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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• +¹(x, s(y)) → +¹(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

• G(s(x)) → G(x) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE