Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(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, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X

Q is empty.


QTRS
  ↳ DirectTerminationProof

Q restricted rewrite system:
The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(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, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X

Q is empty.

We use [23] with the following order to prove termination.

Recursive path order with status [2].
Quasi-Precedence:
2nd1 > head1
2nd1 > [activate1, take2] > from1 > cons2
2nd1 > [activate1, take2] > from1 > nfrom1
2nd1 > [activate1, take2] > from1 > s1
2nd1 > [activate1, take2] > nil
2nd1 > [activate1, take2] > ntake2
sel2 > [activate1, take2] > from1 > cons2
sel2 > [activate1, take2] > from1 > nfrom1
sel2 > [activate1, take2] > from1 > s1
sel2 > [activate1, take2] > nil
sel2 > [activate1, take2] > ntake2

Status:
sel2: [1,2]
activate1: [1]
take2: [2,1]