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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

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


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
QDP

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

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

The TRS R consists of the following rules:

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

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