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.