Term Rewriting System R:
[X, Y, Z, X1]
2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
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:

2ND(cons(X, X1)) -> 2ND(cons1(X, activate(X1)))
2ND(cons(X, X1)) -> ACTIVATE(X1)
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 one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

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


Rules:


2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
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 2
             ↳Polynomial Ordering


Dependency Pair:

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


Rules:


2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
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 2
             ↳Polo
             ...
               →DP Problem 3
                 ↳Dependency Graph


Dependency Pair:


Rules:


2nd(cons1(X, cons(Y, Z))) -> Y
2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
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