Termination w.r.t. Q proof of TRS/AG01#3.35.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:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

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:

G₁(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

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:

G₁(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

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(x)) -> G₁(x)

The remaining pairs can at least be oriented weakly.

G₁(s₁(x)) -> F₁(x)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₁(x₁) ) = 2x₁ + 2

POL( s₁(x₁) ) = 2x₁ + 2

POL( G₁(x₁) ) = max{0, 2x₁ - 2}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.