Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.