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:

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

B(c(x1)) → A(x1)
B(c(x1)) → B(a(x1))
A(b(x1)) → A(x1)
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ Narrowing
  ↳ QTRS Reverse

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

B(c(x1)) → A(x1)
B(c(x1)) → B(a(x1))
A(b(x1)) → A(x1)
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x1)) → B(a(x1)) at position [0] we obtained the following new rules:

B(c(a(x0))) → B(b(x0))
B(c(b(x0))) → B(c(a(x0)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
QDP
          ↳ QDPToSRSProof
  ↳ QTRS Reverse

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

B(c(a(x0))) → B(b(x0))
B(c(x1)) → A(x1)
B(c(b(x0))) → B(c(a(x0)))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
QTRS
              ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
B(c(a(x0))) → B(b(x0))
B(c(x1)) → A(x1)
B(c(b(x0))) → B(c(a(x0)))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
B(c(a(x0))) → B(b(x0))
B(c(x1)) → A(x1)
B(c(b(x0))) → B(c(a(x0)))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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:

C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(c(B(x))) → A1(c(B(x)))
A1(c(B(x))) → B1(B(x))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
QDP
                      ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(c(B(x))) → A1(c(B(x)))
A1(c(B(x))) → B1(B(x))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
QDP
                          ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(c(B(x))) → A1(c(B(x)))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(c(B(x))) → A1(c(B(x))) at position [0] we obtained the following new rules:

B1(c(B(x0))) → A1(A(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
QDP
                              ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
C(b(x)) → A1(b(c(x)))
B1(c(B(x0))) → A1(A(x0))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
QDP
                                  ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x)) → A1(c(x)) at position [0] we obtained the following new rules:

B1(a(B(x0))) → A1(A(x0))
B1(a(b(x0))) → A1(a(b(c(x0))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
QDP
                                      ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
C(b(x)) → A1(b(c(x)))
B1(a(B(x0))) → A1(A(x0))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
QDP
                                          ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
C(b(x)) → A1(b(c(x)))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x)) → A1(b(c(x))) at position [0] we obtained the following new rules:

C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → A1(a(c(B(x0))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ Narrowing
QDP
                                              ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(b(x0))) → A1(b(a(b(c(x0)))))
A1(a(x)) → B1(x)
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → A1(a(c(B(x0))))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x)) → B1(c(x)) at position [0] we obtained the following new rules:

C(b(b(x0))) → B1(a(b(c(x0))))
C(b(B(x0))) → B1(A(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
QDP
                                                  ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → B1(A(x0))
B1(a(x)) → C(x)
C(b(B(x0))) → A1(a(c(B(x0))))
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
QDP
                                                      ↳ Narrowing
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → A1(a(c(B(x0))))
B1(a(x)) → C(x)
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(B(x0))) → A1(b(A(x0))) at position [0] we obtained the following new rules:

C(b(B(x0))) → A1(A(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
QDP
                                                          ↳ DependencyGraphProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

C(b(B(x0))) → A1(A(x0))
C(b(b(x0))) → A1(b(a(b(c(x0)))))
A1(a(x)) → B1(x)
B1(a(x)) → C(x)
C(b(B(x0))) → A1(a(c(B(x0))))
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                    ↳ QDP
                      ↳ DependencyGraphProof
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ Narrowing
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
QDP
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

A1(a(x)) → B1(x)
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(a(c(B(x0))))
B1(a(x)) → C(x)
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)

The TRS R consists of the following rules:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
QTRS
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ Narrowing
        ↳ QDP
          ↳ QDPToSRSProof
            ↳ QTRS
              ↳ QTRS Reverse
                ↳ QTRS
                  ↳ DependencyPairsProof
                  ↳ QTRS Reverse
                  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))

The set Q is empty.
We have obtained the following QTRS:

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))

Q is empty.