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

Q is empty.


QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(a(y0)))) → A(a(b(b(b(y0))))) at position [0] we obtained the following new rules:

A(b(b(a(y0)))) → A(b(b(b(y0))))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(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
      ↳ Narrowing
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ 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(x1) → x1
a(b(b(a(x1)))) → a(a(b(a(b(b(x1))))))
A(b(b(a(a(x0))))) → A(a(b(a(a(b(a(b(b(x0)))))))))
A(b(b(a(x1)))) → A(b(b(x1)))
A(b(b(a(a(x0))))) → A(b(a(a(b(a(b(b(x0))))))))

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(a(b(A(y0))))
A1(a(b(b(A(y0))))) → A1(b(a(A(y0))))



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

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

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

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(b(A(y0)))



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

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

A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(y0))))) → A1(b(A(y0)))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(a(b(b(A(x))))) → A1(a(b(A(x))))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(b(A(y0)))



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

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

A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(x)))) → A1(b(a(a(x))))
A1(b(b(a(x)))) → A1(a(x))
A1(a(b(b(A(y0))))) → A1(b(A(y0)))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

A1(b(b(a(b(b(A(x0))))))) → A1(b(b(b(a(b(a(a(b(a(A(x0)))))))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(b(b(a(b(a(a(b(A(x0))))))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(a(b(b(A(x0))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))



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

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

A1(b(b(a(b(b(A(x0))))))) → A1(b(b(b(a(b(a(a(b(A(x0))))))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(b(b(a(b(a(a(b(a(A(x0)))))))))))
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(a(b(b(A(x0))))))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(a(b(b(A(x0))))))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(a(b(b(A(x))))) → A1(b(a(a(b(a(A(x)))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(b(a(a(b(A(y0))))))
A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))



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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))
A1(a(b(b(A(x))))) → A1(b(a(a(b(A(x))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(a(b(b(A(x0))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(A(y0)))))



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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(A(y0)))))
A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))
A1(b(b(a(b(b(A(x0))))))) → A1(b(a(b(b(A(x0))))))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

A1(b(b(a(b(b(A(y0))))))) → A1(b(b(b(A(y0)))))



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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(A(y0)))))
A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(A(y0))))))) → A1(b(b(b(A(y0)))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(A(y0)))))
A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(b(b(A(y0))))) → A1(b(a(b(A(y0))))) at position [0,0] we obtained the following new rules:

A1(a(b(b(A(y0))))) → A1(b(b(A(y0))))



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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))
A1(a(b(b(A(y0))))) → A1(b(b(A(y0))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

A1(a(b(b(A(y0))))) → A1(b(a(b(a(A(y0))))))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(a(A(x0))))))))))))
A1(b(b(a(y0)))) → A1(b(a(y0)))
A1(b(b(a(x)))) → A1(a(x))
A1(b(b(a(a(b(b(A(x0)))))))) → A1(b(a(b(b(a(b(a(a(b(A(x0)))))))))))
A1(b(b(a(b(b(a(x0))))))) → A1(b(a(b(b(a(b(a(a(x0)))))))))

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.