Termination w.r.t. Q proof of /tmp/tmpdLKvEW/09.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(b(a(x1))) → b(b(b(a(x1))))
b(b(b(a(x1)))) → a(a(a(b(x1))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ DependencyPairsProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(a(x))) → a(b(b(b(x))))
a(b(b(b(x)))) → b(a(a(a(x))))
a(b(b(B(x)))) → b(A(x))
a(a(b(b(B(x))))) → a(b(b(b(A(x)))))
a(b(b(a(b(b(B(x))))))) → b(a(a(a(a(A(x))))))
a(b(b(B(x)))) → B(x)
a(b(A(x))) → a(b(b(B(x))))
a(b(a(b(A(x))))) → a(b(b(b(b(B(x))))))
a(a(b(b(B(x))))) → a(b(b(b(a(A(x))))))
a(b(b(a(b(b(B(x))))))) → b(a(a(a(a(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 1 SCC with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.A¹: 0
B: 0
a: 0
A: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

A^1.0(b.0(b.0(b.1(x)))) → A^1.0(a.1(x))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.0(a.1(A.0(x))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.0(a.1(A.1(x))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(a.0(x))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(a.1(A.0(x0))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.1(A.1(x)))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(x)
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.0(a.1(A.0(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.0(a.1(A.1(x0)))))))
A^1.0(b.0(a.1(x))) → A^1.0(b.0(b.0(b.1(x))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.1(A.0(x)))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.0(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.0(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.1(A.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.0(a.1(A.0(x))))))
a.0(b.0(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.0(a.1(A.1(x))))))
a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.0(a.0(a.0(a.0(a.0(a.1(A.0(x)))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(b.1(x)))) → b.0(a.0(a.0(a.1(x))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.0(a.0(a.0(a.0(a.1(A.1(x))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.0(a.0(a.0(a.0(a.1(A.0(x))))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.0(a.0(a.0(a.0(a.0(a.1(A.1(x)))))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(x))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                          ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A^1.0(b.0(b.0(b.1(x)))) → A^1.0(a.1(x))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.0(a.1(A.0(x))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.0(a.1(A.1(x))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(a.0(x))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(a.1(A.0(x0))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.1(A.1(x)))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(x)
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.0(a.1(A.0(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.0(a.1(A.1(x0)))))))
A^1.0(b.0(a.1(x))) → A^1.0(b.0(b.0(b.1(x))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.1(A.0(x)))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.0(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.0(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.1(A.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.0(a.1(A.0(x))))))
a.0(b.0(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.0(a.1(A.1(x))))))
a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.0(a.0(a.0(a.0(a.0(a.1(A.0(x)))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(b.1(x)))) → b.0(a.0(a.0(a.1(x))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.0(a.0(a.0(a.0(a.1(A.1(x))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.0(a.0(a.0(a.0(a.1(A.0(x))))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.0(a.0(a.0(a.0(a.0(a.1(A.1(x)))))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(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

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                          ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.0(a.1(A.0(x))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.0(a.1(A.1(x))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(a.0(x))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(a.1(A.0(x0))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.0(a.1(A.0(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.0(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.0(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.0(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(a.1(A.0(x0))))))))

The TRS R consists of the following rules:

a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.0(a.1(A.0(x))))))
a.0(b.0(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.0(a.1(A.1(x))))))
a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.0(a.0(a.0(a.0(a.0(a.1(A.0(x)))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(b.1(x)))) → b.0(a.0(a.0(a.1(x))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.0(a.0(a.0(a.0(a.1(A.1(x))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.0(a.0(a.0(a.0(a.1(A.0(x))))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.0(a.0(a.0(a.0(a.0(a.1(A.1(x)))))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used. Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.A¹: 0
B: 0
a: x₀
A: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.1(a.1(A.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(a.1(x))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(a.0(x))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0)))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.1(a.1(A.0(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.1(a.1(x))) → A^1.0(b.0(b.0(b.1(x))))
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(x)
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.1(x0))))))))

