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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

A(x1) → C(x1)
A(b(x1)) → A(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2 + 2·x1   
POL(C(x1)) = 2·x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
QDP
          ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
QDP
              ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(x1) → b(c(x1))
a(b(x1)) → c(a(x1))
c(c(c(x1))) → a(b(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 1 less node.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPToSRSProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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

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


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
QTRS
                      ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
QTRS
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
QTRS
                          ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

C1(c(c(x))) → A1(x)
A1(x) → C1(b(x))
A1(x) → B(x)
C1(c(C(x))) → B(A(x))
B(a(x)) → C1(x)
B(b(A(x))) → A1(c(C(x)))
C1(c(c(x))) → B(a(x))
B(a(x)) → A1(c(x))
B(b(A(x))) → C1(C(x))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
QDP
                              ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

C1(c(c(x))) → A1(x)
A1(x) → C1(b(x))
A1(x) → B(x)
C1(c(C(x))) → B(A(x))
B(a(x)) → C1(x)
B(b(A(x))) → A1(c(C(x)))
C1(c(c(x))) → B(a(x))
B(a(x)) → A1(c(x))
B(b(A(x))) → C1(C(x))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
QDP
                                  ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

C1(c(c(x))) → A1(x)
A1(x) → C1(b(x))
A1(x) → B(x)
B(a(x)) → C1(x)
B(b(A(x))) → A1(c(C(x)))
C1(c(c(x))) → B(a(x))
B(a(x)) → A1(c(x))

The TRS R consists of the following rules:

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

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

C1(c(c(x))) → A1(x)
A1(x) → B(x)
B(a(x)) → C1(x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 1 + x1   
POL(A1(x1)) = 1 + x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(C1(x1)) = x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
QDP
                                      ↳ Narrowing
  ↳ QTRS Reverse

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

A1(x) → C1(b(x))
C1(c(c(x))) → B(a(x))
B(b(A(x))) → A1(c(C(x)))
B(a(x)) → A1(c(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(x) → C1(b(x)) at position [0] we obtained the following new rules:

A1(b(A(x0))) → C1(a(c(C(x0))))
A1(a(x0)) → C1(a(c(x0)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ Narrowing
QDP
                                          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

B(b(A(x))) → A1(c(C(x)))
C1(c(c(x))) → B(a(x))
A1(b(A(x0))) → C1(a(c(C(x0))))
B(a(x)) → A1(c(x))
A1(a(x0)) → C1(a(c(x0)))

The TRS R consists of the following rules:

a(x) → c(b(x))
b(a(x)) → a(c(x))
c(c(c(x))) → b(a(x))
c(c(C(x))) → b(A(x))
b(b(A(x))) → a(c(C(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
QDP
                                              ↳ Narrowing
  ↳ QTRS Reverse

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

C1(c(c(x))) → B(a(x))
A1(b(A(x0))) → C1(a(c(C(x0))))
B(a(x)) → A1(c(x))
A1(a(x0)) → C1(a(c(x0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(A(x0))) → C1(a(c(C(x0)))) at position [0] we obtained the following new rules:

A1(b(A(y0))) → C1(c(b(c(C(y0)))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
QDP
                                                  ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

A1(b(A(y0))) → C1(c(b(c(C(y0)))))
C1(c(c(x))) → B(a(x))
B(a(x)) → A1(c(x))
A1(a(x0)) → C1(a(c(x0)))

The TRS R consists of the following rules:

a(x) → c(b(x))
b(a(x)) → a(c(x))
c(c(c(x))) → b(a(x))
c(c(C(x))) → b(A(x))
b(b(A(x))) → a(c(C(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
QDP
                                                      ↳ Narrowing
  ↳ QTRS Reverse

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

C1(c(c(x))) → B(a(x))
B(a(x)) → A1(c(x))
A1(a(x0)) → C1(a(c(x0)))

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
QDP
                                                          ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

B(a(c(c(x0)))) → A1(b(a(x0)))
C1(c(c(x))) → B(a(x))
B(a(c(C(x0)))) → A1(b(A(x0)))
A1(a(x0)) → C1(a(c(x0)))

The TRS R consists of the following rules:

a(x) → c(b(x))
b(a(x)) → a(c(x))
c(c(c(x))) → b(a(x))
c(c(C(x))) → b(A(x))
b(b(A(x))) → a(c(C(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ QTRS Reverse
                        ↳ QTRS
                          ↳ QTRS Reverse
                          ↳ QTRS Reverse
                          ↳ DependencyPairsProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ RuleRemovalProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP
                                                      ↳ Narrowing
                                                        ↳ QDP
                                                          ↳ DependencyGraphProof
QDP
  ↳ QTRS Reverse

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

B(a(c(c(x0)))) → A1(b(a(x0)))
C1(c(c(x))) → B(a(x))
A1(a(x0)) → C1(a(c(x0)))

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
QTRS

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

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

Q is empty.