Termination w.r.t. Q proof of /tmp/tmpxpn0PB/size-12-alpha-3-num-545.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)) → a(b(b(c(c(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)) → a(b(b(c(c(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)) → A(x1)
A(c(x1)) → B(c(c(a(x1))))
A(c(x1)) → B(b(c(c(a(x1)))))
A(c(x1)) → A(b(b(c(c(a(x1))))))

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

POL( A(x₁) ) = x₁ + 1

POL( c(x₁) ) = x₁ + 1

POL( b(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = x₁ + 1

The following usable rules [17] were oriented:

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

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → a(b(b(c(c(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(b(b(c(c(a(x1)))))) at position [0] we obtained the following new rules:

A(c(y0)) → A(b(c(a(y0))))
A(c(b(x0))) → A(b(b(c(c(x0)))))
A(c(c(x0))) → A(b(b(c(c(a(b(b(c(c(a(x0)))))))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(c(b(x0))) → A(b(b(c(c(x0)))))
A(c(c(x0))) → A(b(b(c(c(a(b(b(c(c(a(x0)))))))))))
A(c(y0)) → A(b(c(a(y0))))

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → a(b(b(c(c(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(b(x0))) → A(b(b(c(c(x0))))) at position [0] we obtained the following new rules:

A(c(b(y0))) → A(b(c(y0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(c(b(y0))) → A(b(c(y0)))
A(c(c(x0))) → A(b(b(c(c(a(b(b(c(c(a(x0)))))))))))
A(c(y0)) → A(b(c(a(y0))))

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → a(b(b(c(c(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(b(y0))) → A(b(c(y0))) at position [0] we obtained the following new rules:

A(c(b(x0))) → A(x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

A(c(c(x0))) → A(b(b(c(c(a(b(b(c(c(a(x0)))))))))))
A(c(b(x0))) → A(x0)
A(c(y0)) → A(b(c(a(y0))))

The TRS R consists of the following rules:

a(b(x1)) → x1
a(c(x1)) → a(b(b(c(c(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

                  ↳ 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(x1)) → x1
a(c(x1)) → a(b(b(c(c(a(x1))))))
b(c(x1)) → x1
A(c(c(x0))) → A(b(b(c(c(a(b(b(c(c(a(x0)))))))))))
A(c(b(x0))) → A(x0)
A(c(y0)) → A(b(c(a(y0))))

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                    ↳ QTRS

                                  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → a(c(c(b(b(a(x))))))
c(b(x)) → x
c(c(A(x))) → a(c(c(b(b(a(c(c(b(b(A(x)))))))))))
b(c(A(x))) → A(x)
c(A(x)) → a(c(b(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 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

C(a(x0)) → C(b(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

C(c(A(y0))) → C(b(A(y0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x)) → C(c(b(b(a(x)))))
C(c(A(x))) → C(b(b(a(c(c(b(b(A(x)))))))))
C(a(x0)) → C(b(x0))
C(c(A(x))) → C(c(b(b(a(c(c(b(b(A(x))))))))))
C(c(A(y0))) → C(b(A(y0)))

The TRS R consists of the following rules:

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

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

                                                            ↳ QDP

                                                              ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

C(a(y0)) → C(b(a(y0)))
C(a(x0)) → C(c(b(x0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(y0)) → C(b(a(y0)))
C(a(a(x0))) → C(x0)
C(c(A(x))) → C(b(b(a(c(c(b(b(A(x)))))))))
C(c(A(x))) → C(c(b(b(a(c(c(b(b(A(x))))))))))
C(a(x0)) → C(c(b(x0)))

The TRS R consists of the following rules:

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

C(c(A(y0))) → C(b(a(c(c(b(b(A(y0))))))))
C(c(A(y0))) → C(c(b(b(a(c(b(A(y0))))))))
C(c(A(y0))) → C(c(b(c(c(b(b(A(y0))))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ Narrowing

                                                                    ↳ QDP

                                                                      ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(y0)) → C(b(a(y0)))
C(a(a(x0))) → C(x0)
C(c(A(x))) → C(b(b(a(c(c(b(b(A(x)))))))))
C(c(A(y0))) → C(b(a(c(c(b(b(A(y0))))))))
C(c(A(y0))) → C(c(b(b(a(c(b(A(y0))))))))
C(c(A(y0))) → C(c(b(c(c(b(b(A(y0))))))))
C(a(x0)) → C(c(b(x0)))

The TRS R consists of the following rules:

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

C(c(A(y0))) → C(b(c(c(b(b(A(y0)))))))
C(c(A(y0))) → C(b(b(a(c(b(A(y0)))))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ Narrowing

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(y0)) → C(b(a(y0)))
C(a(a(x0))) → C(x0)
C(c(A(y0))) → C(b(c(c(b(b(A(y0)))))))
C(c(A(y0))) → C(b(b(a(c(b(A(y0)))))))
C(c(A(y0))) → C(b(a(c(c(b(b(A(y0))))))))
C(c(A(y0))) → C(c(b(b(a(c(b(A(y0))))))))
C(a(x0)) → C(c(b(x0)))
C(c(A(y0))) → C(c(b(c(c(b(b(A(y0))))))))

The TRS R consists of the following rules:

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

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

C(a(x0)) → C(x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ Narrowing

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ Narrowing

                        ↳ QDP

                          ↳ QDPToSRSProof

                            ↳ QTRS

                              ↳ QTRS Reverse

                                ↳ QTRS

                                  ↳ QTRS Reverse

                                  ↳ QTRS Reverse

                                  ↳ DependencyPairsProof

                                    ↳ QDP

                                      ↳ DependencyGraphProof

                                        ↳ QDP

                                          ↳ Narrowing

                                            ↳ QDP

                                              ↳ Narrowing

                                                ↳ QDP

                                                  ↳ DependencyGraphProof

                                                    ↳ QDP

                                                      ↳ Narrowing

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

                                                            ↳ QDP

                                                              ↳ Narrowing

                                                                ↳ QDP

                                                                  ↳ Narrowing

                                                                    ↳ QDP

                                                                      ↳ Narrowing

                                                                        ↳ QDP

                                                                          ↳ Narrowing

                                                                            ↳ QDP

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(a(x0)) → C(x0)
C(a(a(x0))) → C(x0)
C(c(A(y0))) → C(b(c(c(b(b(A(y0)))))))
C(c(A(y0))) → C(b(a(c(c(b(b(A(y0))))))))
C(c(A(y0))) → C(b(b(a(c(b(A(y0)))))))
C(c(A(y0))) → C(c(b(b(a(c(b(A(y0))))))))
C(c(A(y0))) → C(c(b(c(c(b(b(A(y0))))))))
C(a(x0)) → C(c(b(x0)))

The TRS R consists of the following rules:

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

Q is empty.