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)), 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.

↳ QTRS

  ↳ DirectTerminationProof

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.

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

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

0 > nil
[first₂, activate_1] > terms₁ > cons₂ > n_{_first₂}
[first₂, activate_1] > terms₁ > recip₁
[first₂, activate_1] > terms₁ > sqr₁ > add₂ > [s₁, half_1]
[first₂, activate_1] > terms₁ > sqr₁ > dbl₁ > [s₁, half_1]
[first₂, activate_1] > terms₁ > n_{_terms₁}
[first₂, activate_1] > nil

Status:

dbl₁: multiset
first₂: [2,1]
0: multiset
nil: multiset
n_{_terms₁}: [1]
cons₂: multiset
activate₁: [1]
n_{_first₂}: multiset
half₁: multiset
sqr₁: multiset
recip₁: [1]
terms₁: multiset
add₂: multiset
s₁: multiset