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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

POL( b ) = 1


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


POL( a ) = 1


POL( f2(x1, x2) ) = x1 + x2 + 1


POL( c ) = 1



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

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

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

The TRS R consists of the following rules:

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

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