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) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(X) → G(X)
G(s(X)) → G(X)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__f(X)) → F(X)

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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(s(X)) → G(X)

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.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(s(X)) → G(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  x1
s(x1)  =  s(x1)
f(x1)  =  f(x1)
cons(x1, x2)  =  cons(x1, x2)
n__f(x1)  =  x1
g(x1)  =  g(x1)
0  =  0
sel(x1, x2)  =  sel(x1, x2)
activate(x1)  =  activate(x1)

Recursive path order with status [RPO].
Precedence:

sel₂ > activate₁ > f₁ > cons₂
sel₂ > activate₁ > f₁ > g₁ > s₁

Status:

sel₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

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.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x1
s(x1)  =  s(x1)
f(x1)  =  f(x1)
cons(x1, x2)  =  cons(x1, x2)
n__f(x1)  =  x1
g(x1)  =  g(x1)
0  =  0
sel(x1, x2)  =  sel(x1, x2)
activate(x1)  =  activate(x1)

Recursive path order with status [RPO].
Precedence:

sel₂ > activate₁ > f₁ > cons₂
sel₂ > activate₁ > f₁ > g₁ > s₁

Status:

sel₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE