Term Rewriting System R:
[X, Z, N, Y]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons(X, Z)) -> ACTIVATE(Z)
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSNEG(s(N), cons(X, Z)) -> ACTIVATE(Z)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> ACTIVATE(Z)
PI(X) -> 2NDSPOS(X, from(0))
PI(X) -> FROM(0)
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(X)

Furthermore, R contains three SCCs.

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

Dependency Pairs:

2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, activate(Z)))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
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, Z)) -> 2NDSPOS(s(N), cons2(X, activate(Z)))
two new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, nfrom(X''))) -> 2NDSPOS(s(N), cons2(X, from(X'')))
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons(X, nfrom(X''))) -> 2NDSPOS(s(N), cons2(X, from(X'')))
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

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

2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, activate(Z))
two new Dependency Pairs are created:

2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(X''))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, nfrom(X''))) -> 2NDSPOS(s(N), cons2(X, from(X'')))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(X''))
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
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, Z)) -> 2NDSNEG(s(N), cons2(X, activate(Z)))
two new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, nfrom(X''))) -> 2NDSNEG(s(N), cons2(X, from(X'')))
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons(X, nfrom(X''))) -> 2NDSNEG(s(N), cons2(X, from(X'')))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(X''))
2NDSPOS(s(N), cons(X, nfrom(X''))) -> 2NDSPOS(s(N), cons2(X, from(X'')))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

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

2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, activate(Z))
two new Dependency Pairs are created:

2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(X''))
2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, nfrom(X''))) -> 2NDSPOS(s(N), cons2(X, from(X'')))
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(X''))
2NDSNEG(s(N), cons(X, nfrom(X''))) -> 2NDSNEG(s(N), cons2(X, from(X'')))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(X''))
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
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, nfrom(X''))) -> 2NDSPOS(s(N), cons2(X, from(X'')))
two new Dependency Pairs are created:

2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, nfrom(X''')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Narrowing Transformation`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(X''))
2NDSNEG(s(N), cons(X, nfrom(X''))) -> 2NDSNEG(s(N), cons2(X, from(X'')))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(X''))
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

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

2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSNEG(N, from(X''))
two new Dependency Pairs are created:

2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(X''', nfrom(s(X'''))))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSNEG(N, nfrom(X'''))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(X''', nfrom(s(X'''))))
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(X''))
2NDSNEG(s(N), cons(X, nfrom(X''))) -> 2NDSNEG(s(N), cons2(X, from(X'')))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, cons(X''', nfrom(s(X''')))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
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, nfrom(X''))) -> 2NDSNEG(s(N), cons2(X, from(X'')))
two new Dependency Pairs are created:

2NDSNEG(s(N), cons(X, nfrom(X'''))) -> 2NDSNEG(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSNEG(s(N), cons(X, nfrom(X'''))) -> 2NDSNEG(s(N), cons2(X, nfrom(X''')))

The transformation is resulting in one new DP problem:

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

Dependency Pairs:

2NDSNEG(s(N), cons(X, nfrom(X'''))) -> 2NDSNEG(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(X''', nfrom(s(X'''))))
2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(X''))
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

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

2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X'')))) -> 2NDSPOS(N, from(X''))
two new Dependency Pairs are created:

2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(X''', nfrom(s(X'''))))
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSPOS(N, nfrom(X'''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Nar`
`           →DP Problem 4`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pairs:

2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(X''', nfrom(s(X'''))))
2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(X''', nfrom(s(X'''))))
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')
2NDSNEG(s(N), cons(X, nfrom(X'''))) -> 2NDSNEG(s(N), cons2(X, cons(X''', nfrom(s(X''')))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

The following dependency pairs can be strictly oriented:

2NDSPOS(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSNEG(N, cons(X''', nfrom(s(X'''))))
2NDSNEG(s(N), cons2(X, cons(Y, nfrom(X''')))) -> 2NDSPOS(N, cons(X''', nfrom(s(X'''))))
2NDSPOS(s(N), cons2(X, cons(Y, Z'))) -> 2NDSNEG(N, Z')
2NDSNEG(s(N), cons2(X, cons(Y, Z'))) -> 2NDSPOS(N, Z')

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(cons(x1, x2)) =  0 POL(2NDSNEG(x1, x2)) =  x1 POL(s(x1)) =  1 + x1 POL(cons2(x1, x2)) =  0 POL(2NDSPOS(x1, x2)) =  x1

resulting in one new DP problem.

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

Dependency Pairs:

2NDSPOS(s(N), cons(X, nfrom(X'''))) -> 2NDSPOS(s(N), cons2(X, cons(X''', nfrom(s(X''')))))
2NDSNEG(s(N), cons(X, Z')) -> 2NDSNEG(s(N), cons2(X, Z'))
2NDSPOS(s(N), cons(X, Z')) -> 2NDSPOS(s(N), cons2(X, Z'))
2NDSNEG(s(N), cons(X, nfrom(X'''))) -> 2NDSNEG(s(N), cons2(X, cons(X''', nfrom(s(X''')))))

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
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`
`         ↳Nar`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 13`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

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

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
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`
`         ↳Nar`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 14`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(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)
activate(nfrom(X)) -> from(X)
activate(X) -> X

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes