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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

F2(f2(f2(a, x), b), b) -> F2(f2(a, f2(f2(x, b), b)), b)
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = 1 + x1 + x2   
POL(a) = 1   
POL(b) = 0   
POL(f2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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

F2(a, f2(a, f2(a, f2(x, b)))) -> F2(a, f2(a, f2(a, x)))
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = 1 + x1 + x2   
POL(a) = 1   
POL(b) = 0   
POL(f2(x1, x2)) = x1 + x2   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP

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

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

The TRS R consists of the following rules:

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

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