Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/secret2005SLASHmb2.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₂(f₂(x, a), y) -> f₂(f₂(a, y), f₂(a, x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(f₂(x, a), y) -> f₂(f₂(a, y), f₂(a, x))

Q is empty.

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

F₂(f₂(x, a), y) -> F₂(f₂(a, y), f₂(a, x))
F₂(f₂(x, a), y) -> F₂(a, x)
F₂(f₂(x, a), y) -> F₂(a, y)

The TRS R consists of the following rules:

f₂(f₂(x, a), y) -> f₂(f₂(a, y), f₂(a, x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

F₂(f₂(x, a), y) -> F₂(f₂(a, y), f₂(a, x))
F₂(f₂(x, a), y) -> F₂(a, x)
F₂(f₂(x, a), y) -> F₂(a, y)

The TRS R consists of the following rules:

f₂(f₂(x, a), y) -> f₂(f₂(a, y), f₂(a, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₂(f₂(x, a), y) -> F₂(f₂(a, y), f₂(a, x))

The TRS R consists of the following rules:

f₂(f₂(x, a), y) -> f₂(f₂(a, y), f₂(a, x))

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