The TRS R consists of the following rules:

a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(A.0(x))))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(b.1(x)))) → b.1(a.1(a.1(a.1(x))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(a.1(A.0(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(A.1(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.1(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(a.1(A.1(x))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.0(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.1(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(x))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                              ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(a.0(b.0(b.0(B.1(x))))) → A^1.0(b.0(b.0(b.1(a.1(A.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(a.1(x))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(a.0(x))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0)))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(a.0(b.0(b.0(B.0(x))))) → A^1.0(b.0(b.0(b.1(a.1(A.0(x))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.1(a.1(x))) → A^1.0(b.0(b.0(b.1(x))))
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(x)
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.1(x0))))))))

The TRS R consists of the following rules:

a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(A.0(x))))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(b.1(x)))) → b.1(a.1(a.1(a.1(x))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(a.1(A.0(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(A.1(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.1(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(a.1(A.1(x))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.0(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.1(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(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

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                              ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(a.0(x))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0)))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))

The TRS R consists of the following rules:

a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(A.0(x))))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(b.1(x)))) → b.1(a.1(a.1(a.1(x))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(a.1(A.0(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(A.1(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.1(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(a.1(A.1(x))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.0(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.1(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(x))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(a(x))) → a(b(b(b(x))))
a(b(b(b(x)))) → b(a(a(a(x))))
a(b(b(B(x)))) → b(A(x))
a(a(b(b(B(x))))) → a(b(b(b(A(x)))))
a(b(b(a(b(b(B(x))))))) → b(a(a(a(a(A(x))))))
a(b(b(B(x)))) → B(x)
a(b(A(x))) → a(b(b(B(x))))
a(b(a(b(A(x))))) → a(b(b(b(b(B(x))))))
a(a(b(b(B(x))))) → a(b(b(b(a(A(x))))))
a(b(b(a(b(b(B(x))))))) → b(a(a(a(a(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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(b(a(x))) → a(b(b(b(x))))
a(b(b(b(x)))) → b(a(a(a(x))))
a(b(b(B(x)))) → b(A(x))
a(a(b(b(B(x))))) → a(b(b(b(A(x)))))
a(b(b(a(b(b(B(x))))))) → b(a(a(a(a(A(x))))))
a(b(b(B(x)))) → B(x)
a(b(A(x))) → a(b(b(B(x))))
a(b(a(b(A(x))))) → a(b(b(b(b(B(x))))))
a(a(b(b(B(x))))) → a(b(b(b(a(A(x))))))
a(b(b(a(b(b(B(x))))))) → b(a(a(a(a(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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                                                                ↳ QDP

                                                                  ↳ SemLabProof

                                                                  ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.A¹: 0
B: 0
a: x₀
A: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(b.0(b.0(b.0(x0)))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.0(b.0(B.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0)))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.1(a.1(x))) → A^1.0(b.0(b.0(b.1(x))))
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(x)
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.0(b.0(B.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(b.1(a.1(a.1(a.1(x0)))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(b.0(b.0(b.1(x0)))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(b.0(a.0(a.0(a.0(x0)))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(A.0(x0)))))))

The TRS R consists of the following rules:

a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(A.0(x))))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(b.1(x)))) → b.1(a.1(a.1(a.1(x))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(a.1(A.0(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(A.1(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.1(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(a.1(A.1(x))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.0(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.1(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(x))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                                                                ↳ QDP

                                                                  ↳ SemLabProof

                                                                    ↳ QDP

                                                                      ↳ DependencyGraphProof

                                                                  ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(b.0(b.0(b.0(x0)))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.0(b.0(B.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0)))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.1(a.1(x))) → A^1.0(b.0(b.0(b.1(x))))
A^1.0(b.0(b.0(b.1(x)))) → A^1.1(x)
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.0(b.0(B.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(b.1(a.1(a.1(a.1(x0)))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(b.0(b.0(b.1(x0)))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(b.0(a.0(a.0(a.0(x0)))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(b.1(a.1(a.1(a.1(a.1(A.0(x0)))))))

The TRS R consists of the following rules:

a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(A.0(x))))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(b.1(x)))) → b.1(a.1(a.1(a.1(x))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(a.1(A.0(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(A.1(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.1(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(a.1(A.1(x))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.0(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.1(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                                                                ↳ QDP

