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(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))
+(x0, 0)
+(x0, s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

FIB(s(s(x))) → +¹(fib(s(x)), fib(x))
FIB(s(s(x))) → FIB(s(x))
FIB(s(s(x))) → FIB(x)
+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(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))
+(x0, 0)
+(x0, s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) 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(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))
+(x0, 0)
+(x0, s(x1))

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

(8) 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: Combined order from the following AFS and order.
+¹(x1, x2)  =  +¹(x2)
s(x1)  =  s(x1)
fib(x1)  =  fib(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [RPO].
Precedence:

fib₁ > 0 > +^1₁
fib₁ > +₂ > s₁ > +^1₁

The following usable rules [FROCOS05] were oriented:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

(9) 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(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))
+(x0, 0)
+(x0, s(x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

FIB(s(s(x))) → FIB(x)
FIB(s(s(x))) → FIB(s(x))

The TRS R consists of the following rules:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))
+(x0, 0)
+(x0, s(x1))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIB(s(s(x))) → FIB(x)
FIB(s(s(x))) → FIB(s(x))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIB(x1)  =  x1
s(x1)  =  s(x1)
fib(x1)  =  fib(x1)
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive Path Order [RPO].
Precedence:

fib₁ > +₂ > s₁
0 > s₁

The following usable rules [FROCOS05] were oriented:

fib(0) → 0
fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

(14) 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(x))) → +(fib(s(x)), fib(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))
+(x0, 0)
+(x0, s(x1))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE