Termination w.r.t. Q proof of /tmp/tmptxl6HU/size-12-alpha-3-num-556.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(x1)) → x1
a(c(x1)) → c(b(b(c(a(a(x1))))))
b(c(x1)) → x1

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

A(c(x1)) → A(x1)

The remaining pairs can at least be oriented weakly.

A(c(x1)) → A(a(x1))

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

POL( A(x₁) ) = x₁

POL( c(x₁) ) = x₁ + 1

POL( b(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(c(x1)) → A(a(x1))

The TRS R consists of the following rules:

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

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

A(c(b(x0))) → A(x0)
A(c(c(x0))) → A(c(b(b(c(a(a(x0)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(c(b(x0))) → A(x0)
A(c(c(x0))) → A(c(b(b(c(a(a(x0)))))))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

Q is empty.

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

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

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

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x)) → C(x)
C(a(x)) → C(b(b(c(x))))
C(c(A(x))) → C(b(b(c(A(x)))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

C(c(A(x0))) → C(b(A(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x)) → C(x)
C(a(x)) → C(b(b(c(x))))
C(c(A(x0))) → C(b(A(x0)))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x)) → C(x)
C(a(x)) → C(b(b(c(x))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

C(a(b(x0))) → C(b(b(x0)))
C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))
C(a(A(x0))) → C(b(A(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(x)) → C(x)
C(a(A(x0))) → C(b(A(x0)))
C(a(b(x0))) → C(b(b(x0)))
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(x)) → C(x)
C(a(b(x0))) → C(b(b(x0)))
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

C(a(b(c(A(x0))))) → C(b(A(x0)))
C(a(b(a(x0)))) → C(b(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(b(c(A(x0))))) → C(b(A(x0)))
C(a(x)) → C(x)
C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(b(a(x0)))) → C(b(x0))
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(x)) → C(x)
C(a(b(a(x0)))) → C(b(x0))
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

C(a(b(a(c(A(x0)))))) → C(A(x0))
C(a(b(a(a(x0))))) → C(x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x)) → C(x)
C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(b(a(a(x0))))) → C(x0)
C(a(b(a(c(A(x0)))))) → C(A(x0))
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ Narrowing

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ DependencyGraphProof

                                                ↳ QDP

                                                  ↳ Narrowing

                                                    ↳ QDP

                                                      ↳ DependencyGraphProof

                                                        ↳ QDP

                                                          ↳ Narrowing

                                                            ↳ QDP

                                                              ↳ DependencyGraphProof

                                                                ↳ QDP

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(a(x0))) → C(b(b(a(a(c(b(b(c(x0)))))))))
C(a(x)) → C(x)
C(a(b(a(a(x0))))) → C(x0)
C(a(c(A(x0)))) → C(b(b(a(a(c(b(b(c(A(x0))))))))))

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(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

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

a(b(x)) → x
a(c(x)) → c(b(b(c(a(a(x))))))
b(c(x)) → x
A(c(b(x))) → A(x)
A(c(c(x))) → A(c(b(b(c(a(a(x)))))))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                          ↳ QTRS Reverse

                            ↳ QTRS

                          ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(b(x)) → x
a(c(x)) → c(b(b(c(a(a(x))))))
b(c(x)) → x
A(c(b(x))) → A(x)
A(c(c(x))) → A(c(b(b(c(a(a(x)))))))

Q is empty.

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

b(a(x)) → x
c(a(x)) → a(a(c(b(b(c(x))))))
c(b(x)) → x
b(c(A(x))) → A(x)
c(c(A(x))) → a(a(c(b(b(c(A(x)))))))

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

a(b(x)) → x
a(c(x)) → c(b(b(c(a(a(x))))))
b(c(x)) → x
A(c(b(x))) → A(x)
A(c(c(x))) → A(c(b(b(c(a(a(x)))))))

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ QDPToSRSProof

                    ↳ QTRS

                      ↳ QTRS Reverse

                        ↳ QTRS

                          ↳ DependencyPairsProof

                          ↳ QTRS Reverse

                          ↳ QTRS Reverse

                            ↳ QTRS

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

a(b(x)) → x
a(c(x)) → c(b(b(c(a(a(x))))))
b(c(x)) → x
A(c(b(x))) → A(x)
A(c(c(x))) → A(c(b(b(c(a(a(x)))))))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

    ↳ QTRS

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

    ↳ QTRS

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

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

Q is empty.