Term Rewriting System R:
[X, Y, X1, X2]
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
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)))
LENGTH(ncons(X, Y)) -> S(length1(activate(Y)))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))
LENGTH(ncons(X, Y)) -> ACTIVATE(Y)
LENGTH1(X) -> LENGTH(activate(X))
LENGTH1(X) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nnil) -> NIL
ACTIVATE(ncons(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳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)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
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 3
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
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 3
             ↳Polo
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pair:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
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 3
             ↳Polo
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
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
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> ns(X)
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X




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