Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule A(x1) → B(x1) we obtained the following new rules:

A(a(y_2)) → B(a(y_2))



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

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

B(a(b(b(x1)))) → A(a(x1))
B(a(b(b(x1)))) → A(x1)
B(a(b(b(x0)))) → B(a(b(x0)))
A(a(y_2)) → B(a(y_2))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(a(b(b(x1)))) → A(x1) we obtained the following new rules:

B(a(b(b(a(y_0))))) → A(a(y_0))



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

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

B(a(b(b(x1)))) → A(a(x1))
B(a(b(b(x0)))) → B(a(b(x0)))
B(a(b(b(a(y_0))))) → A(a(y_0))
A(a(y_2)) → B(a(y_2))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


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

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

a(x1) → b(x1)
b(a(b(b(x1)))) → b(b(b(a(a(x1)))))
B(a(b(b(x1)))) → A(a(x1))
B(a(b(b(x0)))) → B(a(b(x0)))
B(a(b(b(a(y_0))))) → A(a(y_0))
A(a(y_2)) → B(a(y_2))

Q is empty.

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

a(x1) → b(x1)
b(a(b(b(x1)))) → b(b(b(a(a(x1)))))
B(a(b(b(x1)))) → A(a(x1))
B(a(b(b(x0)))) → B(a(b(x0)))
B(a(b(b(a(y_0))))) → A(a(y_0))
A(a(y_2)) → B(a(y_2))

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ 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) → b(x)
b(b(a(b(x)))) → a(a(b(b(b(x)))))
b(b(a(B(x)))) → a(A(x))
b(b(a(B(x)))) → b(a(B(x)))
a(b(b(a(B(x))))) → a(A(x))
a(A(x)) → 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(B(x)))) → A1(A(x))
B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(x)))) → B1(b(x))
A1(x) → B1(x)
A1(A(x)) → A1(B(x))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(b(x)))) → A1(b(b(b(x))))
A1(b(b(a(B(x))))) → A1(A(x))

The TRS R consists of the following rules:

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

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

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

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

B1(b(a(B(x)))) → A1(A(x))
B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(x)))) → B1(b(x))
A1(x) → B1(x)
A1(A(x)) → A1(B(x))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(b(x)))) → A1(b(b(b(x))))
A1(b(b(a(B(x))))) → A1(A(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B1(b(a(B(x)))) → A1(A(x))
A1(b(b(a(B(x))))) → A1(A(x))
The remaining pairs can at least be oriented weakly.

B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(x)))) → B1(b(x))
A1(x) → B1(x)
A1(A(x)) → A1(B(x))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(b(x)))) → A1(b(b(b(x))))
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( B1(x1) ) = x1 + 1


POL( A(x1) ) = 0


POL( b(x1) ) = 1


POL( A1(x1) ) = x1 + 1


POL( B(x1) ) = max{0, -1}


POL( a(x1) ) = 1



The following usable rules [17] were oriented:

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



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

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

B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
A1(x) → B1(x)
B1(b(a(b(x)))) → B1(b(x))
A1(A(x)) → A1(B(x))
B1(b(a(b(x)))) → A1(b(b(b(x))))
B1(b(a(b(x)))) → B1(b(b(x)))

The TRS R consists of the following rules:

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

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

B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(x)))) → B1(b(x))
A1(x) → B1(x)
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(b(x)))) → A1(b(b(b(x))))

The TRS R consists of the following rules:

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



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

B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(a(a(b(b(b(x0)))))))
B1(b(a(b(a(b(x0)))))) → B1(a(a(b(b(b(x0))))))
A1(x) → B1(x)
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(b(a(B(x0))))))) → B1(b(a(A(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(x)))) → A1(b(b(b(x))))
B1(b(a(b(a(B(x0)))))) → B1(a(A(x0)))

The TRS R consists of the following rules:

a(x) → b(x)
b(b(a(b(x)))) → a(a(b(b(b(x)))))
b(b(a(B(x)))) → a(A(x))
b(b(a(B(x)))) → b(a(B(x)))
a(b(b(a(B(x))))) → a(A(x))
a(A(x)) → 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(a(B(x0)))))) → B1(b(a(B(x0)))) at position [0] we obtained the following new rules:

B1(b(a(b(a(B(y0)))))) → B1(b(b(B(y0))))



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

B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(a(a(b(b(b(x0)))))))
B1(b(a(b(x)))) → B1(b(x))
A1(x) → B1(x)
B1(b(a(b(a(b(x0)))))) → B1(a(a(b(b(b(x0))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(a(A(x0))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(x)))) → A1(b(b(b(x))))
B1(b(a(b(a(B(y0)))))) → B1(b(b(B(y0))))
B1(b(a(b(a(B(x0)))))) → B1(a(A(x0)))

The TRS R consists of the following rules:

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

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

B1(b(a(b(x)))) → A1(a(b(b(b(x)))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(a(a(b(b(b(x0)))))))
B1(b(a(b(a(b(x0)))))) → B1(a(a(b(b(b(x0))))))
B1(b(a(b(x)))) → B1(b(x))
A1(x) → B1(x)
B1(b(a(b(b(a(B(x0))))))) → B1(b(a(A(x0))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(x)))) → A1(b(b(b(x))))
B1(b(a(b(a(B(x0)))))) → B1(a(A(x0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ 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) → b(x)
b(a(b(b(x)))) → b(b(b(a(a(x)))))
B(a(b(b(x)))) → A(a(x))
B(a(b(b(x)))) → B(a(b(x)))
B(a(b(b(a(x))))) → A(a(x))
A(a(x)) → B(a(x))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ 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) → b(x)
b(a(b(b(x)))) → b(b(b(a(a(x)))))
B(a(b(b(x)))) → A(a(x))
B(a(b(b(x)))) → B(a(b(x)))
B(a(b(b(a(x))))) → A(a(x))
A(a(x)) → B(a(x))

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.