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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → c(b(c(a(x1))))
c(c(x1)) → c(b(a(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ Narrowing
  ↳ QTRS Reverse

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

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

The TRS R consists of the following rules:

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



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

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

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

The TRS R consists of the following rules:

a(x1) → x1
a(b(x1)) → c(b(c(a(x1))))
c(c(x1)) → c(b(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
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
QTRS
                  ↳ QTRS Reverse
  ↳ QTRS Reverse

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

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

The set Q is empty.

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

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

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

The set Q is empty.

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

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

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

The set Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ 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:

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

B(b(A(x0))) → B(A(x0))



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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(c(x))))
c(c(x)) → a(b(c(x)))
b(b(A(x))) → a(c(b(c(C(x)))))
c(C(x)) → A(x)
b(A(x)) → C(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ 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(x)) → C1(b(c(x)))
B(b(A(x))) → C1(b(c(C(x))))
C1(c(c(x0))) → B(a(b(c(x0))))
C1(c(C(x0))) → B(A(x0))
B(a(x)) → B(c(x))
B(a(x)) → C1(x)

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(c(x))))
c(c(x)) → a(b(c(x)))
b(b(A(x))) → a(c(b(c(C(x)))))
c(C(x)) → A(x)
b(A(x)) → C(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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



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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(c(x))))
c(c(x)) → a(b(c(x)))
b(b(A(x))) → a(c(b(c(C(x)))))
c(C(x)) → A(x)
b(A(x)) → C(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 1 less node.

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

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

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

The TRS R consists of the following rules:

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

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

B(b(A(x0))) → C1(b(A(x0)))



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

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

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

The TRS R consists of the following rules:

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

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

B(a(C(x0))) → C1(A(x0))
B(a(C(x0))) → C1(C(x0))



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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(c(x))))
c(c(x)) → a(b(c(x)))
b(b(A(x))) → a(c(b(c(C(x)))))
c(C(x)) → A(x)
b(A(x)) → C(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 2 less nodes.

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

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

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

The TRS R consists of the following rules:

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

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

B(b(A(x0))) → C1(C(x0))
B(b(A(x0))) → C1(A(x0))



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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(c(x))))
c(c(x)) → a(b(c(x)))
b(b(A(x))) → a(c(b(c(C(x)))))
c(C(x)) → A(x)
b(A(x)) → C(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 2 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Narrowing
            ↳ QDP
              ↳ QDPToSRSProof
                ↳ QTRS
                  ↳ QTRS Reverse
                    ↳ QTRS
                      ↳ QTRS Reverse
                      ↳ QTRS Reverse
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ DependencyGraphProof
                            ↳ QDP
                              ↳ Narrowing
                                ↳ QDP
                                  ↳ DependencyGraphProof
                                    ↳ QDP
                                      ↳ Narrowing
                                        ↳ QDP
                                          ↳ DependencyGraphProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ Narrowing
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
                                                        ↳ QDP
                                                          ↳ Narrowing
                                                            ↳ 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(x0))) → B(a(b(c(x0))))
B(a(c(x0))) → C1(b(a(b(c(x0)))))
C1(c(c(x0))) → B(a(b(c(x0))))
B(a(x)) → C1(x)

The TRS R consists of the following rules:

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

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

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

Q is empty.