Term Rewriting System R:
[Z, X, Y, X1, X2]
fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ADD(s(X), Y) -> S(nadd(activate(X), Y))
ADD(s(X), Y) -> ACTIVATE(X)
LEN(cons(X, Z)) -> S(nlen(activate(Z)))
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(nadd(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nlen(X)) -> LEN(activate(X))
ACTIVATE(nlen(X)) -> ACTIVATE(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

ACTIVATE(nlen(X)) -> ACTIVATE(X)
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nlen(X)) -> LEN(activate(X))
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





The following dependency pairs can be strictly oriented:

ACTIVATE(nlen(X)) -> ACTIVATE(X)
ACTIVATE(nlen(X)) -> LEN(activate(X))


Additionally, the following rules can be oriented:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(activate(x1))=  x1  
  POL(FST(x1, x2))=  x1 + x2  
  POL(len(x1))=  1 + x1  
  POL(n__fst(x1, x2))=  x1 + x2  
  POL(LEN(x1))=  x1  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(ADD(x1, x2))=  x1  
  POL(add(x1, x2))=  x1 + x2  
  POL(n__from(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(fst(x1, x2))=  x1 + x2  
  POL(n__len(x1))=  1 + x1  
  POL(n__add(x1, x2))=  x1 + x2  

resulting in one new DP problem. 



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


Dependency Pairs:

LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





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


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


Dependency Pairs:

FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X1)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nadd(X1, X2)) -> ADD(activate(X1), activate(X2))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





The following dependency pairs can be strictly oriented:

ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nadd(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nadd(X1, X2)) -> ADD(activate(X1), activate(X2))


Additionally, the following rules can be oriented:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  x1  
  POL(activate(x1))=  x1  
  POL(FST(x1, x2))=  x1 + x2  
  POL(len(x1))=  0  
  POL(n__fst(x1, x2))=  x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(ADD(x1, x2))=  x1  
  POL(add(x1, x2))=  1 + x1 + x2  
  POL(n__from(x1))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(fst(x1, x2))=  x1 + x2  
  POL(n__len(x1))=  0  
  POL(n__add(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



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


Dependency Pairs:

FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ADD(s(X), Y) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





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


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


Dependency Pairs:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(activate(x1))=  x1  
  POL(FST(x1, x2))=  x1 + x2  
  POL(len(x1))=  0  
  POL(n__fst(x1, x2))=  x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(add(x1, x2))=  x2  
  POL(n__from(x1))=  1 + x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(fst(x1, x2))=  x1 + x2  
  POL(n__len(x1))=  0  
  POL(n__add(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pairs:

ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





The following dependency pairs can be strictly oriented:

ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfst(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nfst(X1, X2)) -> FST(activate(X1), activate(X2))


Additionally, the following rules can be oriented:

fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  0  
  POL(activate(x1))=  x1  
  POL(FST(x1, x2))=  x1 + x2  
  POL(len(x1))=  0  
  POL(n__fst(x1, x2))=  1 + x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(add(x1, x2))=  x2  
  POL(n__from(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  x2  
  POL(nil)=  0  
  POL(s(x1))=  x1  
  POL(fst(x1, x2))=  1 + x1 + x2  
  POL(n__len(x1))=  0  
  POL(n__add(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pairs:

FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


fst(0, Z) -> nil
fst(s(X), cons(Y, Z)) -> cons(Y, nfst(activate(X), activate(Z)))
fst(X1, X2) -> nfst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
add(0, X) -> X
add(s(X), Y) -> s(nadd(activate(X), Y))
add(X1, X2) -> nadd(X1, X2)
len(nil) -> 0
len(cons(X, Z)) -> s(nlen(activate(Z)))
len(X) -> nlen(X)
s(X) -> ns(X)
activate(nfst(X1, X2)) -> fst(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(X)
activate(nadd(X1, X2)) -> add(activate(X1), activate(X2))
activate(nlen(X)) -> len(activate(X))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes