Termination w.r.t. Q proof of /tmp/tmpMO4DD7/06.srs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A(a(a(x1))) → B(a(x1))

Used ordering: POLO with Polynomial interpretation [25]:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

      ↳ RuleRemovalProof

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

The TRS R consists of the following rules:

a(a(a(x))) → a(b(b(x)))
a(b(a(x))) → a(b(b(x)))
b(a(b(x))) → b(a(a(x)))
b(a(b(a(B(x))))) → b(a(a(a(A(x)))))
a(b(a(b(A(x))))) → a(b(b(b(B(x)))))
b(a(b(A(x)))) → b(a(a(B(x))))
b(a(a(A(x)))) → b(a(a(B(x))))
a(b(a(B(x)))) → a(b(b(A(x))))
a(b(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(b(A(x))))) → a(b(b(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 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x₁)) = 1 + x₁
POL(A¹(x₁)) = 2 + 2·x₁
POL(B(x₁)) = 1 + x₁
POL(B¹(x₁)) = 2·x₁
POL(a(x₁)) = 2 + x₁
POL(b(x₁)) = 2 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                              ↳ 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(x)))
A¹(a(a(x))) → A¹(b(b(x)))

The TRS R consists of the following rules:

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

A¹(a(a(x))) → A¹(b(b(x)))

The remaining pairs can at least be oriented weakly.

A¹(b(a(x))) → A¹(b(b(x)))

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

POL( A(x₁) ) = x₁

POL( b(x₁) ) = max{0, -1}

POL( A¹(x₁) ) = max{0, x₁ - 1}

POL( B(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                              ↳ 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(x)))

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                              ↳ QDP

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ QDPOrderProof

                                              ↳ QDP

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(A¹(x₁)) = (1/2)x_1
POL(B(x₁)) = 0
POL(a(x₁)) = 3/4
POL(A(x₁)) = 1/2
POL(b(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 9/128.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ QDPOrderProof

                                                              ↳ QDP

                                                                ↳ PisEmptyProof

                                              ↳ QDP

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ Narrowing

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

B¹(a(b(x))) → B¹(a(a(x)))

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ SemLabProof

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(a(a(x))) → a(b(b(x)))
a(b(a(x))) → a(b(b(x)))
b(a(b(x))) → b(a(a(x)))
b(a(b(a(B(x))))) → b(a(a(a(A(x)))))
a(b(a(b(A(x))))) → a(b(b(b(B(x)))))
b(a(b(A(x)))) → b(a(a(B(x))))
b(a(a(A(x)))) → b(a(a(B(x))))
a(b(a(B(x)))) → a(b(b(A(x))))
a(b(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(a(A(x))))) → a(b(b(b(B(x)))))
a(a(a(b(A(x))))) → a(b(b(b(B(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.B: 0
a: 0
A: 0
B¹: 0
b: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ QDPToSRSProof

                        ↳ QTRS

                          ↳ QTRS Reverse

                            ↳ QTRS

                              ↳ QTRS Reverse

                              ↳ DependencyPairsProof

                                ↳ QDP

                                  ↳ DependencyGraphProof

                                    ↳ QDP

                                      ↳ RuleRemovalProof

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ AND

                                              ↳ QDP

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ SemLabProof

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                              ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a.0(a.0(a.1(b.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
b.0(a.1(b.0(A.1(x)))) → b.0(a.0(a.0(B.1(x))))
a.1(b.0(a.0(B.0(x)))) → a.1(b.1(b.0(A.0(x))))
a.0(a.0(a.0(x))) → a.1(b.1(b.0(x)))
b.0(a.0(a.0(A.1(x)))) → b.0(a.0(a.0(B.1(x))))
a.1(b.0(a.0(a.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
b.0(a.1(b.0(a.0(B.0(x))))) → b.0(a.0(a.0(a.0(A.0(x)))))
b.0(a.1(b.0(x))) → b.0(a.0(a.0(x)))
a.1(b.0(a.0(x))) → a.1(b.1(b.0(x)))
a.0(a.0(a.0(a.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.1(b.0(a.1(b.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
a.0(a.0(a.1(x))) → a.1(b.1(b.1(x)))
a.1(b.0(a.1(b.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.1(b.0(a.0(a.0(A.0(x))))) → a.1(b.1(b.1(b.0(B.0(x)))))
a.0(a.0(a.0(a.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(x)))))
b.0(a.1(b.0(A.0(x)))) → b.0(a.0(a.0(B.0(x))))
b.0(a.1(b.1(x))) → b.0(a.0(a.1(x)))
a.1(b.0(a.1(x))) → a.1(b.1(b.1(x)))
a.1(b.0(a.0(B.1(x)))) → a.1(b.1(b.0(A.1(x))))
b.0(a.0(a.0(A.0(x)))) → b.0(a.0(a.0(B.0(x))))
b.0(a.1(b.0(a.0(B.1(x))))) → b.0(a.0(a.0(a.0(A.1(x)))))
a.0(a.0(a.1(b.0(A.1(x))))) → a.1(b.1(b.1(b.0(B.1(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 6 less nodes.
We have reversed the following QTRS:
The set of rules R is

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ RuleRemovalProof

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.