Termination w.r.t. Q proof of TRS/TRCSREx8_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:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(X1, X2)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DirectTerminationProof

Q restricted rewrite system:
The TRS R consists of the following rules:

fib(N) → sel(N, fib1(s(0), s(0)))
fib1(X, Y) → cons(X, n__fib1(Y, add(X, Y)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
fib1(X1, X2) → n__fib1(X1, X2)
activate(n__fib1(X1, X2)) → fib1(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:

fib₁ > sel₂ > activate₁ > fib1₂ > add₂ > s₁ > [cons₂, n_{_{fib1_2]}}
fib₁ > 0 > [cons₂, n_{_{fib1_2]}}

Status:

sel₂: [1,2]
add₂: [1,2]
fib₁: [1]
s₁: [1]
0: multiset
cons₂: [1,2]
fib1₂: [1,2]
n_{_fib1₂}: [2,1]
activate₁: [1]