Term Rewriting System R:
[X, Z, N, Y, X1, X2]
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, nfrom(ns(X)))
2NDSPOS(s(N), cons(X, ncons(Y, Z))) -> ACTIVATE(Y)
2NDSPOS(s(N), cons(X, ncons(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons(X, ncons(Y, Z))) -> ACTIVATE(Z)
2NDSNEG(s(N), cons(X, ncons(Y, Z))) -> ACTIVATE(Y)
2NDSNEG(s(N), cons(X, ncons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons(X, ncons(Y, Z))) -> ACTIVATE(Z)
PI(X) -> 2NDSPOS(X, from(0))
PI(X) -> FROM(0)
PLUS(s(X), Y) -> S(plus(X, Y))
PLUS(s(X), Y) -> PLUS(X, Y)
TIMES(s(X), Y) -> PLUS(Y, times(X, Y))
TIMES(s(X), Y) -> TIMES(X, Y)
SQUARE(X) -> TIMES(X, X)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(ncons(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)

Furthermore, R contains four SCCs.

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

Dependency Pairs:

ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(ncons(X1, X2)) -> ACTIVATE(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(n__cons(x1, x2)) =  1 + x1 POL(n__from(x1)) =  x1 POL(n__s(x1)) =  x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

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

Dependency Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(ns(X)) -> ACTIVATE(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(n__from(x1)) =  x1 POL(n__s(x1)) =  1 + x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

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

Dependency Pair:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

ACTIVATE(nfrom(X)) -> ACTIVATE(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(n__from(x1)) =  1 + x1 POL(ACTIVATE(x1)) =  x1

resulting in one new DP problem.

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

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> 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`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pair:

PLUS(s(X), Y) -> PLUS(X, Y)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

PLUS(s(X), Y) -> PLUS(X, Y)

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(PLUS(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 8`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pairs:

2NDSNEG(s(N), cons(X, ncons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSPOS(s(N), cons(X, ncons(Y, Z))) -> 2NDSNEG(N, activate(Z))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

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

2NDSPOS(s(N), cons(X, ncons(Y, Z))) -> 2NDSNEG(N, activate(Z))
four new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSNEG(N, s(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pairs:

2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSNEG(N, s(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(activate(X'')))
2NDSNEG(s(N), cons(X, ncons(Y, Z))) -> 2NDSPOS(N, activate(Z))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

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

2NDSNEG(s(N), cons(X, ncons(Y, Z))) -> 2NDSPOS(N, activate(Z))
four new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(activate(X'')))
2NDSNEG(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSPOS(N, s(activate(X'')))
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Nar`
`             ...`
`               →DP Problem 10`
`                 ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pairs:

2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSNEG(N, s(activate(X'')))
2NDSNEG(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSPOS(N, s(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(activate(X'')))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

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

2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(activate(X'')))
six new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, nfrom(activate(X''')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSNEG(N, from(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSNEG(N, from(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSNEG(N, from(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, from(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pairs:

2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, from(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSNEG(N, from(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSNEG(N, from(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSNEG(N, from(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSPOS(N, s(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSNEG(N, s(activate(X'')))
2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

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

2NDSPOS(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSNEG(N, s(activate(X'')))
five new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSNEG(N, ns(activate(X''')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSNEG(N, s(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSNEG(N, s(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSNEG(N, s(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSNEG(N, s(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pairs:

2NDSPOS(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSNEG(N, s(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSNEG(N, s(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSNEG(N, s(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSNEG(N, s(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSNEG(N, from(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSNEG(N, from(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSNEG(N, from(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSPOS(N, s(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, from(X'''))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

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

2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(activate(X'')))
six new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, nfrom(activate(X''')))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSPOS(N, from(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSPOS(N, from(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSPOS(N, from(cons(activate(X1'), X2')))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, from(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Narrowing Transformation`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pairs:

2NDSPOS(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSNEG(N, s(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSNEG(N, s(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSNEG(N, s(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, from(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, from(X'''))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSPOS(N, from(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSNEG(N, from(cons(activate(X1'), X2')))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSPOS(N, from(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSNEG(N, from(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSPOS(N, from(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSNEG(N, from(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSPOS(N, s(activate(X'')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSNEG(N, s(X'''))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

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

2NDSNEG(s(N), cons(X, ncons(Y, ns(X'')))) -> 2NDSPOS(N, s(activate(X'')))
five new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSPOS(N, ns(activate(X''')))
2NDSNEG(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSPOS(N, s(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSPOS(N, s(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSPOS(N, s(cons(activate(X1'), X2')))
2NDSNEG(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSPOS(N, s(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pairs:

2NDSNEG(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSPOS(N, s(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSNEG(N, s(X'''))
2NDSNEG(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSPOS(N, s(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSNEG(N, s(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSPOS(N, s(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSNEG(N, s(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSPOS(N, s(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, from(X'''))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, from(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSNEG(N, from(cons(activate(X1'), X2')))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSPOS(N, from(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSNEG(N, from(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSPOS(N, from(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSNEG(N, from(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSPOS(N, from(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSNEG(N, s(cons(activate(X1'), X2')))

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

The following dependency pairs can be strictly oriented:

2NDSNEG(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSPOS(N, s(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, ns(X''')))) -> 2NDSNEG(N, s(X'''))
2NDSNEG(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSPOS(N, s(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSNEG(N, s(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, ns(ns(X'''))))) -> 2NDSPOS(N, s(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSNEG(N, s(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, ns(nfrom(X'''))))) -> 2NDSPOS(N, s(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, from(X'''))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, from(X'''))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSNEG(N, from(cons(activate(X1'), X2')))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ncons(X1', X2'))))) -> 2NDSPOS(N, from(cons(activate(X1'), X2')))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSNEG(N, from(s(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(ns(X'''))))) -> 2NDSPOS(N, from(s(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSNEG(N, from(from(activate(X'''))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(nfrom(X'''))))) -> 2NDSPOS(N, from(from(activate(X'''))))
2NDSPOS(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSNEG(s(N), cons(X, ncons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(activate(X'''), nfrom(ns(activate(X''')))))
2NDSPOS(s(N), cons(X, ncons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons(X, ncons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSPOS(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSNEG(N, cons(activate(X1'), X2'))
2NDSNEG(s(N), cons(X, ncons(Y, ncons(X1', X2')))) -> 2NDSPOS(N, cons(activate(X1'), X2'))
2NDSPOS(s(N), cons(X, ncons(Y, ns(ncons(X1', X2'))))) -> 2NDSNEG(N, s(cons(activate(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(n__from(x1)) =  0 POL(from(x1)) =  0 POL(n__cons(x1, x2)) =  0 POL(activate(x1)) =  0 POL(cons(x1, x2)) =  0 POL(2NDSNEG(x1, x2)) =  x1 POL(n__s(x1)) =  0 POL(s(x1)) =  1 + x1 POL(2NDSPOS(x1, x2)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Nar`
`           →DP Problem 9`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Polo`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

TIMES(s(X), Y) -> TIMES(X, Y)

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

The following dependency pair can be strictly oriented:

TIMES(s(X), Y) -> TIMES(X, Y)

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(TIMES(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`
`         ↳Nar`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 16`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, ncons(Y, Z))) -> rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, ncons(Y, Z))) -> rcons(negrecip(activate(Y)), 2ndspos(N, activate(Z)))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
s(X) -> ns(X)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:04 minutes