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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(y0))) → B(c(b(b(y0))))



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

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

B(b(c(x1))) → B(x1)
B(b(c(y0))) → B(c(b(b(y0))))
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)

The TRS R consists of the following rules:

a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → 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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
QTRS
                  ↳ QTRS Reverse

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

a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
B(b(c(x1))) → B(x1)
B(b(c(y0))) → B(c(b(b(y0))))
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)

Q is empty.

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

a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
B(b(c(x1))) → B(x1)
B(b(c(y0))) → B(c(b(b(y0))))
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

a(x) → b(x)
b(b(c(x))) → c(b(c(a(b(x)))))
c(c(x)) → x
B(b(c(x))) → B(x)
B(b(c(x))) → B(c(b(b(x))))
B(b(c(b(c(x))))) → B(c(a(c(b(c(a(b(x))))))))
B(b(c(x))) → A(b(x))
A(x) → B(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(b(x))) → C(b(c(x)))
C(b(c(b(B(x))))) → A1(c(B(x)))
C(b(c(b(B(x))))) → C(b(c(a(c(B(x))))))
C(b(c(b(B(x))))) → C(B(x))
C(b(b(x))) → C(x)
C(b(B(x))) → A2(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
C(b(B(x))) → C(B(x))
C(b(c(b(B(x))))) → A1(c(b(c(a(c(B(x)))))))
C(b(b(x))) → A1(c(b(c(x))))

The TRS R consists of the following rules:

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

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

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

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

C(b(b(x))) → C(b(c(x)))
C(b(c(b(B(x))))) → A1(c(B(x)))
C(b(c(b(B(x))))) → C(b(c(a(c(B(x))))))
C(b(c(b(B(x))))) → C(B(x))
C(b(b(x))) → C(x)
C(b(B(x))) → A2(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
C(b(B(x))) → C(B(x))
C(b(c(b(B(x))))) → A1(c(b(c(a(c(B(x)))))))
C(b(b(x))) → A1(c(b(c(x))))

The TRS R consists of the following rules:

a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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 6 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

C(b(c(b(B(y0))))) → C(b(c(b(c(B(y0))))))



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

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

C(b(c(b(B(y0))))) → C(b(c(b(c(B(y0))))))
C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))

The TRS R consists of the following rules:

a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
                                ↳ QDP
                                  ↳ DependencyGraphProof
QDP
                                      ↳ Narrowing
                      ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

C(b(c(b(B(y0))))) → C(b(c(B(y0))))



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

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

C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(c(b(B(y0))))) → C(b(c(B(y0))))

The TRS R consists of the following rules:

a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
QDP
                                              ↳ Narrowing
                      ↳ QTRS Reverse

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

C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
QDP
                      ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

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

Q is empty.