Termination w.r.t. Q proof of TRS/SK902.24.trs

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, 3x₁ - 2}

POL( s₁(x₁) ) = 3x₁ + 3

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.