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.

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(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
      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.

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)
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  x1
f2(x1, x2)  =  x1
a  =  a
b  =  b
c  =  c

Lexicographic Path Order [19].
Precedence:
[a, c] > b


The following usable rules [14] were oriented:

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))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

F2(f2(f2(a, b), c), x) -> F2(c, f2(b, x))
F2(f2(f2(a, b), c), x) -> F2(a, f2(c, f2(b, x)))
F2(x, f2(y, z)) -> F2(x, y)
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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

F2(f2(f2(a, b), c), x) -> F2(a, f2(c, f2(b, x)))
F2(x, f2(y, z)) -> F2(x, y)
F2(x, f2(y, z)) -> F2(f2(x, y), z)
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x1)
f2(x1, x2)  =  x1
a  =  a
b  =  b
c  =  c

Lexicographic Path Order [19].
Precedence:
F1 > [a, b] > c


The following usable rules [14] were oriented:

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))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.