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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F2(f2(a, f2(x, a)), a) -> F2(f2(x, a), a)
Used argument filtering: F2(x1, x2)  =  x1
f2(x1, x2)  =  f2(x1, x2)
a  =  a
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.