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(c(a(x1)))
c(c(x1)) → b(a(c(a(x1))))

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS
  ↳ DependencyPairsProof

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

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

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

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

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

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



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

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

C(c(x1)) → A(x1)
C(c(x1)) → A(c(a(x1)))
A(b(b(x0))) → C(b(c(a(x0))))
C(c(x1)) → C(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)) → b(c(a(x1)))
c(c(x1)) → b(a(c(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 1 less node.

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

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

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

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

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



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

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

C(c(b(x0))) → A(c(b(c(a(x0)))))
C(c(x1)) → A(x1)
C(c(x0)) → A(c(x0))
C(c(x1)) → C(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)) → b(c(a(x1)))
c(c(x1)) → b(a(c(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 1 less node.

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

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

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

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

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

The TRS R consists of the following rules:

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

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

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

The set Q is empty.

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(x)))
c(c(x)) → a(c(a(b(x))))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(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)) → C1(a(b(x))) at position [0] we obtained the following new rules:

C1(c(a(x0))) → C1(a(a(c(b(x0)))))
C1(c(A(x0))) → C1(a(C(x0)))
C1(c(A(x0))) → C1(a(A(x0)))
C1(c(y0)) → C1(b(y0))



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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(x)))
c(c(x)) → a(c(a(b(x))))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(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(A(x0))) → C1(a(C(x0))) at position [0] we obtained the following new rules:

C1(c(A(y0))) → C1(C(y0))



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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

a(x) → x
b(a(x)) → a(c(b(x)))
c(c(x)) → a(c(a(b(x))))
c(C(x)) → C(x)
c(C(x)) → A(x)
c(C(x)) → c(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(A(x0))) → C1(a(A(x0))) at position [0] we obtained the following new rules:

C1(c(A(y0))) → C1(A(y0))



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

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

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

The TRS R consists of the following rules:

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

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

C1(c(a(x0))) → C1(a(a(c(b(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
C1(c(x)) → B(x)
B(a(x)) → B(x)
C1(c(y0)) → C1(b(y0))

The TRS R consists of the following rules:

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

The set Q is empty.

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

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

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

The set Q is empty.

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

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

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

Q is empty.