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:

f(a, f(a, x)) → f(c, f(b, x))
f(b, f(b, x)) → f(a, f(c, x))
f(c, f(c, x)) → f(b, f(a, x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(a, f(a, x)) → f(c, f(b, x))
f(b, f(b, x)) → f(a, f(c, x))
f(c, f(c, x)) → f(b, f(a, x))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(a, f(a, x)) → F(c, f(b, x))
F(a, f(a, x)) → F(b, x)
F(c, f(c, x)) → F(b, f(a, x))
F(b, f(b, x)) → F(c, x)
F(b, f(b, x)) → F(a, f(c, x))
F(c, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, f(a, x)) → f(c, f(b, x))
f(b, f(b, x)) → f(a, f(c, x))
f(c, f(c, x)) → f(b, f(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:

F(a, f(a, x)) → F(c, f(b, x))
F(a, f(a, x)) → F(b, x)
F(c, f(c, x)) → F(b, f(a, x))
F(b, f(b, x)) → F(c, x)
F(b, f(b, x)) → F(a, f(c, x))
F(c, f(c, x)) → F(a, x)

The TRS R consists of the following rules:

f(a, f(a, x)) → f(c, f(b, x))
f(b, f(b, x)) → f(a, f(c, x))
f(c, f(c, x)) → f(b, f(a, x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(a, f(a, x)) → F(b, x)
F(b, f(b, x)) → F(c, x)
F(c, f(c, x)) → F(a, x)
The remaining pairs can at least be oriented weakly.

F(a, f(a, x)) → F(c, f(b, x))
F(c, f(c, x)) → F(b, f(a, x))
F(b, f(b, x)) → F(a, f(c, x))
Used ordering: Polynomial interpretation [25,35]:

POL(c) = 0   
POL(a) = 0   
POL(f(x1, x2)) = 1/4 + (1/2)x_1 + (2)x_2   
POL(b) = 0   
POL(F(x1, x2)) = (2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

f(b, f(b, x)) → f(a, f(c, x))
f(c, f(c, x)) → f(b, f(a, x))
f(a, f(a, x)) → f(c, f(b, x))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP

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

F(a, f(a, x)) → F(c, f(b, x))
F(b, f(b, x)) → F(a, f(c, x))
F(c, f(c, x)) → F(b, f(a, x))

The TRS R consists of the following rules:

f(a, f(a, x)) → f(c, f(b, x))
f(b, f(b, x)) → f(a, f(c, x))
f(c, f(c, x)) → f(b, f(a, x))

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