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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.

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

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

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

The TRS R consists of the following rules:

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

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


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

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

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

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.

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

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

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

The TRS R consists of the following rules:

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

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


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

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

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

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.