Termination w.r.t. Q proof of TRS/TRCSRExSec11_1_Luc02a

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.

↳ QTRS

  ↳ DirectTerminationProof

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.

We use [23] with the following order to prove termination.

Recursive path order with status [2].
Quasi-Precedence:

terms₁ > cons₁
terms₁ > recip₁
terms₁ > sqr₁ > 0 > nil
terms₁ > sqr₁ > add₂ > [s₁, half_1]
terms₁ > sqr₁ > dbl₁ > [s₁, half_1]

Status:

dbl₁: multiset
half₁: multiset
first₂: [1,2]
terms₁: multiset
recip₁: multiset
sqr₁: multiset
add₂: multiset
0: multiset
s₁: multiset
cons₁: [1]
nil: multiset