Term Rewriting System R:
[X, Y, Z, X1, X2]
active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

ACTIVE(from(X)) -> CONS(X, from(s(X)))
ACTIVE(from(X)) -> FROM(s(X))
ACTIVE(from(X)) -> S(X)
ACTIVE(2nd(X)) -> 2ND(active(X))
ACTIVE(2nd(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
2ND(mark(X)) -> 2ND(X)
2ND(ok(X)) -> 2ND(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
PROPER(2nd(X)) -> 2ND(proper(X))
PROPER(2nd(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(from(X)) -> FROM(proper(X))
PROPER(from(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains seven SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

CONS(ok(X1), ok(X2)) -> CONS(X1, X2)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(mark(x1)) =  0 POL(ok(x1)) =  1 + x1 POL(CONS(x1, x2)) =  x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 8`
`             ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

CONS(mark(X1), X2) -> CONS(X1, X2)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

CONS(mark(X1), X2) -> CONS(X1, X2)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(mark(x1)) =  1 + x1 POL(CONS(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 8`
`             ↳Polo`
`             ...`
`               →DP Problem 9`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

FROM(ok(X)) -> FROM(X)
FROM(mark(X)) -> FROM(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

FROM(ok(X)) -> FROM(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(FROM(x1)) =  x1 POL(mark(x1)) =  x1 POL(ok(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 10`
`             ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

FROM(mark(X)) -> FROM(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

FROM(mark(X)) -> FROM(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(FROM(x1)) =  x1 POL(mark(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 10`
`             ↳Polo`
`             ...`
`               →DP Problem 11`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

S(ok(X)) -> S(X)
S(mark(X)) -> S(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

S(ok(X)) -> S(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(S(x1)) =  x1 POL(mark(x1)) =  x1 POL(ok(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 12`
`             ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

S(mark(X)) -> S(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

S(mark(X)) -> S(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(S(x1)) =  x1 POL(mark(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 12`
`             ↳Polo`
`             ...`
`               →DP Problem 13`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

2ND(ok(X)) -> 2ND(X)
2ND(mark(X)) -> 2ND(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

2ND(ok(X)) -> 2ND(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(2ND(x1)) =  x1 POL(mark(x1)) =  x1 POL(ok(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 14`
`             ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

2ND(mark(X)) -> 2ND(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

2ND(mark(X)) -> 2ND(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(2ND(x1)) =  x1 POL(mark(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 14`
`             ↳Polo`
`             ...`
`               →DP Problem 15`
`                 ↳Dependency Graph`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polynomial Ordering`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(2nd(X)) -> ACTIVE(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(2nd(X)) -> ACTIVE(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  x1 POL(ACTIVE(x1)) =  x1 POL(2nd(x1)) =  1 + x1 POL(cons(x1, x2)) =  x1 POL(s(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 16`
`             ↳Polynomial Ordering`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(s(X)) -> ACTIVE(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  x1 POL(ACTIVE(x1)) =  x1 POL(cons(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 16`
`             ↳Polo`
`             ...`
`               →DP Problem 17`
`                 ↳Polynomial Ordering`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(from(X)) -> ACTIVE(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  1 + x1 POL(ACTIVE(x1)) =  x1 POL(cons(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 16`
`             ↳Polo`
`             ...`
`               →DP Problem 18`
`                 ↳Polynomial Ordering`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

ACTIVE(cons(X1, X2)) -> ACTIVE(X1)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(ACTIVE(x1)) =  x1 POL(cons(x1, x2)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 16`
`             ↳Polo`
`             ...`
`               →DP Problem 19`
`                 ↳Dependency Graph`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polynomial Ordering`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(2nd(X)) -> PROPER(X)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(2nd(X)) -> PROPER(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  x1 POL(2nd(x1)) =  1 + x1 POL(PROPER(x1)) =  x1 POL(cons(x1, x2)) =  x1 + x2 POL(s(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`           →DP Problem 20`
`             ↳Polynomial Ordering`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

PROPER(s(X)) -> PROPER(X)
PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(s(X)) -> PROPER(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  x1 POL(PROPER(x1)) =  x1 POL(cons(x1, x2)) =  x1 + x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`           →DP Problem 20`
`             ↳Polo`
`             ...`
`               →DP Problem 21`
`                 ↳Polynomial Ordering`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

PROPER(from(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pair can be strictly oriented:

