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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

f2(f2(a, b), x) -> f2(a, f2(a, x))
f2(f2(b, a), x) -> f2(b, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(x, f2(y, z)) -> F2(x, y)
F2(f2(b, a), x) -> F2(b, x)
F2(f2(a, b), x) -> F2(a, x)
The remaining pairs can at least be oriented weakly.

F2(x, f2(y, z)) -> F2(f2(x, y), z)
F2(f2(b, a), x) -> F2(b, f2(b, x))
F2(f2(a, b), x) -> F2(a, f2(a, x))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( b ) = max{0, -1}


POL( a ) = 0


POL( F2(x1, x2) ) = max{0, 2x1 + 2x2 - 1}


POL( f2(x1, x2) ) = x1 + x2 + 2



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

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

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

The TRS R consists of the following rules:

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