Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

f(X) → cons(X, n__f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
f(x1)  =  f(x1)
cons(x1, x2)  =  cons(x1, x2)
n__f(x1)  =  x1
g(x1)  =  g(x1)
0  =  0
s(x1)  =  s(x1)
sel(x1, x2)  =  sel(x1, x2)
activate(x1)  =  activate(x1)

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

sel₂ > [f₁, activate₁] > cons₂ > s₁
sel₂ > [f₁, activate₁] > g₁ > 0 > s₁

Status:

sel₂: [1,2]
cons₂: [2,1]
f₁: multiset
g₁: [1]
activate₁: multiset
s₁: multiset
0: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(X) → cons(X, n__f(g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
activate(X) → X

(2) Obligation:

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

activate(n__f(X)) → f(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(activate(x₁)) = 1 + x₁   
POL(f(x₁)) = x₁   
POL(n__f(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

activate(n__f(X)) → f(X)

(4) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(5) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(6) TRUE

(7) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(8) TRUE

(9) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(10) TRUE