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
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


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 pairs can be strictly oriented:

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


There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
ACTIVATE(x1) -> ACTIVATE(x1)
ncons(x1, x2) -> ncons(x1, x2)
ns(x1) -> ns(x1)
nfrom(x1) -> nfrom(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


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
AFS
       →DP Problem 2
Argument Filtering and Ordering
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


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 using the Ce-refinement that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 6
Dependency Graph
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


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
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:

Termination of R could not be shown.
Duration:
0:00 minutes