PROPER(from(X)) -> PROPER(X)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  1 + x1 POL(PROPER(x1)) =  x1 POL(cons(x1, x2)) =  x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`           →DP Problem 20`
`             ↳Polo`
`             ...`
`               →DP Problem 22`
`                 ↳Polynomial Ordering`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pairs:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(cons(X1, X2)) -> PROPER(X1)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(PROPER(x1)) =  x1 POL(cons(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`           →DP Problem 20`
`             ↳Polo`
`             ...`
`               →DP Problem 23`
`                 ↳Dependency Graph`
`       →DP Problem 7`
`         ↳Nar`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Narrowing Transformation`

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(X)) -> TOP(proper(X))
four new Dependency Pairs are created:

TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Narrowing Transformation`

Dependency Pairs:

TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
TOP(ok(X)) -> TOP(active(X))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(X)) -> TOP(active(X))
six new Dependency Pairs are created:

TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 25`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(2nd(X''))) -> TOP(2nd(proper(X'')))
four new Dependency Pairs are created:

TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 26`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
TOP(ok(s(X''))) -> TOP(s(active(X'')))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(cons(X1', X2'))) -> TOP(cons(proper(X1'), proper(X2')))
eight new Dependency Pairs are created:

TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 27`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(from(X''))) -> TOP(from(proper(X'')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(from(X''))) -> TOP(from(proper(X'')))
four new Dependency Pairs are created:

TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 28`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(X''))) -> TOP(s(proper(X'')))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(mark(s(X''))) -> TOP(s(proper(X'')))
four new Dependency Pairs are created:

TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 29`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(2nd(X''))) -> TOP(2nd(active(X'')))
six new Dependency Pairs are created:

TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 30`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(cons(X1', X2'))) -> TOP(cons(active(X1'), X2'))
six new Dependency Pairs are created:

TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 31`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(ok(from(X''))) -> TOP(from(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(from(X''))) -> TOP(from(active(X'')))
six new Dependency Pairs are created:

TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 32`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(ok(s(X''))) -> TOP(s(active(X'')))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP(ok(s(X''))) -> TOP(s(active(X'')))
six new Dependency Pairs are created:

TOP(ok(s(2nd(cons(X', cons(Y', Z')))))) -> TOP(s(mark(Y')))
TOP(ok(s(from(X')))) -> TOP(s(mark(cons(X', from(s(X'))))))
TOP(ok(s(2nd(X')))) -> TOP(s(2nd(active(X'))))
TOP(ok(s(cons(X1', X2')))) -> TOP(s(cons(active(X1'), X2')))
TOP(ok(s(from(X')))) -> TOP(s(from(active(X'))))
TOP(ok(s(s(X')))) -> TOP(s(s(active(X'))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 33`
`                 ↳Polynomial Ordering`

Dependency Pairs:

TOP(ok(s(s(X')))) -> TOP(s(s(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(from(active(X'))))
TOP(ok(s(cons(X1', X2')))) -> TOP(s(cons(active(X1'), X2')))
TOP(ok(s(2nd(X')))) -> TOP(s(2nd(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(mark(cons(X', from(s(X'))))))
TOP(ok(s(2nd(cons(X', cons(Y', Z')))))) -> TOP(s(mark(Y')))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

