Termination w.r.t. Q proof of /tmp/tmplgK9cJ/size-12-alpha-3-num-222.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(x1) → x1
a(b(x1)) → b(b(c(a(x1))))
b(c(c(x1))) → c(a(x1))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ 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:

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ 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:

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(b(x))))
c(c(b(x))) → a(c(x))
b(A(x)) → c(b(B(x)))
b(A(x)) → c(B(x))
c(c(B(x))) → A(x)
b(b(A(x))) → a(c(b(b(c(B(x))))))
b(b(A(x))) → a(c(b(b(c(b(B(x)))))))
b(A(x)) → 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 2 SCCs with 17 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                ↳ UsableRulesProof

                              ↳ QDP

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(c(b(x))) → C(x)

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                ↳ UsableRulesProof

                              ↳ QDP

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(c(b(x))) → C(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

C(c(b(x))) → C(x)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(C(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ UsableRulesReductionPairsProof

                                      ↳ QDP

                                        ↳ PisEmptyProof

                                ↳ UsableRulesProof

                              ↳ QDP

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                                ↳ UsableRulesProof

                                ↳ UsableRulesProof

                                  ↳ QDP

                              ↳ QDP

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

C(c(b(x))) → C(x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ Narrowing

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

B¹(a(x)) → B¹(b(x))
B¹(a(x)) → B¹(x)

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

B¹(a(x)) → B¹(x)
B¹(a(a(x0))) → B¹(a(c(b(b(x0)))))

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

POL( B¹(x₁) ) = x₁ + 1

POL( A(x₁) ) = 0

POL( c(x₁) ) = max{0, -1}

POL( b(x₁) ) = 1

POL( B(x₁) ) = max{0, -1}

POL( a(x₁) ) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ QDPToSRSProof

                ↳ QTRS

                  ↳ QTRS Reverse

                    ↳ QTRS

                      ↳ DependencyPairsProof

                        ↳ QDP

                          ↳ DependencyGraphProof

                            ↳ AND

                              ↳ QDP

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                      ↳ QTRS Reverse

                      ↳ QTRS Reverse

  ↳ QTRS Reverse

  ↳ QTRS Reverse

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

B¹(a(x)) → B¹(x)
B¹(a(a(x0))) → B¹(a(c(b(b(x0)))))

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ Narrowing

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

Q is empty.

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

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

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

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

Q is empty.

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

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

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

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

Q is empty.