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(c(x1))) → a(c(c(a(a(a(x1))))))

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

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:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(b(c(x1))) → a(c(c(a(a(a(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(b(c(x1))) → a(c(c(a(a(a(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(a(x1)) → b(x1)
a(b(c(x1))) → a(c(c(a(a(a(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ 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)
a(b(c(x))) → a(c(c(a(a(a(x))))))
A(b(c(x))) → A(b(x))
A(b(c(b(c(x))))) → A(a(a(c(c(a(a(a(x))))))))
A(b(c(a(x)))) → A(b(x))
A(b(c(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(c(b(A(x))))) → A1(a(a(c(c(a(a(A(x))))))))
C(b(a(x))) → C(a(x))
C(b(a(x))) → A1(a(a(c(c(a(x))))))
C(b(c(b(A(x))))) → A1(a(A(x)))
C(b(c(b(A(x))))) → A1(a(c(c(a(a(A(x)))))))
C(b(c(b(A(x))))) → C(a(a(A(x))))
C(b(a(x))) → A1(a(c(c(a(x)))))
C(b(a(x))) → A1(c(c(a(x))))
C(b(c(b(A(x))))) → C(c(a(a(A(x)))))
C(b(a(x))) → C(c(a(x)))
C(b(c(b(A(x))))) → A1(A(x))
C(b(c(b(A(x))))) → A1(c(c(a(a(A(x))))))

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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 8 less nodes.

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

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

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

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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(a(x))) → C(a(x)) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ QTRS Reverse
                                      ↳ DependencyPairsProof
                                        ↳ 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(a(a(x0)))) → C(b(x0))
C(b(c(b(A(x))))) → C(a(a(A(x))))
C(b(c(b(A(x))))) → C(c(a(a(A(x)))))
C(b(a(x))) → C(c(a(x)))
C(b(a(c(b(A(x0)))))) → C(b(A(x0)))

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ QTRS Reverse
                                      ↳ 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(a(a(x0)))) → C(b(x0))
C(b(c(b(A(x))))) → C(a(a(A(x))))
C(b(c(b(A(x))))) → C(c(a(a(A(x)))))
C(b(a(x))) → C(c(a(x)))

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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(c(b(A(x))))) → C(c(a(a(A(x))))) at position [0] we obtained the following new rules:

C(b(c(b(A(y0))))) → C(c(b(A(y0))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ QTRS Reverse
                                      ↳ DependencyPairsProof
                                        ↳ 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(a(a(x0)))) → C(b(x0))
C(b(c(b(A(x))))) → C(a(a(A(x))))
C(b(c(b(A(y0))))) → C(c(b(A(y0))))
C(b(a(x))) → C(c(a(x)))

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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(a(x))) → C(c(a(x))) at position [0] we obtained the following new rules:

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



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

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

C(b(a(a(x0)))) → C(b(x0))
C(b(c(b(A(x))))) → C(a(a(A(x))))
C(b(a(c(b(A(x0)))))) → C(c(b(A(x0))))
C(b(c(b(A(y0))))) → C(c(b(A(y0))))
C(b(a(a(x0)))) → C(c(b(x0)))

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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(c(b(A(x))))) → C(a(a(A(x)))) at position [0] we obtained the following new rules:

C(b(c(b(A(y0))))) → C(b(A(y0)))



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

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

C(b(a(a(x0)))) → C(b(x0))
C(b(a(c(b(A(x0)))))) → C(c(b(A(x0))))
C(b(c(b(A(y0))))) → C(c(b(A(y0))))
C(b(c(b(A(y0))))) → C(b(A(y0)))
C(b(a(a(x0)))) → C(c(b(x0)))

The TRS R consists of the following rules:

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

C(b(a(a(x0)))) → C(b(x0))
C(b(a(c(b(A(x0)))))) → C(c(b(A(x0))))
C(b(c(b(A(y0))))) → C(c(b(A(y0))))
C(b(a(a(x0)))) → C(c(b(x0)))

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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(c(b(A(y0))))) → C(c(b(A(y0)))) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ Narrowing
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ QDPToSRSProof
                                ↳ QTRS
                                  ↳ QTRS Reverse
                                    ↳ QTRS
                                      ↳ QTRS Reverse
                                      ↳ DependencyPairsProof
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ 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(c(b(A(x0))))) → C(A(x0))
C(b(a(a(x0)))) → C(b(x0))
C(b(a(c(b(A(x0)))))) → C(c(b(A(x0))))
C(b(c(b(A(x0))))) → C(b(A(x0)))
C(b(a(a(x0)))) → C(c(b(x0)))

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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 2 less nodes.

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

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

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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(a(c(b(A(x0)))))) → C(c(b(A(x0)))) at position [0] we obtained the following new rules:

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



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

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

The TRS R consists of the following rules:

a(a(x)) → b(x)
c(b(a(x))) → a(a(a(c(c(a(x))))))
c(b(A(x))) → b(A(x))
c(b(c(b(A(x))))) → a(a(a(c(c(a(a(A(x))))))))
a(c(b(A(x)))) → b(A(x))
c(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 2 less nodes.

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.