Term Rewriting System R:
[X, Z, Y, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





The following dependency pair can be strictly oriented:

ACTIVATE(nfirst(X1, X2)) -> FIRST(X1, X2)


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(cons(x1, x2))=  x2  
  POL(FIRST(x1, x2))=  x2  
  POL(s(x1))=  0  
  POL(ACTIVATE(x1))=  x1  
  POL(n__first(x1, x2))=  1 + x2  

resulting in one new DP problem. 



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


Dependency Pair:

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


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, Z)) -> SEL(X, activate(Z))


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))=  0  
  POL(from(x1))=  0  
  POL(activate(x1))=  0  
  POL(first(x1, x2))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(SEL(x1, x2))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(n__first(x1, x2))=  0  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
sel(0, cons(X, Z)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
activate(nfrom(X)) -> from(X)
activate(nfirst(X1, X2)) -> first(X1, X2)
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

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