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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

f2(a, f2(a, f2(b, f2(a, f2(a, f2(b, f2(a, x))))))) -> f2(a, f2(b, f2(a, f2(a, f2(b, f2(a, f2(a, f2(a, 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 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP

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

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

The TRS R consists of the following rules:

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

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