Termination w.r.t. Q proof of TRS/TRCSREx1_Luc02b

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:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(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:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(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.

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

0 > [from₁, first₂, nil, activate_1] > s₁ > [cons₂, n_{_from₁}, n_{_{first_2]}}
sel₂ > [from₁, first₂, nil, activate_1] > s₁ > [cons₂, n_{_from₁}, n_{_{first_2]}}

Status:

from₁: [1]
sel₂: [1,2]
first₂: [2,1]
n_{_from₁}: [1]
s₁: [1]
0: multiset
nil: multiset
cons₂: [1,2]
n_{_first₂}: [2,1]
activate₁: [1]