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:

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.

(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:

FIB(N) → SEL(N, fib1(s(0), s(0)))
FIB(N) → FIB1(s(0), s(0))
FIB1(X, Y) → ADD(X, Y)
ADD(s(X), Y) → ADD(X, Y)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
ACTIVATE(n__fib1(X1, X2)) → FIB1(X1, X2)

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 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 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

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 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.

ADD(s(X), Y) → ADD(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  ADD(x1)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > ADD₁

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
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 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(N), cons(X, XS)) → SEL(N, activate(XS))

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 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(N), cons(X, XS)) → SEL(N, activate(XS))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  x2
activate(x1)  =  activate(x1)
n__fib1(x1, x2)  =  x2
fib1(x1, x2)  =  fib1(x2)
add(x1, x2)  =  add(x1, x2)
0  =  0

Recursive Path Order [RPO].
Precedence:

activate₁ > fib1₁ > SEL₁
add₂ > s₁ > SEL₁
0 > SEL₁

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

Q DP problem:
P is empty.
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 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