Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

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

Recursive Path Order [RPO].
Precedence:
+^11 > s1
fib1 > sp1 > +2 > s1
fib1 > g1 > 0 > pair2 > s1
fib1 > g1 > np1 > pair2 > s1
fib1 > g1 > np1 > +2 > s1

The following usable rules [FROCOS05] were oriented:

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))

(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: Recursive Path Order [RPO].
Precedence:
fib1 > sp1 > +2 > s1 > 0
fib1 > g1 > np1 > pair2
fib1 > g1 > np1 > +2 > s1 > 0

The following usable rules [FROCOS05] were oriented:

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))

(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