Termination w.r.t. Q of the following Term Rewriting System could be proven:

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))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


FIB(s(s(x))) → FIB(s(x))
FIB(s(s(x))) → FIB(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FIB(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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))

The set Q consists of the following terms:

fib(0)
fib(s(0))
fib(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.