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(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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(s(s(x))) → SP(g(x))
FIB(s(s(x))) → G(x)
G(s(x)) → NP(g(x))
G(s(x)) → G(x)
SP(pair(x, y)) → +¹(x, y)
NP(pair(x, y)) → +¹(x, y)
+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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:

+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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.

+¹(x, s(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(+¹(x₁, x₂)) = x₂   
POL(s(x₁)) = 1 + x₁   

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(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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:

G(s(x)) → G(x)

The TRS R consists of the following rules:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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.

G(s(x)) → G(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(G(x₁)) = x₁   
POL(s(x₁)) = 1 + x₁   

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(0) → 0
fib(s(0)) → s(0)
fib(s(s(0))) → s(0)
fib(s(s(x))) → sp(g(x))
g(0) → pair(s(0), 0)
g(s(0)) → pair(s(0), s(0))
g(s(x)) → np(g(x))
sp(pair(x, y)) → +(x, y)
np(pair(x, y)) → pair(+(x, y), x)
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

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