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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

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

Q is empty.

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

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

B(b(b(b(A(x0))))) → A1(a(A(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

B(b(b(b(A(x0))))) → A1(A(x0))



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

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

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

The TRS R consists of the following rules:

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

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

B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
A1(b(a(x))) → B(x)
B(b(a(x))) → A1(b(b(x)))
A1(b(a(x))) → B(b(x))
B(b(a(x))) → A1(a(b(b(x))))

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

B(b(a(b(A(x0))))) → A1(A(x0))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

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

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

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

Q is empty.