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 strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x2)
b  =  b
f2(x1, x2)  =  f2(x1, x2)
a  =  a

Lexicographic Path Order [19].
Precedence:
f2 > [F1, b]
a > [F1, b]


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
                ↳ 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 strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x2)
a  =  a
f2(x1, x2)  =  f2(x1, x2)
b  =  b

Lexicographic Path Order [19].
Precedence:
F1 > [a, f2]
b > [a, f2]


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.