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