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:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic Path Order [LPO].
Precedence:

fib₁ > sel₂ > activate₁ > fib1₂ > cons₂ > s₁
fib₁ > sel₂ > activate₁ > fib1₂ > n_fib12 > s₁
fib₁ > sel₂ > activate₁ > fib1₂ > n_add2 > s₁
fib₁ > sel₂ > activate₁ > add₂ > n_add2 > s₁
fib₁ > 0 > s₁

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

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, n__add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
add(X1, X2) → n__add(X1, X2)
activate(n__fib1(X1, X2)) → fib1(activate(X1), activate(X2))
activate(n__add(X1, X2)) → add(activate(X1), activate(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