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(s(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)
activate(nfst(X1, X2)) -> fst(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(nlen(X)) -> len(X)
activate(X) -> X

Innermost 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) -> ACTIVATE(X)
LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(X1, X2)) -> FST(X1, X2)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(nadd(X1, X2)) -> ADD(X1, X2)
ACTIVATE(nlen(X)) -> LEN(X)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FST(x1, x2))=  x1 + x2  
  POL(cons(x1, x2))=  x2  
  POL(n__fst(x1, x2))=  x1 + x2  
  POL(LEN(x1))=  x1  
  POL(s(x1))=  x1  
  POL(n__len(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(ADD(x1, x2))=  x1  
  POL(n__add(x1, x2))=  1 + x1  

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(nlen(X)) -> LEN(X)
ADD(s(X), Y) -> ACTIVATE(X)
FST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfst(X1, X2)) -> FST(X1, 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(s(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)
activate(nfst(X1, X2)) -> fst(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(nlen(X)) -> len(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 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pairs:

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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FST(x1, x2))=  x1 + x2  
  POL(cons(x1, x2))=  x2  
  POL(n__fst(x1, x2))=  1 + x1 + x2  
  POL(LEN(x1))=  x1  
  POL(s(x1))=  x1  
  POL(n__len(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  

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)
ACTIVATE(nlen(X)) -> LEN(X)
FST(s(X), cons(Y, Z)) -> ACTIVATE(X)
LEN(cons(X, 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(s(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)
activate(nfst(X1, X2)) -> fst(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(nlen(X)) -> len(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 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

LEN(cons(X, Z)) -> ACTIVATE(Z)
ACTIVATE(nlen(X)) -> LEN(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(s(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)
activate(nfst(X1, X2)) -> fst(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(nlen(X)) -> len(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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

resulting in one new DP problem. 



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


Dependency Pair:

LEN(cons(X, 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(s(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)
activate(nfst(X1, X2)) -> fst(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nadd(X1, X2)) -> add(X1, X2)
activate(nlen(X)) -> len(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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