Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

sel₂ > [from₁, 2nd₁, activate₁, take₂, 0, nil] > n_from1 > n_take2
sel₂ > [from₁, 2nd₁, activate₁, take₂, 0, nil] > s₁ > cons₂ > n_take2
sel₂ > [from₁, 2nd₁, activate₁, take₂, 0, nil] > head₁ > n_take2

Status:

sel₂: [1,2]
from₁: multiset
head₁: multiset
activate₁: multiset
take₂: multiset
0: multiset
cons₂: multiset
n_from1: multiset
2nd₁: multiset
n_take2: [2,1]
s₁: multiset
nil: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE