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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(f2(f2(a, f2(a, a)), a), x) -> F2(x, a)
The remaining pairs can at least be oriented weakly.

F2(f2(f2(a, f2(a, a)), a), x) -> F2(x, f2(x, a))
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = 3·x1 + 3·x2   
POL(a) = 0   
POL(f2(x1, x2)) = 3   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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