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:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

[fst₂, activate₁, add₂, len₁] > [0, nil]
[fst₂, activate₁, add₂, len₁] > from₁ > cons₂ > [s₁, n_fst2, n_add2]
[fst₂, activate₁, add₂, len₁] > from₁ > cons₂ > n_len1
[fst₂, activate₁, add₂, len₁] > from₁ > n_from1

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(X1, X2)
activate(n__from(X)) → from(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__len(X)) → len(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