Term Rewriting System R:
[X, XS, N, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

2ND(cons(X, XS)) -> HEAD(activate(XS))
2ND(cons(X, XS)) -> ACTIVATE(XS)
TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)
ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X





The following dependency pair can be strictly oriented:

TAKE(s(N), cons(X, XS)) -> ACTIVATE(XS)


The following rules can be oriented:

from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(from(x1))=  1 + x1  
  POL(n__from(x1))=  x1  
  POL(activate(x1))=  1 + x1  
  POL(2nd(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(take(x1))=  1 + x1  
  POL(nil)=  0  
  POL(s)=  0  
  POL(head(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem.
Used Argument Filtering System:
TAKE(x1, x2) -> x2
cons(x1, x2) -> cons(x1, x2)
ACTIVATE(x1) -> ACTIVATE(x1)
ntake(x1, x2) -> x2
from(x1) -> from(x1)
nfrom(x1) -> nfrom(x1)
s(x1) -> s
head(x1) -> head(x1)
2nd(x1) -> 2nd(x1)
activate(x1) -> activate(x1)
take(x1, x2) -> take(x2)
sel(x1, x2) -> x2


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:

ACTIVATE(ntake(X1, X2)) -> TAKE(X1, X2)


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

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


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
head(cons(X, XS)) -> X
2nd(cons(X, XS)) -> head(activate(XS))
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
activate(nfrom(X)) -> from(X)
activate(ntake(X1, X2)) -> take(X1, X2)
activate(X) -> X




Termination of R could not be shown.
Duration:
0:13 minutes