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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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