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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

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

The TRS R consists of the following rules:

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

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