Termination w.r.t. Q proof of TRS/AG01#3.24.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:

f₁(0) -> s₁(0)
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(0) -> s₁(0)
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(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:

F₁(s₁(s₁(x))) -> F₁(f₁(s₁(x)))
F₁(s₁(s₁(x))) -> F₁(s₁(x))

The TRS R consists of the following rules:

f₁(0) -> s₁(0)
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(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:

F₁(s₁(s₁(x))) -> F₁(f₁(s₁(x)))
F₁(s₁(s₁(x))) -> F₁(s₁(x))

The TRS R consists of the following rules:

f₁(0) -> s₁(0)
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(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.

F₁(s₁(s₁(x))) -> F₁(f₁(s₁(x)))
F₁(s₁(s₁(x))) -> F₁(s₁(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

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

POL( s₁(x₁) ) = 2x₁ + 2

POL( F₁(x₁) ) = max{0, 2x₁ - 2}

POL( f₁(x₁) ) = 2

The following usable rules [14] were oriented:

f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(x)))
f₁(0) -> s₁(0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

f₁(0) -> s₁(0)
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> f₁(f₁(s₁(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.