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

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

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

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f2(f2(a, x0), a)


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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f2(f2(a, x0), a)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f2(f2(a, x0), a)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f2(f2(a, x0), a)

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, x), a) -> F2(x, a)
Used argument filtering: F2(x1, x2)  =  x1
f2(x1, x2)  =  f1(x2)
a  =  a
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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, x), a) -> f2(f2(f2(x, a), f2(a, a)), a)

The set Q consists of the following terms:

f2(f2(a, x0), a)

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