Term Rewriting System R:
[X, Y, L, X1, X2]
eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(s(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(nlength(X)) -> length(X)
activate(X) -> X

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

EQ(ns(X), ns(Y)) -> EQ(activate(X), activate(Y))
EQ(ns(X), ns(Y)) -> ACTIVATE(X)
EQ(ns(X), ns(Y)) -> ACTIVATE(Y)
INF(X) -> S(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(Y)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(X)
TAKE(s(X), cons(Y, L)) -> ACTIVATE(L)
LENGTH(nil) -> 0'
LENGTH(cons(X, L)) -> S(nlength(activate(L)))
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(n0) -> 0'
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(ninf(X)) -> INF(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)
ACTIVATE(nlength(X)) -> LENGTH(X)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(X)


Rules:


eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(s(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(nlength(X)) -> length(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

LENGTH(cons(X, L)) -> ACTIVATE(L)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__length(x1))=  x1  
  POL(cons(x1, x2))=  1 + x2  
  POL(ACTIVATE(x1))=  x1  
  POL(LENGTH(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

ACTIVATE(nlength(X)) -> LENGTH(X)


Rules:


eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(s(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(nlength(X)) -> length(X)
activate(X) -> X


Strategy:

innermost




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 Pair:

EQ(ns(X), ns(Y)) -> EQ(activate(X), activate(Y))


Rules:


eq(n0, n0) -> true
eq(ns(X), ns(Y)) -> eq(activate(X), activate(Y))
eq(X, Y) -> false
inf(X) -> cons(X, ninf(s(X)))
inf(X) -> ninf(X)
take(0, X) -> nil
take(s(X), cons(Y, L)) -> cons(activate(Y), ntake(activate(X), activate(L)))
take(X1, X2) -> ntake(X1, X2)
length(nil) -> 0
length(cons(X, L)) -> s(nlength(activate(L)))
length(X) -> nlength(X)
0 -> n0
s(X) -> ns(X)
activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(nlength(X)) -> length(X)
activate(X) -> X


Strategy:

innermost



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