Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/ZantemaSLASHz22.trs

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

Used argument filtering: F₂(x1, x2) = x1
f₂(x1, x2) = x1
a = a
c = c
b = b
Used ordering: Quasi Precedence: [a, b] > c

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

          ↳ QDPAfsSolverProof

            ↳ QDP

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

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

The TRS R consists of the following rules:

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

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