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.

Lexicographic path order with status [19].
Quasi-Precedence:

terms₁ > cons₁ > [0, half_1]
terms₁ > recip₁ > [0, half_1]
terms₁ > sqr₁ > add₂ > s₁ > [0, half_1]
terms₁ > sqr₁ > dbl₁ > s₁ > [0, half_1]
first₂ > nil > [0, half_1]

Status:

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