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:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive Path Order [RPO].
Precedence:

[first₂, activate₁, from₁] > s₁ > [nil, n_first2]
[first₂, activate₁, from₁] > cons₂ > [nil, n_first2]
[first₂, activate₁, from₁] > n_from1 > [nil, n_first2]
0 > [nil, n_first2]

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

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
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