Termination w.r.t. Q proof of /tmp/tmpfPESV-/13.srs

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

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = 2·x₁
POL(B(x₁)) = 12 + 8·x₁
POL(a(x₁)) = 2 + 2·x₁
POL(b(x₁)) = 14 + 8·x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

b(b(a(a(a(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
a(a(a(x))) → a(a(x))
b(b(x)) → a(b(a(x)))
b(a(a(a(b(b(a(a(A(x))))))))) → a(a(a(b(a(a(a(b(B(x)))))))))
b(b(a(a(A(x))))) → a(b(a(B(x))))
b(b(b(a(a(a(B(x))))))) → b(b(b(b(a(a(A(x)))))))
b(b(a(a(b(a(a(a(B(x))))))))) → b(b(b(a(a(b(a(a(A(x)))))))))
b(b(a(a(a(B(x)))))) → a(a(a(b(a(a(a(a(a(A(x))))))))))
a(a(b(a(a(a(B(x))))))) → a(a(a(a(b(a(a(A(x))))))))
b(a(a(a(B(x))))) → a(a(b(a(a(A(x))))))
b(b(a(b(a(a(a(B(x)))))))) → b(b(b(a(b(a(a(A(x))))))))
a(b(a(a(a(B(x)))))) → a(a(a(b(a(a(A(x)))))))
b(b(b(a(a(A(x)))))) → a(b(a(b(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 2 SCCs with 61 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                  ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(b(a(a(a(B(x))))))) → a(a(a(a(b(a(a(A(x))))))))
a(b(a(a(a(B(x)))))) → a(a(a(b(a(a(A(x)))))))
a(a(a(x))) → a(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 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25]:

POL(A(x₁)) = x₁
POL(B(x₁)) = 8 + x₁
POL(B¹(x₁)) = 8·x₁
POL(a(x₁)) = 4 + x₁
POL(b(x₁)) = 12 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

b(b(a(a(a(x))))) → b(b(b(x)))
b(a(a(a(b(x))))) → a(a(a(b(a(a(a(x)))))))
a(a(a(x))) → a(a(x))
b(b(x)) → a(b(a(x)))
b(a(a(a(b(b(a(a(A(x))))))))) → a(a(a(b(a(a(a(b(B(x)))))))))
b(b(a(a(A(x))))) → a(b(a(B(x))))
b(b(b(a(a(a(B(x))))))) → b(b(b(b(a(a(A(x)))))))
b(b(a(a(b(a(a(a(B(x))))))))) → b(b(b(a(a(b(a(a(A(x)))))))))
b(b(a(a(a(B(x)))))) → a(a(a(b(a(a(a(a(a(A(x))))))))))
a(a(b(a(a(a(B(x))))))) → a(a(a(a(b(a(a(A(x))))))))
b(a(a(a(B(x))))) → a(a(b(a(a(A(x))))))
b(b(a(b(a(a(a(B(x)))))))) → b(b(b(a(b(a(a(A(x))))))))
a(b(a(a(a(B(x)))))) → a(a(a(b(a(a(A(x)))))))
b(b(b(a(a(A(x)))))) → a(b(a(b(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ 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:

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B¹(b(b(a(a(a(B(y0))))))) → B¹(b(a(b(a(a(a(A(y0)))))))) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ Narrowing

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ Narrowing

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

                                                                  ↳ QDP

                                                                    ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ 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:

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

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ 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:

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ 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:

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ Narrowing

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

                                                                  ↳ QDP

                                                                    ↳ Narrowing

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

                                                                          ↳ QDP

                                                                            ↳ Narrowing

                                                                              ↳ QDP

                                                                                ↳ DependencyGraphProof

                                                                                  ↳ QDP

                                                                                    ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [21,25] with Interpretation:

POL( B¹(x₁) ) = x₁

POL( A(x₁) ) = max{0, -1}

POL( b(x₁) ) = 1

POL( B(x₁) ) = 0

POL( a(x₁) ) = max{0, -1}

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ AND

                                  ↳ QDP

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ Narrowing

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

                                                                  ↳ QDP

                                                                    ↳ Narrowing

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

                                                                          ↳ QDP

                                                                            ↳ Narrowing

                                                                              ↳ QDP

                                                                                ↳ DependencyGraphProof

                                                                                  ↳ QDP

                                                                                    ↳ QDPOrderProof

                                                                                      ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

Q is empty.

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

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

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

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

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

Q is empty.