Termination w.r.t. Q proof of TRS/TRCSREx7_BLR02

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)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(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)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(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:

2nd₁ > head₁
2nd₁ > [activate₁, take_2] > from₁ > cons₂
2nd₁ > [activate₁, take_2] > from₁ > n_{_from₁}
2nd₁ > [activate₁, take_2] > from₁ > s₁
2nd₁ > [activate₁, take_2] > nil
2nd₁ > [activate₁, take_2] > n_{_take₂}
sel₂ > [activate₁, take_2] > from₁ > cons₂
sel₂ > [activate₁, take_2] > from₁ > n_{_from₁}
sel₂ > [activate₁, take_2] > from₁ > s₁
sel₂ > [activate₁, take_2] > nil
sel₂ > [activate₁, take_2] > n_{_take₂}

Status:

sel₂: [1,2]
activate₁: [1]
take₂: [2,1]