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:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic path order with status [LPO].
Quasi-Precedence:

[first₂, activate₁] > terms₁ > cons₂ > n_first2 > [n_terms1, s₁]
[first₂, activate₁] > terms₁ > recip₁ > [n_terms1, s₁]
[first₂, activate₁] > terms₁ > sqr₁ > add₂ > [n_terms1, s₁]
[first₂, activate₁] > terms₁ > sqr₁ > dbl₁ > [n_terms1, s₁]
half₁ > [0, nil] > [n_terms1, s₁]

Status:

n_first2: [2,1]
sqr₁: [1]
activate₁: [1]
first₂: [2,1]
0: []
terms₁: [1]
add₂: [2,1]
cons₂: [2,1]
half₁: [1]
dbl₁: [1]
s₁: [1]
n_terms1: [1]
nil: []
recip₁: [1]

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__first(X1, X2)) → first(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