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:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(x))) -> +2(fib1(s1(x)), fib1(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(x))) -> +2(fib1(s1(x)), fib1(x))

Q is empty.

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

FIB1(s1(s1(x))) -> FIB1(s1(x))
FIB1(s1(s1(x))) -> FIB1(x)

The TRS R consists of the following rules:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(x))) -> +2(fib1(s1(x)), fib1(x))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

FIB1(s1(s1(x))) -> FIB1(s1(x))
FIB1(s1(s1(x))) -> FIB1(x)

The TRS R consists of the following rules:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(x))) -> +2(fib1(s1(x)), fib1(x))

Q is empty.
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.


FIB1(s1(s1(x))) -> FIB1(s1(x))
FIB1(s1(s1(x))) -> FIB1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FIB1(x1) ) = max{0, x1 - 3}


POL( s1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

fib1(0) -> 0
fib1(s1(0)) -> s1(0)
fib1(s1(s1(x))) -> +2(fib1(s1(x)), fib1(x))

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