TOP(ok(s(s(X')))) -> TOP(s(s(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(from(active(X'))))
TOP(ok(s(cons(X1', X2')))) -> TOP(s(cons(active(X1'), X2')))
TOP(ok(s(2nd(X')))) -> TOP(s(2nd(active(X'))))
TOP(ok(s(from(X')))) -> TOP(s(mark(cons(X', from(s(X'))))))
TOP(ok(s(2nd(cons(X', cons(Y', Z')))))) -> TOP(s(mark(Y')))
TOP(ok(from(from(X')))) -> TOP(from(from(active(X'))))
TOP(ok(from(cons(X1', X2')))) -> TOP(from(cons(active(X1'), X2')))
TOP(ok(from(2nd(X')))) -> TOP(from(2nd(active(X'))))
TOP(ok(from(from(X')))) -> TOP(from(mark(cons(X', from(s(X'))))))
TOP(ok(from(2nd(cons(X', cons(Y', Z')))))) -> TOP(from(mark(Y')))
TOP(ok(cons(s(X'), X2'))) -> TOP(cons(s(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(from(active(X')), X2'))
TOP(ok(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(active(X1''), X2''), X2'))
TOP(ok(cons(2nd(X'), X2'))) -> TOP(cons(2nd(active(X')), X2'))
TOP(ok(cons(from(X'), X2'))) -> TOP(cons(mark(cons(X', from(s(X')))), X2'))
TOP(ok(cons(2nd(cons(X', cons(Y', Z'))), X2'))) -> TOP(cons(mark(Y'), X2'))
TOP(ok(2nd(s(X')))) -> TOP(2nd(s(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(from(active(X'))))
TOP(ok(2nd(cons(X1', X2')))) -> TOP(2nd(cons(active(X1'), X2')))
TOP(ok(2nd(2nd(X')))) -> TOP(2nd(2nd(active(X'))))
TOP(ok(2nd(from(X')))) -> TOP(2nd(mark(cons(X', from(s(X'))))))
TOP(ok(2nd(2nd(cons(X', cons(Y', Z')))))) -> TOP(2nd(mark(Y')))
TOP(ok(from(X''))) -> TOP(mark(cons(X'', from(s(X'')))))
TOP(ok(2nd(cons(X'', cons(Y', Z'))))) -> TOP(mark(Y'))
TOP(ok(from(s(X')))) -> TOP(from(s(active(X'))))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  x1 POL(active(x1)) =  0 POL(proper(x1)) =  0 POL(2nd(x1)) =  x1 POL(cons(x1, x2)) =  x1 POL(s(x1)) =  x1 POL(mark(x1)) =  0 POL(ok(x1)) =  1 POL(TOP(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 34`
`                 ↳Polynomial Ordering`

Dependency Pairs:

TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

The following dependency pairs can be strictly oriented:

TOP(mark(s(s(X')))) -> TOP(s(s(proper(X'))))
TOP(mark(s(from(X')))) -> TOP(s(from(proper(X'))))
TOP(mark(s(cons(X1', X2')))) -> TOP(s(cons(proper(X1'), proper(X2'))))
TOP(mark(s(2nd(X')))) -> TOP(s(2nd(proper(X'))))
TOP(mark(from(s(X')))) -> TOP(from(s(proper(X'))))
TOP(mark(from(from(X')))) -> TOP(from(from(proper(X'))))
TOP(mark(from(cons(X1', X2')))) -> TOP(from(cons(proper(X1'), proper(X2'))))
TOP(mark(from(2nd(X')))) -> TOP(from(2nd(proper(X'))))
TOP(mark(cons(X1', s(X')))) -> TOP(cons(proper(X1'), s(proper(X'))))
TOP(mark(cons(X1', from(X')))) -> TOP(cons(proper(X1'), from(proper(X'))))
TOP(mark(cons(X1', cons(X1'', X2'')))) -> TOP(cons(proper(X1'), cons(proper(X1''), proper(X2''))))
TOP(mark(cons(X1', 2nd(X')))) -> TOP(cons(proper(X1'), 2nd(proper(X'))))
TOP(mark(cons(s(X'), X2'))) -> TOP(cons(s(proper(X')), proper(X2')))
TOP(mark(cons(from(X'), X2'))) -> TOP(cons(from(proper(X')), proper(X2')))
TOP(mark(cons(cons(X1'', X2''), X2'))) -> TOP(cons(cons(proper(X1''), proper(X2'')), proper(X2')))
TOP(mark(cons(2nd(X'), X2'))) -> TOP(cons(2nd(proper(X')), proper(X2')))
TOP(mark(2nd(s(X')))) -> TOP(2nd(s(proper(X'))))
TOP(mark(2nd(from(X')))) -> TOP(2nd(from(proper(X'))))
TOP(mark(2nd(cons(X1', X2')))) -> TOP(2nd(cons(proper(X1'), proper(X2'))))
TOP(mark(2nd(2nd(X')))) -> TOP(2nd(2nd(proper(X'))))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(from(x1)) =  x1 POL(proper(x1)) =  0 POL(2nd(x1)) =  x1 POL(cons(x1, x2)) =  x1 POL(s(x1)) =  x1 POL(mark(x1)) =  1 POL(ok(x1)) =  0 POL(TOP(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`       →DP Problem 6`
`         ↳Polo`
`       →DP Problem 7`
`         ↳Nar`
`           →DP Problem 24`
`             ↳Nar`
`             ...`
`               →DP Problem 35`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

active(2nd(cons(X, cons(Y, Z)))) -> mark(Y)
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:40 minutes