Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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.

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

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

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

The TRS R consists of the following rules:

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

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

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

POL(F2(x1, x2)) = 2·x1·x2 + 2·x2   
POL(a) = 3   
POL(b) = 0   
POL(f2(x1, x2)) = 2 + 2·x1 + 2·x1·x2 + 3·x2   

The following usable rules [14] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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(b, f2(a, x)) -> F2(b, f2(b, f2(b, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x2   
POL(a) = 3   
POL(b) = 0   
POL(f2(x1, x2)) = 3 + 3·x1 + 3·x1·x2 + 2·x2   

The following usable rules [14] were oriented:

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



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

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

The TRS R consists of the following rules:

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

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(b, x)) -> F2(a, x)
F2(a, f2(b, x)) -> F2(a, f2(a, x))
F2(a, f2(b, x)) -> F2(a, f2(a, f2(a, x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = 3·x1·x2 + x2   
POL(a) = 0   
POL(b) = 1   
POL(f2(x1, x2)) = 2 + 3·x1 + 2·x1·x2   

The following usable rules [14] were oriented:

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



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.