Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/D33SLASH04.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(f₁(x)) -> g₁(f₁(x))
g₁(g₁(x)) -> f₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(f₁(x)) -> g₁(f₁(x))
g₁(g₁(x)) -> f₁(x)

Q is empty.

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

F₁(f₁(x)) -> G₁(f₁(x))
G₁(g₁(x)) -> F₁(x)

The TRS R consists of the following rules:

f₁(f₁(x)) -> g₁(f₁(x))
g₁(g₁(x)) -> f₁(x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

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

F₁(f₁(x)) -> G₁(f₁(x))
G₁(g₁(x)) -> F₁(x)

The TRS R consists of the following rules:

f₁(f₁(x)) -> g₁(f₁(x))
g₁(g₁(x)) -> f₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

G₁(g₁(x)) -> F₁(x)

Used argument filtering: F₁(x1) = x1
f₁(x1) = f₁(x1)
G₁(x1) = x1
g₁(x1) = g₁(x1)
Used ordering: Quasi Precedence: [f_1, g_1]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F₁(f₁(x)) -> G₁(f₁(x))

The TRS R consists of the following rules:

f₁(f₁(x)) -> g₁(f₁(x))
g₁(g₁(x)) -> f₁(x)

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