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) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(Y)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

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

Status:

add₂: [2,1]
cons₁: multiset
sqr₁: [1]
dbl₁: [1]
s: multiset
0: multiset
first₂: multiset
terms₁: multiset
nil: multiset
recip₁: multiset

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) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(Y)

(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