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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP

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

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

The TRS R consists of the following rules:

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

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