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 [1,13] 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(b, x)
F(c, f(c, x)) → F(a, x)
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))
F(b, f(b, x)) → F(c, 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
      ↳ EdgeDeletionProof

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

F(a, f(a, x)) → F(b, x)
F(c, f(c, x)) → F(a, x)
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))
F(b, f(b, x)) → F(c, 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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof

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

F(a, f(a, x)) → F(b, x)
F(a, f(a, x)) → F(c, f(b, x))
F(c, f(c, x)) → F(a, x)
F(b, f(b, x)) → F(a, f(c, x))
F(b, f(b, x)) → 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.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

B1(B(x)) → A1(C(x))
C1(C(x)) → B1(A(x))
A1(A(x)) → C1(B(x))
A1(A(x)) → B1(x)
B1(B(x)) → C1(x)
C1(C(x)) → A1(x)

The TRS R consists of the following rules:

A(A(x)) → C(B(x))
B(B(x)) → A(C(x))
C(C(x)) → B(A(x))

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


The following pairs can be oriented strictly and are deleted.


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

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))
Used ordering: Combined order from the following AFS and order.
B1(x1)  =  x1
B(x1)  =  B(x1)
A1(x1)  =  x1
C(x1)  =  C(x1)
C1(x1)  =  x1
A(x1)  =  A(x1)

Recursive path order with status [2].
Quasi-Precedence:
[B1, C1, A1]

Status:
trivial


The following usable rules [14] were oriented:

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))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.