Term Rewriting System R:
[X, XS, N, X1, X2]
from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(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(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(ntake(X1, X2)) -> TAKE(activate(X1), activate(X2))
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ntake(X1, X2)) -> ACTIVATE(X2)

Furthermore, R contains two SCCs.


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


Dependency Pairs:

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


Rules:


from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__from(x1))=  1 + x1  
  POL(from(x1))=  1 + x1  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(n__take(x1, x2))=  x1 + x2  
  POL(cons(x1, x2))=  x2  
  POL(take(x1, x2))=  x1 + x2  
  POL(TAKE(x1, x2))=  x2  
  POL(n__s(x1))=  x1  
  POL(nil)=  0  
  POL(s(x1))=  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 Pairs:

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


Rules:


from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
s(X) -> ns(X)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__from(x1))=  0  
  POL(from(x1))=  0  
  POL(activate(x1))=  x1  
  POL(0)=  0  
  POL(n__take(x1, x2))=  x1 + x2  
  POL(cons(x1, x2))=  x2  
  POL(take(x1, x2))=  x1 + x2  
  POL(TAKE(x1, x2))=  x2  
  POL(n__s(x1))=  1 + x1  
  POL(nil)=  0  
  POL(s(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
Polynomial Ordering
       →DP Problem 2
Polo


Dependency Pairs:

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


Rules:


from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X
take(0, XS) -> nil
take(s(N), cons(X, XS)) -> cons(X, ntake(N, activate(XS)))
take(X1, X2) -> ntake(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(X)
s(X) -> ns(X)


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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 3
Polo
             ...
               →DP Problem 5
Dependency Graph
       →DP Problem 2
Polo


Dependency Pair:

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


Rules:


from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(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(N), cons(X, XS)) -> SEL(N, activate(XS))


Rules:


from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X





The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, XS)) -> SEL(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(0)=  0  
  POL(n__take(x1, x2))=  0  
  POL(cons(x1, x2))=  0  
  POL(SEL(x1, x2))=  x1  
  POL(take(x1, x2))=  0  
  POL(n__s(x1))=  0  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 6
Dependency Graph


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(ns(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))
s(X) -> ns(X)
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(ntake(X1, X2)) -> take(activate(X1), activate(X2))
activate(X) -> X





Using the Dependency Graph resulted in no new DP problems.

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