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