(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) =
x2
s(
x1) =
s(
x1)
fib(
x1) =
fib(
x1)
0 =
0
+(
x1,
x2) =
+(
x1,
x2)
Recursive Path Order [RPO].
Precedence:
[fib1, 0] > +2 > 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))
(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: Recursive Path Order [RPO].
Precedence:
[fib1, 0] > +2 > [FIB1, 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