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)))
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)) → cons(Y)
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

terms₁ > cons₁ > [s₁, nil]
terms₁ > recip₁ > [s₁, nil]
terms₁ > [sqr₁, 0] > add₂ > [s₁, nil]
terms₁ > [sqr₁, 0] > dbl₁ > [s₁, nil]
first₂ > [s₁, nil]
half₁ > [s₁, nil]

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)))
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)) → cons(Y)
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(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