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(ns(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(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(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)
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(activate(X))
ACTIVATE(ninf(X)) -> ACTIVATE(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(activate(X1), activate(X2))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nlength(X)) -> LENGTH(activate(X))
ACTIVATE(nlength(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

ACTIVATE(nlength(X)) -> ACTIVATE(X)
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(activate(X))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ninf(X)) -> ACTIVATE(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(ns(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(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(X))
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(ninf(X)) -> ACTIVATE(X)


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(X))
activate(X) -> 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)
inf(X) -> cons(X, ninf(ns(X)))
inf(X) -> ninf(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(activate(x1))=  x1  
  POL(n__length(x1))=  x1  
  POL(n__take(x1, x2))=  x1 + x2  
  POL(take(x1, x2))=  x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(n__inf(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(inf(x1))=  1 + x1  
  POL(nil)=  0  
  POL(n__0)=  0  
  POL(s(x1))=  x1  
  POL(LENGTH(x1))=  x1  
  POL(length(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(nlength(X)) -> ACTIVATE(X)
LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(nlength(X)) -> LENGTH(activate(X))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

activate(n0) -> 0
activate(ns(X)) -> s(X)
activate(ninf(X)) -> inf(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(nlength(X)) -> length(activate(X))
activate(X) -> 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)
inf(X) -> cons(X, ninf(ns(X)))
inf(X) -> ninf(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(activate(x1))=  x1  
  POL(n__length(x1))=  1 + x1  
  POL(n__take(x1, x2))=  x1 + x2  
  POL(take(x1, x2))=  x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(n__inf(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(inf(x1))=  0  
  POL(nil)=  0  
  POL(n__0)=  0  
  POL(s(x1))=  x1  
  POL(LENGTH(x1))=  x1  
  POL(length(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pairs:

LENGTH(cons(X, L)) -> ACTIVATE(L)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)


Rules:


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


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 5
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polo
             ...
               →DP Problem 6
                 ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


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


Strategy:

innermost



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