                                                                  ↳ SemLabProof

                                                                    ↳ QDP

                                                                      ↳ DependencyGraphProof

                                                                        ↳ QDP

                                                                  ↳ SemLabProof2

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(a.0(b.1(A.0(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.0(x))))))
A^1.0(b.0(b.0(b.0(b.1(A.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.0(x0))))))
A^1.0(b.0(b.0(b.0(b.1(A.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(B.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.1(A.0(x0))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(b.0(b.0(b.0(x0)))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.1(x0))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.1(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(x)))) → A^1.0(x)
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.0(b.0(B.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.0(x0)))))))))
A^1.0(b.0(a.0(x))) → A^1.0(b.0(b.0(b.0(x))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(a.1(A.1(x0)))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(a.0(b.0(a.0(a.0(a.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.1(x0))))))))
A^1.0(b.0(a.0(b.1(A.1(x))))) → A^1.0(b.0(b.0(b.0(b.0(B.1(x))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(b.0(b.0(b.0(b.0(B.1(x0)))))))
A^1.0(b.0(b.0(b.0(b.0(a.0(x0)))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(b.1(a.1(x0)))))) → A^1.0(a.0(b.0(b.0(b.1(x0)))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))))) → A^1.0(a.0(b.1(a.1(a.1(a.1(a.1(A.0(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.1(A.1(x0))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.0(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(b.0(x0))))))) → A^1.0(b.0(a.0(a.0(a.0(x0)))))
A^1.0(b.0(b.0(b.0(b.0(a.0(b.1(A.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.0(b.0(B.1(x0))))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.1(x0)))))))) → A^1.0(a.0(a.0(b.0(b.0(b.1(a.1(A.1(x0))))))))
A^1.0(b.0(b.0(b.0(b.0(b.0(B.1(x0))))))) → A^1.0(a.0(b.0(b.0(b.1(A.1(x0))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(a.1(A.0(x0)))))))
A^1.0(b.0(b.0(b.0(a.0(b.0(b.0(B.0(x0)))))))) → A^1.0(a.0(b.0(b.0(b.1(A.0(x0))))))

The TRS R consists of the following rules:

a.0(b.0(b.0(b.0(x)))) → b.0(a.0(a.0(a.0(x))))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(A.0(x))))))
a.0(b.0(b.0(B.0(x)))) → b.1(A.0(x))
a.0(b.0(b.0(b.1(x)))) → b.1(a.1(a.1(a.1(x))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(a.1(A.0(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(A.1(x))))))
a.0(b.0(b.0(a.0(b.0(b.0(B.1(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.1(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(a.1(A.1(x))))))
a.0(b.0(a.0(b.1(A.0(x))))) → a.0(b.0(b.0(b.0(b.0(B.0(x))))))
a.0(a.0(b.0(b.0(B.0(x))))) → a.0(b.0(b.0(b.1(A.0(x)))))
a.0(b.0(b.0(B.1(x)))) → b.1(A.1(x))
a.0(b.0(b.0(a.0(b.0(b.0(B.0(x))))))) → b.1(a.1(a.1(a.1(a.1(a.1(A.0(x)))))))
a.0(a.0(b.0(b.0(B.1(x))))) → a.0(b.0(b.0(b.1(A.1(x)))))
a.0(b.0(b.0(B.0(x)))) → B.0(x)
a.0(b.0(b.0(B.1(x)))) → B.1(x)
a.0(b.1(a.1(x))) → a.0(b.0(b.0(b.1(x))))
a.0(b.1(A.1(x))) → a.0(b.0(b.0(B.1(x))))
a.0(b.0(a.0(x))) → a.0(b.0(b.0(b.0(x))))
a.0(b.0(a.0(b.1(A.1(x))))) → a.0(b.0(b.0(b.0(b.0(B.1(x))))))
a.0(b.1(A.0(x))) → a.0(b.0(b.0(B.0(x))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ SemLabProof

                                          ↳ SemLabProof2

                                            ↳ QDP

                                              ↳ SemLabProof

                                              ↳ SemLabProof2

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                                                                ↳ QDP

                                                                  ↳ SemLabProof

                                                                  ↳ SemLabProof2

                                                                    ↳ QDP

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ QTRS Reverse

                          ↳ DependencyPairsProof

                          ↳ QTRS Reverse

                            ↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.