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:

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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(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 3 SCCs with 6 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPToSRSProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

B(a(b(b(x1)))) → B(a(b(x1)))

The TRS R consists of the following rules:

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

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPToSRSProof
              ↳ QTRS
                ↳ QTRS Reverse
QTRS
                    ↳ QTRS Reverse
                    ↳ QTRS Reverse
                    ↳ DependencyPairsProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPToSRSProof
              ↳ QTRS
                ↳ QTRS Reverse
                  ↳ QTRS
                    ↳ QTRS Reverse
QTRS
                    ↳ QTRS Reverse
                    ↳ DependencyPairsProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPToSRSProof
              ↳ QTRS
                ↳ QTRS Reverse
                  ↳ QTRS
                    ↳ QTRS Reverse
                    ↳ QTRS Reverse
QTRS
                    ↳ DependencyPairsProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

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

B1(b(a(a(b(x))))) → B1(a(b(x)))
B1(b(a(a(b(x))))) → B1(a(a(b(b(a(b(x)))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(a(b(x)))))) → B1(a(a(a(b(b(a(a(b(x)))))))))
B1(b(a(b(x)))) → B1(a(b(b(b(x)))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(a(a(b(x)))))) → B1(a(a(b(x))))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPToSRSProof
              ↳ QTRS
                ↳ QTRS Reverse
                  ↳ QTRS
                    ↳ QTRS Reverse
                    ↳ QTRS Reverse
                    ↳ DependencyPairsProof
QDP
                        ↳ DependencyGraphProof
          ↳ QDP
          ↳ QDP
  ↳ QTRS Reverse

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

B1(b(a(a(b(x))))) → B1(a(b(x)))
B1(b(a(a(b(x))))) → B1(a(a(b(b(a(b(x)))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(a(b(x)))))) → B1(a(a(a(b(b(a(a(b(x)))))))))
B1(b(a(b(x)))) → B1(a(b(b(b(x)))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(a(a(b(x)))))) → B1(a(a(b(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

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

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

B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(b(b(x)))

The TRS R consists of the following rules:

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

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

B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(a(B(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))



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

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

B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(a(B(x0))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))

The TRS R consists of the following rules:

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

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

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

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

B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0))))) at position [0] we obtained the following new rules:

B1(b(a(b(b(a(B(x0))))))) → B1(b(a(B(x0))))



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

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

B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(a(B(x0))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))

The TRS R consists of the following rules:

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

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

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

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

B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))

The TRS R consists of the following rules:

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

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

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

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

B(a(a(b(b(x1))))) → B(a(a(b(x1))))

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.