> X = nu(cons#(X, Y)) nu(cons#(ok(X), ok(Y))) = ok(X) |> X = nu(cons#(X, Y)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_3, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(incr#) = 1 Thus, we can orient the dependency pairs as follows: nu(incr#(mark(X))) = mark(X) |> X = nu(incr#(X)) nu(incr#(ok(X))) = ok(X) |> X = nu(incr#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(active#) = 1 Thus, we can orient the dependency pairs as follows: nu(active#(incr(X))) = incr(X) |> X = nu(active#(X)) nu(active#(cons(X, Y))) = cons(X, Y) |> X = nu(active#(X)) nu(active#(s(X))) = s(X) |> X = nu(active#(X)) nu(active#(adx(X))) = adx(X) |> X = nu(active#(X)) nu(active#(head(X))) = head(X) |> X = nu(active#(X)) nu(active#(tail(X))) = tail(X) |> X = nu(active#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009.