Term Rewriting System R:
[X, XS, N]
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

AFTER(s(N), cons(X, XS)) -> AFTER(N, activate(XS))
AFTER(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

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


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X





The following dependency pair can be strictly oriented:

ACTIVATE(ns(X)) -> ACTIVATE(X)


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))=  x1  
  POL(n__s(x1))=  1 + x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

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


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


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))=  1 + x1  
  POL(ACTIVATE(x1))=  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
         ↳Polo


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
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:

AFTER(s(N), cons(X, XS)) -> AFTER(N, activate(XS))


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X





The following dependency pair can be strictly oriented:

AFTER(s(N), cons(X, XS)) -> AFTER(N, activate(XS))


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(AFTER(x1, x2))=  x1  
  POL(cons(x1, x2))=  0  
  POL(n__s(x1))=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
after(0, XS) -> XS
after(s(N), cons(X, XS)) -> after(N, activate(XS))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

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