Term Rewriting System R:
[X, Z, N, Y, X1, X2]
from(X) -> cons(X, nfrom(s(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)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ncons(X1, X2)) -> cons(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(s(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) -> 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)
ACTIVATE(ncons(X1, X2)) -> CONS(X1, X2)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


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(s(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)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(X) -> X





The following dependency pairs can be strictly oriented:

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


Additionally, the following rules can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1  
  POL(activate(x1))=  1 + x1  
  POL(plus(x1, x2))=  x2  
  POL(square(x1))=  0  
  POL(2NDSNEG(x1, x2))=  x2  
  POL(negrecip(x1))=  0  
  POL(posrecip(x1))=  0  
  POL(rcons(x1, x2))=  0  
  POL(2ndspos(x1, x2))=  0  
  POL(2ndsneg(x1, x2))=  0  
  POL(n__from(x1))=  0  
  POL(n__cons(x1, x2))=  1 + x2  
  POL(rnil)=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(times(x1, x2))=  0  
  POL(s(x1))=  0  
  POL(pi(x1))=  0  
  POL(2NDSPOS(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Remaining
       →DP Problem 3
         ↳Remaining


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(s(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)
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(X)
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Remaining Obligation(s)
       →DP Problem 3
         ↳Remaining Obligation(s)




The following remains to be proven:

Termination of R could not be shown.
Duration:
0:15 minutes