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
         ↳AFS


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 pairs can be strictly oriented:

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


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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{0, nil}
{take, ntake} > {activate, from, 2nd} > nfrom
{take, ntake} > {activate, from, 2nd} > cons
{take, ntake} > {activate, from, 2nd} > head
{take, ntake} > {activate, from, 2nd} > s
{take, ntake} > TAKE > ACTIVATE
sel > {activate, from, 2nd} > nfrom
sel > {activate, from, 2nd} > cons
sel > {activate, from, 2nd} > head
sel > {activate, from, 2nd} > s

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


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


Dependency Pair:


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
         ↳Argument Filtering and Ordering


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





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{take, ntake} > {activate, from} > nfrom
{take, ntake} > {activate, from} > cons
{take, ntake} > {activate, from} > s
{take, ntake} > nil
2nd > {activate, from} > nfrom
2nd > {activate, from} > cons
2nd > {activate, from} > s
2nd > head
SEL > {activate, from} > nfrom
SEL > {activate, from} > cons
SEL > {activate, from} > s
sel > {activate, from} > nfrom
sel > {activate, from} > cons
sel > {activate, from} > s

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


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


Dependency Pair:


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.

Termination of R successfully shown.
Duration:
0:02 minutes